SSC Advance Maths (English) by Rakesh Yadav - by WWW - Learnengineering.in

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¶'Digital India' ds Concept dks viuk;s¸ vkSj ?kj cSBs Hkkjr ds loZJs"B Teacher 'Rakesh Yadav Sir (Wizard of Maths)' ls i 6 8 , so 3 3  2 Ex.23: Which one is smaller out of

3 3m  2 3

2n 2 n

1/6 2/3  1  .  . 9 .  0.81 27   is –2 1 2/3 –1/2 –1/4  0.9 . 3 .   .243  3

4 1 5 2– – 3 2 4

Clearly,

9n  3 2  (3 – n /2 )–2 –  27 



1/3



 69

–1

9n  3 2  (3 – n /2 )–2 –  27 

2n

0.3

 10

3

n

Sol.

1/6

 32/6   9 

–1

then the value of (m – n) is:

5m –1m n –2–2m –n 2–2×30×5–3

1 1 3 8  –  3 4 3

1 2  2

 1  1–    10   3–1  23   33  2–3   –31

Ex.22: If

22n 2m –22m –2n –4m × 3m –n m n –2–2m 2 ×

33

–  0.1

1/6

 23/6   23 

 68

 31/3  33

n

1 1 1    4 125 500 Ex.20: The value of Expression



1/6

 8

1 – 10 –9 –9 –3    = 3–1 0 3 .2 – 3 9 – 3 6 2

×3m n –25m n –2 4m 2 52m n 32m –2

3/4

is

–1

and

4n 20m –112m –n 15m n –2 16m52m n 9m –1

1/3

0

 3  3  1  3     –  2  2  3

22n 22m –25m –122m –2n 3m –n

Sol.

3

–1

m n –2

4 20 12 15 16m52m n 9m –1



–1

geisnh eeYa ridna gv.i Sni

Sol.

Ex.19: The value of expression



–  0.1

1 3  3

2  21/2  22

 3  3  1  3     –  2  2  3

7 48



0

–1

1 24  2z 7

Sol.

2 3

3 × 10–1 =

 0.6

m –1

E.g.: Which is larger 2 of 3 3 ? Sol. Given surds are of order 2 & 3 respectively whose L.C.M is 6. Convert each into a surd of order 6, as show below :

r

1 24 3  6z 7

n

–

5 3

3  0.3 10 Ex.21: The value of expression 

1 1 1 24    6z 6z 6z 7

z

19

312  10

–

1 4 × 4

, (3)3

6

1 3  3

, (4)4

4

3

(2)12 , (3)12 , (4)12 1

1

1

(26 )12 , (34 )12 , (43 )12 1

1

1

(64)12 , (81)12 , (64)12 1

1

1

clearly (81)12 > (64)12 = (64)12 1

so,

1

1

(3)3 > (2)2 = (4)4

B>A=C

62

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Ex.26: Arrange the following in descending order.



3 – 2, 4 – 3, 5 – 4, 2 – 1

3– 2 3 2  1 3 2

3–2  3 2



1 , 4 3

4– 3 1 5 4

5– 4

 

2 –1 

 

 

3– 2 

4– 3 

 

8  5 or



13+2 22 23  21

2 23  21



21  19 ×

1

1

4,

6

6,

1

12

12

1

3 3 , 4 4 , 6 6 , 1212 LCM of 3,4,6 and 12 = 12 Then 1 2

1 3

1

, 4 4 3 , 6 6 2 ,1212 1

1

1

21  19

It is cleared that 1

1

1

1

1

1



So, 1

1

1

3 3 > 4 4 > 6 2 > 1212

23  21

TYPE – IV

 19 

21 

1

(81) 12 >(64) 12 >(36) 12 >1212



21  19

1

(81) 12 ,(64) 12 ,(36) 12 ,1212

23  21

  

1

(34) 12 ,(43) 12 ,(6²) 12 , 1212



=

4

1 4  4

Sol. Rationalisation of Surds:

9 4 ,



3,

33

> 8 5> 6 7 Ex.30: Which is greater than

= 13+2 40

1

Rationalisation of Surds: If the product of two surds is rational, then each of them is called the (R.F.) rationalising factor of the other.

E.g.: 5 7 × 7 = 5 7  7 = 5 × 7 = 35

2



11  2

2

2

2





13-2 40 >13-2 42 Then



6 7 ,



= 13-2 36

It is cleared that

11  2 , 10  3 Sol. In this type questions we use square method

So,

2



11  2 > 10  3 > 9  4

Ex.28: Arrange the following in ascending order. 8 5 ,

2

6  7 = 13-2 42

13-2 22 >13-2 30 >13-2 36 >

3 3 > 8 5 13  10







So, It is clear that 

2

8  5 = 13-2 40



21  19

2 23  21

>

Ex.31: Arrange the following in decendinding order. 2350,5200,3300,4250 Sol. 2350,5200,3300,4250 Power in same form (27)50, (54)50,(36)50,(45)50 (128)50,(625)50, (729)50,(1024)50 So, 4250 > 3300>5200>2350 Ex.32: Arrange the following in descending order. 272,536,448,360 Power in same form (26)12,(53)12,(44)12,(35)12 (64)12,(125)12,(256)12,(243)12 Then (256)12 >(243)12 >(125)12 >(64)12  448>360>536 >272 Ex.33: Arrange the following in descending order.

ERna

  8  5    13  10 = 10  ×  13  10

8– 5 





wwM wa. th Les aBryn



8 5

9 4,

11  2 , 10  3 Sol. We use square method

Sol. Rationalisation of Surds:



2

So, It is cleared that

11  2 < 10  3 < 9  4

Ex .27: Whic h is g re ater than



= 13+2 30

8 5, 6 7,

1 and 2 – 1  2 1 As we know, if the numerator is same then the fraction whose de nominator is lar ge r the fraction will be lower. Hence the correct order of descending is.



It is cleared that

2

13+2 40 1) are whole numbers such that mn =121, the value of (m – 1)n+1 is (a) 1 (b) 10 (c) 121 (d) 1000 54.

1

–

1 

3– 8

8– 7

1

is equal to

6– 5 5–2 (a) 5 (b) 4

  55.  

–

1 +

3

7– 6

2

1

(c) 3

 7 6

(d) 2



1

 62

=?

8– 7

  2 – 2 2 

is

equal to

(b)

(c) 4 16

(d) 6 80

Rakesh Yadav Readers Publication Pvt. Ltd.

2– 3

(d)

3

9, 3, 4 16, 6 80 is

(a) 4.899 (c) 1.414

3 4

  x  

(d) 0

(c) 0

(a) 0

3

(c)

3 =1.732, the value of 56.

5 3 – 2 12 – 32  50

(d) 11

2

1

2 3

3

(b)

(a) 1  2 3



(b) 2

50.

0.25

(c) 1

3 1

t he v alue of

5 1 5 –1 , b= , the value of 5 –1 5 1  a 2  ab  b 2    a 2 – ab  b 2  is  

 5  3

    

2 6

–

(d) 2

49. If a =

16

3 6

2– 1

(c) 13

5 3

(a) 3 9

97 – 96

(b) 9

5– 3

1  is equal to 2 

(b) 8

46. Given that

1

1 +.... +

2 2.2

(a) 3

5– 3

99 – 98

–

98 – 97



45. Greatest among the numbers

3 3

1

100 – 99



3 3

6 3

3

(c) 3 – 3 36. Simplify %

7– 5

1



3

3 2

44.

t hen

2 3

5– 3

   3  3

1

   9   44 8 –    2     



4 5

(d)

(a) 32

1



3 4

(d)

(a) 1

 2  2

2 34. The simplified form of

+

1

wwM wa. th Les aBryn

1

   

(b)

ERna

2  7

(c)

(a)

and y =

2 15

  3  

,

3

13 – 11

(d)

5 3

(c) 2.5 42.

2

   

(b)

(a)

41. 0.75 × 0.75 – 2 × 0.75 × 0.25 + 0.25 × 0.25 is equal to (a) 250 (b) 2500

 is presented in the form of

2

(b)

7– 5

(c)

2

(c) 6

1  a + 1 – a is :



19 – 17

is simplified to

perfect square it will be equal to :

(a)

2

13 – 11 ,

5– 3 (x+y) equals: (a) 8

(b) 0.95 (d) 0.97



(a)

19 – 17 ,

40. If x =

3

48. If a =

–2 –2     is equal to :

(c)

:

33. When 4  7

(b) 4

(d) 0

–2

(a)

3

(a) 1.06 (c) 0.86

(c)

2

 and  5 – 3 ?

is equal to:

(d) 2 6

0.96 – 0.1 2 2 0.96  0.096  0.1

35.

(b)

(b) 2 15

(c) 2 10

32.

 

is

2 2

+

12  12  12  ....... is equal to (a) 3

(a) 16 (b) 8 (c) –8 (d) –1 39. Which is the greatest among

(a) 189 (b) 180 (c) 108 (d) 198 31.

47.

2 – 3 – 5 

1

1 5

–3

 

1 

(a) 1

   is :

(a) 5.375 (c) 6 30. The value of

1 2 3– 5

2

– 3 8.75  3 2.5



   

in simplified form equals to :

3 3.5  3 2.5

29. The value of

3 3.5

37.

kgei snhe eYari dnag v.iSn ir

28.

(b) (d)

is : 2.551 1.732

2

–2    1 2    –      2         (a)

1 16

(b)

2 2

(d)

2 7

–1

(b) 16

is equal to :

(c)

–

1 16

(d) –16

69

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3 3 3..... is equal to

(a)

(b)

3

(c) 2 3 (d) 3 3 59. The greatest among the numbers

69. 2 

a a a a

0.09

(c) 0.5

(b)

3

(d)

3 5

0.16,

0.16 ,

0.16



(a) 0.16

8,

4

(a)

4

(b)

(c)

10

13 41

x

y

–z

62. If 2 = 3 =6 then

5

(d)

2

16

74 3

(a) 5 is equal

(c) 6

(b)

3 2

(b) (d)

340

(d) –

1 2

0

111





75 

75 –

50

24

660

5

(c) 2 5 66. Evaluate

(c) 0

2

78. If

3 2 4 23 4 ................... is

n 5

243 n

9

–

(d)

2 2

–

3

5 5 5  5 value of a is (a) 4 (b) 5 79. The value of

81

3

3

5 : 6, then one of

3 y 3 y

1

(a)

  

2

is

1

(b) 5

25

 32n 1

 3n –1

2

(c)

(d) 25

5

5



67. If a, b are rationals and a 2 + b 3 = 98 + 108 – 48 – 72 , then the values of a, b are respectively



32 2

–3

 

3

3 2

(d) 12

6

1





7

 8

7

1



(c) 0

8 9

(d) 1

72  .... is

72 

(a) 9

+

4

(b) 18

(c) 8

(d) 12

1 87. The

value

of

1

2



3

1 1

2



is:

3

(b)

2

(c) 1

(d)

3 4



3



2



88. The value of the expression a 2

5

, then t he

6

(c) 6

–2

(d) 8 –3

2



is

(a) 198 (b) 180 (c) 108 (d) 189 80. A tap is dripping at a constant rate into a container. The level (L cm) of the water in the container is given by the equation L = 2 – 2t, where t is time taken in hours . Then the level of water in the container at the level of water in the container at the start is (a) 0cm (b) 1cm (c) 2cm (d) 4cm

Rakesh Yadav Readers Publication Pvt. Ltd.

6



3

is (a) 2 (b) 4 86. The value of

72  is

3

1



1 4 5

1





2

1



1



1

3 10 7

(d)

(a) 3 (b) 9 (c) 6 76. The simplified value of

 = 1, then the

(b)

3x +  3  x –

the value of

(a)

5 –

(a)

(3x –2y) : (2x +3y) =

(a) 16 (b) 88 (c) 176 (d) 132 77. If 2 n – 1 + 2 n+1 = 320, then the value of n is (a) 6 (b) 8 (c) 5 (d) 7

4

is equal to@

366

+

20  12  3 729 –

3

355

 3  110  12 12 – 2 5 – 3 is

3 5

2

4

(d)

11

(d)

2

(d) 0

3

(b)

9

5

–

3

305

(d)

(b)

4

1

(b)

value of x is (a)

(c)

2

(c

84.

2

– 2– 3

(a) 330

(b)

50

2

2 2

–

 7²  8²  9²  10²

7

30  30  ....

75. The value of

40  9 81 is

(c) 7

x –

3

 2 3

2 13

(a) 2 (b) 22 (c) 23 (d) 25 74. 553 +173 –723 +201960 is equal to (a) – 1 (b) 0 (c) 1 (d) 17

wwM wa. th Les B aryn 3

64. The value of

65. If

= A + B , then B – A is

R Enak

–2 3 340

(a)

(a)

70. If

8

63. 2 3 40 – 4 3 320  3 3 625 – 3 3 5 is equal to

(c)

2



83.

–2

6

–

85. The value of

to (c)



(d)

30 

1 1 1   x  y  z    

(b) 1

6²

(c) 1

73. The value of

(a) 0

3 –2

– 2 3

2

(a) 2 6 (b)

equals to

(c) 13 (d) 3 3 – 7 71. The smallest among t he number s 2250,3150,5100 and 4200 (a) 4200 (b) 5100 (c) 3150 (d) 2250 72. Find the value of

13 , 5 16 , 10 41 is : (b)

2

(b)

8

(a) 8 (b) 4 (c) 1/2 (d) 2 The Simplified value of 6

(a)  2 3

(a) – 13

0.16 (c) 0.04 (d) (0.16) 2 61. The greatest of the numbers 2

3

43 3

, 0.04 is



2 3

=?

10  25  108  154  225

82.

1



 

60.The greatest of the following numbers

2

1



3

0.064

81.

< 729 but a > 216 < 216 > 729 =729

6

3 0.09, 3 0.064 , 0.5 and 5

(a)

1, 3 2,3

68. Let 3 a  3 26  3 7  3 63 then (a) (b) (c) (d)

3

(b) (d)

geisnh eeYa ridna gv.i Sni

58.

(a) 1, 2 (c) 2, 1

r

57. The greatest number among 260,348,436 and 524 is (a) 260 (b) 348 (c) 436 (d) 524



6

6  ...  upto is

(a) 30

(b)

5

(c) 3 89. The value of

(d)

2

3 7 5  (a) 1 (c)

 2

5 5 2

 7

(b) (d)

2 3

90. If 11 n of n is: (a) 3



112



(b) 11

2 2



7

 5

is:

0

7

343 , then the value (c) 13

(d) 7

70

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ANSWER KEY (a) (c) (b) (c) (c) (b) (a) (b) (d)

10. 11. 12. 13. 14. 15. 16. 17. 18.

(b) (c) (a) (c) (c) (b) (a) (a) (c)

19. 20. 21. 22. 23. 24. 25. 26. 27.

(c) (a) (c) (a) (a) (b) (b) (a) (b)

28. 29. 30. 31. 32. 33. 34. 35. 36.

(c) (c) (d) (a) (c) (c) (d) (c) (d)

37. 38. 39. 40. 41. 42. 43. 44. 45.

(c) (a) (d) (a) (d) (b) (d) (c) (a)

46. 47. 48. 49. 50. 51. 52. 53. 54.

(d) (b) (a) (b) (c) (c) (a) (d) (a)

55. 56. 57. 58. 59. 60. 61. 62. 63.

(d) (a) (b) (b) (d) (b) (d) (a) (b)

64. 65. 66. 67. 68. 69. 70. 71. 72.

(c) (b) (c) (a) (a) (d) (c) (b) (c)

73. 74. 75. 76. 77. 78. 79. 80. 81.

(a) (b) (b) (c) (d) (a) (a) (b) (d)

kgei snhe eYari dnag v.iSn ir

1. 2. 3. 4. 5. 6. 7. 8. 9.

82. 83. 84. 85. 86. 87. 88. 89. 90.

(d) (a) (d) (a) (a) (c) (c) (b) (d)

SOLUTION

1.

(a)

4. (c)

1

52 6 –

 12 256 , 12 216 , 12 225 , 12 245

0.06  0.06  0.06 – 0.05  0.05  0.05

52 6

 Biggest = 3 4

0.06  0.06  0.06  0.05  0.05  0.05

1

0.06³ – 0.05³

3  2 –    3  2



0.06²  0.06  0.05  0.05²

8.

4

(b)

3 6 9   5   



 911  5 6 3    



 1 5 2      

3 6 9   5   

4

a³ – b³ 

  2  52 6  3 2  3  2     a² +b² +2ab = (a+b)²   3 2





 –  

1



3 2

   

3 – 2 3 –

3 – 2



3 2– 3 2  2 2

0.05  0.05  0.002  0.04  0.04



    2  32 – 3

1



6–3 3 –4–2 3





2–5 3

2–5 3 3.(b)(243)0.16  (243)

2–

3



2–5 3



1



3 2 3 2 , , , 36 26 26 4 6

 6 27 , 6 4 , 6 8 , 6 16

0.16+0.04

 3 2 (Least)

×(243)0.04

a m  a n  a m n   

20 243100

1  5 243 = 3 243 5

7.

(a) 3 4 , 4 6 , 6 15 , 12 245 



1 1 1 1 , 4 , , 6 12 3 6 15 245 4 (take LCM of 3, 4, 12 & 6)



 911  5 3 6    

 1 5 2      

4

4

 3 16 2   

–3   16 2 

  

 3  1   22 4   3  ²    4 2

 3   2 1  16  3       16 2 



4³ 

1 4³

4097



64 1

11. (c)

(0.01024)5

1  (0.45 )5 

12. (a)

3 2 1 4 , 12 , , 6 1512 24512 412

 12 44 , 12 63 , 1215² , 12 245

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10. (b)

1

 6 33 , 6 2² , 6 23 , 6 4²

=1

 2430.20



1

4

4

 5² × 5²  52+2 = 54 9. (d) 3x + 8 = 272x+1 3x + 8 = (3³)2x+1 3x + 8 = 36x+3 x + 8 = 6x + 3 5x = 5, x = 1

2 , 34

 32 , 23 , 22 , 43 (take LCM of 3 &2)

2–5 3

2 3

3 , 32 ,

6. (b)

3 2– 3 – 2 2 3

=

0.05²  0.002  0.04²

(Description: same as above question) a = 0.05, b = 0.04  a–b  0.05 – 0.04  0.01

2

– 2. (c) 2  3 2 – 3 2–5 3



0.05 ³ – 0.04 ³

wwM wa. th Les aBryn 3

b = 0.05  0.01

5. (c) 0.05  0.05  0.05 – 0.04  0.04  0.04

  





 a–b So, a = 0.06,  0.06 – 0.05

  2

3 2–

3–2



a – ba ²  ab  b ² a ²  ab  b ²

ERna



a ²  ab  b ²



51 = 0.4 0.4 5

–2  1 –2 64 3      4  –2  1 –2 4³ 3      4 

 

71

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 

Similarly:

 1 2  1 –2         4  4 



 1 2–2     4 

=

13. (c)



1

2

 

  

5 –

3  1–

5 

3

2

5 –

 16. (a)

 

5

3



4 – 15

2

6  5  3 – 10 – 2 2 5 – 3



2

6

17. (a)



 2–

 

3

2

3

 3 1

 4 –3



  – 1 

3 – 1

3 –1 

3 43–4 3

3

2



3 –1

3–1

 14   

31– 2 3

14 +

   





3



15. (b)

14 + 2 –

   

3

5 –

3   3



 5  32  5  3





1

1

–



12 –



16

  

4  35

8–

4  15

2

–

 

 12 – 8 3  2

1



36  24 – 24 – 16

1 

2

5  24



4 – 15



5

24

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2





  3

–2 5 3

2 2





2



25 – 24

2

 7   3



2

2

3  2    



– 2 3  2



1 2



2



 5 – 3

 7 – 5

2

 3 – 2 2  3 – 2

2 7 3

1 

2

2

6 –2

–2 7 5

1

5 – 24

2

4  15



2

5 – 24 



8 – 2 15

2

 7   5

2 5–2 6



2

6–4



8 – 2 15

10  2 21

5  24



1

–

12 – 2 35

–

5  24



5  3 – 2 15

8  2 15



 8 – 60

10  84

 3  3 5     2     



140

1

–

2

–

4

4 =2

625

2 5–

5  3  2 15



5      2



3

2

2

5

12 –

     125 

125 3      8 

=

1

20. (a)

5  24



= 16 –

 1 2 16     4 

10  4  21

18. (c)

2



2   1  

1

–4 8 3

4



  

2

2 2–



3 – 1   3 1

2

   

 4  3  4   

 

1



R Enak



2

3



 2 1   4² ×  1  4  

(10)²  100

4

2  3 2 – 3  2 – 3  2  3  

2–



(a +b)² =(7.5 + 2.5 )²

wwM wa. th Les B aryn



4  15

a² + 2ab + b²

2



3

8

2  2 1  4 33           2  

a × a +2ab + b × b



 5 – 6

2

2

0

1

62 a = 7.5 and b = 2.5 

    2  5 – 6 2 5 –2 6 



19. (c)

15

16 – 15 

14. (c)

    

4

 2 64 3  2 –2   

16  15  8 15  16  15 – 8 15

3  10 –



1

15

4 – 15



5 –



4 – =



 

5

4  15 

 1 2 1– 2     5  3  5 – 3  



  3

Thus, the expression.

 1 0     =1 4 



2 5 – 3

   

r



 1 –2   4 

–2

geisnh eeYa ridna gv.i Sni

4

2

 7  3 1 

7 –

– 5

1 5 –

 3

2 7 3

72

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Rationalizing in above equation.

3 0.0 04096

25. (b) 

×

×

7 – 5

–

7 5 2

5 3

–

5 3

5 –

7 – 3

2

7 5 –

26. (a)



3 –

7

2.3  2.3  2.3 – 1 2.3  2.3  2.3  1 a = 2.3 b=1

 1   3  2



a –b

 3

0.49,  0.7

0.008,  0.2

0.23  0.23



least

2.89

3

4

4

12

5

12

64 6

2

0.13

24. (b)



50.20 3 0.250.1500.6



7

1 3 3 1 =7 7

– 3 38 74 3

– 3  3  8 4  3  22 3

2







– 3 38

2  3



– 3  38 2 3

1.7

  3

28. (c)

2.890.5

x=

2

x² = 2 +

– 3

4  3



– 34 3



4 =2

3  3 – 2 2

 

x 





3 – 2 2 + 3  2 2

+

3  2 2     9–8    3

3

2– (0.5)²

x 

 (3 – 2 2 + 3 + 2 2 )(17 + 17 – 1)

2  2  .....

 (6)(33)  198

2  2  ......

x ........ x

Then factor the

n1

5 31. (a)

–

5 



   

3 2

3 3



5 2

5 3

 3 – 2 – 3– 2

5 – 2

+ 2 2 

3–2 2 2

+

2 2

+

5 2

5 – 2



n2

5 3

3 3 –



×

 5 – 3 5 – 3

5 –

2



5–2

 5 – 3 5–3

2 22

2 × 1

3 3.5  3 2.5 3 3.52 – 3 8.75  3 2.52

= 15 – 10 –  

32. (c)

 15 – 6 +

10 – 6

15 – 10 – 15  6  10 – 0

6

0.96³ – 0.1³ 0.96²  0.096  0.1²  a = 0.96  b = 0.1

x = 3 3.5 y = 3 2.5

3 3

3 2

15 – 10 n1 > n2

mi.diff.

So n1 is answer

Rakesh Yadav Readers Publication Pvt. Ltd.

3

3 – 2 2     9–8   

b=3+ 2 2

1.75

x² = 2 + x x² –x – 2 = 0 x² – 2x + x – 2 = 0 x(x–2) +1 (x – 2) = 0 (x + 1) (x – 2) = 0 x=2 Shortcut Method When the question is in t his for m i.e

+

3 – 2 2

3  2 2

29. (c)

2



1.732

 4 ²  24 3





 

3 – 2 2

a³ + b3  (a + b)(a² + b² – ab)

3 0.008



– 3  3  16  8 3

– 3

1

3  3 – 2 2

a=3– 2 2

Assending order : 0.2

3

3

1.732

2



3 0.008

3

1.75

wwM wa. th Les aBryn





3  2 2

–3

   2

1

3 – 2

3

3

0.130.07 243 0.25 2.075 30.2 7 7 7



    

2– 0.25

1.7

 

  

1



2–(0.5)2

2

3 – 2 2

+

3

0.2

9

0.07  243 23. (a) 0.25 0.075 0.2 7  49  343

243





2.89

125



ERna

12

a ²  ab  b ²

2.3 –1 = 1.3

1

–3

   2

a ²  ab  b ²

 27. (b)

22. (a) Descending order:

256

    

a ²  ab  b ²

(2.89)0.5

12



a³ – b³ 

 0 21. (c)

 0.25

3.5 + 2.5 = 6

3

3  2 2

3

2

0.5²,



30. (d)

2

5 –

3 3.5  3 2.5

 0.4

 5  3 –  7 – 3 –

2



3

0.4  0.4



x³ + y³

(16³ = 4096)

0.16



3

7 – 3

+

7 3

7 5 

1

7 5

(x + y) (x² –xy + y²)



kgei snhe eYari dnag v.iSn ir

1



a³ – b³ 

a ²  ab  b ²

73

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

a – ba ²  b ²  ab a ²  ab  b ²



a–b

1 

=

7 8



82 7

8 –



2

4–

3 

3–

1

     

2

   

1

36. (d)

100 –

 7  1

2

7

2 +

12 –

5

99

3

12 – 5

5

–

12 –

Similarly

7

×

7 – 5 7 – 5



99 –

+

12 –



39. (d)

7

12 





98

  7

–2



 

100  99

 19  17 



19  17 ×

98  97 ....and



5 3

5 3



2

 5  3 

5 3



5 3

98  98  97

2



Largest + (Because, Same Numerator is dividided by Smallest denominator)

40. (a) x 

100  99 – 99 –

13  11

99  98

3





7 – 5



7 +



12  5

5      

7

100  1



37. (c)

 12  7  5

   

 

7 – 5  12  5 – 12 – 7

2

5 3

Similarly



 2

2

 10 + 1 = 11

1 

2 3– 5

2– 3 –



  5 



 2  3   5  × 3 –  5   2  3   5 

232 6 – 5

2 3 5





1

4– 3



4 –



3



2–



5  3  2 15  5  3  2 15 2

2 6

3 –

16



41. (d) 





 5  

2– 2–

  5 3  5

8 2 0.75 = a, 0.25 = b a×a–2×a×b+b×b



a² – 2ab + b²



(0.75 – 0.25)² (0.50)² = 0.2500



4– 3

4 3

3 

1

1 42. (b)

3+



3+



3

3

4 – 3



1 Similarly



4 5

=

5 –

1 

5 6





6– 5

1 3

4

 2  3  5 –  2 – 3  5 2 6 2 6

Rakesh Yadav Readers Publication Pvt. Ltd.

 3

+

1 3

1 +

3 3

3–3

1 +

3

–2 6

Now put the value in question

(a–b)²



1 +

1

2– 3 5



2

x+y

1

Similarly

1

 5  3 



2 3 5

3 4

y=

5 3 5 3

1

0

1

35. (c)

–

wwM wa. th Les B aryn

2

19  17 2

 13  11 

5 

2



19  17

.. + 2  1

2

–4

19  17

19  17

Similarly

98 – 97



2

19  17 

soon Now : expression:

12  7 

1 

100  99



1

5 5

 12  5 –  12  5





(–2)4 =16



100  99

×

R Enak



–2

1 



1 7



–2

2

–2 –2 

1

2 7 

–2   

38. (a)

100  99 

   

2 6

7  9– 8

9–



2

 7  12   2

8

4 6– 5 7 –

3 5 –

6 8 –

 1  2. 7.1 2

7 5

9

9–



r



 7



1

2 3

8 – 7

Now put values

2



2 6

geisnh eeYa ridna gv.i Sni



34. (d)



1

4 7



2 3 5 – 2 3 – 5

7 – 6

1 

 0.96 – 0.1 = 0.86

33. (c)



6 7

–

1

3 3

3 –    3 3

–

3–

3

3 – 3 – 3

 

9–3

 3

1 3

1 –

3

=3

74

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46. (d)

2 2  2 2   2

      

2

5 3 – 2 12 – 32  50

3 6 5 3 – 4 3 –4 25 2

3

3 6

5 –1



3 2



3 2





3 2





5 1

48. (a)

3 2

    

2 6

3 2

 3 1

( 6 

      

3

6 

2 6



3

3 1

 



3  1

6  2

62

6

3

2 3 



 6  3

3 2



2 3





 45.(a)

39 

91/3 

2    

6

 3  1 2

    



4





3



31/2 

36/12





12 27²

6 80



3 1

2

801/6





163/12

5–3

51. (c)

4



12 16³

12 80² 3

9

=

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6– 5

6–3

6 –

5

6–5

5 – 3 – 6  – 3  6 – 5 –2 3



2  7 – 2 10

2

 2

 2

2

5



2

 5 – 2



2 5 – 2 



–2 5 2

2

3 –1 2

3 –1

52. (a) x =



2

x

5

1 2  3 1 x –1

 1 2  3 

1 1 2  3 –1



1 2  3 



1 2 3

2 3

6– 5

×



2 3



×

6– 3

 5 – 3  – 3  6  3 +

2

80 2/12



3 –

1

6 5

  5 

2

3 1 3 –1



6

5 – 3



161/4

Square of 81 is largest . So Ans

2 

1

5 – 3

+

6 3

4

 3  1

put values :

4 16





1–a =

+

2

a  b – ab 2 a  b – 3ab

6– 3

×

6 3

2

2

12  6  18  6  18  2 3

3 1

=1

5 1



5 3



Similarly,

2 32 3 0

5 –1

3

–

5 3

3 1

4

1 a =

12  2 3

2

 3  1

a 1 =



=

a ² – ab  b ²

2



2

6 2

94/12 12 94



a+1=

   2

2



   

+1

42 3

3  1

wwM wa. th Les aBryn

  

62

3 )



6 

2

2 3

 6 3

2

ERna

44. (c)

3

32 

=3

50. (c)

2

a+1=

[8  8] = 0 Ans.





4

2 4 3 –1 9 –1 = = 2 9–3 3 3 –3

3

a=



×

5 –1

a ²  ab  b ²

=

  61  8  2 2           



12  12  12  .......

4 × 3



2 5 1

=

put value in expression



1/2   8  212/2    

 1

5 –1

3 = 1.732



47. (b)

 

 5

a.b =

1/2      9/2 3/2      2 .2   8     1     2      2     

5 1 2

2

 

2 



5 –1 

 5  1   5 – 1   5 – 1 5  1

5 3 – 22 3 – 4 2  5 2



5 1

a+b=

2

3 6 

1/2       29    2 4 212   8       1     2     4   



49. (b)

kgei snhe eYari dnag v.iSn ir

43. (d)

3 6

1

   9  4 4 8     2   

3 –

2

3–2

75

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n=2

 2   1   –      2   



 (m – 1)n+1  (11 –1)

2+1

 10³  1000

3–

8

3–

 1 4  –     2

3 8

×

8

3 8

16



3 8

 348  (34)12  (81)12 (Greatest) 36

9–8

4

3 12

 (4 )  (64)



3 3 625



3× 3 5  5  5  5  3×5 3 5



15 3 5

Now put the value in question

1

57. (b) 260  (25)12  (32)12

1 

 1 22   –    2



1 54. (a)

–2–1



43 5 – 16 3 5 + 15 3 5 – 3 3 5



19 3 5 – 19 3 5



0

12

64. (c)

 3 8 Similarly,

58. (b)

3 3 3 ............. Shortcut method  When the question in from

1 

8 –

= 8 7

7



1 

7 –



6

= 6 5



1 =

5 –2

60. (b) 0.16

52

 3 8 – 8 – 7  7  6 – 6 – 5  5 2 3+2=5 55. (d)

2

7 6

6–2





62

6–2

1

+



+

1

+

2





8 7

+

6 –2

6–4

7 – 6

7 6



7

+2 – 2 2

7 – 6

+2– 2 2

7–6

20

20

5 16





20



20

5 64

0.04

4 13

5 13 5 13

k

+



x

1/y

×k

1



411/10 412/20





4 16

20

x – 24 75  50 = 1

 20

63. (b)



6 –2 7 – 6 8  7 2–2 2



6 –2 7 – 6 2 2  7 2– 2 2  2 7

–1

Rakesh Yadav Readers Publication Pvt. Ltd.





x – 24



4 16

=

75 – 50 75  50 2

75 – 50  = 75 – 50

75  50 – 2 75 50



x – 24 =



x – 24 =



x – 24

=



x – 24

=

25

 20

125 – 2  5 3  5 2 25 125 – 50 6 25

2 41

28



x –2 6



x=5



25 5 – 2 6

= 5–2 6

–1/z

66. (c)

20 +

12 + 3 729 –

1  2 5 +2 3+ 9 –

z 1



   

4



5 – 3

4 5 –

3

– 81

5  3

– 9 

5  3

1 

y



25

=k

 –

x

x – 24

–1/z

y

1





2 41

8 7 8–7

7

49

1041

161/5 164/20

62. (a) 2x = 3y = 6–z = k  2 = k1/x; 3 = k1/y = 6 = k 1/x





0.0256



Greatest =

+2– 2 2

 2  –2   1   –    2       





1

6

40  9

0.04

 2×3=6

7 –

56. (a)

2

0.16

131/4 135/20



8 – 7



0.6

81/2 810/20

1

8 7

8–



61. (d) 2 8

wwM wa. th Les B aryn +

62



0.5



20 10 8

1



R Enak

3  8 –  8  7   7  6 –  6  5    5  2

3/5

0.4

0.40

0.16



0.5

0.16



Put value in question

2



75 – 50

3 0.064

0.09

0.3 

40  9  9

65. (b)

n n n .................

59. (d)

6– 5



 So n is answer 3

7 6

1 

40  9 81

geisnh eeYa ridna gv.i Sni

 524  (52)12  (25)12

r

mn = 121 = 112  m = 11

53. (d)

z

=0

23 40  2  3 2  2  2  5

4    

 5  3  – 9



2 5 + 2 3+9–



2 5+2 3+ 9 – 2 5 – 2 3– 9  0



2 × 2 35



43 5



43 320



4 × 34445



4 ×4 3 5





16 3 5

– 22223 –

2

 

67. (a) a 2 + b 3 =

98 + 108 – 48 – 72 772 + 33322 33222

76

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7 2 +6 3 –4 3 –6 2

= 1 22 3 a 2b 3 a =1 b =2

250

3 a < 3 27 + 3 8 + 3 64



3 a 1

a b2  1 = x 2b

4  16 – 4 2

=

2ab b2  1

1 x 1 y 1 z × × =1 1– x 1– y 1– z TYPE – V

Recurring number

ax  a –x

Find

=

So,

3mx2+m = x2+3x x2+3x–3mx2–m = 0 Ex.40. If x =

......(III)

a b c × × =1 b c a

m 1 m –1

x 3  3x m = 3x 2  1 1

1 a+ = 4 a a2–4a+1 = 0 ax2+bx+c we use,

– 4 

2

  1

Again use C & D

1 b= a Now, a +b = 4 Put the value b



1 z c = 1– z a (I)×(II)×(III)

 3x  3x 2  1

eYrai dnag

Sol.

ab = 4:1 find the value a a +b = 4,

3.232323....  3.23 =3+ ......(I)

1 y b = .....(II) 1– y c

23 23 = 3 99 99

5.564564564  5.564  5

564 999

85

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*

expanding the meaning Step - 1 Subtract the non-recurring initial part of the decimal (in this case, it is 4) from the number formed by writing down the starting digits of the decimal value upto the digit where the recurring decimals are written for the first time; (456 - 4) Step - 2 As many 9's as the number of digits in the recurring part of the decimal. (in this case, since the recurring part '56' has 2 digits, we write down 2 9's.) These nines have to be followed by as many zeroes as the number of digits in the non recurring part of the decimal value. (In this case, the non recurring part of the decimal value is '4'. Since, 4 has 1 digit, attach one zeroes to the two nines to get the number to divide the result of the first step.)

(xi) (xii) Sol. (i)

705 47 = 900 60

=5

5 (ii) 0.555...=0. 5 = 9

783  78 900

=9+

116 99

71 71 =4 99 99

0.567 567  56 511 = 900 900

Sol. 0.037 

3

3

0.037

37 1 = 999 27

0.037 =

3

I. 

× 5 4 7 7

× 6 7 5 9

× 5 4 6 7

× 6 7 6 0

× 5 4 6 7

× 6 7 6 0

× 5 4 6 5

= 12.47970 Where maximum digits non recurring

Type – VI

43542  43 43499 = 99900 99900

Ex.45 Find the value of

Rakesh Yadav Readers Publication Pvt. Ltd.

× 2. 8 3. 7 5. 8 12.4

Number of Digits

0.43542 

2.856  3.74  5.8756 I III II

[where ( 5.8756 ) has 3 max. non recurring digit (875)] II.  LCM of recurring digits of number (where 2,2 and 1 recurring digits) III.  atleast 3 digits

0.43542542......

456 152 = 999 333

Sol.

873 – 8 990

Ex.44 Sol

2.856  3.74  5.8756

73  00 73 =5+ 9900 9900

865 173 = 900 180

17 99

= 10 + 0.17 = 10.17 Ex.47 Find the value of

73 9900

=

47 (iii) 0.474747...=0. 47 = 99

(iv) 0.456456....=0. 456 =

37  56  23 99

Ex. 43 0.56777........ Sol

37 56 23 + + 99 99 99

=9+

67  6 61 = 90 90

(xii) 0.8737373....=0.8 73 = =



86 99

=9+1+

4.7171...= 4. 71 =4+

(xi) 5.00 73 =5+



=8+1+

(ix) 0.6777...=0.6 7 = (x)

0.37  8.56  1.23 37 56 23 +8 +1  99 99 99

654 109 = 900 150

(viii)0.783333...0.78 3 =

5.00 73 0.87373.........

2 0.222....= 0 . 2 = 9

726  72 900

(vii) 9.868686...9+0. 86 = 9+

=

0.37  8.56  1.23



(vi) 0.7266....=0.72 6 = =

Ex.46 Find the value of

Sol.

516 43 = 900 75

wwM wa.th Les B aryn

456  4  990 42. Convert to fraction. (i) 0.22222......... (ii) 0.5555 ........ (iii) 0.444747........ (iv) 0.456456456......... (v) 0.57333333.......... (vi) 0.72666666.......... (vii) 9.868686 (viii) 0.783333...... (ix) 0.67777....... (x) 4.717171........

=

573 – 57 900

r

456  4 990

0.57333...=0.57 3 =

eeYa ridn agv .iSn i

0.4565656......... = 0.456 

(v)

R Enak geisnh

the number of of 9's in this group equals the number of digits in the recurring part of the decimal. Impure recurring decimal number convert into fraction

1 1 = 27 3

= 0.333.....  0.3

Ex.48 How many digits are required to write the counting number from 1 to 50 ? Sol. No. × digit = total digit From 1 to 9 =9 × 1 = 9 From 10 to 50 = 41 × 2 = 82 total digit = 9 + 82 = 91 * There are 9 numbers from 1 to 9, and each number has 1 digit hence, the total number of digits are (1×9) = 9. In the same way there are 41 number (50 – 10 + 1) from 10 upto 50, and the number of digits are (41 ×2) = 82.

86

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Digit

No.

2 digit 3 digit 4 digit

Ex. 52 Calculate the number of digits in the prod uct of 8232 × 25348 Sol. 8232 × 25348 (23)232 × (52)348 (By equalising power) (2)696 × (5)696 = (10)696 number of digits = 696 + 1 = 697

v.iSn ir

From 10 to 99 = 90 × 2 = 180 From100 to 999 = 900 × 3 = 2700 From1000 to 8756 = 7757 × 4 = 31028 total digit = 9 + 180 + 2700 + 31028 = 33917 Ex.51Calculate the number of digits in the product of 411111 × 522222 Sol. 4 = 22 = (22)11111 × (5)22222 (am × bm = (ab)m) (By equalising power) = (2)22222 × (5)22222 = (10)22222 So, Number of digits = 22222 + 1 = 22223

eYrai dnag

Hence, The total number of digits (9 + 82) = 91 Ex.49 How many digits are requried to write the counting number from 1 to 672? Sol. No.× digit = total digit From 1 to 9 =9 × 1 = 9 From 10 to 99 = 90 × 2 = 180 From 100 to 672 = 573 × 3 = 1719 total digits s = 1908 Ex.50How many digits are requried to write the counting number from 1 to 8756? Sol. No. × digit = total digit From 1 to 9 = 9 × 1 = 9

SOLUTION Simplify :

5.

1

1

On simplification of 2

2

1 2

1

0.288

4

(a) 1 (b) 4 (c) 5 (d) 6 What is the square root of 0.09 ?

5

6.

(c) 1

2.

11

(b) 1

17 6

(d) 1

17

5 7 21 17

7.

2

Simplify : 1 

3

0.75

3

1 – 0.75

wwM wa.th Les aBryn

1

1

4 5

8.

(a)

7 4

(b)

4

(c)

7

5

7



3.

Simplify :

3 2 9

(a) 4.

(a) 0.3 (b) 0.03 (c) 0.003 (d) 3.0 Find the value of

ERna

(a) 1

1

2

(b) 4



51 5

7

(d)

5

of

of

7

(c) 2

17

5 28 5

–

7

9.

3 2 3

(d)

1

1 3 0 = 11.40 (approximately)

1.3  1300

1

–1

5 

is equal to : (a) 1.6 (b) 0.8 (c) 1.0 (d) 0

1

1

13. When   –

2

 

by   –

 1

5

(d) 0.25 (a) 5

4

38

10.

(b) 36.304 (d) 37.164



1

5

4 3

 9

1 

2

(a) 1.64 (b) 2.64 (c) 1.764 (d) 0.1764 By which s mallest numb er should 5808 be multiplied so that it bec omes a p erfe ct square ? (a) 2 (b) 7 (c) 11 (d) 3 

16

13 = 3.605 (approximately)

 0.013 : (a) 36.164 (c) 37.304



2

1

1002 0.0013  0.00164  30  4

5

1

5

–

–

1

  is divided

6

   , the result is: 18  7

(b) 2

10

1 18

x  441 = 0.02

1

–

 

 0.75  0.75

(a) 4 (b) 1 (c) 2 Find the value of

3

Assume that

find the value of

2

2.644 – 2.356

3

12. When simplified, the expression

kgei snhe

1.

(a) 7

100 49

 3 125 is equal to :

(b) 1

3 4

(c)

7 100

(d)

4 7

11. By which smallest number 1323 must be multiplied, so that it becomes a perfect cube ? (a) 2 (b) 3 (c) 5 (d) 7

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(c) 3

1

(d) 3

6

3 10

14. The square root of (2722–1282) is: (a) 256 (b) 200 (c) 240 (d) 144 15. One-third of the square root of which number is 0.001? (a) 0.0009

(b) 0.000001

(c) 0.00009 (d) None of the above 72.9 16. 3

0.4096

is equal to :

(a) 0.5625

(b) 5.625

(c) 182

(d) 13.6

87

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1

3

17



22

2

7 9

22

22

(b)

5

(c)

5

1

x  1

4

(b) 4

25.

1

2

is :

:

1

2

0.03



1

1 4

43

38

(d)

19

43

1 1   1 1 1 1 1 8 – 3  1 –  1 – –   2   4 2 2 3 6  4 

2

2

(b) 4

21. If 50  *

*

12

1

1

6

(c) 9

1

9

, then the value of *

2

is: (a)

25 2

(b)

4

(c) 4

25

(d) 25

22. Find the sum of the following : 1

1



9

(a)

1



6

1

2

1



12

1



20

1



30

(b) 0 (c)

1



42

1 9

1



56

72

2

 0.21

2

 0.065

(c)

34.

2

3

0.004096 +

0.00004096 up to two place of decimals is: (a) 7.09 (b) 7.10 (c) 7.11 (d) 7.12 30. The least number that must be subtracted from 63522 to make the result a perfect square is : (a) 18 (b) 20 (c) 24 (d) 30 31. By which s mallest numb er should 20184 be multiplied so that it bec omes a p erfe ct square? (a) 2 (b) 3 (c) 5 (d) 6

1

If 2=x + 1

, the n the

1 3

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(d)

17

12 17

20

(d)

101 1



1



42

(a)

2

1

101

10 1 10 0 1



56

90

(b)

10 1

1

1



72

 90

2

(b)

27 5

(d)

27

110

=?

1 9 6 55

35. The value of 1 ÷ [1 + 1 ÷ { 1 + 1 ÷( 1 +1 ÷ 2)}] (a) 1

(b)

5

(c) 2

8

(b)

1 2

36. The simplification of 3.36 – 2.05  1.33 equals :

(a) 2.60

(b) 2 .6 1

(c) 2.64

(d) 2.64

37. (0.2 × 0.2 +0.01) (0.1 × 0.1 + 0.02 )–1 (a)

5 3

(b)

41 12

(c)

41 4

(d)

9 5

38. The value of 5  11  19  29  49

(a) 3 39.

(b) 9

The value of 3

1 4

value of x is :

10 0

30

(c)

4096  64 , then the value of

(d) 1

23. The value of 25–5[2+3 (2–2 (5– 3 )+5)–10 ]÷4 (a) 5 (b) 23.25 (c) 23.75 (d) 25

(a)

2

(b) 10 (d) 0.01

40.96  0.4096 +

32.

13

(c)

        1 1  1   1 –     1   1    10      10     10   10   

10

of 104.04  1.0404  0.010404 is equal to (a) 0.306 (b) 0.0306 (c) 11.122 (d) 11.322

2

(d)

2

21 17

            1 1 1 1 – 1– 1 1– 1   1  1  1  1    10     10      10    10      10  10  10  10 

(a) 0.1 (b) 10 (c) 10 (d) 10 28. If (102)2 = 10404, then the value

29. If

(b)

17

9

2

wwM wa.th Les B aryn 1

(d)

20

2

20. Simplify:

(a) 4

2

     2 2 2 0.003  0.021  0.0065

1

3

43

9

2

(a) 102 (c) 0.1 27. The value of

(d) 6

1

(c)

1

0.1  0.01  0.009 0.01  0.001  0.0009



19

5

(c)

2

19

(b)

(b) 1

26. The value of

(c) 5

43

(a) 1

2

(d)

R Enak geisnh

7

3

(c)

1  1 5 1 1   is equal to –    –    20  4 6 3 2  5 

2

19. Simplify :

5

9

(a) 0

1

(a) 3

5

(b)

1

1

value of 2x +

2

18

33. Simplify:

, the n the

1

1

(a) (d) 1

22

(a)

eeYa ridn agv .iSn i

18. If

12

1

then x  ?

1 2

1

2 –

(a)

1

24. If x =

r

1

17. The value of

(a)

1 3

(b)

(c) 7 7 875

1 15

(c)

(d) 5

is equal to

1 4

(d)

1 5

88

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2

is simplified to

2

 0.39

2

3

3

1 1 1 1 1 1         3.5 5.7 7.9 9.11 11.13 13.15  

(a)

1

(b) 2

3

(c) 6 –

41.

1

(d) None of these 2

4

1

1

3

–  –   2 3 5 3 5 4 1 2 4 1 1 4  –  – – 2 3 3 3 5 5

(a) –

10

is simplified to

(b) –

3

3 10

(c) 1 (d) – 2 42. The simplification of



y ield s



0.63  0.37  0.80

the

2

(b)

45

2 25

(c)

45

x

(b) 1.81

(c) 1.79

(d) 1.80

(a) 0.000196 (c) 0.0196

43. The square root of 0.4 is :

(c) 0.7

(d) 0.9

(a) 1010 (c) 101

7  3 57 – 3 5 (b)

5

52.

to:

13

48

(a)

(c) 3

(d) 4

38

(b)

109

109

(c) 1 (d)

38

116 109

8  57  38  108  169  ?

(a) 4 (b) 6 (c) 8 (d) 10 56. If the number p is 5 more than q and the sum of the squares of p and q is 55, then the product of p and q is (a) 10 (b) – 10 (c) 15 (d) – 15 2 1 – 7 2 1 1 3 1 2 7 4

57.

(b) 0.00196 (d) 0.196

÷

1

is equal to

1

2

1

2

(a) 1

(b)

1 2

1 5

(c)2

(d)

58. If

1 3

part of a th

          5   4 –   1  1   1    3 1   2    4  journey takes ten minutes,

(b) 110 (d) 100

100  99 98 ... 3 21

100

3

then to complete

99  98  97  ..  3  2  1

(b) –

(b) 2

55.

is equal to

(c) 3 5 (d) 2

 11 15 2  11 15 2 45.  4    – 4 –   is equal  15 71  15 71 (a) 1

(a)

100 – 1100 – 2100 – 3...100 – 200

wwM wa.th Les aBryn

44. The square root of

1 3

5–

2 2  0.1 – 0.01    + 1 is equal to 51.  0.0001   

ERna

(b) 0.6

1 3

15

kgei snhe

(a) 1.80

(a) 4

11 1

3 

50. If 2 0.014  0.14x =0.014× 0.14 2 y . find the value of y ,

(a) 0.8

(d)

2

48. The number of digits in the square root of 625686734489 is (a) 4 (b) 5 (c) 6 (d) 7 49. There are some boys and girls in a room. The square of the number of the girls is less than the square of the number of boys by 28. If there were two more girls, the number of boys would have been the same as that of the girls. The total number of the boys and girls in the room are (a) 56 (b) 14 (c) 10 (d) 7

result is:

46.

7

2

–1

3

3

(a)

1 3

54. The value of

is equal to

2

2

v.iSn ir

2

 47.  

eYrai dnag

40.

th

of that

5

journey. it will take (a) 80 minutes (b) 50 minutes (c) 48 minutes (d) 60 minutes

1

99  98  97  ...  3  2  1

(c) 0

59.

is equal to

(d) –

(b)

1

3

1

1

2

1

1

1

1

1

1 8

1

1

(a)

4

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21 13

1

(b)

17 3

(c)

34 21

1

2

1

2

(d)

1 5

1

(a) 1

8 5

(b) 4

(c) 3

(d) 2

2 3

2

1

1

1

1 1

4 1

5–

1

3

1

is equal to

1

1

1

1

(d)

1

3

1

–2

3 1 2 7

1

53. The value of 1 

1 7 1

99  98  97  ...  3  2  1

8

16

(c)

4

1

1

(a)

2

60.

1 1  a –b b –a 1 2 1 2

(a) a – b (c) 1

is equal to

(b) b – a (d) 0

89

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61. Find the sum of 1   2   3  n    1 –   1 –   1 –   .....  1 –  n  1  n  1  n  1 n  1  

(d)

  (c)   7 – 2 5 9 4

(b)

3 2

1 2

n  1

  (d)  2 7 – 5 



a  b –

13

b – a

1 2



(a) 2 (b) 2 (c) 4 (d) –2 66. The simplified value of

1. 2. 3. 4. 5. 6. 7. 8.

4 3 6 2

+

(a) (a) (c) (c) (c) (a) (a) (d)

9. 10. 11. 12. 13. 14. 15. 16.

(d) (b) (d) (a) (a) (c) (d) (b)

a b

is close

is equal to

(a) 27.8

(b) 27.7

(c) 27.08

(d) 27.07

71. Find the value of 6. 74 +7. 32

25. 26. 27. 28. 29. 30. 31. 32.

(a) (b) (b) (d) (c) (a) (d) (b)

Rakesh Yadav Readers Publication Pvt. Ltd.

0. 9

(c)

0.0 9

(d)

0. 09

(a)

1 1000

(b)

1 999

(c)

1 99

(d)

1 9

(a)

127 100

(b)

73 100

(c)

14 11

(d)

11 14

(a)

8. 912

(b)

8.9 12

(c)

8.97 9

(d)

8.9 79

76. The difference of 5. 76 & 2. 3

(b) 14.07

(a)

2. 54

(b)

3. 73

(c)

3. 46

(d)

3. 43

77. The value of (0. 63 + 0. 37 )

0.9, 0. 9 , 0.0 9 , 0. 09

(d) (c) (d) (a) (d) (a) (c) (b)

(b)

75. 8.3 1 +0. 6 +0.00 2 is equal to

(c) 13.06 (d) 13.07 72. Which of the following number greatest of all ?

3 2

0.9

p 74. 1. 27 in the form is equal to q

5. 6 + 7. 3 +8. 7 +6. 1

(a) 14.06

6

17. 18. 19. 20. 21. 22. 23. 24.

4  0.4

(a) 0.0025 (b) 0.025 (c) 0.25 (d) 0.00025 69. The value of (1001)3 is (a) 1003003001 (b) 100303001 (c) 100300301 (d) 103003001 70. Find the Value of

wwM wa.th Les B aryn

3 6

–

is

4 – 0.04

0.05  0.5  a = 0.5 × 0.05 × b , then

(d) 16

2

1/2

3 2

68. If

(a)

73. 0. 001 is equal to

to (a) 0.4 (b) 0.8 (c) 1.0 (d) 1.4

72 5

64. If x [–2{–4(–a)}]+5 [–2{–2 (–a)}] =4a, then x = ? (a) – 2 (b) – 3 (c) – 4 (d) – 5 65. If a = 64 and b =289, then the value of

2

67. The value of

(b) 

(c)

(d) 0

2

63. What number must be added to the expression 16a2 –12a to make it a perfect square ? (a)

3 –

n

62. The square root of 33 – 4 3 5 is : (a)  2 7  5

(c)

2

r

(c) (n+1)

1

(b)

eeYa ridn agv .iSn i

(b)

2

R Enak geisnh

(a) n

1

(a)

(a)

1

(b)

100 99

(c)

99 100

(d)

100 33

ANSWER KEY 33. 34. 35. 36. 37. 38. 39. 40.

(c) (d) (b) (d) (a) (a) (d) (d)

41. 42. 43. 44. 45. 46. 47. 48.

(b) (b) (b) (d) (d) (d) (d) (c)

49. 50. 51. 52. 53. 54. 55. 56.

(b) (b) (d) (c) (c) (a) (a) (c)

57. 58. 59. 60. 61. 62. 63. 64.

(c) (c) (a) (c) (b) (d) (a) (b)

65. 66. 67. 68. 69. 70. 71. 72.

(a) (d) (b) (b) (a) (a) (b) (b)

73. 74. 75. 76. 77.

(b) (c) (c) (d) (b)

90

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SOLUTION According to the question, 1

 2

1

3

2

1

1 2 2

1

–



3 51 5 3 2 5 28 2  × – 9 7 5 3 5

 1

17

×

8. (d)

3

7

1

7

 – 3 15 3 2 2 4– 9 3



1

9



9 8–6

4



2



18  15 9

130

11.4 10

18

5. (c)

33 1

1

 1

33  18

33 51





51

 1

2



3

1

1

4

1



35

54

5

6. (a)



1

1

15

1



3. (c)

24  18 24

9  15



7

of

x

=

21

x =

42



17

–

2

100

42

100

Squaring both sides.

24



4



3 51 5 3 2 5 28 2  of – 9 7 5 3

0.288

Rakesh Yadav Readers Publication Pvt. Ltd.

10. (b) According to the question, 38

=5





According to the question,



3

3

0.75 

1 – 0.75



3 3 1 –  0.75  1 – 0.75

a³–b³ =  a – b

0.75 

 a² +b² + ab

3  1 –  0.75  0.25



100

2



4 2 4 7 4

 3 125

49

16

According to the question,

3



Factors are:

a²–b² =  a+b  a–b 



5  0.288



1

According to the question,

3 5808 2 1936 2 968 2 484 2 242 11 121 11 11 1

0.288

2  0.75   0.75   1   1 – 0.75 

7

9. (d)

Smallest number is = 3

2.644  2.356  2.644 – 2.356

 0.75 

According to the question, 5

= 0.02

441

3, 2, 2, 2, 2, 11, 11

18

9



2

Square root of 0.09 = 0.09 = 0.3

7. (a)

2

–  2.356 

17

2

1

2

11

According to the question,

1

11.4

0.288



84

51

2. (a)

 36.05 

According to the question,

wwM wa.th Les aBryn

33

51  33

10000

100

2.644



130

 1300 

100

ERna

1



 1.14 + 36.05 + 0.114  37.304

1

1

2

2

kgei snhe





3

x



 x = 0.1764

1 1

–

3 2

1.3  1300  0.013

35 9

 1

 4

According to the question,

4. (c) According to the question,

2 2

–

9 8



9

1

0.25

1

7–3

54 5

 1

1

x  441 = 0.02

4 5

 1

7



v.iSn ir

1

5

eYrai dnag

1. (a)

10

5

7 

7

5

10

=1

3 4

11. (d) According to the question, 3 3 3 7 7

1323 441 147 49 7 1

Factors are: 3, 3, 3, 7, 7, 7

 Smallest number is = 7

91

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12. (a) According to the question,

 10 × 0.1 – 0.2 + 2

 1–  

4 5



10

729

9

17. (d) According to the question, 1

–

2 2

–

5

19

7

1

19

4 5

1



5 3



9

–

3

1







7

3 

17

19





17 60



90 5

272 

5

= 5

19





– 128 

2

If

17





a²–b² =  a+b  a–b 

1

1

0.4096 3

729  10000 4096  10

1

1



  



13 5 1  9  2  1          4 4 2  6  



13 5 1   4  4  2  1    



13  4



13  4



2

2 17

1

1

1



3



2

5  2 4



3 4 

5

 x=



5 8



13 8

 Value of 2x +

Rakesh Yadav Readers Publication Pvt. Ltd.

13 5 1  3 1 1          4 4 2  2 3 6 

2

17

2 3

x= 1

 2×

43

1

1

x = 0.003

72.9

2

17





1

x 1

3

3

17



17

x =0.001

38

1  1  1 1  1 1 1   3  1  1    2  4  4 2  2 3 6   

2

x =

Squaring both sides x = 0.000009  Number is = 0.000009 16. (b) According to the question,

8

1

1

400  144

1



=1



1

1





43

20. (a) According to the question,

22

x = 1

 20×12 = 240 15. (d) According to the question,





22

22

 272  128   272 – 128 



19

22

18. (c) According to the question,

1 10

wwM wa.th Les B aryn





15  7

22

14. (c) According to the question, 2

 

90

17

R Enak geisnh



60 36 – 50  54 – 35

5

5

2

43

30 – 15  12 – 10

1



19

22

5

18

4

5

43

6 7

–

5

7

1

4

1

2

22 9

1

1



17



1 2 –

1 1

43

1

1 3



3

1 2

1 9



=5



43

22

2 –

5



17



1

3 

= 1.6

1

4

7 4

19. (d) According to the question,

5

5

20

16



+

4



 10

5 –1 4

8

13

4096

13. (a) According to the question,





 1000

 5.625

4



3

r

4  10 × 0.1 – 0.2 × 1 + 5



eeYa ridn agv .iSn i

–1 1 1 1 5 1002  0.001 3 – 0.0016 4  30     4

13 8

+

7 4

7 4

 is

17



2

16 4

34  16 4 9

=4

2

=

18 4

1 2

21. (d) According to the question, 

50 x



x 12

1 2

92

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 x²



2

 x² = 25 × 25  x = 25





7



8

1



8

9

23. (c) According to the question.  25–5 [2 +3 (2 –2 (5–3 )+5)–1 0]÷4

 25 – 5 [2 + 3 (2 – 4 + 5) – 10] ÷ 4  25 – 5 [2+3 × 3– 10] ÷ 4  25 – 5 [11– 10] ÷ 4

100 = 10

13

0.000009  0.000441  0.00004225 0.049225



100 = 10



0.00049225

28. (d) According to the question,



104.04  1.0404  0.010404

10404



4

 

1

0.0009  0.0441  0.004225

100  5



100

5

 25 –

0.00010181

(0.003)²  (0.021)²  (0.0065)²



 25–5 [2 +3 (2 –2 ×2 +5 )–1 0 ]÷4



102 10

10404



10000



102

100

1

24. (b) x 

2

1

=

2 5

1



=

x

5 2

25. (a) According to the question

1   5

20

1  5   4  6



1

3



1 

 2 

 1  1  5 5    20   5   4   6  6       9

9





20





1 5 

20

1  5

9

9

9





20



17

21

x=

17

10404

                1 1 1  1   1  1 1  1         1 1 1 1    10      10      10      10     10   10   10   10   

1000000

        1 1 1    1 –      1 1   10     10       10   10   

÷





64

10



64

100



64



1000

20

=0

64

1

1

10000

b

101

10

 

a ²  b²

[ a ²  b ²  (a  b )(a  b )]

a b

(a  b )(a  b ) a b

 (a – b)  

111



101

91 101

20 101

34. (d) According to the Question

Factors are 2, 2 3 , 2 , 29, 29  It should be multiplied by = 6 32. (b) According to the question, 1

If 2  x 

Rakesh Yadav Readers Publication Pvt. Ltd.

a

101

91



1

10 

4 20184 3 5046 2 1682 29 841 29 29 1

(0.01)²  (0.001)²  (0.0009)²

111

10

30. (a) According to the question, As we know that the square of 252 is which is near the value of 63522  63522 – x = 63504 x = 18 31. (d) According to the question,

(0.1)²  (0.01)²  (0.009)²



1

10 

 6.4+0.64+0.064+0.0064 = 7.11

1 4 

1

Let 1 

4096 4096 4096 4096    100 10000 1000000 100000000

1     0 4 

26. (b) According to the question,



17

33. (c) According to the question

1000

wwM wa.th Les aBryn

2



13

34  13

 x 

40.96  0.4096  0.004096  0.00004096



9

17

 x 2

102



ERna

= 23.75

4

2x



29. (c) According to the question,



4

 10.2+1.02+0.102 = 11.322

4 95

1

2  x 

(0.03)²  (0.21)²  (0.065)²



1  2

1



27. (b) According to the question,

1 1 1 1 1 1 1 1 1 1 1           9 2 3 3 4 4 5 5 6 6 7 1

0.010181

v.iSn ir



1

12  1

eYrai dnag



4

1

13

1 1 1 1 1 1 1 1        9 6 12 20 30 42 56 72 2 33 4 4 5 5 6 6 7 7 8 8 9

1

 2x 

0.0001  0.000001  0.00000081



22. (a) According to the question,



0.01  0.0001  0.000081

kgei snhe

25

 50 

1



1 3



1 4



1

1



30

42



1 56



1



72

1 90

1



110

1 1 1 1 1 1 1 1 1 1 1 1            5 6 6 7 7 8 8 9 9 10 10 11 1 5



1 11



11  5 55



6 55

93

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



3



3



 1 

1 

1

 

 1  1  1 

  1  1  1   

3 



2 

7

1 125

 

1 11



3 

3

1

2 22 2  0.39 39





200

 0.2  0.2  0.01 0.1  0.1  0.021 0.2  0.2  0.01 0.1  0.1  0.02

222

100

wwM wa.th Les B aryn

60 15  20 – 40  10 – 6 – 24 30



15



–30

60

25



–

2 15 



71 

15

 

2  4

 

11  

15   2   15   71 

71

2 

 2

15 

  4

71 

15

1

1



1

3

1

2

1

3

1

1

1

4 9

4

1



9

3

13



48

13

Satisfied

5  11  19 

29  7

 

36

0.63  0.37  0.80

0.6363...+0.3737....+0.8080...

43. (b) According to the question,



47. (d) According to the question,  1

1

1

1

1

1



  3.5  5.7  7.9  9.11  11.13  13.15 

–3 10

 1.81



11

1

42. (b) According to the question,

5  11  19 

 

– 4

11 15 11 15  11 15 11 15    4 – 4  – 4   15 71 15 71 15 71 15 71

111

2 4 1 1 3 –  –   2 3 5 3 5 4  1 2 4 1 1 4  –  – – 2 3 3 3 5 5 –

3

5  11  19  29  49



15



5

38. (a) According to the question,



 71 

41. (b) According to the question,



0.01  0.02

=

2 15 



–30 – 40  48 – 20  12  45

0.04  0.01

0.03

100 

11

46. (d) According to the question,

22

R Enak geisnh

2



37. (a) According to the question,



6 × 0.39 11

2

 2.64

0.05

 

 4

2

2

1.333333...  2.646464...



2 3

2



3.36 – 2.05  1.33



× 0.39

2 3

 3.363636 ...– 2.050505... +

 

4

2

2

5

8  5   5  8 36. (d) According to question,



45. (d) According to the question,

5

2

3 

4 =2



49 – 45

40. (d) According to the question,

2 

 3  1  1  5   



1



3



7  3 5 7 – 3 5 



875

 1     1  1  1  1  1   1  2      



44. (d) According to the question,

r



1  1  1  1  1  1  1  2

39. (d) According to the question,

eeYa ridn agv .iSn i

35. (b) According to the question,

0.4

1 1

1

1

1

1

1

1

1

1

1

1

1



 2  3 – 5  5 – 7  7 – 9  9 – 11  11 – 13  13 – 15  



1 1 2  3 1 2



–

4 15

  15  1



1  5 – 1 2  15



2 15

48. (c) According to the question,

5  11  19  6



5  11  25



5  11  5



5  16



9 =3





4 9

2 3

 0.66666.....  0.6

Rakesh Yadav Readers Publication Pvt. Ltd.



625686734489 =6

NOTE: For counting the digits of square root we make pairs first. Then the digits will be equal to number of pairs. 49. (b) Let the number of boys = x the number of girls = y

94

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Second part

According to the question,

 x² – y² = 28 .........(i)

1

 1

1

1

 x = y+2

1

1

1

 x – y = 2 ..........(ii)

1

1

 1

1

1 3

1

2

5

1

2

1

2

8

5

1 

1

5–

1

5

1

 (x + y)×2 = 28

50. (b) According to the question,

–1

3

=

11 1

3

2 0.014  0.14x =0.014×0.14 2 y

3

1

77 – 39 33 10 3 33



38 33  33 109



= 0.014 × 0.14

53 130

=

=

53

53



65

130

53

130

=



65

53

y

= 0.00196

55. (a) According to the question

51. (d) According to the question

1

3

0.1  0.010.1 – 0.01   +1    0.0001  0.11  0.09 0.0001



52. (c) According to the question



99  98  97  .......1  0  –1  –2 ....... –100 100  99  98  .....3  2 1

=0

88



16

8  64

=4 ......(i) .......(ii)

2pq = 30  pq = 15

4

1

1 1

2 3

Rakesh Yadav Readers Publication Pvt. Ltd.

–

1

30

1

5  31

= 4–

40

1

4

31

9

9 5

=4–

40 31

=4–

31

=

8

1 8

According to question 1 8

part = 10 minutes

1 part = 80 minutes 3 5

part = 80 ×

3 5

= 48 minutes

59. (a) Take first part

60 – 7

– 53 14 2 7 2 = = = 49  16 65 1 7 8 3 1  14 2 7 2 7 7 1

1 1

2

1

1

31

 (p–q)² = p² + q² – 2pq (5)² = 55 – 2pq 25 = 55 – 2pq

1

1



9

5

= 4–

1

5

= 4–

57.(c) Take a first part

1

5

4 –

3

p–q=5 p² + q² = 55

53. (c) According to the question  1

=

56. (c) According to the question

100  99 98 ....... 3 21

9

1

8  57  7



 100 –1100 – 2100 – 3............100 – 200

1

3

8  57  49



 99 + 1 = 100

1

1

8  57  38  11



+1

5

 4

4

8  57  38  121



1 4

8  57  38  108  13



wwM wa.th Les aBryn

0.1 ² – 0.01 ²   +1    0.0001 



2

8  57  38  108  169



=2

65

1

1

 109

ERna

x

130

5

58. (c) 4 –

38



1 24 2 53

According to question,

=

kgei snhe

y



3

 0.014×0.14x = (0.014)² ×

x

7 13 – 3 11 1 3 3 3 10

1

3

Squaring both sides (0.14)² × y

2

1 130 53

eYrai dnag

1

=

1 2 53 24

5 24

2

21

1

=

1

2

 21

1

=

=

2



 1

34

54. (a) According to the question

 x + y = 14



8 13

 (x + y) (x – y) = 28

13

v.iSn ir

 1

24

5

From eq. (i)

 x² – y² = 28

1

2

1

4

= 3

1 7 1 2

– 2 1

29

1 4 1 7

=

7 7 2

– 

9 4 8 7

95

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62. (d) 33 – 4 3 5 =

28

= 49  16

53

=

28

14

×

=

65

=

33 – 2 × 2 35

13 0

=

33 – 2 × 2 ×

=

28 + 5 – 2 × 2 7  5

Take second part 1

1

33 – 4 35 = 

1

=

=

1

2

2

5

2

53

24 1

=

24

2

=



  2 7 5

1

=

1 06  24

53

53 13 0

53

53 

130

60. (c)

=

=

130 53

1 1  a –b – a –b  1 2 1 2  a –b 1 2  a –b a –b 1 2 2 1

=

  2    +  1– +  n 1  n   .........+  1 –   n  1

=

= 1 =1

 130



1



n  1

a–b 1 2 a–b 1 2

=1

 3    1– +..  n 1

n 1–1 n 1–2 n 1– 3      +   +    ..........  n 1   n 1   n 1 

=

n 1 1

n 1



n –1 n 1



n –2 n 1

.......+





=

+

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3 2

3 6

3 2

–

a

5 5

10  100

a

1 

b

40

= 0.025

69. (a) (1001)3 = 1001 × 1001 × 1001 = 1002001 × 1001

6 3 7 1 +7+ +8+ +6+ 9 9 9 9

 6  3  7  1  = 26 +   9  

1 2



= 26 +

1 2

17 9

= 26 + 1 +

6 2

6– 3

–

3– 2

+

6 3 2

8 8 = 27 = 27. 8 9 9

4 3

6– 2 

6 2

6– 2

= 6+

74 32 +7+ 99 99

=13+

74  32 106 =13+ 99 99

=13+1+

 3 2

= 0.8



b

= 5+

= 5 31/2 = 2

4 3

6– 3

6

23

71. (b) 6. 74 +7. 32



6 3

19

a = 0.5 × 0.05 × b



25 – 9

46

=

70. (a). 5. 6 + 7. 3 + 8. 7 +6. 1

1 8  17 – 17 – 8 2 1 2

38

0.05  0.5  a = 0.5×0.05× b

b = 17

b – a

4.6

=

= 1003003001

8a

a  b –

3.8

0.05×0.5×a = 0.5 × 0.5 × 0.05 × 0.05 × b

n 1

(n+(n–1)+(n – 2)...........1)

 n n  1  n    n 1  2  2 1

1



66. (d)

 n n  1    1+2+3.....n =  2    





=

Squaring both sides

x=–3 65. (a) a = 64 , b = 289 a =8,

4  0.6

68. (b)

24a

x= –

=

=



4  0.4

4 – 0.2

=

64. (b) x [–2{–4 (–a)}]+5 [–2{–2(–a)}] = 4a x [– 2{4a}] + 5[– 2{2a}] = 4a x [– 8a] + 5[– 4a] = 4a – 8ax – 20a = 4a – 8ax = 24a

=

n  1 – n   +     n 1  n



 3 2 9 Number be added =     = 2  4

1 1  a –b b –a 1 2 1 2

61. (b)  1–



 3 2 3 = (4a)² – 2×4a× +     2 2 

wwM wa.th Les B aryn

=

130

53

4 – 0.04

2

2 7 5

53

According to the question

= 2 3 – 6 –3 2+ 6 +3 2–2 3 =0 67. (b)

63. (a) (a – b)² = a² + b² – 2ab 16a²–12a

24 1

1

– 22 7  5

1

 6 – 2

 3 – 2

6

+

2

5

5

4 3

4

eeYa ridn agv .iSn i

5–

=

25 – 1

5

R Enak geisnh

2

1

2

1

7

2 7   5 2 2 7  5



1

2

1

2

2

1

=

 6 – 3 – 3

53

14

3 2

r

116 – 63

3– 2

7 7 =14 99 99

= 14. 07

96

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0. 9 = 0.999 ........ 0.0 9 = 0.09999....... 0. 09 = 0.090909........ 0.9 is the greatest of all

73. (b) 0. 001 =

1 999

3 14 = 11 11

=5+

75. (c) 8.3 1 + 0. 6 +0.00 2

xx 8. 31 0. 66 0. 00 8.97

x 1 6 2 9

xx 11 66 22 99

x 1 6 2 9

=8.979999 = 8.979

 76 3  =3+   99  9   

= 3. 43 77. (b) 0. 63 +0. 37

76. (d) (5. 76 ) – (2. 3 )  76   5  99   

76 3 -299 9

=

 3  2  9   

63 37 100 + = 99 99 99

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ERna

kgei snhe

eYrai dnag

27 74. (c) 1. 27 = 1 + 99

=1+

v.iSn ir

72. (b) 0.9 = 0.9

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CHAPTER

09

LINER EQUATIONS IN TWO VARIABLES

(iii)

3x  7y  2

1. 2.

3.

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The condition a  0, b  0, is often denoted by a2 + b2  0 Note: The graph of a linear equation ax + by + c = 0, is a straight line. Solution of linear equation : Any pair of values of x and y which satisfy the equation ax + by + c = 0, is called its solution. E.g.: show that x = 2 and y = 1 is a solution of 2x + 5y = 9 Sol: Substituting x = 2 and y = 1 in the given equation, we get LHS = 2  2 + 5  1 = 9 = RHS

If 11x -13 = -2x + 78, then x =? (a) 7 (b) 8 (c) 6 (d) 4 If 2x + 3y = 29 and y = x + 3, what is the value of x ? (a) 5 (b) 6 (c) 4 (d) 7 If 2x + 3y = 5 and x = -2, then the value of y is :

1 (b) 3 (c) 1 (d) 9 3 The value of x + y in the solution x y 5 + = of the equations and 4 3 12 x + y =1 2 1 5 3 (a) (b) 2 (c) (d) 2 2 2

(i)

5.

6.

(a) 4.

(ii)

an infinite number of solutions, if

a1 b1 c 1   a 2 b2 c 2

r

 x =2 , y = 1 is a solution of 2x+5y =9 • System of Linear Equatio ns : Consistent System :- A system consisting of two simultaneous linear equations is said to be consistent, if it has at least one solutions. Inco ns is tent Sys tem : A system consisting of two simultaneous linear equations is said to be inconsistent, if it has no solution at all. E.g.: Consider the system of equations: x + y = 9 & 3x + 3y = 5. Clearly, there are no values of x and y whic h m ay simulatneously satisfy the given equations. So, the system given above is inconsistent. Conditions for Solvability : The system of equations a1x + b1y + c1 = 0, a2x + b2y + c2 = 0 has :

geisnh eeYa ridna gv.i Sni

Linear Equations in Two Variables: An equation of the form ax + by + c = 0 where a, b, c  R(real numbers) and a  0, b0 and x, y are variables is called a linear equation in two variables. Examples : Each of the following equations is a linear equation : (i) 4x + 7y = 13 (ii) 2x - 5y = 36

R Enak

•

7.

a unique solution, if

a1 b1  a 2 b2

(i)

(ii)

•

(i)

(ii) (iii)

no solution, if

a1 b1 c1   a 2 b2 c 2

Homogenous System of Equations: The system of equations a1x + b1y = 0; a2x + b2y = 0 has only solution x = 0, y = 0, when a1 b1  a 2 b2 an infinite number of solutions a1 b1  when a 2 b2 The graphs of a1x + b1y + c1 = 0, a2x + b2y + C2 = 0 will be : Parallel, if the system has no Solution ; Coincident, if the system has infinite number of solutions ; Intersecting, if the system has a unique solution.

EXERCISE

If 2x + 3y = 12 and 3x - 2y = 5, then x and y must have the values : (a) 2 and 3 (b) 2 and -3 (c) 3 and -2 (d) 3 and 2 The equations ax + b = 0 and cx + d = 0 are consistent, if : (a) ad = bc (b) ad + bc = 0 (c) ab - cd = 0 (d) ab + cd = 0 The equations 2x + y = 5 and x + 2y = 4 are (a) consistent and have infinitely many solutions (b) consistent and have a unique solution. (c) inconsistent (d) none of these

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(iii)

8.

The cost of 2 sarees and 4 shirts is Rs. 16000 while 1 saree and 6 shirts cost the same. The cost of 12 shirts is : (a) Rs. 12,000 (b) Rs. 24,000 (c) Rs. 48,000 (d) Can’t be determined 9. The system of equations kx - y = 2 and 6x - 2y = 3 has a unique solution when : (a) K = 0 (b) K  0 (c) K = 3 (d) K  3 10. The value of y in the solution of the equation 2x+y = 2x-y  8 is : (a) 0

(b)

1 4

(c)

1 2

(d)

3 4

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3x  y 1 2x  y  2 3x  2y 1   is 3 5 6 (a) x = 2, y = 1 (b) x = 1, y = 1 (c) x = -1, y = -1 (d) x = 1, y = 2 12. If x + 2y  3, x > 0 and y > 0, then one of the solutions is : (a) x = -1, y = 2 (b) x = 2, y = 1 (c) x = 1, y = 1 (d) x = 0, y = 0 13. A purse contains 25 paise and 10 paise coins. The total amount in the purse is ` 8.25. If the number of 25 paise coins is one- third of the number of 10 paise coins in the purse, then the total number of coins in the purse: (a) 30 (b) 40 (c) 45 (d) 60

14. The value of k for which the system of equations x + 2y = 5, 3x + ky + 15 = 0 has no solution, is: (a) 6 (b) -6 (c) 2 (d) 4 15. The equations 2x - 5y = 9 and 8x - 20y = 36 have : (a) no common solution (b) exactly one common solution (c) exactly two common solutions (d) more than two common solutions 16. The difference between two numbers is 5 and the difference between their squares is 65. The larger number is : (a) 9 (b) 10 (c) 11 (d) 12

17. The number of solutions of the equations x 

1  2 and 2xy - 3y y

= -2 is : (a) 0 (b) 1 (c) 2 (d) None of these 18. If 2a + 3b = 17 and 2a+2 - 3b+1 = 5, then: (a) a = 2, b = 3 (b) a= -2, b = 3 (c) a = 2, b = -3(d) a = 3, b = 2 19. The solution to the system of equations |x + y|= 1 and x - y = 0 is given by:

kgei snhe eYari dnag v.iSn ir

11. The solutions of the equations

(a)

x y 

1 2

(b) x  y  

1 2

1 1 or x  y   2 2 (d) x = 1, y = 0 (c)

x y 

ANSWER KEY 1. 2.

(a) (c)

3. 4.

(b) (d)

5. 6.

(d) (a)

7. 8.

(b) (b)

9. (d) 10. (a)

11. (b) 12. (c)

13. (d) 14. (a)

15. (d) 16. (a)

17. (d) 18. (d)

19. (c)

1. (a) 11x - 13 = -2x + 78  11x + 2x = 78 + 13

1 3 x  y 1  2 2 5.(d) 2x + 3y = 12 (i) 3x - 2y = 5 (ii) (i)  2 + (ii)  3, we get ; x = 3 putting x = 3 in (i), we get 2  3+3y = 12  3y = 6  y = 2  x = 3 and y = 2 6.(a) The equations are consistent if

wwM wa. th Les aBryn

 13x = 91

ERna

SOLUTION

91 7 13 2.(c) Putting y = x + 3 in 2x + 3y = 29, we get, 2x + 3(x + 3) = 29  2x + 3x + 9 = 29

 x

20 4 5 3.(b) Putting x = -2 in 2x + 3y = 5, we get ; -4 + 3y = 5  3y = 5 + 4 = 9

 5x = 29 - 9 = 20  x 

9 y  3 3 4.(d) Given equations are : 3x + 4y = 5 (i) and x + 2y = 2 (ii) (i) - 2  (ii): x = 5 - 4 = 1  from (ii) 2y = 2 - x = 2 - 1 =1 1 y  2

a b  c d i.e. ad = bc 7.(b) 2x + y = 5 (i) x + 2y = 4 (ii) On solving we get, x = 2, y = 1 Thus (b) is true 8.(b) Let cost of 1 saree = Rs. x & cost of 1 shirt = Rs. y ........(i)  2x + 4y = 16000 and x + 6y = 16000 .......(ii) Multiplying (ii) by 2 and substracting (i) from it, we get, 8y = 16000  y = 2000

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 cost of 12 shirts = (Rs.2000  12) = Rs. 24000 9.(d) For a unique solution, we must have

a1 b1  a 2 b2

 10.(a)

k 1  1   k   6   k  3 6 2  2

2x+y = 2x-y =

x+y=

8 =2

3/2

3 2

and x - y =

.......(i)

3 2

........ (ii)

(i) – (ii) 2y = 0  y = 0 11.(b)

3x - y +1 2x + y + 2 = 3 5

 5(3x - y +1) = 3(2x + y +2)

99

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= 2(3x - y + 1) = (3x + 2y + 1)

 3x - 4y = -1 ........(ii) (i) - 2  (ii):(9 - 6)x - 8y + 8y = 1 - (-2)  x =1 putting x = 1 in (i) we get, 9  1 - 8y =1  8y = 8  y = 1 x =1, y = 1.  12. (c) Here we will go through options. in option (a) x < 0 and in option (d) x = 0 hence (a) and (d) can’t be the required answer because both does not satisfy the given condition i.e. x > 0. Now option (b) x = 2, y = 1, then x + 2y = 2 + 2(1) = 4 which is > 3 clearly, values of option (b) do not satisfy x + 2y  3

25 10 x y  8.25 100 100



putting y = 3x in (i), we get : 5x + 6x = 165  11x = 165  x = 15  from (ii), y = 3x = 3  15 = 45  Total number of coins in the purse = x + y = 15 + 45 = 60 14.(a) a1x + b1y + c1 = 0 a2x + b2y + c2 = 0 will have no solution if

a1 b1 c 1 1 2     k  6 a 2 b2 c 2 3 k

15.(d) The given equations are 2x - 5y = 9 and 8x - 20y = 36  2x - 5y =9 Thus, there is one equation in two variables. So, the given equations have an infinite number of solutions. 16.(a) Let the numbers be x and y. Then, x - y = 5 and x2 - y2 = 65 

x 2  y 2 65   x  y  13 x y 5

solving x - y = 5 and x + y = 13, we get; x = 9 and y = 4  larger number = 9 17.(d)

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option (c) x = 1, y = 1, then x + 2y = 1+ 2 = 3  3 So, x =1, y = 1 is one of the solutions. 13. (d) Let the number of 25 paise coins be x & that of 10 paise coins be y, then:

5x + 2y = 165

x

(i)

 y(2x - 3) = -2 ......(ii) (ii)

1 1 1  2  2x y  ........( i ) y y 2x and 2xy - 3y = -2

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putting y =

1 in (ii) 2x

2x  3  2  2x - 3 = -4 + 2x. this 2x gives 1=0 This is impossible So, there is no solution. 18.(b) 2a + 3b = 17 and 2a+2 - 3b+1 = 5  22.2a - 3.3b = 5  4.2a - 3.3b = 5 let 2a = x & 3b = y then x + y = 17 ...........(i) 4x - 3y = 5 ............(ii) 3  (i) + (ii), we get 7x = 56  x = 8  2a = 8 = 23 

r

3x – y  1 3x  2y  1 = 3 6

1 and x  y  y  3x 3

geisnh eeYa ridna gv.i Sni

and

.........(i)

R Enak

9x - 8y = 1



a=3

putting x = 8 in (i), we get y = 17 - 8 = 9  3b = 9 = 32 

b=2

 a = 3 and b = 2. 19.(c) Note that |a| = 1 means a = 1 or a = -1 So, |x + y|=1  x + y = 1 or -(x + y) =1 (x + y) = – 1  solving x + y = 1, x - y = 0, we get

1 1 and y = 2 2 solving x + y = – 1, x – y = 0, we get x = –1/2 and y = – 1/2 x=

 x=y= 

1 2

100

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CHAPTER

10

POLYNOMIALS

(ii)

(iii)

(iv)

3 2 u - 8u + 7 is a poly7 nomial in u of degree 3.

3u3 +

5t4 –

2 3 3 t  3t2  is a poly7 8

Nature of Roots The value of x at which value of equation will be zero. 1. Roots are imaginary : b² – 4ac  0 2. Roots are real: b² – 4ac  0 b² –4ac  0 different  b²–4ac>0

 & , then

–b  b² – 4ac & = 2a



1 5 etc. x 5 , , 2 x  3 x 3x 1

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

are not polynomials.

• (1)

Polynomials of Various Degrees : Linear Polynomial : A polynomial of degree 1 is called a linear polynomial. A linear polynomial is of the form p(x) = ax + b, where a  0 e.g.

3x  7 , 

(2)

 5  etc. 2



2

 7y 

–b – b² – 4ac 2a

Sum of root:–b a Product of root:

 =

 =

c a

then, ax² + bx + c = 0 can be written as:



E.g. (2x +7x - 9), 3x  2x  7 ,

y

=

7  2x  5 ,  x   etc. 3 

Quadratic Polynomial : A polynomial of degree 2 is called a quadratic polynomial. It is of the form p(x) = ax2 + bx + c, where a  0 2

equal  b²–4ac=0

rational irrational b²–4ac  perfect square a + b a– b Sum & product of root:Let there are two roots named

nomial in t of degree 4. (v)

•

If the two roots α &β be equal in magnitude and opposite in sign, then b = 0 If a,b,c are rational number and

kgei snhe eYari dnag v.iSn ir

(i)

Polynomials : An expression of the form p(x ) = a0 + a1x + a2 x 2 + ......+ an xn , where an  0, is called a polynomial in x of degree n. Here a0, a1, a2, .....an are real numbers and each power of x is a non-negative integer. e.g. 2x + 7 is a polynomial in x of degree 1. 2y2 - 5y + 7 is a polynomial in y of degree 2.

ERna

•

b c x+ =0 a a



x² +



 –b  c x² –  a  x + =0   a

x²–(sum of root)x + product of root=0

•

Rakesh Yadav Readers Publication Pvt. Ltd.

If the roots  &  be reciprocal to each other then a = c.

•

a  b is one r oot of the quadratic equation, then the other root must be conjugate

a  b and viceversa Ex.1Find the Quadratic equation

whose one root is 3  3

Sol. If one root is 3  3 then second root will be 3  3 Sum of root



 



= 3  3 + 3 – 3 =6 Product of root



= 3 3

 3 – 3  =6

using, x2 (sum of root) x + (product of root) = 0  x2 –6x + 6 = 0 Ex.2: Two roots of equation 2x2 – 7x + 12 = are

α β α &β then, find β + =? α Sol. 2x2 – 7x + 12 = 0 On comparing with standard equation ax2 + bx + c = 0 a =2, b = –7, & c = 12 –b 7 α +β =  α +β = a 2 c α β = α β = 6 a α β α2  β2  β + = αβ α =

αβ2 – 2αβ αβ 2

7    –26 = 2 6

101

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Ex.3:

(2)

zeros are  and β is given by p(x ) = {x2 - (  + β )x +  β }

Find the product of the root

of the equation x 2 – 3  0 Sol. On comparing this equation with ax2 + bx + c = 0

A quadratic polynomial whose

(3) If  , β and γ are the zeros of p(x) = ax3 + bx2 + cx + d, then, b a

(i)

 +β + γ = 

(ii)

( β +β γ + γ  ) 

a = 1, b = 0 & c = – 3

•

Byquadratic Polynomial : A polynomial of degree 4 is called a biquadratic polynomial. It is of the form P(x) = ax4 + bx3 + cx2 + dx + e where a  0 E.g. (3x4 + 7x3 - 4x2 + 6x + 11), (4t4 - 7t3 + 6t2 - 11t + 9) etc. (3) Cubic Polynomial : A polynomial of degree 3 is called a cubic polynomial. It is of the form P(x) = ax3 + bx2 + cx + d, where a  0 E.g. (4x 3 – 2x 2 + 7x + 9) ,

(4)

p(x)

= {x 3 - (  + β + γ )x 2 + (  β



2y 3 – 5y 2 – 8 etc.

+β γ + γ  ) x - β γ } Factor Theorem : The Condition that (x - a) is a factor of a polynomial f(x), if and only if f (a) = 0 Thus, (x - a) i s a factor of f (x)  f (a) = 0. Remarks : (i) (x + a) is a factor of polynomiual p(x) if and only if p(-a) = 0 (ax - b) is a factor of a polyno-

(ii)

b mial p(x), if p   0 a

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Value of a Polynomial at a given point: If P(x) is a polynomial in x and if  is any real number, then the value obtained by putting x =  in P(x) is called the value of P(x) at x =  The value of P (x) at x =  is denoted by p(  ). e.g. Let p(x) = 3x2 - 2x + 7. then p(2) = (3  22 - 2  2 +7) = (12 - 4 + 7) = 15 p(-1) = [3  (-1)2 - 2(-1) + 7] = (3+2+7) = 12 Zeros of a Polynomial : A real number  is called a zero of the polynomial p(x), if p(  )= 0

d a A cubic polynomial whose zeros are  , β and γ is given by

 βγ = 

R Enak

2

(iii)

Note : 1. If  and β are the

zeros of p(x) = ax 2 + bx + c, a  0, then. (i)

 β 

(ii)

 β

c a

b a

c a

geisnh eeYa ridna gv.i Sni

c Product of root αβ = = – 3 a

(iii)

(ax + b) is a factor of a polyno-

 b mial p(x), if p    0  a

(iv)

(x - a) (x - b) are factors of a polynomial p(x) if p(a) = 0 and p(b) = 0. • Remainder Theorem : If a polynomial f (x) of degree n  1, is divided by (x - a), then the remainder is f (a). e.g. Let f (x ) = x 3 + 3x 2 - 5x + 4 be divided by (x -1). Find the remainder. Sol. Remainder = f (1) = 13 + 3  12 - 5  1+ 4 = 3 Important Results : (i) (x n - an) is divisible by (x - a) for all values of n. (ii) (x n + an) is divisible by (x + a) only when n is odd.

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(x n - an) is divisible by (x + a) only for even values of n. (iv) (x n + an) is never divisible by (x - a) • H.C.F & L.C.M of Polynomials : Divisor : A polynomial p(x) is called a divisor of another polynomial f (x ) = p(x ).g(x ) for some polynomial g(x ). • H.C.F. or (G.C.D.) of Polynomials : A polynomial h(x ) is called the H.C.F. or G.C.D of two or more given polynomials, if h(x ) is a polynomial of heighest degree dividing each one of the given polynomials. • Remark : The coefficient of heighest degree term in H.C.F is always taken as positive. e.g. What is the HCF of (x + 3)2 (x 2)3 and (x - 1) (x + 3)(x - 2)2 ? Sol. p(x ) = (x + 3)2 (x - 2)3 q(x ) = (x - 1) (x + 3) (x - 2)2 We see that (x + 3) (x - 2)2 is such a polynomial that is a common divisor and whose degree is heighest among all common divisors. • L.C.M. of Polynomials : A polynomial p(x ) is called the L.C.M. of two or more given polynomials, if it is a polynomial of smallest degree which is divided by each one of the given polynomials. e.g. Find the L.C.M of (x - 3) (x + 4)2 and (x - 3)3 (x + 4) : Sol : p(x ) = (x - 3) (x + 4)2 q(x ) = (x - 3)3 (x + 4) we make a polynomial by taking each factor of p(x ) and q(x). If a factor is common in both, then we take the factor which has highest degree in p(x ) and q(x ).  LCM = (x - 3)3 (x + 4)2 Note : For any two polynomials p(x ) and q(x ) p(x)  q(x ) = (Their H.C.F.)  (Their L.C.M.) (iii)

r

49 – 12 49 – 48 1 = 4 = = 46 24 6

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• (i) (ii)

Factorisation of Polynomials : To express a given polynomial as the product of polynomials, each of degree less than that of the given polynomial such that no such a factor has a factor of lower degree, is called factorisation. Formulae for Factorisation: (x + y)2 = x 2 + y2 + 2x y (x - y)2 = x 2 + y2 - 2x y

(x + y)2 + (x - y)2 = 2 (x 2 + y2) (x + y)2 - (x - y)2 = 4 xy (x + y)3 = x 3 + y3 + 3x y (x + y) (x - y)3 = x 3 - y3 - 3x y (x - y) x 2 - y2 = (x + y) (x - y) (x 3 + y3 )= (x + y) (x 2 + y2 - x y) (x 3 - y3 )= (x - y) (x 2 + y2 + x y) (x + y + z)2 = (x2 + y2 + z2 + 2 (xy + yz + zx)] (x 3+ y 3 + z 3 - 3xyz = (x + y + z) (x 2 + y 2 + z 2 - x y - yz - z x)

(iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi)

=

1 (x + y + z) [(x - y)2 + (y - z)2 2

+(z -x)2] (xii) x 2 + y2 + z2 - x y - yz - zx =

1 [(x - y)2 + (y - z)2 + (z - x )2] 2

(xiii) x 4 + x 2y2 + y4 = (x 2 + x y + y2)(x 2 - x y + y2)

kgei snhe eYari dnag v.iSn ir

•

EXERCISE

If f (x) is divided by (3x + 5), the remainder is :

3 (a) f   5

3.

4.

5.

6.

7.

8.

 3 f    5

5  5 (c) f   (d) f    3  3 11 If (x + 1) is divided by (x + 1), the remainder is : (a) 0 (b) 2 (c) 11 (d) 12 4 3 When (x - 3x + 2x 2 - 5x + 7) is divided by (x - 2) , the remainder is : (a) 3 (b) -3 (c) 2 (d) 0 If (x -2) is a factor of (x 2 + 3qx 2q), then the value of q is : (a) 2 (b) - 2 (c) 1 (d) - 1 The value of  for which the expression x3 + x2 -5x +  will be divisible by (x -2) is : (a) 2 (b) - 2 (c) - 3 (d) 4 If (x + 1) and (x - 2) be the factors of x 3 +(a + 1)x 2 - (b - 2)x - 6, then the value of a and b will be : (a) 2 and 8 (b) 1 and 7 (c) 5 and 3 (d) 3 and 7 The polynomial (x 4 - 5x 3 + 5x2 10x + 24) has a factor as : (a) x + 4 (b) x - 2 (c) x + 2 (d) None of these (x 29 - x 25 + x13 -1) is divisible by: (a) both (x -1) & (x + 1) (b) (x - 1) but not by (x + 1) (c) (x + 1) but not by (x - 1)

10.

4 12x + 7) for x   is : 3 (a) 7 (b) 0 (c) - 7 (d) 18 When (x 3 - 2x 2 + px - q) is divided by (x 2 - 2x - 3), the remainder is (x - 6). The values of P and q are : (a) p = - 2, q = -6 (b) p = 2, q = -6 (c) p = - 2, q = 6 (d) p = 2, q = 6 If (x - a) is a factor of (x 3 - 3x 2 a + 2a 2x + b), then the value of b is : (a) 0 (b) 2 (c) 1 (d) 3 If x100 + 2x 99 + K is divisible by (x + 1), then the value of K is : (a) – 3 (b) 2 (c) – 2 (d) 1 If the polynomial f (x ) is such that f (-1) = 0, then a factor of f(x) is : (a) – 1 (b) x – 1 (c) x + 1 (d) –1 – x If x 3 + 5x 2 + 10K leaves remainder -2x when divided by x 2 + 2, then the value of k is : (a) - 2 (b) 1 (c) - 1 (d) 2 Which of the following is a polynomial ?

11.

wwM wa. th Les aBryn

2.

(b)

(d) neither (x - 1) nor (x + 1) The value of expression (9x 2 +

9.

ERna

1.

12.

13.

14.

15.

(a) x 2 - 3x + 2 x +7 (b)

x 

1 x

(c) x 7/2 - x + x 3/2 (d) None of these

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16. If  and β are the zeros of x 2 + 3x + 7, then the vaue of (  + β ) is : (a) -3 (b) 3 (c) 7 (d) -7

17. If  and β are the zeros of 2x 2 + 3x -10, then the value of  β is:

5 3 (b) 5 (c) - 5 (d)  2 2 2 18. If common factor of x + bx + c and x 2 + mx + n is (x + a), then the value of a is : (a) 

cn cn (b) bm bm cn c 1 (c) (d) mb bm (x 4 + 5x 3 + 6x 2) is equal to : (a) x (x + 3) (x 2 + 2) (b) x 2 (x + 3) (x + 2) (c) x 2 (x -2) (x - 3) (d) x (x 2 + 3) (x + 2) The factors of (x 4 + 625) are : (a) (x 2 - 25) (x 2 + 25) (b) (x 2 + 25) (x 2 + 25) (c) (x 2 -10x + 25)(x 2 + 5x + 24) (d) do not exist The factors of (x 4 + 4) are : (a) (x 2 + 2)2 (b) (x 2 + 2) (x 2 - 2) (c) (x 2 + 2x + 2)(x 2 - 2x + 2) (d) None of these (x + y)3 - (x - y)3 can be factorized as : (a) 2y (3x 2 + y 2) (b) 2x (3x 2 + y 2) (c) 2y (3y 2 + x 2) (d) 2x (x 2 + 3y 2)

(a)

19.

20.

21.

22.

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(c) a = 6, b = 8 (d) a = 5, b = 8 33. Factories : (x 8 + x 4 y 4 + y 8) (a) (x 2 + x y + y2) (x 2 - x y + y2) (x 4 -x 2 y2 + y4) (b) (x 2 + x y - y2) (x 4 - x 4 y 4 + y 4) (c) (x 2 + x y + y2)2 (x 4 - x 2 y 2 + y 4) (d) (x 2 - x y + y2)2 (x 4 - x 4 y 4 - y 4)

r

 6 y6  34. Factorise :  x  27   

geisnh eeYa ridna gv.i Sni

28. The H.C.F of 2(x 2 - y2) and 5(x 3 - y3) is : (a) 2(x 2 - y2) (b) (x - y) (c) (x + y) (d) (x 2 + y2) 29. The L.C.M of (2x 2 - 3x + 2) and (x 3 - 4x 2 + 4x ) is : (a) x (2x 2 + 1) (x 2 + 2) (b) x (2x + 1) (x - 2)2 (c) x (2x 2 + 1) (x - 1)2 (d) x (2x + 1) (x 2 - 1) 30. The L.C.M of (a3 + b3) and (a4 b4) is : (a) (a3 + b3) (a2 + b2) (a - b) (b) (a3 + b3) (a + b) (a2 + b2) (c) (a + b) (a2 + ab + b2) (a3 + b3) (d) (a 3 + b3) (a2 - b2) (a - b) 31. If Polynomials 2x 3 + ax 2 + 3x 5 and x3 + x2 - 2x + a are divided by (x – 2), the same remainder are obtained. Find the value of a : (a) 3 (b) - 9 (c) - 3 (d) - 5 32. If the polynomial f (x) = x 4 - 2x 3 + 3x 2 - ax + b is divided by (x - 1) and (x + 1), the remainders are 5 and 19 respectively. The values of a and b are: (a) a = 8, b = 5 (b) a = 5, b = 6

R Enak

23. The H.C.F. of x 2 - x y - 2y2 and 2x 2 - x y - y2 is : (a) (x + y) (b) (x - y) (c) (2x - 3y) (d) None of these 24. The H.C.F. of (x 3 + x 2 + x + 1) and (x 4 - 1) is: (a) (x 2 – 1) (x 2 + 1) (b) (x + 1) (x 2 - 1) (c) (x + 1) (x 2 + 1) (d) (x 2 + 1) (x + 1) (x 3 + 1) 25. The L.C.M of the polynomials X and Y, where X = (x + 3)2 (x - 2) (x + 1)2 and Y = (x + 1)2 (x + 3) (x + 4) is given by : (a) (x - 2) (x + 4) (x + 3)2 (x + 1)2 (b) (x + 1) (x - 2) (x + 3) (x + 4) (c) (x - 2) (x + 1) (x + 3)2 (x + 4) (d) (x - 2) (x + 1)2 (x + 3) (x + 4) 26. The L.C.M of (x + 2)2 (x - 2) and (x 2 - 4x - 12) is : (a) (x + 2)(x – 2) (b) (x + 2) (x - 2) (x - 6) (c) (x + 2) (x -2)2 (d) (x + 2)2 (x - 2) (x - 6) 27. The H.C.F. of (x 2 - 4), (x 2 - 5x 6) and (x 2 + x -6) is : (a) 1 (b) (x - 2) (c) (x + 2) (d) (x 2 + x - 6)

 2 y2   4 x 2y2 x 2y6 (a)  x  3   x  3  9   

 2 y2  (b)  x  3   

 2 y2  (c)  x  3   

   

 4 x 2 y2 y4   x –  3 9  

 4 x 2y2 x 2y 4  x     3 9  

 2 y2  (d)  x  3   

 4 x 2y2 y 4 x    3 9 

   

35. Factorise : (x 4 + x 2 + 25) (a) (x 2 + 3x + 5)(x 2 + 3x – 5) (b) (x 2 + 5 + 3x) (x 2 + 5 – 3x) (c) (x 2 + x + 5) (x 2 – x + 5) (d) None of these

ANSWER KEY

(d) (a) (b) (d)

5. 6. 7. 8.

(b) (b) (b) (b)

9. 10. 11. 12.

(a) (c) (a) (d)

13. 14. 15. 16.

(c) (b) (d) (a)

wwM wa. th Les B aryn

1. 2. 3. 4.

1.(d) 3x + 5 = 0  x  

5 3

 5 So, remainder is f     3 2.(a) Remainder = f (-1) = (-1)11 + 1 = -1 + 1 = 0 3.(b) Remainder = f (2) = 24 - 3 (2)3 + 2(2)2 -5  2 +7 = 16 - 24 + 8 - 10 + 7 = - 3 4.(d) Since (x - 2) is a factor of f (x ) = x2 + 3qx - 2q  f (2) = 0  22 + 3q  2 - 2q = 0

17. 18. 19. 20.

(c) (a) (b) (d)

21. (c) 22. (a) 23. (d)

24. (c) 25. (a) 26. (d)

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30. (a) 31. (c) 32. (d)

33. (a) 34. (b) 35. (b)

SOLUTION

4q = - 4  q = -1 (x - 2) is a factor of polynomial f (x ) = x 3 + x 2 - 5x +    f (2) = 0  23 + 22 - 5  2 + =0  12 – 10 +  = 0   = – 2 6.(b) Since (x + 1) & (x - 2) are the factors of f (x ) = x3 + (a + 1)x 2 - (b - 2) x-6  f (-1) = 0 and f (2) = 0 or -1 + (a + 1) + (b - 2)-6 = 0 and 8 + 4(a + 1) - (b - 2)  2 5.(b)

27. (a) 28. (b) 29. (b)

6= 0 or a + b = 8 ......(i) and 2a - b = -5 ......(ii) (i) + (ii) 3a = 3  a = 1 From equation (i) b=8-1=7  a=1&b=7 7.(b) Since x = 2 makes the given expression zero, so, (x - 2) is its factor. 8.(b) Since x = 1 makes x 29 - x 25 + x 13 - 1 zero, so (x - 1) is its factor. And x = -1 does not make it zero so (x + 1) is not its factor.

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x x2  2x  3 x3  2x2  px  q x3  2x2  3x     p  3 x  q  remainder

 (p + 3)x - q = x - 6  p + 3 = 1 and q = 6 or p = - 2 and q = 6 11.(a) let f (x ) = x 3 - 3x 2a + 2a2x + b  (x - a) is a factor of f (x )  f (a) = 0  a3 - 3a3 + 2a3 + b = 0  b=0 12.(d)  x 100 + 2x 99 + k = f (x) (let) is divisible by (x + 1)  f (-1) = 0  1-2+k=0  k=1 13.(c) Since x = -1 makes f (x ) zero, So (x + 1) is its factor. 14.(b) x 5 x2  2 x3  5x2 10k x3  2x  

wwM wa. th Les aBryn

5x2  2x 10k 5x2 .  10    2x  10k  10  Remainder but given, remainder = - 2x  - 2x + 10k - 10 = -2x  10k = 10  k=1 15.(d) For polynomial, each power of x must be a non-negative integer.

16.(a)   β   17.(c) 18.(a)



 or



cn bm 19. (b) x 4 + 5x 3 + 6 x 2 = x 2 (x 2 + 5x + 6) = x 2 (x 2 + 3x + 2x + 6) = x 2 [x (x + 3) + 2(x + 3)] = x 2 (x + 3) (x + 2) 20.(b) Do not exist 21. (c) x 4 + 4 = (x 2)2 + (2)2 + 4x 2 - 4x2 = (x + 2)2 - (2x)2 = ( x 2 + 2x + 2) (x 2 - 2x + 2) 22. (a) Using formuale , a3 - b3 = (a - b) (a2 + b2 + ab)  (x + y)3 - (x - y)3 = [(x + y) - (x - y)] + [(x + y)2 + (x - y)2 + (x + y) (x – y)] = 2y [2(x 2 + y2) + (x 2 - y2)] = 2y (3x 2 + y 2) 23.(d) x 2 - x y - 2y2 = (x 2 - y2) - (x y + y2) = (x + y) (x - y) - y (x + y) = (x + y) (x - y - y) = (x + y) (x - 2y) 2x 2 - x y - y2 = (x 2 -x y) + (x 2 - y2) = x(x - y) + (x + y) (x - y) = (x - y) (x + x + y) = (x - y) (2x + y) Clearly, no factor is common, So, H.C.F = 1 24.(c) x 3 + x 2 + x + 1 = x2 (x + 1) +1 (x + 1) = (x + 1) (x2 + 1) x 4 - 1 = (x 2 - 1) (x 2 + 1) = (x + 1) (x - 1) (x 2 + 1)  Required H.C.F = (x + 1) (x 2 + 1) 25.(a) X = (x + 3)² (x – 2) (x+1)² Y = (x + 1)² (x + 3)(x + 4) So, LCM = (x – 2) (x + 4) (x +3)² (x + 1)² 26.(d) x 2 - 4x - 12 = x 2 - 6x + 2x - 12 = x (x - 6) + 2(x - 6) = (x + 2) (x - 6) and other is (x + 2)2 (x -2)  L.C.M = (x + 2)2 (x – 2) (x - 6) 27.(a) x 2 - 4 = (x + 2) (x -2) x 2 - 5x - 6 = x 2 - 6x + x - 6 = (x -6) (x + 1) and x 2 + x - 6 = x 2 + 3x - 2x - 6 = (x + 3) (x - 2) Clearly, ther is no common factor. So, H.C.F = 1. 28.(b) 2(x 2 - y 2) = 2(x - y) (x + y) and 5(x 3 - y3) = 5(x - y) (x 2 + y 2 + x y)  a (b  m)  c  n or a 

b 3    3 a 1

c 10   5 a 2 Let f (x ) = x 2 + bx + c and g (x) = x 2 + mx + n (x + a) is a common factor of f (x ) and g(x) f (-a) = 0 and g (–a) = 0 a2 – ba + c =0 and a2 – ma + n=0 a2 = ab - c...(i) and a2 = ma n .....(ii)

 β

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 H.C.F. = (x - y) 29.(b) 2x 2 - 3x + 2 = 2x 2 - 4x + x 2 = 2x (x - 2) + 1 (x - 2) = (x 2) (2x + 1) x 3 - 4x 2 + 4x = x (x 2 - 4x + 4) = x (x - 2)2  L.C.M. = x (x - 2)2 (2x + 1) 30.(a) a3 + b3 = (a + b) (a2 - ab + b2) a4 - b4 = (a - b) (a +b) (a2 + b2)  L.C.M. = (a - b) (a + b) (a2 -ab + b2) (a2 + b2) = (a - b) (a3 + b3) (a2 + b2) 31.(c) f (x ) = 2x 3 + ax 2 + 3x - 5 g(x ) = x 3 + x 2 - 2x + a By remainder theorem, f (2) = 2(2)3 + a(2)2 + 3  2 - 5 = 17 + 4a and, g(2) = 23 + (2)2 - 2  2 + a =8+a 17 + 4a = 8+a   3a = - 9 or a = - 3 32.(d) By remainder theorem, f (1) = 5 .......(i) [ x - 1 = 0  x = 1] and f (-1) = 19 .....(ii) [  x + 1 = 0  x = -1] Now, from (i) 1 - 2 + 3 - a + b =5 or b - a = 3 ...... (iii) from (ii) 1 + 2 + 3 + a + b = 19 or a + b = 13 ......(iv) (iii) + (iv) 2b = 16 or b = 8 Now from (iv), a = 13 - 8 = 5  a = 5, b = 8 33.(a) x 8 + x 4y4 + y8 = x 8 + 2x 4 y4 + y8 – x 4y 4 = (x 4 + y 4)2 - (x 2y2)2 = (x 4 + y 4 + x 2y2) (x 4 + y4 - x 2 2 y) = [(x 2 + y2)2 - (x y)2] (x 4 - x 2y2 + y4) = (x 2 + x y + y2) (x 2 - x y + y2) (x 4 - x 2 2 y + y4)

kgei snhe eYari dnag v.iSn ir

2

 4  4  4  f     9     12     7  3  3  3 = 16 - 16 + 7 = 7 10.(c)

 from (i) and (ii) ab - c = ma - n

ERna

f (x) = 9x 2 + 12x + 7

9.(a)

34.(b)

y6 x   x2 27 6

 

3

 y2       3 

3

 2 y2   4 x 2y2 y 4  =  x  3   x  3  9      35.(b) x 4 + x 2 + 25 = (x 2)2 + (5)2 + 10x 2 - 9x2 = (x 2 + 5)2 - (3x )2 = (x 2 + 5 + 3x) (x 2 + 5 - 3x)

105

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CHAPTER

11

ALGEBRIC IDENTITIES

12. a²+b²+c²–ab–bc–ca =

(ii) x4+

e.g. x –

3.

(ii) x–

1 =3, then x

1 = a, then x4

1 = a  2 =b x2

1 = b 2 x

 x²+

If x +

wher e

1 = 3, x

Then, 1 1 =3²–2 = 7, and x4 + 4 x² x = 49–2=47

1 =1, then x³ = –1 x

1 = 119  2 =11 x2

If x+

1 =–1, then x³ =1, x

1 1 = 3 then x³+ = 0  x6 x x³ = –1 or x6+1 = 0 11. If ax + by = m and bx –ay = n then, (a²+b²) (x² + y²) = m² + n2 Note : If the sum of squares of real numbers be zero, then each number is equal to zero i.e. if (x – a)² + (y – b)² + (z – c)² = 0, then x – a = 0  x = a, y – b = 0  y = b and z–c=0  z=c or if x ²+ y² + z² = 0, then x = 0, y =0&z=0 Based on Increasing power Ex.1 If x  x² 

1 = b–2 x

1 e.g. x4+ 4 =119 x

–b  b² – 4ac 2a Some Important Results

b = a²– 2 0 e.g. x +

10. If x+

1 = 11²–2= 119. x4

(ii) x+

1 2

2

1 = b²–2, where b = a²+2 x4

x4+

0, then, x =

1 = a, then x 1 (i) x² + 2 = a²–2 x 1 (ii) x 4+ 4 = b 2 – x

9.

1 =3² + 2 = 11 and x2

(i) x²+

If x+

1 =a2+2 x2

x²+

If x4+

8.

geisnh eeYa ridna gv.i Sni

(i) x²+

[(a–b)²+(b–c)²+(c–a)2] 13. If ax2 +bx +c =

1.

1 – a, then x

If x –

wwM wa. th Les B aryn

1 = ( a + b + c ) [ ( a – b ) 2+ ( b – c ) 2 2 + (c–a)²] Note:a³+b³+c³–3abc = 0, If {(i) a + b + c = 0 (a  b  c) or (ii) a²+b²+c²–ab–bc–ca = 0

2.

R Enak

An alg ebraic identity is an algebraic equation which is true for all values of the variable (s). Important Formulae: 1. (a+b)2 = a2 + b2 + 2ab 2. (a–b)2 = a2 + b2 – 2ab 3. (a+b)2 = (a – b)2 + 4ab 4. (a–b)2 = (a + b)2 – 4ab 5. a2–b2 = (a + b) (a–b) 6. (a+b)3 = a3 + b3 + 3ab(a+b) 7. (a–b)3 = a3 – b3  – 3ab(a–b) 8. (a³+b³) = (a + b) (a2–ab+b2) 9. (a³–b³) = (a – b) (a²+ab+b²) 10. (a+b+c)² = a² + b ² +c ² + 2(ab+bc+ca) 11. a³+b³+c³–3abc = (a+b+c) (a²+b²+c²–ab–bc–ca)

r

•

Sol.

1  3 , find the value of x

1 ? x²

1 3 x Squaring both sides, x

2

1 2    x     3 x 

x+

1 = 11  2 = 13 x

 x² 

1 1  2 x   9 x² x

x–

1 = 11 – 2 =3 x

 x² 

1 9–27 x²

4.

If x+

1 =2, then x =1 x

(If x 

5.

If x+

1 = –2, then x = –1 x

Same x 4 

6.

x³+

3  1 1 1  = x   –3  x    x x x³   

7.

x³–

3   1 1 1 = x –  +3  x –   x x x³  

x2+

Rakesh Yadav Readers Publication Pvt. Ltd.

1 1 = a²–2)  a then x ²  x x² 1 = (a² – 2)2 – 2 x4

Alternate:-

x

1 3 x

x² 

1 = 3² – 2 = 7 x²

106

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1 1  5 , find the (i) x ²  x x²

(ii) x 4  Sol.(i) x 

1 x4

1 5 x

1 x²  = 5² – 2 = 23 x²

1 = 23 x² Again squaring both sides

Sol.

2

2  2 1   x  2    23  x  

x4  x4 

1 = 34² – 2 = 1154 x4

x8 

1 = 1154²– 2 = 1331714 x8

1 1  4 , find the (i) x ²  x ² x 1 (ii) x 4  4 x

x4 

(If x –

1 4 x

1  – 2 x   9 x² x

1 = 14 x² Again squaring both sides

1  121 – 2  119 x4

1 1  a²  2 = a then x ²  x² x

Ex.6 If x –

1  4 , Find the value of x

(i) x ² 

1 x  4  196 – 2 x

Sol.

1 x  4  194 x 4

x

x³ 

1  27 – 9  18 x³

x–

x4 

1 x4

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x³ 

x³ 

1  18 x³

1  3 , Find the value of x

1  4 , Find the value of x

1 x³

1 4 x

x



1 = 4³ – 3×4 = 64 – 12 x³ = 52 Same as:



If x 

x³ 

1 x4

1 = 18² – 2 = 322 x4

1 x³

1  3³ – 3  3 x³

Sol.

x ³

x³ – Sol.

1  3, 4, 5, 6 , then x

1 18, 52, 110, 198 x³

Ex.9 If x –

1 = 4² + 2 = 18 x²

Ex.7 If x 

1 6 x

4 (ii) x 

1 4 x

x² 

1  6 , Find the value of x

1 x²

x³ 

x³ 

1 2 4 and x  4  a ²  2  – 2 ) x

4

Sol.

1  3  3 = 27 x³

Ex.8 If x 

1 1  2 x²   121 x4 x²

x² 

1 x8

x³ 

1 3 x

1 (If x 1 a then x ³  = a³ – 3a) +x = x³ Alternate: Here, a = 3

1 2   x²    11 x²  

wwM wa. th Les aBryn

1 = 4² – 2 = 14 x²

8 (iii) x 

1  1  3  x   x    27 x³ x x

 Put the value of x 

1 4 (ii) x  4 x

2

x4 

4 (ii) x 

1  3 , Find the value of x

ERna

1 Sol.(i) x   4 x

1 x2

x³ 

1  9  2 = 11 x² squaring both sides, x² 

1  527 x4

2 (i) x 

3

1  3 x    3 x 

1 x – 3 x squaring both side,

x² 

1  529 – 2 x4

Ex.4 If x 

1 3 x Cube both sides, x

2

1 1  2  x 4  4  529 x4 x

(ii) x 

Sol.

1  2 x –   3 x 

Ex.3 If x 

x² 

x4 

1 (i) x ²  x²

x² 

x4 

1 = 6² – 2 = 34 x2

Ex.5 If x –

1 5 x

(ii) x 

x2 

kgei snhe eYari dnag v.iSn ir

Ex.2 If x 

1  4 , then the value of x

1 x³

1 4 x Cube both sides, x–

3

1  3 x –   4 x  

107

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Put the value of x – x³ – x³ – x³ –

1 4 x

1 – 3  4  64 x³

Sol.

1  76 x³

x³+

*

1 = 4³+3×4 = 64+12 x³ = 76 1 = 3,4,5,6 x

Sol.

1 = 18×76 – x5 = 1368–4 = 1364 x5 –

x5 +

*

*

1 1 +x + 5 = 2530 x x 1 =5 x

1 =2530–5 = 2525 x5

x5–

1 = (a²–2)(a³–3a)–a x5

Sol.

Rakesh Yadav Readers Publication Pvt. Ltd.

1 3 x

1  33 – 3  3 = 18 x3 Squaring both side,



1 = (a²+2) (a³+3a)–a x5

1 x5

1 x – =5 x

x6 

1  1   2  x 3   3   324 6 x x 

x6 

1  324 – 2 x6

x6 

1  322 x6

1 1  a ,then x 6  6 x x = (a³ – 3a)² – 2 x

Ex.15. If x 

1  x –  x 

1 Ex.13 If x– =5 then find the value of x

1 If x+ = a x Then x5 +

1 =4 x

1  If  x –  = a, then x 

x5 –

x

1 . x6

x3 

1 1 +x – 5 =18×76 x x

Put the value of x –

 2 1   3 1   x  2   x  3  = 23×110 x   x  

Put the value of x +

Sol.

.....(ii)

   x³ –  = 18×76 x ³ 

....(i)

...(ii)

1  3 , Find the value of x

 3 1  2  x  3   18 x  

1 x³– = 4³+3×4 = 76 x³ (i) × (ii)

wwM wa. th Les B aryn 1 = 5³–3×5 = 110 x³ (i) × (ii) x³+

x6 

1 x²+ 2 = 4²+2 = 18 x ......(i)

x5 –

1 = 52 – 2 = 23 x²

.....(ii)

2

 2 1  x  2  x  

1 x+ = 5 x

...(ii)

1 = 4 find the value of x

R Enak

1 x+ 5 x 5

x5 +

Ex.14 If x 

1 =4 x

x–

......(i)

1 = 27×140 – 5 = 3775 x5

....(i)

1 x5

x5 –

1 Ex.10 If x+ = 5 find the vlaue of x

x²+

x5 –

1 = 14×52 – 4 = 724 x5

Ex.12 If x –

1 Then x³– = 36, 76, 140, 234 x³

Sol.

1 = 5³+3×5 = 140 x³ (i) × (ii)

1 = 4²–2 = 14 x²

x5 +

1 = 5²+2 = 27 x2

x³–

1 = 4³–3×4 = 52 x³ (i) × (ii)

1 1 =a then x³– = a³+3a) x x³

Same as if x–

1 x5

 =4 x

x+ x²+

Alternate: Here, a = 4 x³–

x² +

Find the value of x5 +

1  64  12 x³

(If x–

 = 4, x

Ex.11 If x+

r

1 1 1 – 3  x   x –   43 x³ x x

geisnh eeYa ridna gv.i Sni

x³ –

x7  Sol.

x

1  3 , Find the value of x

1 . x7

1 3 x

x2 

1 = 3² – 2 = 7 x2

x4 

1 = 7² – 2 = 47 x4

....(i)

1 = 3³ – 3 × 3 = 18 ....(ii) x3 Multiply (i) and (ii) x3 

 3 1   4 1  x  3  ×x  4  x   x  

= 47 × 18

108

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x7  x 

1 1  = 18 × 47 x x7



x =

x2 

1  843 x7

 a



x2 – x2 –

1 x7 – 7 . x

+

1 = 4² + 2 = 18 x2

Sol. x – ...(i)

1  43  3  4  76 ...(ii) x3 Multiply (i) and (ii) x3 –

x7 – x7 –



x3 

1  17 20 x3

2



x2 –

1 x2

x2 –

1  x  1  x – 1  =   x  x x2 

When x –

Rakesh Yadav Readers Publication Pvt. Ltd.

– 3  20

...(i)

1 =4 x

1 = 4³ + 3 × 4 = 76 ...(ii) x3 Multiply (i) and (ii)  3 1   3 1  x  3  ×x – 3  x   x  

= 17 20 × 76 x6 –

1 = 1292 20 x6

of x 

13



13

Sol.

3



–3×

1 = – x ³ 13 13 3 13

x³ +

1 = x ³ 10 13

13

1  4 , Find the value of x

1 x6

1 4 x

1  27 , Find the value x2

2 Ex.20 If x 

x³ +

x–

3



2

Ex.19 If x –

Sol:

20

x3 –

1 x³ + = x³

x6 –



Then,

1 = x Now,

3

1  4 , find the value of x

1 = 3 find the value of x³ x

x+

 2  – 2  a  3a   a

Ex.17. If x 

Sol.

1 1 – = 24472 x x7

1 = 24472 + 4 = 24476 x7

 a

1 = 4 12 x2

1   x   = 13 x 

1 1 x –  a , then x 7 – 7 x x

=

1  20 20 – 3 20 x3

1 + 2 = 11+ 2 x2

x² +

1 1  = 24472 +  x –  x7 x 

2

x3 

1 = 3² + 2 = 11 x2 Adding 2 both sides

x² +

= 322 × 76 x7 – x 

1  1 1  x   x –  2 = x x   x

1 =3 x

wwM wa. th Les aBryn

 4 1   3 1  x  4  ×x – 3  x   x  

1 x3

1  20 x 1  x3

ERna

1 x 4  4  182 – 2  322 x

x

x3 

Ex.18. If x –

1 x– 4 x x2 

2

1   x    20 x 

1  12 x We know that, x–

1 Ex.16 If x –  4 , Find the value of x

Sol.

1 – 2 = 14 – 2 x2

1  2  18  2 x2

2

– 2  – 2  a 3 – 3a  – a

2

x2 

1   x –  = 12 x 

1 1  a,then x 7  7 x x 2

1  4²  2  18 x2 Adding 2 both sides, x2 

x2 

1  18  47 – 3 x7

x7 

1 4 x

1 = 4² – 2 = 14 x2 Subtract 2 both side

1 3 x

Put the value of x 

x7 

x

kgei snhe eYari dnag v.iSn ir



1 x

1  27 x2 Adding 2 both sides, x2 

x2 

1  2  27  2 x2 2

1   x    29 x 

x

1  29 x

2 Ex21. If x 

of x –

1  31 , find the value x2

1 x

109

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1  31 x2 Substract 2 both sides,

1  14 x2 Again adding (2) both side We get,

x2 

2

1   x –   29 x 

x2 

1 x –  29 x

x3 

1 x  4  23 x By adding (2) both side, We get, 4

x4 

1  2  23  2 x4

1 252 x2

r

1  43 – 3  4 x3

Ex.24 If x 

2

Sol.

x

1 4 x Now, x3 –

1  43  3  4 x3

1 =5 x

x3 –

1  76 x3

1 2 5 – 2 = 23 x2 Subtract 2 both sides, x2

Ex.26

3 of x 

Sol.

1 x –  21 x We know that,

1  194 , find the value x4 1 . x3

1  194 x4 By adding (2) both side, We get,

2

 1  x –   21 x 

x2 –

1  1  1 x – x   2 =  x  x x 

= 5 21

x4 

Ex.25 If x 4  of x 3 –

1 x 4  4  2  194  2 x 1 x 4  4  2  196 x

Sol.

2

2  2 1   x  2   14  x  

Rakesh Yadav Readers Publication Pvt. Ltd.

1  4 , Find the 7x

Sol.

2x 

1 4x 2

1 4 7x

Multiply by

7 both sides 2

7 1  7  2x    4 2 7x  2

1  14 2x Squaring both sides, 7x 

2

1  322 , Find the value x4

1  2   7x    14  2x  

1 x3

49x 2 

1  322 x4 By adding (2) both side, We get, x4 

x4 

If 2x 

2 value of 49x 

1 x  2 – 223 – 2 x

1  7 x

4 Ex.23 If x 

1 – 2  16 x2

1 x2

x2 –

wwM wa. th Les B aryn x

x2 

x–

2

1  x    7 x 

1 – 2  18 – 2 x2

1 2   x –    4 x 

1 =5, find the value of x

R Enak

x2 

x2 

2

2

1  5 ....(i) x2 Adding 2 both side We get,

1  18 ....(i) x2 Again subtracting (2) from both side, We get, x2 

1 x  3  52 x 3

 2 1   x  2  = (5)2 x  

x2 

1  2  14  2 x2

1 4 x Now,

1 of x  x

Sol.

2

2

x

1  2  324 x4

 2 1   x  2  = (18)2 x  

1 2   x     4 x 

1  23 , find the value x4

4 Ex.22 If x 

x4 

x2 

geisnh eeYa ridna gv.i Sni

Sol.

1  2  322  2 x4

1 1  2  7x  4x 2 2x

= 196 49x 2 

1  196 – 7 4x 2

49x 2 

1  189 4x 2

110

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1  5 , Find the 3x

1 2 value of 9x  16x 2 4x 

Sol.

Find xy + yz + zx = ? Sol.

 7

x

2

1    15   3x     4 x    4 

2

1 1 225  2  3x  = 2 16x 4x 16

9x 2 

1 225 3  – 2 16x 16 2

 2   2x  2

2

2 2 a 4 – 2a 2b 2  b 4 = a – b 

wwM wa. th Les aBryn a + b = 2x

.....(i)

2 ....(ii) x From (i) and (ii)

Ex .31

2 2

a – 2a b  b

Sol.

4

a 4 – 2a 2b 2  b 4

....(i)

2

a

2

b

2 2



= (4 – ab)2

a 4  a 2b 2  b 4  = 24 – 2  22  0  0 4 = 16 Option c is correct.

Rakesh Yadav Readers Publication Pvt. Ltd.

a

2

 ab  b

2

a

2

– ab  b

8 = 4 × a – ab  b 2

– ab  b 2  = 4

...(ii)

–1 2

2ab , find

(a) x + y (b) x – y (c) xy (d) 2xy x = a² + b² Squaring both side, 2

...(i)

y ²  2a ²b ²

...(ii)

From (i) and (ii) x2  a4  b4  y2

2



Now, Put the value of a4  b4 a 2 – ab 2  b 2

a4 + b4 + 2a2b2 = 16+a2b2– 8ab From equation (i), 8–a²b²+2a2b2 = 16+a2b2– 8ab 8 = 8ab ab = 1 Alternate:We use formula,

Then,

2

x2 – y2  a 4  b4

a  b  4 – ab Squaring both sides,

2  =  2x   = 16 x 

Alternate: Let x = 1 Then, a 2, b = 0 Put the value of a and b in equation,

and

a 4  a 2b 2  b 4  8 2

a

y  2ab

a 4  b 4  8 – a 2b 2

2

...(i)

x ²  a 4  b 4  2a ²b ²

a 4  a 2b 2  b 4  8

If

 ab  b 2  = 3

x ²  a ²  b ² 

a 2  ab  b 2  4 find the value of ab .

a –b 

4

Sol.

4 x4 =1+ 2 2 = 3

2

2

a4  b4 a 2 – ab 2  b 2

x2  x2 2

 a  b a – b  

a

Ex.33If x = a² + b², y =

x2 2 x4 – x3 + x2 + 5 Put the value of of x4 = –4 –4 – x2(x – 1)+5 1 – x2(x – 1) Put the value of x – 1

1–

 ab  b 2 a 2 – ab  b 2 

ab =

x 2  2x – 2

1–

2

(i) – (ii) 2ab = –1

ERna

(d) 8

 ab  b 2  = 4( As Given)..(ii)

12 = 4 × a 2  ab  b 2 

x –1=

(c) 16

2

....(i)

a 4  a 2b 2  b 4 

a

2

x 2  2  x – 1

the value of a 4 – 2a 2b 2  b 4

=

Sol.

x 4  –4 x² + 2 = 2x

1 201  16x 2 16

– ab  b 2  = 2

2

x  4  4x  4x

1 1 Ex.28 If a  x  and b  x – , find x x

Sol.

2

4

9x 2 

(b) 4

 19  2  xy  yz  zx 

2

(ii) – (i) 2ab = 2 ab = 1 Ex.32 If a4 + a2b2 + b4 = 12, a² – ab + b² = 4, Find the value of ab.

15 = xy  yz  zx Ex.30 If x² + 2 = 2x, Find x4 – x3 + x2 + 5 Sol. x² + 2 = 2x Squaring both sides,

1 15  4x 4 Squaring both side,

(a) 10

2

30 = 2  xy  yz  zx 

3x 

2



49 – 19 = 2  xy  yz  zx 

3 both sides 4

3 1  3  4x    5 4 3x  4

9x 2 

x  y  z 

2

x ²  y ²  z ²  2  xy  yz  zx 

1 5 3x

Multiply by

a a

Ex.29If x  y  z  7 , x 2  y2  z 2  19

kgei snhe eYari dnag v.iSn ir

If 4x 

Ex.27

2



 x – y   a ²  b ² – 2 ab   

=

x 2 – y2 x –y

 x  y  x – y  x – y 

= x  y 

Ex.34 If x + y =1, x4 + y4 = –1, Find x² y²– 2xy

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Ex.35If x  y  z  3 ,

Sol.

x+y=

1 = 0 find (x – a)³ – x – a 3 = ?  

1 1 1    2, x y z

Sol.

1 1 1   2 x y z

2



x ²  y ²  z ²  2  xy  yz  zx 

9 = 6 + 2 (2xyz) 3 = 4xyz 3 xyz = 4 p q r Ex.36 If x  y  z  1 &

M – 5=

1 =5 M Now,

p2 q 2 r 2   ? x2 y2 z 2

Sol.

M³ –

q p r , b = y and c = x z Then, a + b + c = 1

Let a =

M³ –



1 1 1   0 a b c ab + bc + ca = 0

a  b  c 

2



p2 q 2 r 2   1 x2 y2 z 2 Ex.37 If x³ + y³ = 0 find x + y = ?

Hence,

3xy

(c) 3xy

1 = 140 M3

1 (x – a)³ – x – a 3 = 140  

a ²  b ²  c ²  2 ab  bc  ca  1 = a² + b² + c² + 2(0) a² + b² + c² = 1

(a)

1 = 5³ + 3 × 5 M3

Put the value of M = x – a So,

&



1 M

M–

wwM wa. th Les B aryn

x y z    0 Find p q r

(x – a) (x – b) = 1 ...(i) a – b + 5 = 0  –b=–a–5 Put the value (– b) In equation (i) (x – a) (x – a – 5) = 1 let (x – a) = M M(M – 5) = 1

R Enak

x  y  z 

(a) 125 (b) –125 (c) 0 (d) 140 (x – a)(x – b) = 1 1  (x – b) =  x – a 

xy  yz  zx  2xyz

Ex.40 If x² + x = 5 find the value of 1 (x + 3)³ + x  3 ³  

Sol.

(b) 2xy (d) 4xy

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   

(m – 3)² + (m – 3) = 5 m² + 9 – 6m + m – 3 = 5 m² – 5m + 1 = 0 m² + 1 = 5m



m+

3xy

Ex.38 If x4 + y4 = x²y² find x6 + y6. Sol. x6 + y6 = (x²)³ + (y²)³ = (x² + y²)(x4 + y4 –x²y²) = (x² + y²) (x²y² – x²y²) =0 Ex.39 If (x –a)(x – b) = 1 and a – b + 5

x ²  y ²  z ²  6 , Find xyz = ? Sol.

(x + y)³ = x³ + y³ + 3xy(x + y)  x³ + y³ = 0 (As given) (x + y)³ = 3xy(x+ y) (x + y)² = 3xy

1 =5 m

Now, 1 = 5³ – 3 × 5 m³

m³ +

r

x+y=1 Squaring both side (x + y)2 = (1)² x2 + y2 + 2xy = 1 x² + y.² = 1 – 2xy Again Squaring both sides, (x² + y²)2 = (1 – 2xy)2 x4+y4+2x2y2 = 1+ 4x2y2–4xy Put the value of x4 + y4 = –1 –1 + 2x² y² = 1+ 4x2y2–4xy –1 = 1 + 2x² y² – 4xy –2 = 2(x² y² – 2xy) x² y² – 2xy = –1

1 = 110 m³

geisnh eeYa ridna gv.i Sni

Sol.

Let (x + 3) = m x+3=m x=m–3 Put the value of 'x' x² + x = 5

m³ +

Put the value of m = x + 3 Then,

1 (x + 3)³ +  x  3  ³ = 110

Ex.41If x(x –3) = –1 find the value of x³(x³ – 18) Sol. x(x –3) = –1 x–3=

x+

x³ +

–1 x

1 =3 x

1 = 3³ – 3 ×3 = 18 x3

x³ – 18 =

–1 x3

......(i)

Now, x³(x³ – 18) From Equation (i) = x³×

–1 =–1 x3

x³(x³ – 18) = – 1 Ex.42 If (a + b)² = 21 + c², (b + c)² = 32 + a² and (c + a)² = 28 + b², find a +b+c=? Sol. (a + b)² – c² = 21  (a + b+ c)(a + b – c) = 21 ...(i) (b + c)² – a² = 32  (b + c + a) (b + c – a) = 32 ..(ii) (c + a)² – b² = 28  (c +a + b)(c + a – b) = 28..(iii) Adding all three equations:(a +b + c) [(a + b + c) + (b + c + a  ) + (c + a – b)] = 81 (a + b + c)² = 81  a+b+c=9 

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3

2

2

2

(a) 4

So. (a = b = c) Ex.45 If a 3 + b 3 + c 3 - 3abc = 0 , a + b + c  0 and a, b & c are natural numb er f ind the possible value of a + b + c (a) 4

(b) 8

(c) 5

(d) 12

Sol.(d) We know that in this condition a = b = c and given a, b and c are natural no. we take option (d) because 12 is divide 3 equal natural part 12 4 3

(b) 8

(c) 5

Ex.50 If a2 + b2 = c2, Find the value of a 6  b6  c 6 a 2b2c 2 a2 + b2 = c2 Cube both side (a2 + b2)3 = (c2)3 a6 + b6 + 3a2 b2 (a²+ b2) = c6 a6 + b6 + 3a2 b2 c2 = c6 a6 + b6 - c6 = -3a2 b2 c2 A.T.Q.

Sol.

(d) 12

Sol.(b) We know that in this condition a = b = c and given a, b and c are natural no. we take option (b) because 8 = 2 × 2 × 2 We can say that 8 is possible value of a × b × c Ex.47 Find the value of (x 2 - y 2 ) 3 + (y 2 - z 2 ) 3 + (z 2 - x 2 ) 3 3

3

3

x  y   y  z   z  x  Sol.

Let, a = x2 – y2 b = y2 – z2 c = z2 – x2 a+b+c=0 Then, a3 + b3 + c3 = 3abc Thus, p = x – y q=y–z r=z–x p+q+r=0 Then, p3 + q3 + r3 = 3pqr A.T.Q.

(x 2 - y 2 ) 3 + (y 2 - z 2 ) 3 + (z 2 - x 2 ) 3 3

3

3

x  y   y  z   z  x 

3(x 2 - y 2 ) (y 2 - z 2 ) (z 2 - x 2 ) = 3 x  y y  z z  x   (a2 - b2) = (a + b) (a - b)

wwM wa. th Les aBryn

= (a + b + c) (a + b + c - ab - bc - ca) (i) If (a + b + c) = 0 3 3 3 then a + b + c - 3abc = 0 3 3 3 a + b + c = 3abc 3 3 3 (ii) If a + b + c - 3abc = 0 a, b and c are distinct no then. a + b + c = 0 3 3 3 (iii) a + b + c - 3abc = 0 a, b and c all are +ve integer no then a= b = c . (iv) a2 + b2 +c2 - ab - bc - ca = 0 2 2 2 a + b + c = ab + bc + ca then a= b = c Ex.43 If a + b + c = 0, then the value of a³ + b³ + c³ is: (a) 0 (b) abc (c) 3abc (d) None of these Sol.  a³ + b³ ++ c³ – 3abc = (a + b c) (a² + b² + c² – ab – bc – ac) or a³ + b³ + c³ – 3abc = 0 a³ + b³ + c³ = 3abc  Hence (c) is the correct option. Ex.44 If a3 + b3 + c3 - 3abc = 0 and a +b+c  0 Which statement is true (a) a>b>c (b) a = b = c (c) a > b < c (d) a < b < c Sol.(b) We know that, a3 + b 3 + c 3 - 3abc = (a + b + c)[(a - b)2] + (b - c)2+ (c - a)2  a3 + b 3 + c 3 - 3abc = 0 ( As given) When a + b + c  0 Then (a - b)2 + (b - c)2 + (c - a)2 =0 (a - b)2 = 0 (a - b) = 0 Then a = b Same b = c c=a

a=b=c=4 Then 12 is possible a + b + c Ex.46 If a 3 + b 3 + c 3 - 3abc = 0, a + b + c  0 and a, b & c are natural number find the possible value of a × b × c

kgei snhe eYari dnag v.iSn ir

3

ERna

3

a + b + c - 3abc 2 2 2 1  (a + b + c)[(a - b) ] + (b - c) + (c - a) 2

(x + y) (x - y) (y + z) (y - z) (z + x) (z - x) x  yy  zz  x

= (x + y) (y + z) (z + x) Ex.48 If a + b + c = 0 Find the value of a 3 + b 3 + c 3 + 3 a b c (a) 0 (b) 1 (c) abc (d) 6abc Sol.(d) (a + b + c) = 0 Then a 3 + b 3 + c 3 = 3 a b c So, a 3 + b 3 + c 3 + 3 a b c = 3abc + 3abc = 6abc Ex.49 If x = 1.235 y = 3.422 z = 4.377 Find x3 + y3 - z3 + 3xyz Sol. x + y = z (x) + (y) + (-z) = 0 (x)3 + (y)3 + (–z)3 = 3(x)(y)(–z) x3 + y3 - z3 = –3xyz x3 + y3 -z3 + 3xyz = 0 

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a 6  b6  c 6 a 2b2c 2

-3a 2 b 2 c 2 a 2b2c 2 = –3 Ex.51 If a1/3 + b1/3 = c1/3 Which statement is true (a) a3 + b3 - c3 = 3abc (b) a3 + b3 - c3 + 3abc = 0 (c) (a + b - c)3 - 27abc = 0 (d) (a + b - c)3 + 27abc = 0 Sol.(d) (a)1/3 + (b)1/3 + (-c)1/3 = 0 If (a + b + c) = 0 Then a3 + b3 + c3 = 3abc (a) + (b) + (-c) = 3(a)1/3 (b)1/3 (-c)1/3 a + b -c = -3a1/3 b1/3 c1/3 Cube both side (a + b - c)3 = (-3a1/3 b1/3 c1/3 )3 (a + b - c)3 = - 27abc  (a + b – c)3 + 27abc = 0 Ex.52 If x + y + z = 2s Find the value of (s - x)3 + (s y) 3 + 3 (s - x) (s - y) z =

(a) y3

(b) x3

(c) z3

(d) 0

Sol.(c) x + y + z = 2s x+y+z=s+s s-x+s-y-z=0 (s - x)+ (s - y) + (-z) = 0 (s - x)3 + (s - y)3 + (-z)3 = 3(s - x) (s - y) (-z) (s - x)3 + (s - y)3 - z3 = -3(s - x) (s - y) z (s - x)3 + (s - y)3 + 3 (s - x) (s - y) z = z3 Ex.53 Find the value of 3

Sol. =

3

3

2  3333  3343  3  3332  334

2  3333  3343  3  3332  334 After describing

3333  3333  3343  3  333  333  334

a3 + b3 + c3 - 3abc 2 2 2 1  (a + b + c)[(a - b) ] + (b - c) + (c - a) 2

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3

a 3  b3  c 3  3abc a  b2  c 2  ab  bc  ca 2

Sol.

a3 + b3 + c 3 - 3abc = (a + b + c) 2 2 2 (a + b + c - ab - bc - ca) 3

3

3

a  b  c  3abc a 2  b2  c 2  ab  bc  ca A.T.Q.

=

(a  b  c) a 2  b2  c 2  ab  bc  ca





a 2  b2  c 2  ab  bc  ca =a+b+c Put the value = 20 + 25 + 15 = 60 Ex.55 If a + b + c = 3 and a2 + b2 + c2 = 6 , 1 1 1    1 , Find abc a b c

Sol.

1 1 1   1 a b c

2

2

2

2

(a + b + c) = a + b + c + 2(ab+bc + ca)

(a + b + c)2 = a2 + b2 + c2 + 2abc (3)2 = 6 + 2abc 9 - 6 = 2abc 3 2 Ex.56 If a + b + c = 15 and a2 + b2 + c2 = 83 find a3 + b3 + c3 - 3abc Sol.

abc =

2

2

2

2

(a + b + c) = a + b + c + 2(ab+bc + ca)

(15)2 = 83 + 2 (ab+bc+ca) 225 - 83 = 2 (ab + bc + ca) ab + bc + ca = 71 a3 + b3 + c3 - 3abc = (a + b + c) [a2 + b2 + c2 -(ab + bc + ca)] a3 + b3 + c3 - 3abc = 15(83-71) = 15 × 12 = 180

1259

(c) –48 (d) None of these Sol.  a + b + c = 0  a³ + b³ + c³ – 3abc = 0 Thus (a) is the correct option. Ex.59 Find the value of a³ + b³ + c³ – 3abc If a + b + c = 12 and ab + bc + ac =47. Sol. a + b + c = 12  (a + b + c)² = a² + b² + c² + 2(ab +bc + ac) = 144  a² + b² + c² + 2 × 47 = 144  a² + b² + c² = 50 Now, since a³ + b³ + c³ – 3ab = (a + b + c) (a² + b² + c² – ab –bc – ac) Then, a³ + b³ + c³ – 3abc = 12(50 – 47) = 12×3 = 36 Ex.60 I f a = 9 9 7 , b = 9 9 9 a n d c = 9 9 6 f in d th e v alu e of a 3 + b3 + c3 - 3abc Sol.



2 2 2 1 99  97  97  96  96  99    2



2 2 1 2 2  1  3      2

1 4  1  9 7 2 Ex.62 If a = 556, b = 558 and c =561 Find the value of a2 + b2 + c2 - ab - bc - ca Sol. a2 + b2 + c2 - ab - bc - ca 

=

1 1 4  9  25   38 = 19  2 2 Ex.63If a2 = b+c, b2 = c+a c2 = a+b 

Sol.

a=

c=

TYPE IV

1  2 2 2 a  b  b  c  c  a    2  When, a² + b² + c² – ab – bc – ca = 0 a² + b² + c² = ab + bc + ca Then a = b = c Ex.61If a = 99, b = 97, and c = 96 Find the value of a2 + b2 + c2 - ab - bc - ca

Rakesh Yadav Readers Publication Pvt. Ltd.

ca b

ab c Now put the value of a, b and c

1  2992 14 2 = 2992 × 7 = 20944

=

bc a

Thus, b =



a2 + b2 + c2 - ab - bc - ca

1 1 1 + + 1 a 1 b 1 c He re a, b and c non ze ro number a2 = b+c Divide by 3 both sides a2 b  c = a a Then,

 999  997 2       2 1  999  997  996 997  996    2  996  999 2      

*

1  2 2 2 a  b  b  c  c  a       2

2 2 2 1   556 558 558 561 561 556      2

a3 + b3 + c3 - 3abc 2 2 2 1  (a + b + c)[(a - b) ] + (b - c) + (c - a) 2

wwM wa. th Les B aryn

ab  bc  ca 1 abc ab + bc + ca = abc

(b)

1  2 2 2 a  b  b  c  c  a       2

r

1 1000  2 = 10 2 Ex.54 If a = 20, b = 25, c = 15, Find

=

(a) 0

=

geisnh eeYa ridna gv.i Sni

=

1 10000  1  1 2

3

R Enak

=

a2 + b2 + c2 - ab - bc - ca

Sol.

=

Ex.57 Find (a - b)3 + (b - c)3 + (c - a)3 =? Sol.  (a - b) + (b - c) + (c - a) = 0  (a - b)3 + (b - c)3 + (c - a)3 = 3 (a - b) (b - c) (c - a) Ex.58 The value of a³ + b³ + c³ – 3abc, Where a = 87, b = –126 and c = 39 is:

1 (333  333  334 )[(333 – 333)² 2 (333 – 334 )2  (334 – 333 )2 ]



 

1 1 1 bc + ca + ab 1 1 1 a b c a b c + + a  b c b c a a  b c

a  b c =1 a  b c Alternate: We put the value a=b=c=2 a2 = b+c = (2)2 = 2+2 4=4

114

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1 1 1 + + =1 3 3 3 x ²  y²  z ² Ex.64 If xy  yz  zx  1 , then find x y y z z  x the value of z  x  y

Sol.

x ²  y²  z ² 1 xy  yz  zx

x² + y² + z² = xy + yz + zx In this condition x = y = z Put the value of x = y = z = 1 Then,

=

Thus, c+a = –b c2+a2–b2 = –2ac Put the value

1 1 1 + + 1 2 1 2 1 2

x y y z z x   z x y

1 1 1  1 1 1   1 1 1

1 1 1 2 2 + 2 2 2 + 2 a b –c b c –a c  a 2 – b2    + + –2ab –2bc –2ab



a  b c O = =0 –2abc –2abc Alternate:We choose the value which we follow the condition. If a+b+c = 0 a = 1, b = 1, c = –2 Put the value a,b,c

1 1 1 2 2+ 2 2 2 2 + 2 a  b – c b c –a c a2 –b2

=

1

=

2

x y z   yz z x yx

=

1 1 1   1 1 1  1 1  1

1 1 1 3 =   = 2 2 2 2

Ex.66 If a + b + c = 0, Find the value of 1 1 1 + + a 2  b2 – c 2 b2  c 2 – a 2 c 2  a 2 – b2

Sol.

2

+

a+b+c = 0 a+b = –c Squaring Both side (a+b)2 = (–c)2 a2+b2+2ab = c2 a2+b2–c2 = –2ab ....(I) Same, b+c = –a b2+c2–a2 = –2bc ...(II)

p2 q2 r2 + + p2 – qr q 2 – pr r 2 – pq

Sol.

pq + pr + rp = 0 pq + rp = – qr Same,pq + qr = – pr qr + rp = – pq

2

2

1  –2 – 1

2

2

 –2  1 – 1

1 1 1 + + =0 2 4 4 Ex.67 If x + y + z = 0 then the value

–

x ²y²  y² z ²  z ² x ² of is: x 4  y4  z 4

(a) 0 (b) 1/2 (c) 1 (d) 2 Sol. (x + y + z) =0  x² + y² + z² + 2(xy + yz + zx) = 0  x² + y² + z² = –2(xy + yz + zx)  (x² + y² + z²)² = 4(xy + yz + zx)²  x4 + y4 + z4 + 2(x²y² + y²z² + z²x²) = 4[x²y² + y²z² + z²x² + 2xyz (x + y + z)] x4 + y4 + z4 = 2(x²y² + y²z² + z²x²)  [ (x + y + z) = 0] 

=

x 2 y 2  y 2z 2  z 2 x 2 x4  y4  z4

r2 r  qr  rp 2

p2 q2 + p p  q  r q p  q  r

+

r2 r p  q  r

=

p q r + + pq  r pq  r pq  r

=

p q r =1 p q r

Alternate: Put the value p = –1, q = 2 r=2 So, pq  qr  rp = –1×2+2×2+2×–1 = 0 =

p2 q2 r2 + 2 + 2 p – pq a – pr r – pq

=

(–1)2 (2)2  2  (–1) – (2  2) (2) – (–1  2)

2

2

(2)2 (2) – (–1  2)

1 x 2 y 2  y 2z 2  z 2 x 2 = 2 x4  y4  z4

Hence (b) is the correct option. Alternate: x+y+z=0 Then, Let x = 1, y = 1, z = – 2

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p2 q2 + 2 p  pq  rp q  pq  qr 2

+

2

1

2

p2 q2 r2 + + p2 – qr q 2  pr r 2 – pq

1 1 1 + + 1 1 – 4 1 4 – 1 4  1 – 1

= =

1 4  4 9 1 = = 1  1  16 18 2 Ex.68If pq+qr+rp = 0, Find the value

1

wwM wa. th Les aBryn

Here, x=y=z Let put the value of x = y = z =1 Now,

+

(1²)  (1²)  (1²)  (–2²)  (–2²)  (1²) (1)4  (1)4  (–2)4



ERna

x ²  y ²  z ²  xy  yz  zx

2

1 1 – –2

E x. 65I f x ²  y ²  z ²  xy  yz  zx , find the v alue of

Sol.

.....(III)

2

= 2+2+2=6

x y z   yz z x yx

Now,

kgei snhe eYari dnag v.iSn ir

Then

2

=

1 4 4   1– 4 4 2 4 2

–1 4 4   3 6 6 –2  4  4 6 = = =1 6 6 =

115

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1 1 + 1  q  r –1 1  r  p –1

1 p 

1 q

+

1 + 1  q  r 1

1 1 1 1  pq p

1 pq q + + q  pq  1 1 q  pq pq  1 q pq  1  q =1 q  pq  1

Alternate p=1=q=r Then =

1 1 1 + + 1 1 1 1 1 1 1 1 1

=

1 1 1 + + =1 3 3 3

x  a 2 x  b2 x  c 2 + + bc ca ab = 4(a+b+c) Find the value of x (a) (a2+b2+c2) (b) (a+b+c)2 (c) (a2+b2+c2–ab–bc–ca) (d) (ab+bc+ca) Sol. We take option (B) Then x = (a+b+c)2

a  b  c – c 2  + a  b 2 2 a –b = (a+b) (a–b) a  b  c – a  a  b  c  a b  c

a  b  c – ba  b  c  b a  c a  b  c – ca  b  c  c + a  b b  c2a  b  c +  b  c a  c2b  a  c + a  c a  b2c  a  b a  b +



=

z+

1 y 1 = + x 1 – y y –1

1 y 1– y  1 – 1– y 1– y 1– y

So, x  y  z 

a 2  b2  c 2 – c 2 a 2  b2

1 1 1   x y z

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1 2 Then y = 2 Let, x =

1 1 + =1 2 2

So,

1 = z

If y = 2 Then z = – 1 2+

1 = 2–1 = 1 –1 1 x

1 = -1+2 = 1 12

Ex.73 If

Find

1 y

1 y + =1 z

1 =1 y

–1+

1 Sol. x + y = 1

y –1 x= y

x+

Put the value z and x

1 Find the value of (i) z + x

x = 1–

(i)

Now, z 

1 1 Ex.72 If x + y = 1, y + =1 z

1 1 1   =3 x y z

Alternate:-

Now, y +

b2  c 2 c2  a 2 a 2  b2 + +  2 b  c 2 c2  a 2 a 2  b2 1+1+1 = 3 L.H.S = R.H.S Then, (x = a2+b2+c2)

(ii) x  y  z 

1 1 1 y  z  y z x

= 1+1+1=3

a2  b2  c2 – a2 a 2  b2  c 2 – b2 + b2  c2 c2  a 2

+

x

(ii)

x  a2 x  b2 x  c2 Ex.71 If 2 2 + 2 2 + b c c a a 2  b2 = 3 Find the value of x (a) a2+b2+c2 (b) a2+b2+c2–ab–bc–ca (c) (a+b+c)2 (d) a2+b2+c2+ab+bc+ca Sol. We take option (A) a2+b2+c2

2

2



(i)

2 x–1= 8 x=9 Option (B) x = (a+b+c)2 = (1+1+1)2 = (3)2 = 9 So, It is proofed  x = (a+b+c)2

2 a  b  c – a + a  b  c – b a  c b  c 2

....(ii)

From (i) & (ii)

x –1 x –1 x –1 + + = 4×3 2 2 2

wwM wa. th Les B aryn

Ex.70 If

1 1– y

z=

x – 1  3 =4×3

1 (Pqr = 1, r = pq )

2

1 = 1–y z

geisnh eeYa ridna gv.i Sni

1

Sol.

2 a + b + c + 2 b +a + c + 2 c + a + b 4 a+ 4 b + 4 c = 4 ( a+ b +c ) L.H.S = R.H.S So, x = (a+b+c)2 Alternate: Let, a = b = c =1 Then

R Enak

+

 

r

1 Ex.69If p×q×r = 1 Find 1  p  q –1

a b c + + =1 bc ca a b a2 b2 c2 + + bc ca a b

Sol. Divide and Multiply a + b + c Now, ....(i) =

a b c + + bc ca a b

1  a  b  c

a  b  c 116

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 a b c        (a+b+c) b  c c  a a  b 

= a+b+c a a  b  c  bc

+



Sol.

c a  b  c ab

a 2  a b  c bc

+

b a  b  c ca

= a+b+c b2  b a  c + + ca

b –c – a ac –b a – bc + + =0 a b c – a – b  c

a2 b2 c2 +a+ +b+ +c bc ca ab =a+b+c 2

Then,

2

a b c + + =0 bc ca a b

a b c + + =1 bc ca a b We break equation small step Let, a = 0, b = 1 Put the value of a and b Then,



=

1 2 +c c Put the value of c² Then,

b – c a c a – b + + = 1 And a c b a-b+c  0 which statement is true

(a)

1 1 1 = – b a c

(b)

1 1 – =2–1=1 12 1

1 1 1 = – a b c

Ex.76 If a 

1  2 , find the value of a

50 (i) a 

1 a 50

1 a 49

49 (ii) a –

(iii) a 3  a 2  a  1 1 2 a a=1

Sol. a 

(i)

a 50 

1  150  50 50 = a 1

= 1

(ii)

a 49 –

 =2 1

1  149 – 49 49 = a 1

= 1– (iii)

 =0 1

a3  a2  a  1 1³ + 1² + 1 + 1 = 4

Ex.77 If p 

  5 , f ind the p–3

value of :-

1 1 1 = – b a c

(i)

 p – 3

(ii)

p² – 3p  4

100

100

TYPE – V 1 = 2 then a = 1 a

1 2 a

a 2  1  2a a² – 2a + 1 = 0 ( a –1)² = 0 So, a = 1 1 Ex.75 If a   2 , find the value of a 1 a100  100 a

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  1+1=2 1

So, L.H.S = R.H.S

Proof a 

Ex.74 If

1

c

1 1 1 = – b a c

When a +

1 +c –1 = 1–1 = 0 c



1 1100

1 2

Option (A)

1 c2 + c  0 0 1

So,

b

ERna a=

a2 b2 c2 + + bc ca a b

0+

1100 

a – b – c + a – b  c

1 –1 a 1 a –1 + + =1 a 1 1 a+1 + a–1 = 1

wwM wa. th Les aBryn

1 + c = 1, c 1+ c² = c c² = c – 1 Again put the value a, b

+

1 1 = – b a Alternate: Let, b = 1 c=1 Then, Put the value of b and c

2

0 1 c + + =1 1 c c  0 0 1



–1 1 1 + + =0 a b c

Alternate:



bc ac a –b + + =1 a b c We adding + 1 in this equation

a =0

1 2 a Then, a = 1 Put the value of a,

Sol. a 

(d) N.O.T

b–c ac a–b –1+ –1+ + 1 = 1–1 a b c

c 2  c a  b = a+b+c ab 

1 1 1 = + c a b

kgei snhe eYari dnag v.iSn ir



(c)

c 1



Sol.(i)

p

      p – 3

 5 p–3

Subtract 3 both sides,

 p – 3   p – 3 

1

 p – 3 1

 p – 3

5–3

2

Let a = (P – 3) Then, a+

1 =2 a

117

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 a = 1 = (P – 3) Now, p–3=1 p=4

(iii)

 p – 3 1100



      p – 3

of

1  100 = 2 1

42

1 When a   –2 , then a = –1 a

(iii) P³ + 4P² + 5P + 1

1  –2 , Find the value a a 200

1 (P – 3) +  p–3 = – 2



1

 –1

= 1+1=2 1  –2 , find the value of a

(– 1)42 +

wwM wa. th Les B aryn

Ex.79 If a  99 (i) a 

1 a 99

32 (ii) a 

1 a 31

(ii)

1 a   –2 a

(–1)

So, a = –1

 –1

99



(iii)

1

 –1

99

(–1) + (–1) = –2 a 32 

 –1

1 a 31

32



(A).

1

 –1

42

1 (P – 3)101 – p–3 101   101

1 a 99

1

=1+1=2

(iii) a 4  a 3  a 2  a  1

a 99 

1 =1 a Then, a³ = – 1

a15 +

 –1

31

1–1=0

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(i) a36 + 101

 –1

= (–1) – (–1) = –1 + 1 = 0 P³ + 4P² + 5P + 1 Put the value of P = 2 2³ + 4 × 2² + 5 × 2 + 1 = 8 + 16 + 10 + 1 = 35

1 = 1, a

or a² – a + 1 = 0 Then a³ = – 1 or a³ + 1 = 0

1 a 36

(ii) a37 +

1 a 37

1 a 38 Sol. a² – a + 1 = 0 or

(iii) a38 +

1 =1 a Then, a³ = – 1 1 a36 + 36 a 1 (a³)12 + a 3 12

a+

TYPE – VI

When a +

1 1 5 + a³ 5 15 = (a³)   a

1 (– 1)5 + –1 5 = (– 1) + (– 1)   =–2 Ex.82 If a² – a + 1 = 0, Find the value of

1

–

1 a15

Sol. a+

1 (i) (P – 3)42 + p–3 42  

200

1 = 1, Find the value a

Ex.81 If a + of a15 +

So, (P – 3) = – 1 P=–1+3=2 P=2

a 200 200

1 +a=0 a a³ + 1 = 0 or a³ = –1

1 (P – 3) + p–3 = 1 – 3  

1

1 =–a a

a³ +

Subtract 3 both sides

1

 –1

a³ +

1 =1 p–3

Sol. p +

R Enak

=

r

(ii) (P – 3)101 – p–3 101  

a 200 

1  = a  a ²  a ²  –1× a  

1

1 Sol. a   –2 a Now, a = –1

(ii)

 p–3

4² – 3×4 + 4 =8

of a 200 

(i)

1 = –1 a² Multiply a both sides

1

(i) (P – 3)42 +

1 = 1² – 2 = – 1 a²

a² +

p² – 3p  4

Ex.78 If a 

Sol.

a² +

geisnh eeYa ridna gv.i Sni

(ii)

1 =1 a Squaring both sides

Proof a +

1 Ex.80 If p + p–3 = 1, Find the value

100

100

a4  a3  a2  a  1 (–1)4 + (–1)3 + (–1)² + (–1) + (1) =1

(i)

 

1 = (–1)12 + –1 12 = 1 + 1 = 2  

118

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(ii)

Then,

1 a 37 Break the power multiple of 3 a37 +

= a² +

1 a+ =1 a We know that a³ = –1 Put the value a

1 a × a + 36 a a 36

Then,

3

= (–1)12 × a + –1 12  a  

(iii)

(B).

a36 × a² +

1 a  a²

1 =–1 a 38

1 = 1find the value of a + a99 + a98 + a97 + a96 + a95

Ex.83 If a + a100

1 =1 a Then, a³ + 1 = 0 a100 + a99 + a98 + a97 + a96 + a95 = a100 +a97 + a99 + a96+ a98+ a95 = a97 (a³ + 1) + a96 (a³ + 1) + a95 (a³ + 1) = a97 × 0 + a96 × 0 + a95 = 0 Note: When difference of the power is 3 then the value of both terms is =0

1 =–1 a

1 =0 a a³ – 1 = 0 a³ = 1

(i) a48 +

1 = – 1 Find the value of a 1 a 48

(ii) a50 +

Sol. a+

Ex.84 If

(iii) a51 +

Sol. (i) a48 +

x 3 + = 1, Find the value 3 x

of x³ Sol. Let a =

x 3

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116 +

1 a 51

1 a 48

1 =1+1=2 116

(ii) a50 +

1 a 50

a48 × a² +

1 =–1 a We know that a³ = 1 a² + 1 = – a ........(i) a + 1 = – a²..........(ii) (a² – a + 1) (a + 1– a² ) From Equation (i) and (ii) (– a – a) (–a² – a²) = – 2a × – 2a² = 4a³ = 4 × 1 = 4 Ex.87 If a² + a + 1 = 0, find the value of a5 + a4 + 1 Sol. a² + a + 1 = 0 or 1 a+ =–1 a Then, a³ = 1 a5 + a4 + 1 a³ × a² + a³ × a + 1 Put the value of a³ = 1 a² + a + 1 = 0 (As Given)

a+

1 =–a a

Ex.85 If a +

1 = 1+1=2 117 Ex.86 If a² + a + 1 = 0 find the value of (a² – a + 1) (a + 1– a² ) Sol. a² + a + 1 = 0 or

= 117 +

a³ + a +

wwM wa. th Les aBryn

a38 +

1 a 50

Ex.88 If x1/4 + of x252 +

1 a 48  a ²

= 116 × a² +

1 a 51

1 (a3)17 + a ³ 17

1   a a ²   a²  = – 1 × a 

a³ +

1 =–1 a 50

(iii) a51 +

ERna

1 = 1² –2 = – 1 a²

a² +

1 = –1 a

1 = (–1)² – 2 = – 1 a² Multiply a both sides.

1 a2

Then,

When a +

a² +

1 = 1 (As Given) a

=a+

 a50 +

Proof :- a +

36

1 = (–1)12 × a² + –1 12  a ²  

= a² +

=–1

or a² + a + 1 = 0 Then, a³ = 1

1 a38 + 38 a Break the power multiple of 3

1 = (–1)² – 2 a²

kgei snhe eYari dnag v.iSn ir

1 a + 37 = 1 a 37

= a² +

x³ =–1 27 x³ = – 27

1 =a+ = 1 (As Given) a So,

1 =–1 a

=a+

x    =–1 3

1

1 a²

1 116  a ²

1 = 1 find the value x 1/4 1

x 252

1 =1 x 1/4 Squaring both sides 1 x1/2 +  1/2 = 1² – 2 = – 1 x

Sol. x1/4 +

119

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Again Squaring both side (ii) x

1 = (–1)² – 2 = – 1 x Then,

96

x+

(x6)16 +

1 =( – 1)16 + –1 16  

1

(x³)

x 252

Ex.90 If x +

+ x³  

84

of

1 (1)84 + 1 84   =1+1=2

1 = x

1 x  x²

–3×

3

x² +

wwM wa. th Les B aryn 1 = x

1 ) x²

1 = x Then, x

3

1 =0 x³ Multiply x³ both sides x6 + 1 = 0 x6 = – 1

x² +

3 (As Given)

1 = x²

 3

x92 +

1 x 90

(ii) x96 +

1 x 96

1 Sol. x + = 3 x Then x6 = – 1

1 x 90

1 =–1 x 92

1 x 93 Break the power multiple of 6

(ii) x93 +

x90 × x³ +

x 

1 =( – 1)15 + –1 15  

= – 1– 1 = – 2

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x6+ 1 + 4 = 4

0 (ii) x102 + x96 + x101 + x95 + x100 , + x94 96 6 x (x + 1) + x95(x6 + 1) + x94(x6 + 1) = x96 × 0 + x95× 0 + x94 × 0 = 0 Note: When difference of the power is 6 then the value of sum of both of terms value is 0. TYPE – VIII

1 x  x3

If x+y = 0 Then x = –y or x = 0, y = 0 If x2+ y2 = 0 Then x2 = 0, x = 0 And y2 = 0, y = 0 If (x–1)2 + (y–2)2 = 0 then we can say x = 1 and y = 2 Ex.92 If (x + 3)2 + (y - 5)2 + (z + 2)2 = 0 find the value x + y + z Sol. (x + 3)2 + (y - 5)2 + (z + 2)2 = 0

90

1 = (–1)15 × x3 + –1 15  x 3  

= –( x3 +

1 ) x³

1 = x

3

x

1 = 3 x Then, x6 + 1 = 0 (i) x6 + 5

Sol. x +

–2

1 =3–2=1 x²

1

6 15

1 = 3 find the value of x (i) x6 + 5 (ii) x102 + x96 + x101 + x95 + x100 , + x94

Ex.91 If x +

2

So,

3 , Find the value

of

(x6)15 +

1 x 93

90

R Enak

  3

= –( x² +

x³ +

(i) x90 +

(ii) x93 +

1 = (–1)15 × x² + –1 15  x ²  

1 = 3 x Cube both sides

(i) x90 +

1 x 92

x90 × x² +

Proof x +

Ex.89 If x +

3 find the value

1 x 92 Break the power multiple of 6

3

– 3× 3

1 =0 x 93

x93 +

Sol. (i) x92 +

Then x6 + 1 = 0 or x6 = – 1

1 x³ + = x³

1 = x

(i) x92 +

TYPE – VII When x +

1 =3 3 –3 3 =0 x³

3

So,

= 1+1=2

1 84

x³ +

6 16

x 

 3

r

+

1 = x³

geisnh eeYa ridna gv.i Sni

x

x3 +

1

1 x+ =–1 x x³ = 1 Now, 252

Then,

1 + 96 x

0 x = -3 x+y+z =

0 y=5

0 z = -2

-3 +5-2

=0 Ex.93 If(a - 4)2 + (b - 5)2 + (c - 3)2 = 0 (As Given)

find the value

a+b c

120

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TYPE – IX

Sol. (a - 4)2 + (b - 5)2 + (c - 3)2 = 0

=

Rationalising factor of the s urd

0 c=3

a+b 4+5 = =3 c 3 Ex.94 If x2 + y2 + z2 + 4x + 2y + 5 = 0 find the value x2 + y3 + z4 Sol. x2 + y2 + z2 + 4x + 2y + 5 = 0 x2 + 4x + 4 + y2 + 2y + 1 + z2 = 0 (x + 2)2 + (y + 1)2 + z2 = 0 0 0 0 x = -2 y = -1 z = 0 x2 + y3 + z4 = (-2)2 + (-1)3 + (0)4 =4-1+0=3 Ex.95 If a2 + b2 + c2 = 2 (a -2b -2c) - 9 find the value of a3 + b4 - c2 Sol. a2 + b2 + c2 = 2 (a -2b -2c) - 9 a2 + b2 + c2 - 2a + 4b - 4c + 9 = 0 2 a - 2a + 1 + b2 + 4b + 4 + c2 - 4c + 4 = 0 (a-1)2 + (b + 2)2 + (c - 2)2 = 0

find

and

Ex.100. If x = 2  3 , find the value

Sol.

1 4 x Then, x

1 1  y x 5–2 6

3 = x 

=

1 1  x 1 1 1 x

y=

Sol. (x+y - z- 1)2 +(y +z -x- 5)2 + (z + x - y - 3)2= 0

0 0 0 x+y-z=1 y+z-x=5 z+x-y=3 Adding all three eqations x + y - z + y + z - x + z + x - y= 1 + 5 + 3 x+y+z=9 9 =3

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

1 1 3 +x + x x Put the value 52 + 4 = 56

1 1  =1 1 x 1 y



3 2



–3



3– 2



–3



1 x ² of x ² – 3x  1 x4 

Sol.

–3

–1



x 52 6

1 5–2 6 x x

Then 1  x

Ex101. If x  5  2 6 Find the value

,

–1

3 2

3– 2



1 1 + 3 x x

= x³ +

Find the value of (x + 1) + (y + 1) Sol: x =

x6  x4  x2 1 x3

= x³ + x +

1 1 x 1 x 1  = 1 x x 1 1  x Note:

x=

1  52 x3

Now,

1 1  = x 1 y 1

If xy = 1 then

1 3   4 – 3  4 3 x

x3 

1 y x Put the value of y in equation

=

x = 2 3 1 1  = 2– 3 x 2 3

1 1  x 1 y 1

1 Sol. x = 52 6

Now,

x6  x4  x2 1 x3

of

Find the value of

x+ y + z

Then x+ y + z =

a b

1 1 ,y= 52 6 5–2 6

Ex.98 If x =

Ex.99

0 0 2x = -y x = –1 x = –1, 2x = –y 2×–1= –y y=2 Ex.97 If (x + y - z - 1)2 + (y + z - x - 5)2 + (z + x - y - 3)2 = 0

=

1  a b. x

wwM wa. th Les aBryn

0 0 0 a=1 b = -2 c=2 a3 + b4 - c2 = (1)3 + (-2)4 - (2)2 = 1 + 16 - 4 = 13 Ex.96 If 5x2 + 4xy + y2 + 1 + 2x = 0 find the value of x & y Sol. 5x2 + 4xy + y2 + 1 + 2x = 0 4x2 + 4xy + y2 + x2 + 2x + 1 = 0 (2x + y)2 + (x + 1)2 = 0

x

kgei snhe eYari dnag v.iSn ir

0 b=5

ERna

0 a=4

1 1 x 1 x  1 = x 1 1  x 1 x

1  10 x

–3

=y

1 y x Now, (x + 1)–1 + (y + 1)–1 1 1 = x 1  y 1

1 1  = x 1 1 1 x

3 Then x 

x3 

1 = (10)3 – 3×10 x3 1 = 970 x3

Now, x divide or nominator and denominator 1 x² x ² – 3x  1 x4 

121

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Sol. x = 3 – 2 2

x4 1  2 x x x = x ² 3x 1 –  x x x

=

x

970 970 = 10 – 3 7

Put the value of y =

x

1 x   14 x Then,

7x x ² – 5x  1 x divide or nominator and denominator 7x x x ² – 5x  1 = x

R Enak

=

Sol. x = 5  2 6

3 1 x ²  3x  5   x x² = 1 x²  x²

1 1   3 x    5 x² x  1 x²  x²

194  3  14  5 241 = 194 194

Ex.103 If x = 3 – 2 2 , y = 3  2 2 Find the value of

1 3   6 – 3  6 3 x

x ²  y ²  2xy Find the value of x ³  y ³  3xy

1 5–2 6 x xy = 1 Then,

wwM wa. th Les B aryn

x 4  3x 3  5x 2  3x  1 x² x4 1 x²

Sol.

x² y²  y x

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Ex.106.

1 2 x² 1 x³  x³  3 Now,

1  10 x Then 1 2 x²   10  – 2 x² x

1  10 x

1 x

(As Given)

2p   , Find the p² – 2p  1 4

value of p 

x ²  y ²  2xy x ³  y ³  3xy

x² 

x –5

7 7 = 10 – 5 5

(As given)

1 x

7

7 1 x  –5 x

x

1 5–2 6 y= x

Put the value of y 

1  10 , Find the value x

7x x ² – 5x  1

of

Ex.104. If x = 5  2 6 and xy = 1,

x 4 3x 3 5x 2 3x 1     x² x² x² x² x² = x4 1  x² x²

=

Ex.105. If x 

1 x  3  198 x 3

1  194 x² x² divide or nominator and denominator

98  2 100 = 970  3 973

1 6 x

x3 

1 2 x²   14  – 2 x²

=

1 x

1 x³ Now, x³ 

1 7–4 3 x

x² 

1  970 x³ Put the value in equation,

2

x 74 3

x² 

1 3  10  – 3  10 x³

x³ 

x²  1  1   1  x  x x

x 4  3x 3  5x 2  3x  1 of x 4 1

=

x³ 

x² y²  y x

1 –3 x

Ex102. If x  7  4 3 , find the value

Sol.

1 y x

r

=

1 x3

And,

geisnh eeYa ridna gv.i Sni

x3 

1  32 2 x

Then,

1  98 x²

x² 

Sol.

1 p

2p   p² – 2p  1 4

8p = p² – 2p + 1 10p = p² + 1 divide by p both sides, p

1  10 p

Ex.107 If x 

a  1 , find the value of x

x²  x  a x ³ – x²

122

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a 1 x x² + a = x x² – x = –a

a b c + + = 13 a –1 b – 2 c – 3

x

....(i) ....(ii)

1 2 3 Ex.110 If + + ...... x 1 x  2 x  3

x²  x  a x³ – x² From equation (i)

1007 = 1249 x  1007

x x x³ – x²

x x x find + ...... x 1 x  2 x  1007

2 = x² – x From equation (ii)

Sol.

Type – X

a2 – bc b2 – ac c2 – ab   a2  bc b2  ac c2  ab = 1 Find 2a 2 2b 2 2c 2  2  2 a  bc b  ac c  ab

c 2 – ab a 2 – bc b 2 – ac 1 2 1+ 2 +  2 a  bc b  ac c  ab 1 =1+3 2

2

2

a – bc  a  bc b – ac  b  ac + a 2  bc b2  ac

+

c 2 – ab  c 2  ab 4 c 2  ab

2a 2 2b 2 2c 2  2  2 4 2 a bc b ac c ab Note: In this type of question when base d is s am e we add or substract 1 

1 2 3 + + = 10 , a –1 b  2 c  3 Find the v alue of

Ex.109 If

a b c + + a –1 b – 2 c – 3

Sol.

x x x + ...... x 1 x  2 x  1007 = –242 Ex.111 If x = 101 find the value of



x x 2 – 3x  3

Sol:



2

x x – 3x  3

Ex.113 : If

1 2 3 + + = 10 a –1 b  2 c  3

1 2 3 +1+ +1+ +1 a –1 b 2 c 3 = 10+1+1+1



a³+b³ =a+b(a²–ab+b²) 1

53  1



3

 1  5 3   13    

3

1 25 – 3 5 1

= A 3 25 + B3 5 + C Find the value of A + B – C

Sol.

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3

1 25 – 3 5 1

= A 3 25 + B3 5 + C

1

1

1 2 3 5 3  1 = A 5 3  B  5 3   c  5 1 =     5 1 6  

1





x³ – 3x² + 3x We add and substract 1 then, x³–3x²+3x–1+1 = (x–1)³+1 Put the value x = (101–1)³ +1 = 100³+1 = 1000001 Ex.112 If x = 102 find the value of x(x²+3x+3) Sol: x(x²+3x+3) x³ + 3x² + 3x We add and substract 1 then, x³+3x²+3x+1–1 = (x+1)³–1 Put the value x = (102 + 1)³ +1 = 103³+1 = 1092727+ 1 = 1092728

wwM wa. th Les aBryn



2



ERna

a2 – bc b2 – ac c2 – ab   =1 a2  bc b2  ac c2  ab Added 1 every terms

1

 1  1   5 3  1     2   1 1 1      5 3  1  5 3  – 5 3  1  (1)2           

x –1 = 1249 – 1007 x  1007 

2

–

1

= A  5 3  B  5 3  C

1 53

 1    5 3  1 multiply and divided    

1 2 3 + + ...... x 1 x  2 x  3

2 1 3 –1+ –1+ –1....... x 1 x 2 x 3



Ex.108 If

Sol.

2 53

1007 = 1249 x  1007

x²  x  a –2 = x³ – x² a

2

1



kgei snhe eYari dnag v.iSn ir

Sol.

2  1 53 1  = A 5  3  B  5 3   c   6 6   Comparison of the terms

A = 0, B =

1 1 and C = 6 6

Then, A+ B – C = 0 + Ex.114 : If

3

1 1 – =0 6 6

1 16  3 4  1

=A 3 16 + B3 4 + C Find the value of A + B + C Sol.





3

1 = A 3 16 + B3 4 + C 16  3 4  1 1

2 43

 1  43  

–

1 43

2

1

= A  4 3  B  4 3  C 1

 – 1 multiply and divided  

 1  1   4 3 – 1     2  1  1   1   4 3 – 1  4 3   4 3  1  (1)2           

123

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–1

 

– 13

1



=

A  4

1

1   x – 1   x  1 x

 1  B  43   c    

1  x² –1 =x x So, x = 7232 120 find the

=

1



1 2 4 3 1 = A 4 3  B  4 3   c   –   3 3  

Comparison of the terms 1 1 A = 0, B = and C = – 3 3

Then, A+ B + C = 0 +

Ex.



4  x² – 4 =x x Then, x = 675 Ex.121 Find the

3

0

=

4

x – 12x³ – x³ + 12x² + x² – 12x + 11 – x³ 0

So, 16 will be added the product of number

Maximum or minimum value of quadratic equation Quadratic equation in general form 1. When a>0 (In the equation ax²+bx+c.) The expression gives minimum value

v alue

of

16   x – 4  x  4  x

0

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6 2 =3

Divided by 2 in difference =

Square of 3 = 3² = 9 (30 × 36) + 9 = 1080 + 9 = 1089 1089 is perfect square of 33 So, 9 will be added the product of number

4ac – b ² 4a Max =  When a 0

=

Then, min value = here, a = 5,

b = –8,

Now, min =

=

4ac – b ² 4a

c = 14

4  5 14 – 8 ² 45

280 – 64 216 = 20 20

Min value =

54 5

Ex.125 Find the minimum value of (x–2) (x–9)

124

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Sol. (x–2) (x–9) = x²–2x–9x+18 = x²–11x+18 Coefficient of x² is 1 which 1>0 Then min value

Relation between A.M & G.M A.M > G.M When a,b,c are +ve Real number

A.M = –49 4 Ex.126 Find the maximum value of 12–7x–x² Sol. 12 – 7x – x² or –x² – 7x +12 Here a 1

1 m

1 =2 m

x4 , where x is Real number x 1 8

ERna

Sol.

1 –3 x² –1 Add and substract 1, Now,

Then, max value =

Let x² –1 = m

(iii)

1 m –2 m 1 =2 m

Then, 2–2 = 0 So, minimum value F(x) = 0 Ex.129 Find the minimum value of x² 1 –4 x² 1 Sol. Add and substract

+

Now, (x²+1) + Let x²+1 = m



1 –4–1 x2 1



1 =2 x4

2 1

wwM wa. th Les aBryn

4 Min value = x 

1 –3+1 x ²  –1

Minimum value of m 

x4 x8 1 componendo and dividendo divide by x4

x4 1 x4 = 1 x8 1 x4  4 4 x x

x² 

(x²–1) +

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

1 7 m Minimum value = 7+2 = 9 Ex.131 Find the maximum value of

1 value of f (x) = x ²  –3 x² – 1

=



1 +5+2 x2 – 2

Now, m 

> 2

1 =2 mn When m is + ve Real number Ex.128 If x >1, find the minimum

–60 – 144 –12

1 5 x² – 2

Let x²–2 = m

mn 

Sol.

1  5 , here x > 2 x² – 2

(x²–2)+

And We can say minimum value

4  –3  5 –  –12 ² = 4  –3

a b 2 and G.M = ab Thus when a, b and c are three numbers Then, A.M = a+b+c G.M = 3 abc

>

Minimum value of m 

4ac – b ² So, max value = 4a

A,M =

x²+

2 When m +ve Real number.

=

(ii)

1 m

2

4  –112 –  –7 ² 4   –1

–204 = 17 –12 a and b are two numbers Then,

1 m

kgei snhe eYari dnag v.iSn ir

m

4ac – b ² 4a = a = –1, b = – 7 and c = 12

=

M

1 =2 m

= 2–5 = –3 So, minimum value of F(x) = –3 Ex.130 Find the minimum value of

A.M > G.M

Max value =

–48 – 49 –97 97 = = –4 –4 4 Ex.127 Find the maximum value of 5–12x–3x² Sol. 5–12x–3x² or –3x²–12x5 Here coefficient of x² is –3 which less than 0, Now, a = –3, b = –12, c = 5

1 m , G.M =

2

min value =

=

M

1 –5 m

Minimum value of m 

1 m and are two number m

4 1 18 –  –11 ² 72 – 121 = = 4 4 1

Then, m 

If x+y will be given then xy

will be maximum When x=y Ex.132 Find maximum value xy if x+y = 20 Sol. x+y = 20 For maximum x = y Then x = y = 10 Max value x y = 10×10 = 100 Ex.133 Find the maximum value xy, if x + y = 25 Sol. x + y = 25 For maximum x = y Then x=y =

25 2

125

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Put the value m Then Max value = (4)³ = 64 Ex.137 If x+y+z = 24, maximum value of (x–1) (y–2)(z+3) Sol. For max value (x–1) = (y–2) = (z– 3) = m x = m +1 y = m +2 z = m –3  x+y+z = 24 Put the value x, y and z m +1 + m+2 + m – 3 = 24 3m = 24 m=8 Max value = (x+1) (y+2)(z –3) = m×m×m = m³ So, max value = (8)³ = 512

r

625 4 (iv) If x y will be given then x + y will be minimum when x = y here x & y (+ve) Real number Ex.134 Find the minimum value of x + y if x y = 16 Sol. x y = 16 For minimum value x = y = z x=y=4 Min value = x + y = 4 + 4 = 8 Ex.135 Find the minimum value of x + y + z if x y z = 216 Sol. xyz = 216 Min value x = y = z x×x×x = x³ = 216 =

x=6 x=y=z=6 Minimum value = x + y + z = 6+6+6 = 18 Ex.136 If x+y +z = 18, Find the Maximum value of (x–1) (y–2)(z–3) Sol. For maximum (x–1)= (y–2)=(z–3) = m x = m+1 y = m+2 z = m+3  x+y+z = 18 (As given) Put the value x, y, and z m+1 + m+2 + m+3 = 18 3m = 12 m=4 So, max value of (x–1) (y–2)(z–3) = m×m×m = m³

geisnh eeYa ridna gv.i Sni

25 25 Max value = x y × 2 2

EXERCISE

1.

The value of

a

 1  1  1  1      1   1  1  1    x  x  1 x  2 x  3

5.

If

is:

b 

3

4

1

(c)

x4

1

(d)

x

6.

x4 x

7.

If x = 7 – 4 3 , t h en t he v alu e of

 1  x     x

8.

(b) 8 3

(c) 14  8 3

(d) 14

9.

3.

If

2a  b

a  4b

 3 , then find the value of

a b

a  2b

10.

5

(a)

10

2

(b)

9

7

(c)

10

(d)

9

4.

If A : B =

5

1

2

:

3

8

B:C=

1 3

:

5 9

(c) 2

and C : D

(d) 3

2



2

x – 3

5 D

: then find the ratio of A : B : C 6 4 :D (a) 6 : 4 : 8 : 10 (b) 6 : 8 : 9 : 10 (c) 8 : 6 : 10 : 9 (d) 4 : 6 : 8 : 10

12.

If

x 

3 1 3 –1

& y 

value of x2+ y2 is : (a) 14 (b) 13 (c) 15

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1

x

   

is :

(b) 2

(c) 2 2 (d) 3 3 If p = 9 99, t hen t he v alue of

3 p p 2  3 p  3  1 is



t he

(d) 37

3 –1 3 1

15.



, t hen

2 xy  y 2 2 x –y

is

16.

If

3

(b)

4

xx

x

17.

4 3

 

= x x

4 (a)

9

(c) x

3

7

(d)

7 3

, then x equals

2 (b)

3

9 (c)

4

3 (d)

2

If a = 7, b = 5 and c = 3, then the value of a2 +b 2 +c 2 –ab –bc –ca is (a) 12 (b) –12 (c) 0 (d) 8

1 1 1 1 1 1 1 1 1 1 1 1 . .  . . –3 . .  . . 3 3 3 4 4 4 3 4 5 5 5 5 18. is 1 1 1 1 1 1 1 1 1 1 1 1  .  .  . – .  .  .    3 3 4 4 5 5 3 4 4 5 5 3  equal to (a)

(d) 10

(b) 999 (d) 1002

If x : y = 7 : 3 then the value of

+ 6E ,

3

=

x –

(a)

1 2 3 4 5 31 1     .....   x 4 6 8 10 12 64 2 (c) 36

14.

is

then the value of 5A + 3B + 6C + D + 3E is (a) 53.6003 (b) 53.603 (c) 153.6003 (d) 213.0003 If 3x+3+7= 250, then x is equal to (a) 5 (b) 3 (c) 2 (d) 1

value of x is (a) 31 (b) 32

   

(a) 1000 (c) 998

If 47.2506 = 4A + 7B + 2C +

If

x =3 + 2 2 , t h en t he v alue of

(a) 1

(a) 1 (b) –2 (c) 3 (d) 2x – 4 Given that 100.48 = x, 100.70 = y and xz = y2 , then the value of z is close to (a) 1.45 (b) 1.88 (c) 2.9 (d) 3.7

7

11.

c

14.4

x – 1

is :

(a) 3 3

(b) 1

If

is equal to

 , then the value of x is 0.144 x (a) 144 (b) 14.4 (c )1.44 (d) 0.0144 If 1 < x < 2 , t hen t he v alu e of If

wwM wa. th Les B aryn

2.

144

(b) x +4

7

13.

a  b c

then

R Enak

(a) 0 (a) 1 

c



2 3

(b)

3 4

(c)

47 60

(d)

49 60

126

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19.

If x = 7 – 4 3 , then

x 

1 x

(a) x

is equal

6



1 6 x

(b) x

1 8

(d) x

8



1 8 x

42.

If 4b

2

1  2, then the value of 2



b

to:

20.

(b) 2

(c) 3

5 1

If a 

& b 

5 –1

(c) x

(d) 4

5 –1 5 1

2

31.

a  ab  b 2 2 is a  –ab  b

22.

23.

24.

25.

3

(b)

If

3a – 5b

(c)

1 4

x a

8

(c) –

1 1 (d) 5 4

a b (b) < ab 2

If x 

3

2 equal to

, then

(a) 1

29.

1 x

1 1 x

+

(b) 2/

1– x

1– 1– x

If x 



(c) 1

If x

2

(b) 12 3

1 3

1 3

46.

47.

 z , then {(x + y – z) +27 (c) 0

48.

41. is

equal to

Rakesh Yadav Readers Publication Pvt. Ltd.

(b) 4

If x –

1 x

t he n

  1– x  

 4 , then

(a) 5 2 (b) 2 5

x 4 x –4

x 4 – x – 4 (a) 2.4 (b) 3.2 If x 

 2  1

(c) 9

(c) 34

 1 x      x

(d) 16 50.

51.

(c) 4 2 (d) 4 5

If

7

5 7

(d)

t he

5 v al ue

of

1 3

is

(c) 1

(d)

3

 2 then x is equal to (c) 4

,

(d) 5

the

v alu e

of

1   3  is x 

x

(b) – 2 (c) + 2 (d) 3 2

2 2

– x 1

3 

 x 1

2

, then the value of

is

(b) – 5

(c) 6

(d) 8

2 1 If x  3  8 , then x  2 is equal to x (a) 38 (b) 36 (c) 34 (d) 30 If x  5  2 6 , t hen t he v alue of

   

(d) 30 is equal to

–

(a) 4

2 1 If x  3  8 , then x  2 is equal to x (b) 36

If

 1 x     x 

49.

If x, y and z are real number such that (x – 3)2+ (y –4 )2+(z –5)2 = 0 then (x + y + z) is equal to (a) – 12 (b) 0 (c) 8 (d) 12

(d) 18

1 x –

x

If 5 x  12 x  13 x , then x is equal to

(a) 38

5

,

2

(c)

1 x  1 – x

(a) 0

25

40.

3

x 

 3  x 

(d) 27

37.

39.

12

(d) 1

3

a b  =1, a  0, b  0 the value of b a

4

12

(b)

then x is

 2 then x is equal to

(a) – 3 (b) – 1

a3 +b3 is (a) 0 (b) 1 (c) – 1 (d) 2 If p = 99, then value of p(p2 +3p +3) is (a) 999 (b) 9999 (c) 99999 (d) 999999

(a)

If

   

1   3  is x 

y

(d) 5

(c) – 1

3 x – 3 – x

5

3  2 , then t he v alue of

If

38.

(c) 2

3 x  3 – x

(a)

45.

(d) 2

(c) 18 3 (d) 24 3 If x + y = 7, then the value of x3 + y3 +21xy is (a) 243 (b) 143 (c) 343 (d) 443 1 3

1 3 is b (b) 1

 3 3  3 –6  3 2x –1  ,        =     5  5  5 

If

44.

36.

3

 1 1  2 1  2 1  x   x   x  2 – 1 x  2  1    x  x  x  x 

1

xyz} equals : (a) –1 (b) 1

(c) 2 – 3 (d) 2 If for non-zero, x, x2 – 4x – 1 = 0, the

1 value of x  2 is x (a) 4 (b) 10 (c) 12 30.

(b)

(a) 6 3

is

2

3

equal to (a) –2 (b) 2

1 1 1  1 and b   1 then c  b c a is equal to If a 

 3 x   

35.

a b (c) > ab (d) All of the above 2 27. If x, y are two positive real number and x1/3= y1/4, then which of the following relations is true? (a) x3 = y 4 (b) x3 = y (c) x = y4 (d) x20 = y15 28.

33.

34.

If a  b,then which of the following statements is true?

a b (a) = ab 2

8b

–4 7 (d) 3

wwM wa. th Les aBryn

26.

1 (b)

1 6

x

4 (b) – 3

(a) 0

2

a perfect square ?

18

32.

 5 , then a : b is equal to

For what value(s) of a is x 

1

–

43.

4 (c) 3

(d)

(a) 2:1 (b) 2:3 (c) 1:3 (d) 5:2 If p : q = r : s = t : u = 2 : 3, then (mp + nr +ot ) : (mq +ns +ou) equals : (a) 3:2 (b) 2:3 (c) 1:3 (d) 1:2 If x : y = 3 : 4 , then (7x + 3y) : (7x –3y) is equal to : (a) 5 : 2 (b) 4 : 3 (c) 11 : 3 (d) 37 : 19

(a) 

72

5

4 3 5 3 If a = 4.36, b = 2.39 and c = 1.97, then the value of a3 – b3 – c3 – 3abc is (a) 3.94 (b) 2.39 (c) 0 (d) 1 3a  5b

6

 a 7  b , then the value of

ERna

21.

7–2

If

11 (a) 3

4

(a)

x

a is

2

3

–

(a) 0

, then the

value of

8

kgei snhe eYari dnag v.iSn ir

(a) 1

   x

1

x 

is.

(a) 2 2

(b) 3 2

(c) 2 3

(d) 3 3

If x 

3  2 , t hen t he v alue of

 2 x    (a) 4

1   2  is : x  (b) 6 (c) 9

(d) 10

127

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9

x

If

 6 , t he n t he v al ue of

x

 2 x   

63.

9   2  is x 

If

a b c   = 1, then the value 1–a 1–b 1–c

of

1 1 1 + + 1– a 1– b 1– c

(a) 1 (b) 9

(c) 10

(d) 12

2x – y

3 1 If 2p+ =4, then value of p  3 is p 8p 1

53.

54.

55.

64.

If

1

 3 , t he n t he v al ue of

x

58.

65.

(b)

2

7 (c)

2

11 (d)

66.

If x  a 

1

67.

1



2 2 x y  xy 3 3 is x –y

– 5 x 3

,

is

76.

12

12 (c) 1

(a) 0

77.

(d)

7

1 3x

2

11 33 3

(d)

,

71.

4

(b)

79.

81.

(d) 9

82.

6

5

1

(b)

4

(d)

2

73

(c)

77

770

(d)

74 77

(b) 1331000 (d) 1330030

1 1 If x2+y2+ 2  2  4 , then the value of x y x2+ y2 is (a) 2 (b) 4 (c) 8 (d) 16 If x2 = y + z, y2 = z +x, z2 = x +y, then the value of

1 +

1 +

y 1

z 1

is

(a) – 1 (b) 1 (c) 2 (d) 4 If a2 +b2 = 2 and c2 +d2 = 1 then the value of (ad –bc)2 +(ac+bd)2 is (a)

9

1 (b)

2

4ab 83.

73

1 , then the value of a² –331a is a 3 11

4

7

If x varies inversely as (y2 – 1) and x is equal to 24 when y = 10, then the value of x when y =5 is (a) 99 (b) 12 (c) 24 (d) 100 If x2 + y2 +2x + 1 =0, then the value of x31 +y35 is (a) – 1 (b) 0 (c) 1 (d) 2

Rakesh Yadav Readers Publication Pvt. Ltd.

If

77

1

.

(d)

(c)

(a) 1331331 (c) 1334331

(d) 64

1

a b c is :

(b) 4

730

x 1

(c)

(c)

If 1.5x = 0.04y, then the value of

(a)

5 15

(c) 4

1

(d) 2 3 If (a – 3)2+(b – 4)2+(c – 9)2= 0, then the

4

80.

1

1

(a)

+

2 2 y –x 2 2 is y 2xy  x

4

,

 5 . Find the value of

 2 0 x 1

2

b – c  a –  bc – a 

+

(b) 3

value of

78.

2a 2  3c 2  4e 2 =? 2b 2  3d 2+ 4 f 2

If 2 x 

(d) – 2

c – a a – bb – c

5x

72.

5

3

(b)

(c) 65

(b) 3

2

, then x equals

2

(b) 61

1

2

3

a c e   b d f =3

6x

2

a  b  b  c c  a 

32 x If 1  = , then the value of x is 961 31

If

(c) –

75.

(a) –4

10 5

, then value of

–b –c)–3 then the value of 2a –3b +4c is (a) – 1 (b) 0 (c) 1 (d) 2 If (3a +1)2 +(b – 1)2 +(2c –3 )² = 0, then the value of (3a +b +2c) is equal to; (a) 3 (b) – 1 (c) 2 (d) 5 The value of the expression

= a b 6 , then the values

15

(a) 2

a b 2a 3b  , then value of is 3 2 3a – 2b

(b)

5

(d)

4

,–

15

then

70.

(a) 1 (b) 37 (c) 324 (d) 361 If 50% of (p –q) = 30% of (p +q), then p : q is equal to (a) 5 : 3 (b) 4 : 1 (c) 3 : 5 (d) 1 : 4

5

3



48  18

(a) 63

69.

– 3 p  3 – 1 is

12

(c) 3

3

4 3 5 2

wwM wa. th Les B aryn

is

(a) 100 (b) 101 (c) 102 (d) 1000 If x = 19 and y = 18, then the value of

(a)

2

(b) 4 (d) (136)1/3

9

(b) 2 (c) (d) 20 2 20 If p = 1 01, t hen t he v alue of

If

x

6

If a, b, c are real and a2 +b2 +c2 = 2 (a

1

100

9

(a)

(c)

1



62.

If

(b)

2

5

=

74.

2x

3x

of a and b are respectively

1

and y  a – , then the a a value of x4+y4–2x2y2 is (a) 24 (b) 18 (c) 16 (d) 12 If a = 11 and b = 9, then the value of

3p p

61.

1 –

If

(a) 2 (c) 16

1

2

 5 , then

(b)

2

68.

60.

x

(a) 5

5

a 2  b 2  ab     a3 – b3   

59.

1

If x 

(b) 1

5

1

 1  0 a0 then the value of a (a4 –a) is: (a) 0 (b) 1 (c) 2 (d) – 1

(a)

3 x y

(a) 2

4 (c)

5

R Enak

57.

If a 

, then value of

3 (b)

5

1

5 x 2

1

3x – y

equal to

3

56.

2

1 (a)

x is : x² – x  1

2

=

2

(d) 4

1

x  2y

1

x³ 

(a)

If

(c) 3

2x

 1  x  x   is:  

is :

(a) 4 (b) 5 (c) 8 (d) 15 If a4 +b4 = a2b2, then (a6 +b6) equals (a) 0 (b) 1 (c) a2 + b 2 (d) a2b4 + a4b2

x

(b) 2

If

geisnh eeYa ridna gv.i Sni

(a) 8

x

73.

r

52.

If x =

a b

(c) 1

(d) 2

a  b , the value of

x  2a x  2b + is x – 2a x – 2b (a) a (b) b (c) 2 ab (d) 2

128

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1

If m 

= 4,

96.

m –2 2

find the value of m – 2





1



m–2

 85.



is

1 (a) 3

(b) – 3

(c)

x y 86.

If x–y =

(a) 18

3

4

(b)

(c)

4

(d)

4

2 

y

2 

z

88.

yz

zx

3

xy

=?

1

1 + a c b a a  b b  c       1 + c a c b    (a) 1 (b) 0 (c) – 1 (d) – 2 If a + b +c = 0, then the value of

2 2 2 a b c is 2 a – bc

(a) 1

94.

95.

(b) 2

a c

b (c) 3

is:

(d) 0

1 If x   3 then the value of x18+x12+x6 x +1 is (a) 0 (b) 1 (c) 2 (d) 3 If for two real constants a and b the expression ax3 +3x2 –8x +b is exactly divisible by (x + 2) and (x –2) (a) a = 2, b =12 (b) a = 12, b = 2 (c) a = 2, b = –12 (d) a = –2, b = 12 If x2 –3x +1 = 0, then the value of 3 1 x  3 is x (a) 9 (b) 18

(d) 16

x

1 y

3

(c) 27

(d) 1



1

108. If x 

2

(c) 123 (d) 125 and x is real , then the

x

(b) 3 xyzabc

xyz

(d)

(b) 1

(c) 3

(c) 13 13

(d) 10 13

1

(a) 118

= 5, t h en t he v al ue of

1

27 x

3 is:

1

(b) 30

2

10

(c)

23

z

+

(a) 1

y –z

is

(b) 3

and its reciprocal is 1

and a  0 , b  0 , then the value of a3 +b3 is (a) 2 (b) –1 (c) 0 (d) 1 104. If x =2 –21/3 + 22/3 then the value of x3– 6x2+18x + 18 is (a) 22 (b) 33 (c) 40 (d) 45 105. If a3 –b3 –c3 –3abc = 0, then (a) a = b = c (b) a +b+ c =0 (c) a + c = b (d) a = b + c

1

(d) 2 2 111. If a3b = abc = 180, a, b, c are positive integers, then the value of b is (a) 110 (b) 180 (c) 4 (d) 25 112. If a, b ar e r ational number and

(c)

a – 1 2  3  b a  b  is

2  a , the value of

(a) – 5 (b) 3 (c) – 3 (d) 5 113. If ax 2 +bx +c =a(x –p)2, then the relation among a, b, c would be (a) abc = 1 (b) b2 = ac (c) b² = 4ac (d) 2b =a +c 114. If a +b + c + d =1, then the maximum value of (1 +a) (1 +b) (1 +c )(1 + d) is 3

1  (b)     2 

(a) 1

3

4

3 5  (c)     (d)   4 4 

115. If a2 +b2 +c2 +3 = 2(a+ b +c) then the value of ( a +b+ c) is (a) 2 (b) 3 (c) 4 (d) 5

a b

7

27

(c) 0 (d) 1 102. If x + y = z, then the expression x³+y3 – z³ +3xyz wil be equal to : (a) 0 (b) 3xyz (c) –3xyz (d) z3 103. If the sum of

=?

(d) –2

(b) 8 13

3 8x 

1 19 is

7 7 (d) 29 17 110. If x +y = 2z t h en t he v al ue of

x –z

(a) 6 13

2x

x

(b)

25

x

3 1 value of x  is 3 x

3x 



7

(a)

abc

1 100. If x  4 = 119 and x >1, then the x 4

17

5 x – 2y 7 x  2y

abc

(a) 0

value of x

(a) 1 (b) 0 (c) 2 (d) –2 109. If x : y = 3 : 4, then the value of

1 3 3 – x – y is: 3 3 x y

Rakesh Yadav Readers Publication Pvt. Ltd.



(b) 7

3

If xy(x +y) = 1, then the value of

101. If

2 107. If p = 124, 3 p p  3 p  3  1  ?

( x0, y0, xy )

3xyz 99.

106. If a = 2.361, b = 3.263 and c =5.624, then the value of a3 +b3 – c3 +3abc is (a) (p – q) (q – r)³ + (r – p)³ (b) 3 (p – q) (q–r) (r–p) (c) 0 (d) 1

(a) 5

3abc

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(a) 0 (b) 1 (c) 2 (d) 3 If x2 +y2 –4x –4y +8 = 0, then the value of x –y is (a) 4 (b) – 4 (c) 0 (d) 8 If x = b +c –2a, y = c +a –2b, z =a +b –2c, then the value of x2+y2–z2+2xy is (a) 0 (b) a +b +c (c) a – b + c (d) a + b – c For real a, b, c if a² +b² +c² = ab +bc +ca, then value of

93.

x y

(c)

2

(a) (xyz)2 (b) x2 +y2 +z2 (c) 9 (d) 3 If a +b+c = 0, then the value of

92.



(c) 24

 x  y   z             =? then  a  b   c 

1

1

If x + y + z = 0, then

91.

of

xyz

87.

90.

1 =

3

(a)

x

89.

v al ue

98.

, the numerical

3

3

(b) 36

ERna

4

t he

then the value of x3–y3 is (a) 0 (b) 1 (c) – 1 (d) 2 If x =a(b–c), y = b(c –a) and z =c(a–b),

xy =

7

3

f in d

If

1

(b) –

,

97.

value of xy is (a)

4x 2

1

a –b a b

3 

1 3 8x  3 . 8x

2

(a) – 2 (b) 0 (c) 2 (d) 4 If a2 +b2 +2b +4a +5 =0, then the value of

If

1

x

kgei snhe eYari dnag v.iSn ir

84.

116. If x –

1 1 2 = 5, then x  2 is : x x

(a) 5

(b) 25

(c) 27

(d) 23

117. If x = 3  2 2 , then the value of

 1     x –  is: x  (a) 1

(b) 2

(c) 2 2

(d) 3 3

129

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118. If a +b +c = 0, then the value of

a 2  b2  c 2 is a 2 – bc (d) 3

119. If n = 7+ 4 3 , then the

 1   n  n 

a6 –

(b) 4

(c) – 4

(d) – 2 3

3  2 , t h en t he v alue of

(a) 2 2 (b) 2 3 (c) 2 (d) 3 121. If p + q =10 and pq = 5, then the p q numerical value of q  p will be (c) 22

x  3xy  y is x 2 – 3xy  y 2

(a)

30 31

; fn

(a)

1 3

(b)

(b)

70 31

x

(c)

35 31

(d)

43 (a)

x y z   gS , rks% b c c a a b

x –y y –z z –x   b –a c –b a –c

(b)

135. If

x

x y z   a b c

x –y y –z z –x   (c) c b c (d) None of the above is true 124. If x –y = 2, xy = 24, then the value of (x2 +y2) is (a) 25 (b) 36 (c) 63 (d) 52

23 x

3

 x  

(a) 1

(b) 2

(c) 0

(d)

3

126. If a = x +y, b = x – y, c = x +2y, then a2 +b2 +c2 – ab – bc – ca is (a) 4y2 (b) 5y2 (c) 6y2 (d) 7y2

1 127. If x   2, x  0 t hen v alue of x 1 2 x  3 is equal to x (a) 1 (b) 2 (c) 3 (d) 4 128. If

a b  =1, a  0, b  0 the value b a

of a3 +b3 is (a) 0 (b) 1

(c) – 1

(d) 2

(c) –

1 3

(d) –

2 3

1  5 , t he n t he v al ue of x

47

(b)

is

21

136. If

r eal,

41

(c)

23

x 

1

45

(d)

0

x

21

a nd

4 1



  x

x 

 5 x  



(a) 322

x

(c) 16

(d) 25

 3 , t he n t he v al ue of

47 (c)

4

a

145. If

b

+

1–a

1–b

60

49 (d)

60

c

+

1

value of

1–c

= 1, t hen the

1

1

is 1–a 1–b 1–c (a) 1 (b) 2 (c) 3 (d) 4 146. If a, b, c are real numbers and a2 +b2 + c2 = 2 (a –b –c) –3 then the value of 2a –3b +4c is (a) – 1 (b) 9 (c) 1 (d) 2 147. The v alue of t he expr e ssion





2

2

a – b + b – c  b – c c – a a – bc – a

+

c – a a – bb – a (d) 2 3 148. If (x –3)2 +(y –5)2 +(z – 4)2 =0 then the value of

x

(b) 3

2

y

1   5  is x  (b) 126

(c) 123 (d) 113

3 (b) 4

(c)

2 (d) 1

2

 9

138. If x + y + z =6, then the value of (x –1)3+(y – 2)3 +(z – 3)3 is (a) 3(x – 1) (y +2) (z –3) (b) 3(x + 1) (y – 2) (z –3) (c) 3(x – 1) (y – 2) (z +3) (d) 3(x – 1) (y – 2) (z –3) 139. If a +b +c =6, a2 +b2 +c2 = 14 and a3 +b3 +c3 = 36, then the value of abc is (a) 3 (b) 6 (c) 9 (d) 12

Rakesh Yadav Readers Publication Pvt. Ltd.

3

(b)

3

1

(b) 9

1

2

(a)

(a) 0

is

1 1 4 137. If m  4 = 119, then m – =? m m (a)

1 1 1 1 1 1 1 1 1 1 1 1 . .  . . –3 . .  . . 3 3 3 4 4 4 3 4 5 5 5 5 144. 1 1 1 1 1 1 1 1 1 1 1 1  .  .  . –  .  .  .   3 3 4 4 5 5 3 4 4 5 5 3 

2

1  3 = 0, t hen t he v alu e of x

(a) 4

x2 y2 125. If the expression 2 + tx + is a pery 4 fect square, then the values of t is

2 3

a b  =? b a

x 4  3x 3  5x 2  3x  1 x4 1

37 31

wwM wa. th Les B aryn

(a)

133. If (a2 +b2)3 = (a3 +b3)2 then

134. If

x y z   , then b c c a a b

123. If

(b)

2

of

1 1  x y will be :

2 5 (c) 6 (d) 6 3 132. If a3 – b3 = 56 and a – b =2 then value of a2 +b2 will be : (a) 48 (b) 20 (c) 22 (d) 5 1 3

(d) 18

122. If x=3+ 2 2 and xy = 1, then the value 2

1  2 will be a6

value of

(a)

(b) 20

1  3 , then the value of a

(a) 1 (b) 2 (c) 3 3 (d) 5 131. If x3 +y3 = 35 and x + y = 5, then the

 1  x    is x 

(a) 16

130. If a 

   is: 

(a) 2 3

120. If x =

value of

r

(c) 2

140. If a +b = 1 and a3 + b3 +3ab = k, then the value of k is (a) 1 (b) 3 (c) 5 (d) 7 141. If a = 34, b = c = 33, then the value of a3 +b3 +c3 – 3abc is (a) 0 (b) 111 (c) 50 (d) 100 142. If (2x )(2y) =8 and (9x )(3y) = 81, then (x, y) is: (a) (1,2) (b) (2,1) (c) (1,1) (d) (2,2) 143. The expression x4 –2x2+ k will be a perfect square when the value of k is (a) 2 (b) 1 (c) – 1 (d) – 2

geisnh eeYa ridna gv.i Sni

(b) 1

R Enak

(a) 0

2  1 x    =3 then the value of (x72 +x66 129. If  x  +x54 + x 24+x6 +1) (a) 0 (b) 1 (c) 84 (d) 206

z

2

 25

(c)

16

is

(a) 12 (b) 9 (c) 3 (d) 1 149. x varies inversely as square of y. Given that y = 2 for x = 1, the value of x for y = 6 will be equal to

1 (d) 3 9 150. If x2 –y2 = 80 and x –y = 8, then the average of x and y is (a) 2 (b) 3 (c) 4 (d) 5 (a) 3

(b) 9

(c)

1

151. The third proportional to x

2

y

x y   y  x    

and

2 is

(a) xy

(b)

(c) 3 xy

(d) 4 xy

xy

130

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43 3

152. The value of

is

74 3

(a) 5 3 – 8

160. If x and y are positive real numbers and xy =8, then the minimum value of 2x + y is (a) 9 (b) 17 (c) 10 (d) 8 161. If the expression x² +x +1 is written in

(b) 5 3  8

(c) 8 3  5

 1 2 2  x     q , then the pos 2

(d) 8 3 – 5 the form ,

5 3

x  20

t he

v alue

of

1

x  12 +

(a) 1

x – 12

(b) 2

154. If x = 5 –

(a) 

is (c)

(d)

3

(c) 

2 1

(b)

2



is

21

3 –

7

1



155. The v al ue of

2 1

(d)

2



7 –



x  x 

c –a

(b) 2

x

1

(c) – 1

(d) 0

(a) –a

(b)

1

x

1

(c) a

a

(d) 

1

x

 99 , f in d t he v al ue of

2x

2

158. If

1

(b)

6

4x – 3 x

1

(c)

2

4y – 3

+

4z – 3

+

y

3

z

; fn

(a) 8

the value of (a) 9

x

xy

= a,

z

(c) 4

xz

is

(d) 6

yz

= b and

(a)

ab  bc – ac

(b)

2abc (c)

y z

ac  bc – ab

2abc ab  ac – bc 2abc

(d)

2



yz

y

ab  bc – ac

2



zx

z

(b) – 3

(c) 9

xy

=?

2

2

a

2

b

2

1 b

 1 and

(c) 1

(d) 2

s

2

Rakesh Yadav Readers Publication Pvt. Ltd.

(c) 32

(b) 16

(c) 23

a 2  b2 



b a

2

 

2

(d) – 23

xy + yz + zx =0 , t h en

r ks

(x,y,z  0)

(a) 3 (b) 1 (c) x + y +z (d) 0 174. If a +b +c = 9 (where a, b, c are real numbers ), then the minimum value of a2 +b2 +c2 is (a) 100 (b) 9 (c) 27 (d) 81 175. If a2 +b 2 + 4c 2 = 2(a +b – 2c) – 3 and a, b, c are real, then the value of (a2 +b 2 +c 2) is (b) 3

(a) 3

1

(d) 2

(c) 2

4

1 4

176. Number of solutions of the two equations 4x – y = 2 and 2x –8y +4 = 0 is (a) zero (b) one (c) two (d) infinitely many

a

 and



b

5

b

15 

c

16

,

then

2 2 18c – 7a 2 2 is equal to 45c  20a

(a)

1

2

(b)

3

5

(c)

3 4

(d)

1 4

1 1 1 1 1 1      2 2 2 xy yz zx then the x y z relation among x, y, z is (a) x + y + z = 0 (b) x + y = z

2 1 169. If x = 3 + 2 2 , the value of x  2 is x

(b) 30

(d) 8

178. If x  0, y  0 and z  0 and

2

(b) 0

(c) 6

(d) 1

2

c

x

 1 1 1      x2 – yz y2 – zx z2 – xy   

177. If

 1 , then the value of abc is :

(a) a2 +b2 +c2

(a) 36

1 is 2



(b) 2

; fn

(d) 0

s – a  s – b  s – c 

=c,

1

2

a +b +c = 0, then the value of

(a) – 1 (b) 3 (c) – 3 168. If a + b +c = 2s, then

where a, b, c are all non -zero numbers, then x equals to

2abc

1 c

xz

x

=0 t hen

1



y

(b) 3

xy

159. If

1



3

167. If a, b, c are non - zero a 

b

1

(d) 4 x –

a b b c c a  a b c             a b  b c c  a a  b c

4

1

172. If a2 +b2 = 5ab, the value of

173. If

(b) 4x + 3

3

(d) 3

1

(d)

(d) 40

(c) 9

 102x  2

1

(a)

(c) 35

(b) x2 + y2 z2

166. If

4 1 4 (b) 2 (c) 3 (d) 3 9 9 9 9 171. If x2 – 3x + 1= 0, then the value of (a) 2

(a) 10

(a) (xyz)2

100 x

1 2 is x



is : (a) 32

165. If x + y + z = 0, then

a

2

, then the value of

2

2 x  3 – 5 , then the value of x – 16x +6 is (a) 0 (b) – 2 (c) 2 (d) 4

wwM wa. th Les aBryn

157. If

(a) 4x +

164. If

1

 – , t hen t he v alue of x a a x – x 2 is :

156. If

(b) 30

(c) 4x – 3

x  (a) 1

1

2 4 3x +5x – 2 3 is : b –c

b c

(a) 25

x

 

x

163. One of the factors of the expression

7  3

a –b

a b

3

x

x2 + x 

1 3 a2+ 2 +3a– is a a

(b)

3

x

3 2

162. If a 2 – 4a – 1 = 0, t h en v alu e of

 7  3

c a

(d) 

3

21 , then the value of

32 – 2x – 1

(b) 

2

5

x

(a)

3

ERna

x – 20

sible values of q are

2



kgei snhe eYari dnag v.iSn ir

4 15

x 

153. If



170. If x  3 –

(d) 34

1

(c)

1 

x

1 

y

z

=0

(d) x = y = z

131

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t  1 , then the value 2

179. If x = 3t, y =

189. If x + y + z = 6 and x2 +y2 +z2 = 20 then the value of x3 +y3 +z3 –3xyz is (a) 64 (b) 70 (c) 72 (d) 76

of t for which x =2y is

3

1

2

1 

(c) – 1

2

x a

(d)

3 (a) – 8

2

is a perfect square,

5 then a is

1 1 1 1 (b)  (c) (d) – 100 10 10 10 181. Find the value of x for which the expression 2–3x –4x2 has the greatest value. (a)

(a) –

41

(b)

3

(c) 

3

(d)

41

16 8 8 16 182. The expression x4 –2x 2 +k will be a perfect square if the value of k is 1 (a) 1

(b) 0

(c)

5x 183. If

4

=

3

   2x 

(a) 15

(b) 10

(c) 20

1 3 3 – x – y is: 3 3 x y

x >1

x

x

a nd

3

–

2



(d) – 2 1  83 2 x

a

(a) 2 3

(b) 2

(c) 3 3

(d) 0

x

187. If

t h en

(b) 750 (d) 760

 1 2  a    = 3, then  a

x

2

– 2x  1

=

1

3

3



1 3 =?

a

(a) –

188. If

(b) 110 (c) 81

1



 4 , 

x

 3 , t h en t he v alue of

9

(b) –

(d) 124

t hen t he v alue of

1 x  4 is : x (a) 64 (b) 194 (c) 81

x 1

=1, then (x+1)5 +

(c) 4

5

x  1

(d) 8

2 b

b –a

then the value of a3 +b3 is

(a) 6 ab (b) 0 (c) 1 (d) 2 201. If p – 2q = 4, then the value of p 3– 8q 3 – 24pq – 64 is (a) 2 (b) 0 (c) 3 (d) – 1 202. If x = – 1, t hen t he v alu e of 1 1 1 1 1 1 1       –1 99 98 97 96 95 94 x x x x x x x

(d) 124

(a) 1

Rakesh Yadav Readers Publication Pvt. Ltd.

(c) 2

(d) 3

3 204. If x  2  3 , t hen the v alue of x

3



1 3 is x

(a) 8

(b) 9

(c) 2

(d) 4

3 205. If x  5  2 , then the value of x3 –6x2 +12x– 13 (a) – 1 (b) 1 (c) 2 (d) 0 206. The simplest form of the expression

2 2 – p p –1 p   3 2 2 2p  p p  3p p  1 p

2

1

(a) 2p2

(b)

(c) p +3

(d)

2p

7

(c)

1 1 1   , then the value of a b a b a3 – b3 is (a) 0 (b) – 1 (c) 1 (d) 2 197. If a +b +c = 0, then a3 +b3 +c3 is equal to (a) a +b + c (d) abc (c) 2 abc (d) 3 abc 198. If x =y = 333 and z =334, then the value of x3 +y3 + z³ – 3xyz is (a) 0 (b) 667 (c) 1000 (d) 2334 199. Out of the given responses one of the factors of (a2 –b2)3+(b2–c2)3+(c2–a2)3 is (a) (a + b)(a–b) (b) (a + b)(a+b) (c) (a – b)(a–b) (d) (b – c)(b–c)

4

(b) 1

207. If

x

1 x

2

1 p3

 2 , t he n t he v al ue of

1

1

195. If x 

200. If a 

x

25

(d) 11 8 8 8 194. If a + b + c = 15 and a2 +b2 +c2 = 83 then the value of a3 +b3 +c3 –3abc (a) 200 (b) 180 (c) 190 (d) 210

, then the value of

1 x  3 is : x

 x  

, then

1  2 is 3 x

3

(a) 64

3– 2

196. If

1 3 is x

(a) 764 (c) 756 186. If

2 193. If 2x+

equals (a) 1 (b) 2

(c) 3

3 2

and y =

the value of x3 +y3 is : (a) 950 (b) 730 (c) 650 (d) 970

wwM wa. th Les B aryn

185. If

(b) 1

3 2

(d) 5

184. If xy (x +y) = 1, then the value of

(a) 0

3– 2

2

1



(a) 0

(b) 8

192. If x =

x3+

, then the value of

is:

(c) 2 2 (b) 1 191. If x = a – b, y = b – c, z =c – a, then the numerical value of the algebraic expression x3 +y3 +z3 –3xyz will be (a) a + b +c (b) 0 (c) 4(a +b+c) (d) 3 abc

1

2x ²  5 x  1

  x 

1 (d)

 1  x –    x

geisnh eeYa ridna gv.i Sni

180. If x

(b)

190. If x = 1– 2 , the value of

R Enak

(a) 1

2

1 3 +c and a, 203. If 3 = 3 4  32  1 a 4  b 2 b, c are rational numbers then a + b + c is equal to

r

1

(b) 0

(c) – 2

(d) – 1

 2 x  

1  3 1  x  3   is 2  x  x  (a) 20 (b) 4 (c) 8 (d) 16 208. If a,b,c be all positive integers then the least positive value of a3 +b3 +c3 – 3abc is. (a) 0 (b) 2 (c) 4 (d) 3 209. When f(x) = 12x3 – 13x2 –5x+7 is divided by (3x +2), then the remainder is (a) 2 (b) 0 (c) – 1 (d) 1 210. If the equation 2x2 – 7x +12 =0 has two 

 roots and , then the value of

   

is (a)

7

(b)

2

1 24

(c)

7 24

(d)

97 24

 3 3 211. If x   4 (a3 + b3) and 3x  4 3 x x (a3 + b3), then a2 – b2 is equal to

(a) 4 (b) 0 (c) 1 (d) 2 212. The term to be added to 121a2+64b2 to make a perfect square is (a) 176 ab (b) 276 a2b (c) 178 ab (d) 188 b2a 213. If a = 2+ 3 , t h en t he v alue of

 2  a 

1   2 a  (a) 12 (b) 14 

(c) 16

(d) 10

132

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214. For what value(s) of k the expression

p

1  4

p k

2

is perfect square ?

1 (b) 

(a) 0

1 (c) 

4

224. If a + b + c +d =4, then find the value of

(d) 

8

1

2

x

x 2

1

(b)

1

1

(c) x –

x

1 – c 1 – d1 – a

x 1

1

1 a

(c) 3

 



1

(a) 0

(b) 5

1

225. If x –

1

 is b c (d) 1

x

x



3 = 1, then the

3x

4

2

218. If

(c) 4

(d) 9

1

1

is

 a

b

x

2

2

(a) 1

(b) 2

b

1

=b 

c

=c 

(a  b  c),

a

then the value of abc is (a)

1

(b)



2

(c) 0

(d)

1

2 220. If (x –2) is a factor of x2 +3Qx –2Q, then the value of Q is (a) 2 (b) – 2 (c) 1 (d) – 1 221. If a +b = 12, ab = 22, then (a2 +b2) is equal to (a) 188 (b) 144 (c) 34 222. If

x 

3 –

1

3

an d

x

then the value of (a)

3

2

y



(d) 100

y 

3

1

3

y

x

x

(d) 4

t he

v alue

of

2

2

2 x

3x

3

(c)

(d) 0

=?

(b) 10, – 6

x2 is

+ 3 xy 3 z 2

x ²  y²  zx  y – 3z

when x = +1, y = – 3, z = – 1,

is

(a) 1

(b) 3 3

232. If

(c) 16 3 (d) 2 3 223. If x² +ax +b is a perfect square, then which one of the following relations between a and b is true (a) a2 = b (b) a2 =4b (c) b = 4a (d) b2 =a

x

 2  x 



(a) 20

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x 3

(b) 0

1 x

(c) – 1

(d)

1 2

 2 , t he n t he v al ue of

1  3 1  x  3   is 2  x  x  (b) 4

(c) 8

1 6 is

x

2

1 will be

3

x 

2 2

5 2,

(d) 16

t h en

t he

v alue

– 3x – 2

– 4x – 3

is equal to

(a) 0.1785 (b) 0.525 (c) 0.625 (d) 0.785 237. If a = 2.234, b= 3.121 and c = –5.355, then the value of a3 +b3+c3–3abc is (a)– 1 (b) 0 (c) 1 (d) 2 238. If x2 +y2 +1 =2x, then the value of x3 +y5 is (a) 2 (b) 0 (c) – 1 (d) 1 239. If 3(a2+b2+c2)= (a +b +c)2 then the relation between a, b and c is (a) a = b = c

 1 , t he n t he v al ue of

1 – x 



numerator and denominator of x is: (a) 3 (b) 4 (c) 5 (d) 7

2x

(a) 1 (b) – 1 (d) 2 (d) – 2 230. If x = –2k and y =1 –3k, then for what value of k, will be x = y ? (a) 0 (b) 1 (c) – 1 (d) 2 231. Find the value of

,

2

x

x

4

3

(c) 6, – 10 (d) 4 228. If a2 +a + 1 =0, then the value of a9 is (a) 2 (b) 3 (c) 1 (d) 0 229. If



6

x  1 – x – 1 2 2 = 2, then the sum of x  1 – x – 1

236. If

(a) 8

1

wwM wa. th Les aBryn

219. If a 

1

(c) 2 3 (d) 4

 5 , then x

(a) 18 (b) 15 (c) 21 (d) 30 235. If x is a r at iona l number and

1 2 =66, then the value of x



– 1  2x x

x

x

1 2 x

(b)

x

227. If

6

+

4 2 4 226. If x +y =15, then (x–10)3+(y –5)3 is (a) 25 (b) 125 (c) 625 (d) 0

2  3 a  2 – 3 b  1 then the

value of

t h en

1

(a)

(b) 1

x

(c) 1

ERna

(a) – 1

+

 5x – 3

1

1 1 + 2 2 a 1 b 1

1 ,

–

value of

x

(a) 12098 (b) 12048 (c) 14062 (d) 12092 234. If x² –3x+1 =0, then t he value of

1 – d1 – a1 – b



then the least value of

217. If a 2  3 = b 2 –

1 – d

1

1

(b) 5



is

x (d) x x 216. If a, b, c are positive and a+b+c =1,

(a) 9

1–c

1

kgei snhe eYari dnag v.iSn ir

(a)



x

+

1 – a1 – b1 – c

1

1–b

215. The reciprocal of x 

233. If x 

1

(b) a = b

c

(c) a < b < c (d) a > b > c 240. If x(x–3) = – 1, then the value of x 3(x 3 – 18) is (a) – 1 (b) 2 (c) 1 (d) 0 241. The factors of (a2+4b 2+4b–4ab–2a– 8) are (a) (a –2b–4) (a –2b+2) (b) (a –b–2) (a + 2b +2) (c) (a +2b–4) (a +2b+2) (d) (a +2b–4) (a –2b+2) 242. The value of 1

1 2ax –  2 2 2 4 2 2 4 a²  ax  x a – ax  x a a x  x

is (a) 2 (b) 1 (c) – 1 (d) 0 243. If x =11, then the value of x5–12x4+12x3– 12x2+12x –1 is (a) 5 (b) 10 (c) 15 (d) 20 244. I f p = 9 9 , t h e n t h e v a l u e o f p (p 2 +3p +3) is (a) 10000000 (b) 999000 (c) 99999 (d) 990000 245. Which one is not an example of an equality relation of two expressions in x: (a) (x +3)2 = x2 +6x +9 (b) (x + 2y)3 = x3 +8y3 +6xy (x +2y) (c) (x +2)2 =x2 +2x + 4 (d) (x +3 ) (x – 3) = x 2 – 9

133

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(b) 1

(d) 6

1

1

247. If a 

(c) 2

3 , then the value of a18 +a12

 a

+a6 +1 is (a) 0 (b) 1 (c) 2 (d) 6 248. If x = 997, y = 998 and z =999 then the value of x2 +y2 +z2 –xy –yz –zx is (a) 0 (b) 1 (c) – 1 (d) 3

x

249. If

1 x

3x x

2

 3 , t he n t he v al ue of

– 4x  3

2

4 (a)

is

– x 1

3 (b)

3

5 (c)

2

2

5 (d)

3

x

x

then

4 x

x 3

2

1 is equal to

a

b

3 (b)

2

3

is

3b

(a)

2

b

(b) b

(c)

255. If x >1 and x 

value of x

4

–

1 x

20736

20736

1

12

(d)

1 4 is

(a) 44

20736

(d) 64

3 2 3 3 2 3 x  a  a b + a – a  b ,

then x3 +3bx is equal to (a) 0 (b) a (c) 2a (d) 1

262. If

24 x

x

1

x

1 12 =7 t h en t he v al ue of

1 36

(a)343 (b) 433 (c) 432 (d) 322 263. If P = 99 then the value of P(P²+3P+3) (a) 989898 (b) 998889 (c) 988899 (d) 999999 264. If x = 2 then the value of x3+ 27x² + 243x + 631 (a) 1321 (b) 1233 (c) 1231 (d) 1211 265. If x² + y² + z² = 2(x + z –1), then the value of; x³ +y³+z³ = ? (a) – 1 (b) 2 (c) 0 (d) 1 266. If

x

1 x

=1, t h en t he v al ue of

x² – x  2

=?

(a) 2/3 (b) 2

5 – 3 5 3

(c) 1 and y=

(d) 4

5 3 5 3

, then

x ²  xy  y ² the value of

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65 63

]

then the value of

+3=?

a³ (a)

7

3

(b)

(c)

21

(d)

21

16 16 64 16 269. If x = z = 225 and y = 226 then the value of: x² + y³ + z³ – 3xyz (a) 765 (b) 676 (c) 674 (d) 576 270. If x²+x=5 then the value of: 1

(b)

3

x ² – xy  y ²

67 65

(c)

=?

69 67

(d)

x  3

(a) 140 (b) 110 (c) 130 (d) 120 271. If m = – 4, n = –2, then the value of m³ – 3m² +3m + 3n + 3n² + n³ is (a) 124 (b) –124 (c) 126 (d) – 126 272. 2x – ky + 7 = 0 and 6x – 12y + 15 =0 has no solution for: (a) k = – 4 (b) k =4 (c) k = 1 (d) k = –1 273. If x = 332, y = 333, z = 335, then the value of x³ + y³ + z³ – 3xyz is (a) 7000 (b) 8000 (c) 9000 (d) 10000

1 , then the simplest 2 3 value of x is: (a) 1 (b) –2 (c) 2 (d) –1 275. If m – 5n = 2, then the value of (m 3 – 125 n³ – 30mn) is : (a) 6 (b) 7 (c) 8 (d) 9 276. If x = a 3 b a 3 b........ , t he n t he value of x is: (a) 5 ab ³

(b) 3 a 5b

(c) 3 a 3b

(d) 5 a 3b

277. If x 

63 61

1 x

is: (a) – 4

x

278. If

x

2

(a)

+3=0

274. If 2+ x 3 =

72

57895 (d)

(c) 52

259. If 3 a  3 b  3 c , then the simplest value of (a+b–c)3 +27abc is (a) – 1 (b) 3 (c) – 3 (d) 0 260. If 4x+5y =83 and 3x : 2y =21 : 22, then (y –x) equals (a) 3 (b) 4 (c) 7 (d) 11

267. If x =

20736

a³ –

4 a

(x + 3)³ +

(b) 48

59825 (b)

(d) – 1

2

1 3 t  3 is: t

3

, then the

x

57985 (c)

2

=2

58975 (a)

2b

5 (c)

2

268. If 4a –

258. If t 2 –4t+1 =0, then the value of

wwM wa. th Les B aryn

254. If 3a 2=b 2  0, t hen t he v alue of

3

2x  16x – 3

is

1 (a)

x

(a) 216 (b) 192 (c) 198 (d) 204 251. If a + b + c = 0, then the value of (a + b – c)2 + (b + c – a)2 + (c + a – b)2 is (a) 0 (b) 8abc (c) 4(a2 +b2+c2) (d) 4(ab+bc+ca) 252. If p3 +3p2 +3p = 7, then the value of p2+2p is (a) 4 (b) 3 (c) 5 (d) 6 253. If x = 2015, y = 2014 and z =2013, then value of x2+y2+z2 –xy–yz–zx is (a) 3 (b) 4 (c) 6 (d) 2

a  b – a – b 2 2 a  b  a – b

when

1 

of

261. If

250. If x  3  2 2 ,

6

–x

x = 9999 is (a) 1111 (b) 2222 (c) 3333 (d) 6666 257. If a3+b3 = 9 and a+b =3, then the value

1  3 is a

(a) 0

256. The value of

r

a

3

=3, then the v alue of

3

R Enak

246. If

4x

geisnh eeYa ridna gv.i Sni

 1 2  a     a

2

12 =2, then the value of x –

(b) 4

1 x

(c) 2

1 x12

(d 0

=1, t h en t he v al ue of

 3x  1

is: x ²  7x  1 (a) 1/2 (b) 3/7 (c) 2 (d) 3 279. If x +(1/x) =2, then the value of x7 + (1/x5) is: (a) 25 (b) 212 (c) 2 (d) 27 280. The term, that should be added to(4x² +8x) so that resulting expression be a perfect square, is: (a) 2x (b) 2 (c) 1 (d) 4 281. If 999x + 888y = 1332 and 888x + 999y = 555 Then the value of x + y is? (a) 888 (b) 1 (c) 555 (d) 999

134

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1

282. If x =

, y= , then t he 2 3 2– 3 value of 8xy(x² + y²) is (a) 112 (b) 194 (c) 290 (d) 196

x2 x –2 x 2 –

x –2

, then the value

of a² –ax is (a) 2 (b) 1 (c) 0 (d) – 1 284. If a + b = 1, find the value of a³ + b³ – ab – (a² – b²)² (a)0 (b)1 (c) – 1 (d)2

1 = 5, then the value of a–3 1 (a–3)³ – (a–3)³ (a) 7 (b) 14 (c) 2 (d) 5

285. If a –

286. (3x –2y) : (2x +3y) = 5 : 6, then one of

3x 3  x

the value of

+ 3y – 3y

1 (a)

  

(b) 5

(c)

y–

3 +

(b) 3

P² – 2P +1

 1  p+ p    (a) 7

(d) 25

2 = 0 , then value of

x

2

(d) 0

(d) a = b = c

1 1 1 1 – – 289. x = a 2 +a 2 , y = a 2 –a 2 ,

t h en

(c) 14 (d) 16

n= 5– 5– 5–......

(b) m+n+1=0

(c) m+n–1=0

(d) m–n–1=0

+

2x

+

2y

2

2

2z

=0, t hen t he

2

+ + is x y x (a) 20 (b) 10 (c) 5 (d) 15 292. If 2s = a + b + c, then the value of s(s – c) + (s –a) (s – b) is (a) ab (b) 0 value of

(c) abc

(d)

x y

is

x –y

(b)

2

x

2

(a)

x y

(c)

307 If

–y

=

x –1

2

2

a+b+c 2

Rakesh Yadav Readers Publication Pvt. Ltd.

3

x a  

abc

x ² –y² , then the value of

a2

2a

(b)

a²+2 2a

(d)

a²+4

x ²+y²

4a a²+4 4a a²+2

a b   2, then the value of a – b is: b a

(a) 2

(b) –1

(c) 0

(d) 1

308. If x( x+y+z)=20, y(x+y+z)= 30, & z(x+y+z)=50, then the value of 2(x+y+z) is: (b) 10

(c) 15

(d) 18

309. If x+y=4, x²+y²=14 and x > y. Then the correct value of x and y is: (a) 2– 2 , 3

(b) 3, 1

(c) 2+

(d) 2+

3 , 2– 3

3,2 2

310. If for non-zero x, x² – 4x – 1= 0 the value of is (a) 4

1–y

a

and

b

x² 

(b) 10

1 : x²

(c) 12

(d) 18

a+c

b



1 y

a

, then the

3– 2

311. If a 

3+ 2

and b =

3+ 2 3– 2

, then

a

2

b

+

b

2

a

value

of : (a) 1030 (b) 970 (c) 1025 (d) 930 312. If (2a – 1)² + (4b – 3)² + (4c + 5)² = 0 then

x –y 1  xy

–b

is:

2 (b)

ab –b

a

2

b

a² + b² + c²

is:

2

2ab

(a) 3

3 8

(b) 2

3 8

(c) 0

(d) 1

3 8

2

3

y  b 

+

a³ + b³ + c³ – 3abc

the value of

2ab (c) (d) 2 2 a b 2ab 2 2 304. If a + b + c²– ab – bc – ca = 0 then a : b : c is: (a) 1 : 2 : 1 (b) 2 : 1 : 1 (c) 1 : 1 : 2 (d) 1 : 1 : 1 305. If x = a(b – c), y = b(c – a), z = c(a–b) then the value of

a

2 xyz

a 2 =

is

x 1

a

(d)

abc

(a) 20

2

is : b (a) 2 (b) 1 (c) 0 (d) 3 301. If p³ – q³ = (p – q) {(p – q)² – xpq} , then find the value of x is: (a) – 1 (b) 3 (c) 1 (d) – 3 302. If x + y + z = 6 and xy + yz + zx = 10, then 3 3 3 the value of x + y + z – 3xyz is: (a) 36 (b) 40 (c) 42 (d) 48

then among the following the relation between m & n holds is

291. If

1

and

a

value of

3–5z

(d) 10

x –a y–a (a) 0 (b) – 1 (c) 1 (d) 2 300. For real a, b, c if a² + b² + c² = ab + bc

303. If

290. If m = 5+ 5+ 5.......

(a) m–n+1=0

3 x yz

(c)

306. If

(b) 0

abc

+ ca, the value of

4 2 2 4 2 2 value of (x –x y –1) + (y –x y +1)

3–5y

2 5

(c)

2x

–y

+

wwM wa. th Les aBryn

(b) a  b  c

xyz

(a)

(a)

–2y 2x (d) 2 2 2 2 x –y y –x 299. If x + y = 2a, then t he value of a

(c) a  b = c

3–5x

is

(b) 1

(c)

(a) a = b  c

(b) 12

, then the value of

296. If x = 2, y = 1 and z = – 3, then x3 + y3 + z3 –3xyz is equal to (a) 6 (b) 0 (c) 2 (d) 8 297. (x3 +y6) (x3 – y6) is equal to (a) x6 – y12 (b) x9 – y16 (c) x6 + y12 (d) x9+ y36

(a)

288. 3(a²+b²+c²) = (a+b+c)² , then the relation between a, b and c is

(a) 13

4

2y

5

(c) 1

1

1

 x ³ – 20 2  –  y³ + 2 2  (a) 2

=

298. The sum of

is

287. If x – 3 – 2=0 and

If

2

1

25

2p 295.

ERna

283. If a =

293. If p + m = 6 and p³ + m³ = 72, then the value of pm is (a) 6 (b) 9 (c) 12 (d) 8 294. When xm is multiplied by xn, product is 1. The relation between m and n is (a) mn = 1 (b) m + n = 1 (c) m = n (d) m = – n

kgei snhe eYari dnag v.iSn ir

1

z  c 

+

3

is:

2

313. If

a  1    a 

 3 , then find the value

of a 3 0 + a 2 4 + a 1 8 + a 1 2 + a 6 + 1 (a) 0 (b) 27 (c) 1 (d) –1

1 1 1   , then the value of a b a b a3–b3 is: (a) 3 (b) 2 (c) 1 (d) 0

314. If

135

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ANSWER KEY

1. (d)

65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. 80. 81. 82. 83. 84. 85. 86. 87. 88. 89. 90. 91. 92. 93. 94. 95. 96.

(b) (b) (d) (a) (d) (d) (a) (a) (b) (c) (a) (b) (c) (b) (b) (a) (b) (d) (d) (c) (c) (a) (d) (b) (c) (c) (a) (b) (a) (c) (b) (a)

97. (a) 98. (c) 99. (c) 100.(d) 101.(b) 102.(a) 103.(c) 104.(c) 105.(d) 106.(c) 107.(d) 108.(c) 109.(c) 110.(a) 111.(b) 112.(d) 113.(c) 114.(d) 115.(b) 116.(c) 117.(b) 118.(c) 119.(b) 120.(b) 121.(d) 122.(d) 123.(a) 124.(a) 125.(a) 126.(d) 127.(b) 128.(a)

129.(a) 130.(b) 131.(b) 132.(b) 133.(b) 134.(a) 135.(b) 136.(c) 137.(a) 138.(d) 139.(b) 140.(a) 141.(d) 142.(a) 143.(b) 144.(c) 145.(d) 146.(c) 147.(b) 148.(c) 149.(d) 150.(d) 151.(a) 152.(a) 153.(b) 154.(b) 155.(a) 156.(c) 157.(c) 158.(c) 159.(c) 160.(d)

1   1   1     1  1  x  1   1  x  2     x  

1   

1

x 

1 x

  3

=

Taking L.C.M of each term.



 x 1  x   

 x  1  1  x 1   

 x  2 1  x 2   



3.

 x  3  1  x 3   



2.

1 x

× (x +4)



x 4 x

x+

=

1 7–4 3

74 3 49 – 48

1 x

×

=7–4

1

7 – 4 3

10 –10b = = –9b 9

–11b+b

Rakesh Yadav Readers Publication Pvt. Ltd.

257.(b) 258.(c) 259.(d) 260.(b) 261.(c) 262.(d) 263.(d) 264.(b) 265.(b) 266.(b) 267.(d) 268.(c) 269.(b) 270.(b) 271.(d) 272.(b) 273.(a) 274.(d) 275.(c) 276.(d) 277.(d) 278.(a) 279.(c) 280.(d) 281.(b) 282.(a) 283.(d) 284.(a) 285.(b)

225.(b) 226.(d) 227.(b) 228.(c) 229.(a) 230.(b) 231.(b) 232.(b) 233.(a) 234.(c) 235.(b) 236.(c) 237.(b) 238.(d) 239.(a) 240.(a) 241.(a) 242.(d) 243.(b) 244.(c) 245.(c) 246.(a) 247.(a) 248.(d) 249.(c) 250.(d) 251.(c) 252.(b) 253.(a) 254.(a) 255.(a) 256.(c)

3

+7+4

(c)

 

3

= 14

(giv en)

286.(d) 287.(d) 288.(d) 289.(d) 290.(d) 291.(b) 292.(a) 293.(d) 294.(d) 295.(d) 296.(b) 297.(a) 298.(b) 299.(a) 300.(a) 301.(d) 302.(a) 303.(d) 304.(d) 305.(c) 306.(b) 307.(c) 308.(a) 309.(c) 310.(d) 311.(b) 312.(c) 313.(a) 314.(d)

4 : 3



9 : 15



C : D



20 : : B : 3 3

(c)



1 3 : 2 8

8 : 6

B : C

8 5.

A : B =



A 4

a+b  a+2b

 –11b+2b

By rationalisation

4.

74 3

2a+b (c) = 3 a+4b 2a + b = 3 (a+4b) 2a + b = 3a + 12b  –a = 11b a = –11b

1



74 3

= 7 + 4 3

x = 7 – 4

3

193.(c) 194.(b) 195.(b) 196.(a) 197.(d) 198.(c) 199.(a) 200.(b) 201.(a) 202.(c) 203.(a) 204.(d) 205.(d) 206.(b) 207.(b) 208.(a) 209.(d) 210.(b) 211.(c) 212.(a) 213.(b) 214.(a) 215.(a) 216.(a) 217.(b) 218.(d) 219.(a) 220.(d) 221.(d) 222.(b) 223.(b) 224.(a)

SOLUTION

(d)

x

161.(b) 162.(b) 163.(d) 164.(c) 165.(d) 166.(c) 167.(a) 168.(c) 169.(d) 170.(b) 171.(a) 172.(c) 173.(d) 174.(c) 175.(d) 176.(b) 177.(d) 178.(d) 179.(b) 180.(c) 181.(d) 182.(a) 183.(d) 184.(c) 185.(c) 186.(d) 187.(b) 188.(b) 189.(c) 190.(d) 191.(b) 192.(d)

r

(c) (c) (c) (a) (d) (b) (d) (c) (b) (a) (c) (b) (d) (d) (c) (b) (c) (c) (d) (c) (b) (a) (c) (a) (c) (a) (a) (a) (b) (a) (d) (b)

geisnh eeYa ridna gv.i Sni

33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64.

R Enak

(d) (d) (c) (c) (c) (d) (d) (a) (c) (c) (c) (a) (b) (a) (a) (c) (a) (c) (d) (b) (c) (d) (b) (c) (b) (c) (d) (b) (d) (d) (b) (c)

wwM wa. th Les B aryn

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32.

1

5

5

3

 3 : 9  3:5

 6 : 4 18  10 : 9

:

6

a 3

=

:

C

:

D

:

5 10 10

: :

9 9

:

b 4

=

c 7

= k

a = 3k b = 4k c = 7k

136

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a+b+c 3k + 4k + 7k = c 7k

6.

144

(d)

=

0.144

x = 7.

(d)

x 8.



x =

=

= 0.0144 =

2

x

+

– 3

=

(square root cancel with square)  x – 1 + x – 3 = 2x – 4 (c) 10 0.48 = x 10 0.70 = y and x z = y 2 0.48 z

 10  0.48z =  10



1.40

48

=

35

47.2506 = 4×10+7×1+2× 0.1000 + 5×0.0100 + 0 + 6×0.0001  A = 10 B = 1 C = 0.1000

1

11.(c)

100  5A + 3B + 6C + D + 3E = 5 × 10 + 3 × 1 + 6 × 0.1 + 100 + 3 × 0.0001 = 50 + 3 + 0.6 + 100 + 0.0003 = 153.6003 (c) 3 x+3 + 7 = 250 3 x+3 = 250 – 7 3 x+ 3 = 243 3 x+ 3 = 3 5 x + 3 = 5 x = 2

1

4

×

2

6 1 = x 2

×

3 8

×

4

10

5

×

12

×.......×

2

1

7 – 4 3

=

 1 306     2 

7 – 4 3

3 1



3 –1

1 2

x

  

–2



x



2    

1 1 or 36 = x 2 2  x = 36

x=

3 – 1





y

–2



2



2  (1)

2

 

2 =

Rakesh Yadav Readers Publication Pvt. Ltd.

=

1

2 + 1

1 =

2 1

   

17.



x

3

= x 2x

=

2 18.

3 x = 2

3 x or 2

 3

2

=

9 4

(a) a² + b² + c² – ab – bc– ca

–2

2 1

x

x =    2

2



 3 x x = x 2 x  

 x x

 

x = 2 + 1 + 2



(If bases are s ame then their power is also same)

b)² + (a–b)² = 2(a² + b²)





x

= x x

xx

2 2

 

= 1000

x :y 7: 3

(c) x x x

3  1

3



3  p  1 3

(a)

1

2





2 –1

2 1– 2 1 = 2 (a) P = 999

2

2

x



2 xy  y 21  9 30 3 = =  2 2 = 49 – 9 40 4 x –y

16.

(b) x = 3 + 2

x

2 1–

3

= (3 + 1)² –2 = 16 – 2 = 14

2x

=

x

3 = 3 10 00 

1

15.

2 1

 (a+

1

=

=

49 – 48



13.

x –

3 = 3  999  1 

7 – 4 3

 3 1  3 –1   –2 =   3 –1  

64

1





7 – 4 3

×

7 4 3

 

31

2 –1

 3 p 3  3 p2  3 p  1

= 4 + 3 + 4 3

7 – 4 3

= 

=



2  3 

=  x+

2 –1

3 p p2  3 p  3  1

x² + y² = x² +

5



2

 x2 + y2 = 7 + 4 3 + 7 – 4 = 14 Alternate:-

=

 1 30  1    × 2   2   

14.

2  3 

wwM wa. th Les aBryn

10.

= 100 , E = 0.0001

1

2 1

=

42 3

=

2

y2 = =

5 (c) 47.2506=4A+7B+2C+ +6E D

D =

3 1

3 1  2 3

y

= 2.9

12

×



y2 = x

2 –1 ×

3 1

= 7 + 4 3



1

2

3 1

0.70 2

= 10

x

2 –1

=

ERna

9.

140



1 =

( gj d k i fj es; d j . k d j usi j )

y

3 –1

x2 =

10 (If a = a , if ba se e qua l pow er a re eq ua l: (x = y)  0.48z = 1.40 z =

3 1

3 –1

1 < x < 2

– 1

x =

3 1

1000  10

2



x  10

144

10000

3 –1

 

144

10 x

144

1

3 –1 and y =

1

144

 1000 

(a) x =

x

144

 x 

12.

14.4

144  1000



3 1

= 2

kgei snhe eYari dnag v.iSn ir



=

1 a – b 2  b – c 2  c – a 2  2

=

1  7 – 5 2   5 – 3 2   3 – 7 2  2

=

24   4  4  16  = = 12 2 2

(c)

1 1 1 1 1 1 1 1 1 1 1 1      – 3      3 3 3 4 4 4 3 4 5 5 5 5 1 1 1 1 1 1 1 1 1 1 1 1      –       3 3 4 4 5 5 3 4 4 5 5 3

137

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47 20  15  12 = 60 60

= 19.

(d) x



= 7 – 4 3

= 4 + 3 – 4 3 2

 

2

3

= (2) +



= 2– 3

– 2 × 2 3

2

1 2

2 – 3 

x =

22.

2

1

1

=

2– 3

= 2

2 3 2 3

3

x 



23.

1

3 + 2 +

5 1 (b)

a =

5 –1

5 –1 b =

5 1



a =

24.





5 –1

+



62 5 6–2 5 5 –1

a

2



 x1/3 

12

5

 ab  b – ab  b

=

12

m  3x  n  3x  o  3x

4

=



2 2

3

2:3



=

42 3 4

1 + x =

1 3  2 3 4

1 2 



3

2



25.

1 a   ab 2 a = 1 2 a  – ab 2 a

1

(b) x +

4

1 



Rakesh Yadav Readers Publication Pvt. Ltd.

=

 x 2

 2

2



x a

3 –1 2



1 x



2



1 3 3 3



3 –1 3–

1 3 3



1– x 1– 1– x

3 –1 2 + 3 –1 1– 2

1 3 2 = 1 3 1 2

[(A + 2BA + B ) = (A + B)²]

2



1 3 2

=

1 x

1– x

=

8

3

Similarly,

x + a2

1

3

4

1 1 x

2

2





4 11 = 21 – 12 = –3 3 4 4

 2 1

4

1 + x

2112

2

2 3 2

=

2 2



3

(c) x : y = 3 : 4

2

3 2

or 1 + x = 1 +

=

2

mp + nr + ot : mq + ns + ou

3

5

 

3 = y

(divides and multiply by 2)

m  2x  n  2 x  o  2x

4 = 21

= 3

12

y1/4 



3 2

(b) x =

mq  ns  ou

21

2

LCM of 3, 4 = 12

=

 x  7 y  3 7  3  3 y 7x  3y   4 =y  7 x – 3  = 3 7x – 3y  y  7 4 – 3

5 1

5 1 2 5  5 1– 2 5 2 2 5 – 1 

a



5 –1

 



28.

2a = 5b

a

5 1

 option (c) is correct. (d) x 1/3 = y 1/4

12a = 30b

 3 x m  n  o 

1

a + b = a +

ab

x 20 = y 15

a : b 5 : 2 (b) p : q = r : s = t : u = 2 : 3



a+b > 2





2 x m  n  o 

b

16  4 = 8

3a + 5b = 15 a – 25b



1

ab =

 x4 

wwM wa. th Les B aryn

20.

and





3 = 4

a+b 16+4 = = 10 2 2

So,

27.

1 8

a =

(c) Given that a  b Let a = 16, b = 4  by options

= 5

3a – 5 b

mp  nr  ot

x

= 2 –

3a  5 b

(d)

R Enak

x







x4 = y3 take power '5' on both sides

  

x 2– 3

26.

(a– b– c)[(a– b) 2+ (b– c) 2+(c– a) 2 ]

= 0

 [(a 2 + b 2 – 2ab = (a–b) 2]



2

(c) a = 4.36 b = 2.39 c = 1.97 a – b – c = 4.36 – 2.39 – 1.97 = 0 a³ –b³ – c³ – 3abc =



(ab = 1)

1  ab 2 4 7 1 8 a = = = 1 7 –1 6 3 2 a  – ab 2 a a

21.

= 9 – 2 = 7

and B = a

x

1 8

B =

1 a2 + 2 a

2 2 2 1 1 1 1 1 1  1 1 1   1  1 1         4    5  – 3  4 – 4  5 – 5  3   3 5 4   3        1 2  1 2  1 2 1 1 1 1 1 1          –  –  –   3 4 4 5 5 3  4 5  3 

Here, A =

= 3

a

3

1 1 1 1 1 1  3   4 – 3 3  4  5   5        2 2 2 1 1 1 1 1 1 1 1 1 3  4  5 – 3  4 – 4  5 – 5  3      



=

3

1

a +

geisnh eeYa ridna gv.i Sni

3



r

A 3 + B 3 + C 3 – 3ABC = (A + B + C) (A 2 + B 2 + C 2 – AB – BC – CA)

3

3 1



3 1



3



3 1

 138

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1 1  3 3

29.

3

(d) x 2 – 4x – 1 = 0 x 2 – 1 = 4x (divide x both sides)

1 = 4 x

x –

x2 +

x2 +

30.

=

1 x2

1 x2

– 2 = 16

= 18



( d)  x  x   

1  x – x   

 2 1   x  2 –1  x  33.

  A + B   A 2 – AB + B2   A – B   A 2 + AB + B2 

1

= A 3 +B3

×

7 2 (Rationalisation) 7 –2 2



=

7 –2

1

x

11

7 + 3

 3 3  2

1





3 2

3 2

 5





39.

1 = x

3– 2

7 and

(c)

169 = 169

(x – 3) 2 = 0

x = 3

(y – 4) 2 = 0 (z – 5) 2 = 0

y = 4 z = 5

x3 

11 3

=

1 1 1 =1, b+ =1, c+ = ? b c a

b = 2

3– 2 =

34.

c = –1 35.

1 1 = –1 +  1  = –1 + 2 = 1 a 2

Rakesh Yadav Readers Publication Pvt. Ltd.

2 3 

3



– 3 2 3

 x +

= 24 3 – 6 3 = 18 3 (c) x + y = 7 (cubing both sides) (x + y) 3 = (7) 3 x 3 + y 3 + 3(x + y)xy = 343 x 3 + y 3 + 21xy = 343 (c) x 1/3 + y 1/3 = z 1/3 (cubing both sides) 1/3 3

 x1/3  y 



= z

1/3 3



8

3

3 8

8

1

= 3+

x 1 x

8

×

= 3 –

9 –8

x+

12

1 =

3 –

3

1 1 1   = x   – 3 x   x x3 x  



4 + 3 +5

1 =

x

= 2 3

4 3



x = 3 +

1

3 2 +

4

 13



x 3 2

4

5 2 + 12 2 = 13 2

(x + y + z) 40.

 12



1

x =

4

hence, x = 4 (d) (x – 3) 2 + (y – 4) 2 + (z – 5) 2 = 0



= 9 3  11 2 + 9 3 – 11 2

x3

1 and x + = x

Put values,

c+

3

 12 x  13 x By option put x = 4

= 18 3

Compare the cofficients of constant term

1 a = 2

2



3



x

Alternate:

= a 7+b = R.H.S

(c) a+

(b) 5

= 9 3 – 11 2

x +

74–4 7 7–4

11 4 4 – 7 = – 3 3 3

32.

3

3

11 – 4 7 3

b =

38.

x 3 = 9 3  11 2

7 –2

2

–  4

a = –

3

3 2

= 3 3 2 2 9 2 6 3

wwM wa. th Les aBryn

=

 7



ERna

7 –2

b 0

37.

3– 2

    3

=

 0,

a 2 + b 2 = ab a 2 + b 2 – ab = 0 (a + b) (a 2 + b 2 – ab) = (a + b) ×0 [(multiply both sides by (a + b)] a3 + b3 = 0 (d) p = 99 p(p 2 + 3p + 3) = p 3 + 3p 2 + 3p + 1 – 1 = (p + 1) 3 – 1 = (100) 3 – 1 = 1000000 – 1 = 999999

3 2

=

 x3 =

= A 3 –B3

a b  1 b a

(a) a

1 x3

x3 +

= a 7 b

L.H.S =

=

x =

= 3 3 2 2 3 6

7 2



(c)

 x

1  3 1  1  3 6 = x  3  x – 3  = x – 6 x  x   x  7 –2

36.

1 1 b  = 1 – b b –1 a

c +

z

(cubing again both sides) (x + y – z) 3 = – 27 xyz (x + y –z) 3 + 27xyz = 0

b –1 1 = b b

1–b 1 b – = = 1 1–b 1–b 1–b

=

1  2 1   1  2 1   =  x  x  x  2 –1 x – x   x  2  1     x   x 

(b)

a = 1 –

1 1 = 1 – b, c  1–b c

 2 1   x  2 1   x

31.



b 1 1 =  b+ c =1 b –1 a

1





1 a + = 1 ..... (i) b

 x + y + 3x 1/3.y 1/3.(x 1/3 +y 1/3) =  x + y – z + 3x 1/3.y 1/3.z 1/3 = 0  x + y – z = –3x 1/3 y 1/3 z 1/3

kgei snhe eYari dnag v.iSn ir

=

Alternate:

2

3 –

8

3 –

8

8

8 + 3–

8 = 6

= 6

squaring both sides x2 +

1 2 x

+ 2 = 36

x2 +

1 2 x

= 34

139

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41.

(b)

1

x –

44.

= 4

x

3x  3–x

(b)

2 1

=

3x – 3–x

x



(by c–d rule ) – 2 = 16

3 x



x2 + 1 2 x

x2 +

= 18

A

B D A +B C +D = A –B C –D

x+

42.

1

= 20

x =

=

x

20

4 5 =2 5

(a)

1 2 b

4b 2 +

1 b  

(2b) 2 +

 2b+  

  b

1

  b

1

45.

2

 

– 4 = 2

=

2

=

b

  b

1

1 x

1 3 b

(c)

   

 2b+  = b 

6 = 6

6 6

=

1 x  1– x

1 3 x

=



x3 –

2



2

1 1– x

2

3



4 × 2

3

y–1

1   

1

y+1

3

2  2 × 2 = 2 = 3 3

(d)

5 5     3–6

2x –1

3 5  

=

x  4 

x – 4

x  4 –

x – 4

x 4



x –4

2x –1

=

3 5  



– 3= 2x – 1 2x = – 2 x = –1

Rakesh Yadav Readers Publication Pvt. Ltd.



2 +1=2

– x 1

 x 1



 

1 x

3

=

Given

2

   – 1   1    1 x  1 x



3 2

 

 y

=

3 2

3

=

  

2

x

=

x4 x –4

2 1 2 –1

2

(c)

3

=

2

2(y – 1) = 3(y + 1) 2y – 2= 3y + 3 y = – 2 – 3 = – 5

1 x

= – 5

x = 3 +

8

x = 9 + 8 + 2 × 3

1

2

x = 17 + 6

1 2 x

3 1  

= 9 again C & D rule

3

2

2

=

=

 x

1 49.

  –4

x 4



3

by C – D rule

–6

1 3 = 2 +1–

x  1  – 1   x   x  1  1   x  

x

1–

2

x  

Let =

2 –1

x

  x    x 

x

2

2

x

x



2 –1

×

2 – 1

x

48. (b)

2

1– x

1

2 1

1



2x

6

= 0

3

3 5  

1

46.

3

2

22 1– x

3 6 )

= (

+ 6

1 8b 3 + 3 b

×

 –

=

1 1 8b3+ 3 +3×2b× b b 8b 3 +

1 x  1– x

1 x  1 – x

1

=

2 –1

6

3

3

2 1

2 + 1

2

1 x 1– x  2 1– x

=

1



1 3

1 x –1 x

= 6

× 4 = 5

8

x

=

5

1 x – 1– x

=

10

x3 =



3

x =

+ 4 – 4 = 2

8

Take cube on both sides

1 x  1– x

Take cube both sides

 2b+  

x =

12

=

10

(d)

= 2

2

1

2b +

43.

24

wwM wa. th Les B aryn

 2b+  

(c)

= 9 3–x 3 + x = 27 – 9x 10x = 24

2

 

x

x =

= 3

R Enak

=

1



47.



3–x Squaring both sides

+ 2 = 20

4



3x

3x

x  

    

C

=

10

=

1

+ 2 – 2 = 18

1 2 x

x2 +

    

=

2 –1

x



3

geisnh eeYa ridna gv.i Sni



1 2 x

=

3– x

(On Squaring)

2 1

9 –1

r

1 2 x

x2 +

9 1

=

4

x2 +

1 2 x

8

= 17 – 6

= 17 + 6

8

8

8

+ 17 – 6

8 = 34

140

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50.

(c)

x = 5 + 2

6

x = 3+ 2 + 2 x = x =

2

3 ×

    3



+

3 

2

2



2

9

x =

56.

x

2 3 ×

2

53.

2

Squaring both sides

3 –

1

2p

2

+

2

= 2 51.

(d)

=

x

3 +

2 +

3 –

p +

3 +

2

x2 = 3 + 2 + 2 2

x = 5 + 2



6

6

6 +5 – 2

6 54.

p3 +

1 3 = 8 – 3 = 5 8p



55.

= 6

9 2 x

x2 +

(c)

= 9 +

x

9 9

x

9 x

3

2



On squaring

 x 



x

 

x

2 9 x

 

a2 + 1 =



a +



a2 + 1 = – a

57.

= 3

x

–1 2 a

1

= – 1

a

(c)



1

a

+ y) = a +

(x – y) = a +

 58.

  2a 



a

x +



x2 +

x

2



x

 1 x



= 3

1 2 x

1 2 x –1

= 9 – 2 = 7

= 59.

=

7 3 –1

=

7

= 2a

1 a

=

2 a

2 a

  

2

2

2  b  ab 3 3 a –b

(a3 – b3) = ((a – b) (a2 + ab + b 2))

a 1

– a +

1 a

= (4) = 16 a = 11 b = 9

(a)



x

1 a

+ a –

2

–1

x 

Rakesh Yadav Readers Publication Pvt. Ltd.

2





= 0

1 x  2 x = 1 x –1 x

1 a

 x 4 + y 4 – 2x 2y 2 = (x 2 – y 2) 2  ((x + y) (x – y)) 2

(Divide by x)

1 2 x

1

.....(ii)

a

y = a –



3

(Given)

1

x = a +

(Given)

1  2 x x 2 x x 1 –  x x x

2

x

.....(i)

– a = 0 × a = 0 (Multiply a both sides)

– x 1

=

81 9 ×x = 36 2 +2 × x x

–



 (x

x

= 36

81 x2 + 2 – 18 = 0 x

x  

1 2 = – 1 a

 a4

1



x 

x 2+

a2 +

+ 2 = 1

= 10

= 6

2 9

1

x +

Alternate:

x +



a 6 + b 6 = (a 2) 3 + (b 2) 3 (a 2 + b 2) (a 4 – a 2b 2 + b 4) a 6 + b 6 = (a 2 + b 2) × 0 = 0

(a) =

3 Prove So, x = 3



1 2 a

–1  – a = 2 fr om equat ion (i) a and (ii) a3 = 1  a³ – 1 = 0

 p+ 1   2p  = 8  

1 3 × 2 = 8 3 + 8p 2

wwM wa. th Les aBryn

9

= (2)

3

ERna

= 6 x Take values of x Let x = 3 3 +

x +

3

p3 +

= 10 9

  2p  1

1 1 3 +3×p× 2p 8p

p 3+

1 x 2+ 2 x

(c)

2p

 p+  

6

1 2 = 5 – 2 x

52.

a2 +

= 2

Take cube on both sides

x =



2

1

2

3

=5+2

4

=

2p

1

x +

= 4

p

= –1

kgei snhe eYari dnag v.iSn ir

=

2p +

1 a

Divide by 2

1 x

(b)

+ 1 = 0

a

a +

9 9 Hence x 2+ 2 =9+ = 10 9 x

1



1

a +

x2 = 9 +2

2

x = 3 + Similarly

(a)

a 1 a –b

2

– b

=

2  b  ab 2 2 a  ab  b



1 11 – 9

=

1 2

(a)

p = 101

=

3 p(p 2 –3p+3)–1

=

3 p 3 –3p 2 +3p–1

2



141

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[(p – 1) 3 = p3 – (1) 3 – 3p (p – 1)]

=

3 (p–1)3

= (a)

p – 1 = 101 – 1 = 100 x = 19 y = 18

2x–y x+2y

19 – 18

12 – 3

65.

= 1

 50%      

(b)

3x

3x

=

4 : 1 

b a  b 2

3a – 2b

=

3 2

2

23 32 33 – 22

(b)



12

9–4

(d)

=

+

1–a

b

1–b

+





 a 1 – a 

 c 1 – c 

 c  



+

1–a

1–c

b

+

1–b

 

 1

c

1–c

+



+3=1 + 3



 b 1 – b 

 

 1

67.

a 1 – a   1–a   

1 1–a

1 x

  

3x 

x

1–

+

b 1 – b   1–b   

+

(d)

 

= 4

+

25

25

–5

68.

=

=

=

x

1–b

+

1 1–c

= 4

Rakesh Yadav Readers Publication Pvt. Ltd.



x

961

x

1+

6 = a + b

6

4

b=

15

32

=

31

1024

=

961

961  x 961

961 =

1024 961

x = 1024 – 961 = 63

5

2

69.

3

a

(d)

b



e

c

=

=

d

3

=

f

1

29  39  49 2 1  3 1  4 1

100 3

=

18  27  36 234

=

81 9

= 9

100

16  100 25

70.

= x³

= x³

4 3 5 2 48  18

2x +



6x 2 + 1 = 15x

= a + b

6x

6

4 35 2

 20 x  1

5x

=

5x 35 x

=

1 7

1 71.

4 35 2

×

2

15 x  20 x

16  3  9  2

4 33 2

= 5

3x

5x



=

4 3 5 2

1

(d)

(a)

x

4 33 2

1

15

5

1

(a)



3 5   x

4

(Squaring both sides)

3

=

3

9

6

3, 4   5 15   

35 – 5

–5

3

100

16

x

2

=

  

3



 16 × 4  4=x

+

 

1–c



100

x

= 1

 1 = 4

1–c

x

5

15

3

a=

5

1–

 1–

c

Add 3 both sides

a

x

3

5

a

2

1



10

wwM wa. th Les B aryn

66

+

5







30

By comparing cofficients of rational and irrational parts.

– 5x  3

5x

–

 3 x 

p : q

=

3



2

66.

=

2

x

1p = 4q

2a  3b

2

30

4

+

5

18

+

3

6 +

3

=

5

= 5

x

x



2p = 8q

=

4 15

2x

5p – 5q = 3p + 3q

a 3

(a)

1

x +

5(p –q) = 3 (p + q)



8 6

=

30

3

=

15

(Divide by x)

2 

3 –3 2

8 6  18

=

343

9

=

12  3

1



 4

48 – 18

=

2x

(p + q)

10

3 5 2

r

=

4

34–3

=

3x  y



3

=

2

63.

3x – y

50% (p – q) = 30% (p + q)

p–q



geisnh eeYa ridna gv.i Sni

x –y

(b)

1

=

(Cross Multiply)

4x – 2y = x + 2y



2

1

1 2

3x = 4y x : y = 4 : 3

2  y  xy 2 2 ( x – y )( x  y  x y)

=

62.



2  y  xy 3 3 x – y

x

=

(b)

2

x

61.

64.

R Enak

60.



4 3 –3 2 4 3 –3 2

x = k ×

 y2 – 1

(Given)

1 2 y –1 (k is constant)

142

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Now x = 24 when y = 10 given

1 2 (10) – 1

24 = k ×



75.

k

24 =

99

k = 24 × 99



x = ? y = 5

x 2 + 2x + 1 + y 2 = 0 (x + 1) 2 + y 2 = 0 Hence both terms are squares and there addition is zero so, it can be possible only when both t er ms are zeros.  x+1=0

x=–1  x 31 + y 35

76.

c =

2

2x



– 1 + 1 + 3 = 3

2 (a–b)

=

(b–c)(c–a)

 5x  2

5x



x

2 =

 

1 6

x

2

×2

1

2x 

5 = 6

x

x  



1



6 – 5 = 2



1 = 2



x +

x

x  

 +5 

80.

x  





1

 x

x

 

1

1 = x 2 a 2 + b 2 + c 2 = 2(a – b – c) – 3

a2 + b2 + c2 = 2a – 2b – 2c – 3

a 2 + b 2 + c 2 – 2a + 2b + 2c + 1 + 1 + 1 = 0

77.

 (a 2 – 2a + 1) + (b 2 + 2b + 1) + (c 2 + 2c + 1) = 0  (a – 1)2 + (b + 1)2 + (c + 1)² = 0 a = 1 b = – 1 c = –1

Rakesh Yadav Readers Publication Pvt. Ltd.

=

15

×

a – b 

+

1

73

=

75  2

77

1 1 2 + y2 = 4 x



1

+

1

Hence x2 + y2 = (b) x 2 = x2 + add x (x

1 y x x +

+ 1 = 2 + z ........(i) = y + z + x on both sides 1) = x + y + z



z2 = y + x



z2 + z add z z (z + x (x +

1

c – a  c – a  a – b  b – c  c – a 



x y z

=

y 1

z

3



a – b  b – c  c – a 

(c)(a – a b c

3) 2 + (b – 4) 2 + (c – 9) 2 = 0 – 3 = 0 a = 3 – 4 = 0 b = 4 – 9 = 0 c = 9



a b c = 16

x 1

y x y z

±4

1

=

x yz

+

349

........(ii)

= x + z + y on both sides 1) = x+ y + z 1) = x + y + z

x

3 a – b  b – c  c – a 

........(ii)

y 2 + y = x + y +z add y on both sides y (y + 1) = (x + y +z)

2

=

= 4

1

y2 = x + z

a –b

a – b 2 a – b  b – c  c – a  a – b 

=

2 75

y – xy  x  y  x 2

75 – 2

x2 + y2 +

(a)

+

Let a – b = x b – c = y c – a = z  x+y+z=0  x 3 + y 3 + z 3 = 3xyz  (a – b) 3 + (b – c) 3 + (c – a) 3 = 3 (a – b) (b – c) (c – a)

1

10

Take x = y = 1

81.

b – c 2 b – c  a – b  b – c  c – a 

= 6

=

×

a 1/3 = 11, a = 11 3 = 1331 a 2 – 331a = a (a – 331) = 1331 (1331 – 331) = 1331× 1000 = 1331000

(c – a) in IIIrd term

wwM wa. th Les aBryn

2

4 100

(b)

1 + 1 +

(b – c) in IInd term



2

 

(a–b)(c–a)

a – b   b –c c –a   

6

yx

Multiply divide by (a – b) in Ist term



(c)

3

+ 1+

2 (b–c)

2

1

1



+

3

ERna

x

–1

= (b)

=

1.5

2

(a–b)(b–c)

x x

3

3a +b+2c =3×

Now

x 2

79.

0.04

y –x



b –1 = 0



y = 0

x

(b)

3

2 (c–a)

(–1) 31 + (0) 35 = –1

=

–1



=

2 2 y –x = 2 2 y  x  2xy



b = 1



x = 99

1.5x = 0.04y

x

(2c – 3) = 0

25 – 1



74.

a =

(b)

 y

3a = –1

24 x 2 + y 2 + 2x + 1 = 0

(a)

2

 

1

1

= 24× 99×

73.

(3a + 1) 2 = 0

(b – 1) 2 = 0

x = 24× 99×

72.

 

78.

kgei snhe eYari dnag v.iSn ir



 2a – 3b + 4c = 2 × 1 – 3 × (–1) + 4× (–1) = 2 + 3 – 4 = 1 (a)(3a + 1) 2 + (b – 1) 2 + (2c – 3)² = 0

1

=

z 1

By adding them

x =

y +

x yz z

x yz

x yz

+

x 1

x yz =

1

1

=

=

y 1

1

+

z 1

1

1 x 1

+

x yz

+

y 1

+

1 z 1

=1

Alternate:x = y = z = 2

 1 3

1 2 1 +

1 3

+

1 2 1

+

1 2 1

=

1 3

+

= 1

143

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a2 + b2 = 2 c2 + d2 = 1 Put values of a, b, c, d Take a = b = 1 c = 1 d = 0 (d)

a b

2a

x  2a

12k 2 = 4k

a b

2b – a – b

=

b–a

x

=

x  2a x – 2a

2a – a – b x  2b

+

x – 2b

=

=

3b  a b –a

+

x

=

b –a

= 2

b – a 

88.

xyz

= 4



Squaring both sides

m – 2

= 2



m – 2

= 4

89.

1

m – 22

= 2

(c) a 2 + b 2 + 2b + 4a + 5 = 0 a 2 + b 2 + 2b + 4a + 4 + 1 = 0 a² + 4a+ 4 + b² + 2b + 1 = 0 (a + 2) 2 + (b + 1) 2 = 0 a + 2 = 0 a = –2 b + 1= 0 b = – 1





–1 –3

=

z

–2  1 –2 – 1

1 3

2

xy

×

z z

=0

3

= 0

92.

a 2 – bc 2a 2 – 2bc a 2 – bc



2 a 2 – bc





a 2 – bc

= 2

(c)x 2 + y 2 – 4x – 4y + 8 = 0 x 2 + 4 – 4x + y 2 + 4 – 4y = 0 (x – 2) 2 + (y – 2) 2 = 0 x – 2 = 0, y – 2 = 0 x = 2 , y = 2  x–y=2–2=0 (a)x = b + c – 2a y = c + a – 2b z = a + b – 2c x + y + z = ( b + c – 2 a ) + (c + a – 2b ) + (a + b – 2c)  Now = x 2 + y 2 + 2xy – z 2 = (x+ y)2 – z2 (A2 – B2 = (A + B) (A –B) (x + y – z) (x + y + z) = As we know (x + y + z) = 0  x2 + y2 – z2 + 2xy = 0 × (x + y – z) = 0 (b)a 2 + b 2 + c 2 = ab + bc + ca take value a = b = c = 2

a c

=

b

= 3

93.

22 2

= 2

+

(a  c )(b  a )

+

x  



=

3 ( t a k e cu b e on b ot h



1 x

  

3

3



 3

1 1 3 x  3 + 3x  x x

(a  b )(a  c )(b  c )



x

3



2(a  b  c )

(a  b )(a  c )(b  c )

2

= 0 x

2 2 b c 2 a – bc

440 4–0



8 4

= 2

3



1 3 x

1 3 +3 x

x  1   x  

 3

= 3

= 3 3

3

= 0

x6 = – 1 x 18 + x 12 + x 6 + 1 = (–1) 3 + (–1) 2 + (–1) + 1 = –1 + 1 – 1 + 1= 0 (c) ax 3 + 3x 2 – 8x + b is divisible by (x +2) and (x–2)  (x + 2) and (x – 2) are factors

 

94.

Alternate:a + b + c = 0 b + c = – a Squaring both sides (b + c ) 2 = a 2 b 2 + c 2 + 2bc = a 2

Rakesh Yadav Readers Publication Pvt. Ltd.

x

1

(a  c )  (b  c )  (a  b )

a

1

sides)

(  a + b + c = 0) (c) a + b + c = 0 Assume values a = 2 b = –2 c = 0 a+ b + c = 2 – 2 + 0 = 0(satisfy)



( a) x +

1

2 + 2 × (m – 2) ×



a b

+

(c  a )(c  b )

1

a –b

a 2  a ² – 2bc



(a  b )(b  c )

1

85.

y

zx  y

z



(b) a + b + c = 0

1

(m – 2) +

a 2 – bc

3xyz



(m – 2) +

2

3

2

= 0

xy

 3xyz)



m – 2

a 2  b2  c 2

then, x 3 + y 3 + z 3

wwM wa. th Les B aryn

m –2

(m – 2) 2 +

x

y

91.

2

1

1

(c) m +

+

z

3

 If x + y + z = 0

a –b

2 b – a  =

zx

x

y

+

xyz

b –a 2b – 2a

3

2

y

+

×

3a  b

3b  a – 3a – b



2

yz

3a  b a –b

2

yz x

= 2a  a  b

=

x – 2b

84.

x

(d)

2a a b

x  2b



87.

4

 xy = 4k =

3

again

2b



1

k =

3b+a

(By C – D rule)



90.

 xy  4k × 3k = 12k²

2b

2b  a  b

=

x – 2a

 

x = 4k y = 3k

4ab

=

..........(i) .........(ii) .........(iii)

x + y = 7k ...(ii)

(–1) 2 + (1) 2 = 2





 x – y = k .....(i)

(0–1)² +(1+ 0)²

x

= k (let)

4

x – y = k x + y = 7k xy = 4k

(ad – bc) 2 + (ac + bd) 2

(d) x =

=

7

r

83.

(a)x – y =

b 2 + c 2 = a 2 – 2bc

xy

geisnh eeYa ridna gv.i Sni

  

86.

x y

R Enak

82.

 x + 2 = 0  x = –2 x – 2 = 0 x = 2 Put x = –2  a (–2)3 + 3 (–2)2 – 8 (–2) + b = 0 = – 8a + 12 +16 + b = 0 – 8a + b + 28 = 0 – 8a + b = – 28............. (I)

144

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and Put x = 2

=

2

a(2) + 3(2) – 8 × 2 + b = 0 8a + 12 – 16 + b = 0

8a + b – 4 = 0 8a + b = 4 From equation (I) & (II)

......(II)

 – 8a + b = –28

98.

8a + b = 4 2b = –24 b = –12 a=2 95.

(b)x 2 – 3x + 1 = 0 x 2 + 1 = 3x Divide by x x

=

2

y

z

x

3

1

x 

Cubing both sides x

3

 x 

a

 = 27

x

= 99.

1 4x



2

1

2x

3

  2x  1

= (3) 3

y +  b  y b

×

z +  c 

3

3



13

3



1 3 + 3 13 = 13 13 x



1 3 = 10 13



x

1

101. (b) 3 x 



2x

= 5

Multiply both sides by

z

2

c

1

2

 3x × 3 + 2 x × 3 = 5 × 2x 

x y

=

2 3 2 3

10 1 = 3x 3

x

1 27x 3

+ 3×2x ×

1  1   2x  3x  3x  

2x  1    2x  

100. (d) x

1 +

3

 10 

1 (x + y) 3 = 3 3 x y

x

4

=    3 

 1   x  y  xy   

4

y

1

 xy Rakesh Yadav Readers Publication Pvt. Ltd.

 2 x 

1 4 = 119 x



1 4 +2 = 119 + 2 = 121 x

x

2

1  2 x

2

  

8x ³ 

x > 1

8x 3 

1  2 = 11 x

 2

1 27 x 3

10 3

=



1000 27

1000 20 – 3 27

10 1000 – 180 820 = = 30 27 27 27

102. (a) x + y = z x + y – z = 0 If a+ b+ c = 0 then a³ +b³ + c³ –3abc = 0  x 3 + y 3 – z 3 = –3xyz x 3 + y 3 – z 3 + 3xyz = 0 3xyz – 3xyz = 0



103. (c) = (11) 2

1 27x ³

=



= 27 – 9 = 18

1

8x 3 

xy

1 3 3 – x³ – y³ = 3 x y



3

1 (a)

x

3

1 x³ + y³ + 3 = 3 3 x y

1

8x

97.

= C

c

1 x³ + y³ + 3xy (x + y) = 3 3 x y

1 3 = 8x  + 3 × 3 = 27 3 8x 

1 3 + 3 13 =



(c)xy (x + y) = 1

= 27

3

x

abc



= 3

1 3 1 = 8x  3 +3×2x× 2x 8x

8x

Taking cube both sides

3 xyz

wwM wa. th Les aBryn

=

13

Cubing both sides

3



Take cube both sides

 2x  

z

×

x + y =

= 18

x3

2x 

=

x

 Taking cube on both side

Multiply by 2 both sides



1

 13

1

1

(a) x 

x

= 3×

1 

1 3 x  +3×3 = 27 3 x

x3 

96.

1 1 3 + 3x  x



x



= B

ERna



y

b

3

x  a   

= 3

x

3

 

x

x 



2 1



x

 A +B+ C = b– c +c– a+a – b = 0  A 3 + B 3 + C 3 = 3 ABC

x

 x 

x

= A

a

= a – b

c

3x



x

= b – c

= c – a

b

1

+

x

x a

 2  11  2

x²

xy = (x + y)² x 2 + y 2 + 2xy = xy x 2 + y 2 + xy = 0  x 3 – y 3 = (x – y) (x 2 + y 2 + xy) (x 3 – y 3) = (x – y) × 0 = 0 (c)x = a (b – c) y = b (c – a) z = c (a – b) Let

1

x² 

xy

kgei snhe eYari dnag v.iSn ir

 

x y

3

  

a b + = 1 b a a 2 + b 2 = ab a 2 + b 2 – ab = 0

a 3 + b 3 = (a + b) (a 2 – ab + b 2) = (a + b)× 0 = 0

145

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112. (d) (a – 1)

x – 2 = 2 2/3 – 2 1/3 ........ (I) Take cube both sides (x – 2) 3 = (2 2/3 – 2 1/3) 3 x 3 – 8 – 6x(x – 2) = (2 2/3) 3 – (2 1/3) 3 – 3×2 2/3.2 1/3 (2 2/3 – 2 1/3) x3 – 8 – 6x2 + 12x = 22 – 2 – 21

(x – 2) 32 3 From equation (I)  x 3 – 8 – 6x 2 + 12x = 4 – 2 – 3 × 2 (x – 2) x³ – 8 – 6x² + 12x = 2 – 6x + 12 x³ + 18x – 6x² – 8 – 14 = 0 x ³ + 18x – 6x 2 – 22 = 0  x3 – 6x² + 18x + 18 = 22 + 18 = 40 105. (d) a 3 – b 3 – c 3 – 3abc = 0  a–b–c=0 a = b + c 106. (c)a = 2.361 b = 3.263 c = 5.624 a + b – c = 0 2.361 + 3.263 – 5.624 = 0

 a3

+ b 3 – c 3 + 3abc

0

Comparing Cofficient of 2 constant terms.  a=3  a–1=b 3 – 1 = b b = 2 a + b = 3 + 2 = 5 113. (c) ax 2 + bx + c = a (x – p)² ax 2 + bx + c = a(x 2 + p 2 – 2px)

 ax

2





21  8

=

x

x –z

=



1 19

1

x2 +

x –z x –z

1 4

2–

3

n

n +

=2 3 2–

n

120. (b) x =

=

x

3 = 4

3 2

1

1

3 2

29

×

3– 2

=

3– 2

– (y – z)

........(i)

3 – 2

x

x –z

–

z

x –z

= 1

111. (b) a³b = abc = 180 or a =1, b = 180 then c = 1  b = 180

Rakesh Yadav Readers Publication Pvt. Ltd.

x =



1  x

2 1



1

x –



1 = x

2 

3 –

2 = 2

3

q

p

2

q pq

2 =

x

2

90 5

=18

xy = 1, y² =

1 x²

1 = 1 x 

32 2

= 3–2 2

= 3+ 2 2 +3– 2 2 = 6 1 2 = 36 – 2 = 34 x

2



2 1

x

2



2 1

1

 x+

2

x =

= 3 

121. (d) p + q = 10 .... (i) and pq = 5 Squaring both sides of equation (i) (p + q) 2 = (10) 2 p 2 + q 2 + 2pq = 100 p 2 + q 2 + 2 × 5 = 100 p 2 + q 2 = 90 Now,

y =



2 1

x

122. (d) x = 3 + 2 2 ,

x = 2 + 1 + 2 2 2 + (1) 2 + 2 × 1 × 2

  x = 

1

p

1 = 27 x2

+

x +

 q+p=

117. (b) x = 3 + 2 2

=

3

1

1 – 2 = 25 x2

 x2

7

y –z

4a



(From (i))

2

1 = 5 x

116. (c) x –

53 – 24

z

+

3

2

1

b2

 1   1  1  1  5 = 1 4  1 4 1 4 1  4  =           4 115. (b) a 2 + b 2 + c 2 + 3 = 2(a + b + c) a 2 + b 2 + c 2 + 3 = 2a + 2b + 2c a 2 – 2a + 1 + b 2 – 2b + 1 + c 2 – 2c + 1 = 0 (a – 1) 2 + (b – 1) 2 + (c – 1)² = 0 a = 1 b = 1 c = 1 (a + b + c)  1 + 1 +1 = 3

110. (a) x + y = 2z x – z = z – y x – z

+ 2 × 2 ×

2  3 

n  2 

114. (d) a + b + c + d = 1 (1 + a) (1 + b) (1 + c) (1 + d)  For maximum value a, b, c, d

 7 x  2y = 73  2 4 =

2

.......(i)

4ac = b2

= 1 + 1 = 2 109. (c) x : y = 3 : 4

15 – 8

 3

n = (2) 2 +

n =

c = a ×

wwM wa. th Les B aryn 1 17 19 = (1) +

5 x – 2y

n = 4 + 3 + 4 3

b = – 2ap

R Enak

= 2

(assume x = 1, so,1 + 1 =2)

x

119. (b) n = 7 + 4 3

4

3 3 = 3  p  1 = 3 125  = 125

x 17 +

12  12   –22 6  2 3 = 2 1 – 1  –2

=

+ bx + c = ax + ap – 2apx

a = b = c = d =

= 3 p3  3 p2  3 p  1

x

a ²  b²  c ² a ² – bc



and c = ap 2



1

118. (c) Put a = b = 1 and c = – 2 we get a + b + c = 1 + 1 – 2 = 0 0 = 0 (satisfy)

2

–b  p = 2a

3 p p2  3 p  3  1

108. (c) x +

2

&

Comparing cofficients of x² and x

107. (d) p = 124



2 + 3 = b 2 + a

r

+ 2 2/3

geisnh eeYa ridna gv.i Sni

21/3

104. (c) x = 2 –

1 3 2 x  3 xy  y x = 2 2 = 2 1 x – 3 xy  y x  –3 2 x

2

2 –1 2 –1

=

2 –1

2  1 – 2 1 = 2

=

2

x

2



37 34  3 = 34 – 3 31

146

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For More Visit : www.LearnEngineering.in

y z x = = (given) b c c a a b x

 b c



x –y x –y x  b  c = b  c – c  a = b – a Similarly

y –z

z

x –y



y –z

x y

2

x y  

4

(giv en)

(giv en)

A =

y

, B =

So, tx = 2 ×

y

2

x y

y 2  

=

=

1

2

1 2 1 2

((2y)² + (–3y)² + y²)

=

x

3

x2 +

1 3 + 3 3 = 3 3 x

1

2

((a –

b

1 = 5 x

x 72 + x 66 + x 54 + x 24 + x 6 + 1 12

11

 6

+ x

1 a + = a

+

x6 

9

 6

+ x

1

3

(14y²) = 7y²

Rakesh Yadav Readers Publication Pvt. Ltd.

 a3

x4 1

=

1  –1 + 2

1 1   3x    5 x x2  1 2 x  2 x

x2 

x y 5 xy = 6

3

– b 3 – 3ab (a – b) = (2) 3 (By cubing) 56 – 3ab × 2 = 8 – 6ab = 8 – 56 6ab = 48



3 1  x x2

1 x2

x2 

3

132. (b)a – b = 56  a–b=2

(4y² + 9y² + y²)

+2 = 25

x 4  3x 3  5x 2  3x  1

=

= –1 + 1 + 2 = 2 131. (b) x 3 + y 3 = 35  x+y=5 Take cube on both sides, (x + y) 3 = (5) 3 x 3 + y 3 + 3xy(x + y) = 125 35 + 3xy(5) = 125 15xy = 125 – 35 15xy = 90 xy = 6

1

1 x2

= 23

x 2  3x  5 

4

6

 x y =

x2 +

x 4 3x 3 5 x 2 3x 1  2  2  2  2 x2 x x x x = x4 1  x2 x2

x6 = – 1

x 6 

1

x2

now,

2 3

=

divided by x 2,

1  a 6 – a 6 + 2 = –1 –

((x + y – x + y)² + (x – y – x –2y)² 2 + (x + 2y – x – y)²) =

1

a = – 1

b) 2 + (b – c) 2 + (c – a) 2) =

= 3

+ x6 + 1  (–1) 12 + (–1) 11 + (–1) 9 + (–1) 4 + – 1 + 1  1–1–1+1–1+1=0

tx = x t =1 126. (d) a = x + y b = x – y c = x + 2y

2 a 2  b2 = 3 ab

 b+a

x6 + 1 = 0

130. (b)

1

 

   

y 2

a² + b² + c² – ab – bc – ca =

x



a

2 1

1 3 x  3 = 0 x

& 2AB = tx

×

a 2b 2 (3a 2 + 3b 2) = 2a 3b 3 3a 2 + 3b 2 = 2ab 3 (a 2 + b 2) = 2ab

x3 +



A² + 2AB + B²

x

 = 1



wwM wa. th Les aBryn

=



x +

2

+ tx +

= 1

a

ERna

=

b

= 1 +1 = 2

134. (a) x +

T o m a k e i t a p er f e c t s q u a r e i t should be in the form A² + 2AB + B² = (A + B)²

2

+

x  

129. (a)

2

y

2 + tx +

= 8 (a – b) = 2 (a – b) 2 = a 2 + b 2 – 2ab = 4 = a 2 + b 2 = 4 + 2ab a 2 + b 2 = 4 + 2 × 8 = 20 133. (b) (a 2 + b 2) 3 = (a 3 + b 3) 2  a 6 + b 6 + 3a 2b 2(a 2 + b 2) = a 6 + b 6 + 2a 3b 3  a 6 + b 6 + 3a 4b 2 + 3a 2b 4 = a 6 + b 6 + 2a 3b 3  3a 4b 2 + 3a 2b 4 = 2a 3b 3

x 0

a 2 + b 2 – ab = 0 a 3 + b 3 = (a + b) (a 2 – ab + b 2) = (a + b) × 0 = 0

z –x

= = b –a c –b a –c 124. (d) x – y = 2, xy = 24 x 2 + y 2 – 2xy = 4 x 2 + y 2 – 2 × 24 = 4 x 2 + y 2 = 4 + 48 = 52

125. (a)

b

x³

ab

z –x z z –x  a  b = a  b – bc = a – c



a

y–z

x b c

=

1

2 2 a +b

 c  a = c  a – a b = c – b again

 a b

x

x2 +

128. (a)

y z = c a a b y

= 2,

put x = 1 1 + 1 = 2

y c a

=

 ab

1

127. (b) x 

kgei snhe eYari dnag v.iSn ir

123. (a)

23  3  5   5 23

135. (b) x 3 +



43 23

1 = 0 x3

3 1 1 1    x  x  – 3x × x  x  x  = 0  

1  x  x   

3



1

– 3 x  x  = 0   3

1 1   x  x  = 3 x   x    2

1  x  x   

= 3

2  1    x    x  

2

= (3) 2

4

1  x  x   

= 9

147

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(Squaring both sides)

1

x2 +

= 7

x2

On cubing both sides 1 1 1  3x   x    27 x³ x x

 x³ 

x³ 

a3 + b3 + c3 – 3abc = (a + b + c) (a 2 + b + c 2 – ab – bc – ca)

1 



1 

  x ³  x ³   x ²  x ²   18  7 1  5 x  5 x 

1      x    126 x  

1  5 x  5 x 

   3  126 

m4

1   2  m + 2   121  m 

= 100 ×

m2

= 11

m² +

1 (1 2

(a) 2x.2 y

= 8





1  m – m   9  

1 m–  3 m

143. (b)

1–a a

2 x+y = 2 3

1–a

4

5

47 60

b

+

1

+

c

+

1–b

1–c

= 1

+

b

+

1–b

+1+

b 1–b

c

+3 = 1 + 3

1–c

+1+

c 1–c

+1 = 4

.....(i) a 1–a



1–a

+

b+1–b 1–b

+

c+1–c 1–c

= 4

2x + y = 4

x+ y = 3 2x + y = 4 – – – –x = –1

x = 1 y = 2  (x, y) = (1, 2) x 4 – 2x 2 + k (A + B)² = A 2 + 2AB + B 2 (A – B) 2 = A 2 – 2AB + B 2



(x 2) 2 – 2 × x 2 + k (A) 2 – 2 × AB + B 2 A = x 2, B = – 1 B2 = K (–1) 2 = K K = 1

1 1 1 1 1 1 1 1 1 1 1 1      – 3      3 4 5 5 5 5 1 1 1 1 1 1 1 1 1 1 1 1      –       3 3 4 4 5 5 3 4 4 5 5 3

144. (c) 3 3 3 4 4 4

Rakesh Yadav Readers Publication Pvt. Ltd.

a

1–a

3

=

60

a





20  15  12

1

+

Adding 3 on both sides

1

.....(ii) from equation (i) and (ii),

2

1

A + B + C =

145. (d)

+ 0 + 1)



 A2 +B2 +C2 –AB–BC–CA 

=A + B + C

=

1 2

x + y = 3 9 x .3 y = 81 3 2x .3 y = 3 4

1 – 2 =11 – 2 m²

138. (d) x + y + z = 6 (x – 1) 3 + (y – 2) 3 + (z – 3) 3  as x + y + z = 6 Take values x = 1, y = 2, z = 3 (1 + 2 + 3) = 6  (1 – 1) 3 + (2 – 2) 3 + (3 – 3) 2 = 0 Now assume values in options. option 'd' satisfies the given relation. Hence 'd' is correct. 139. (b) a + b + c = 6 = a 2 + b 2 + c 2 = 14 a 3 + b 3 + c 3 = 36 Put values as a = 1, b = 2, c = 3 1 + 2 + 3 = 6



= 100 × 1 = 100

142.

1

 A+B+C  (A 2 +B2 +C2 –AB–BC –CA)

 34 – 33 2   33 – 33 2   33 – 34 2   

wwM wa. th Les B aryn

m2 +

= (34 + 33 + 33) ×

+ 2 = 119 + 2 2

1 = C 5

1  a – b 2  b – c 2  c – a 2  2 

1 = 119 m4

1

1 = B 4

36 – 3abc = 6 × 3

R Enak

m4 +

36 – 3abc = 6 (14 – 11)

–3abc = 18 – 36 3abc = 18 abc = 6 140. (a) a + b = 1 By cubing a³+b³ +3ab (a + b) = 1³ a³ + b³ + 3ab = 1 (a + b=1) a³ +b³ + 3ab = k From above both equations k = 1 141. (d) a = 34, b = 33, c = 33  a 3 + b 3 + c 3 – 3abc = (a + b + c) ×

1   5  x  5   123 x   137. (a) m 4 +

3

1 = A 3

Let

2

 

3

 A 3 + B 3 + C 3 – 3ABC = (A 2 + B2 + C2 – AB – BC – CA) (A + B + C)



1  3  3  27 x³

1  18 x³

3

1 1 1 1 1 1  3  4  5  – 3 3  4  5       = 2 2 2 1  1  1 – 1  1  1  1  1  1 3 4 5 3 4 4 5 5 3        

geisnh eeYa ridna gv.i Sni

x³ 

1 + 4 + 9 = 14 1 + 8 + 27 = 36  abc = 1 × 2 × 3 = 6 Alternate:  (a + b + c) 2 = a 2 + b 2 + c 2 + 2(ab + bc + ca) 36 = 14 + 2(ab + bc + ca) (ab + bc + ca) = 11

r

1 = 3 x

136. (c) x +

1–a

+

1 1–b

+

1 1–c

= 4

146. (c)a 2 + b 2 + c 2 = 2 (a – b – c) – 3



a 2 + b 2 + c 2 – 2a + 2b + 2c + 1 + 1 + 1 = 0



a 2 – 2a + 1 + b 2 + 2b + 1 + c 2 + 2c + 1 = 0



(a – 1) 2 + (b + 1) 2 + (c + 1) 2 = 0 a = 1, b = – 1, c = –1

 2a – 3b + 4c = 2 × 1 – 3 × –1 + 4 ×– 1 = 2+ 3 – 4 = 1 2

147. (b)

2

a – b  b – c  + b – c  c – a  a – b  c – a 

+

2

c – a  a  – b  b – c  148

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2



a – b  b – c  c – a 

a – b  a – b 

×

2

b – c  a  – b  c – a 

151. (a) Third proportional of a and b =

Third proportion of

b – c  b – c 

×

+

+



x²  y²

y

a – b  b – c  c – a  a – b 

3

3

3) 2 + (y – 5) 2 + (z – 4) 2 = 0 3) 2 = 0 x = 3 5) 2 = 0 y = 5 4) 2 = 0 z = 4



25 16 9 + + = 3 9 25 16

x =

x =



1 =

 22

x =

4

 6

2

=



=

2

=

......(II)

7– 3

2











32 – 10 – 2 21 – 21





2

22  2 21 – 21



2  5  12 5 3

7– 3

=

=

12

2 5

 2 

=

=





21  1

 – 21  

2

2 5– 5– 3

x  20 x – 20

5 3 =

21  1 – 21

7– 3





155. (a) (x b+c) b–c (x c+a ) c–a (x a+b) a–b (x

2 3

=





2

2 5 5 3

=

2

5 3

5– 3

20

7– 3

2

7– 3 =

3 5 3

x

......(I)





7– 3

x

x – 12

150. (d) x 2 – y 2 = 80

x y

=

= ?

7– 3





5 3

x  12

1 1 = = 9 9

(x – y) (x + y) = 80 x – y = 8  (x + y) × 8 = 80 (x + y) = 10 Now average of x and y

5 3

by C – D rule

k = 4 For y = 6



x – 12



2

 7  3 

7– 3

2  20  3

(Given)

k  1= 4

x  12

+

7– 3

1

– 2

x 32 – 2x – 21

2

4 5  3

x =

or

k

1 2

5 3

x – 20

y2

(y = 2) for (x = 1)

x =

=

4 15

x  20

k





2

2



1 2

=



153. (b)

2x =

......(i)

 7 +  3 

2x =

x

– 16 3– 12 × 3 28 + 21 3 28 – 36 + 5 3 = 5 3 –8

(Inversely proportional)

y2

7–4 3

=

1

149. (d)x 



7–4 3

×

21

2



ERna

x2 c2 y2 + + 25 9 16

x = 5 –

2x = 10 – 2 21

4+ 3 3 ×7–4 3

wwM wa. th Les aBryn



= 2

5– 3

154. (b)



3 a – b  b – c  c – a 



2 5 –2 3

=

49 – 48

(If x + y + z = 0 then x³ + y³ + z³ = 3xy)

– – – –

3– 5

5 – 3

xy

(4  3 3 ) 7 – 4 3 =

= 3

3 3 5

+

5– 3

3 5  3 –3 3 – 5

 xy

= x²  y²

74 3

74 3

a – b   b – c   c – a   b – c  c – a  a – b 

b – c  c – a  a – b 

x²  y²

x

43 3

c – a  a – b  b – a  c – a  3

2

x – 20

3 5 3

=

(By Rationalization of denominator)

3

(x (x (y (z

and

43 3

152. (a) +



y



+

3

b – c  a – b   c – a  b – c 

148. (c)



x

x  20

+

kgei snhe eYari dnag v.iSn ir

x

3

=

y

x  12

 x – 12

=

c – a  c – a  × a – b  b – a  c – a 

=



a

x²  y²

2



x y 

b²

156. (c)

=

xb

xb

2

x a

2 3 5 3

x

2 3– 5– 3

a

2

2

–c .

xc

2

2

–a .

xa

 0) 2

–b2

–c 2 c 2 –a 2 a 2 –b 2 = x 0 = 1

=

–

1 a 1 a

–

1 x

= –

1 x

10 2

=

= 5

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3 3 5 3– 5

 x – 1  a   

= –

1 x

149

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a

x

x (1 – x) = a x – x2 = a 157. (c)

1

x +

x

= 99



x 2 + 1 = 99x 2(x 2 + 1) = 2 × 99x 2x 2 + 2 = 198x

=

2

100 x 2x

 2  102 x

100 x

=

198 x  102 x 4x – 3

158. (c)

x



4x

= 4 –

x

–

=

100 x 300 x

4y–3 + 3 x

+

y 4y +

y

=

4z–3 z

3 –

y

+

4z z

1 3

= 0

–

3 z

= 0

160. (d) xy = 8 Given So, (x, y) = (1, 8) We have to Questions the options and check them (8, 1) (2, 4) (4, 2)  2x + y = 2 × 1 + 8 = 10 2 × 8 + 1 = 11 2 × 2 + 4 = 8 minimum 2 × 4 + 2 = 10 Hence in this question we have all the options. So, take all positive factor otherwise we should have to take –ve values also. (x, y) = (1, 8) (8, 1) (2, 4) (4, 2) (–1, –8) (–8, –1) (–2, –4) (–4, –2) 161. (b) x 2 + x + 1 = 0 .....(i)

1  x  2   

3 3 3 + 4 – y + 4 – = 0 x z

1

1

1

12 – 3  x  y  z  = 0  

1

1

1 1 1   x y z = 4

xy yz xz 159. (c) x  y =a, x  z = b, y  z = c

x2 +

 

1 c

 +





x + y + z = 0

x 3 + y 3 + z 3 – 3xyz = 0 x + y 3 + z 3 = 3xyz

1 4  = 0

......(ii)

166. (c) a + b + c= 0 Have values a = 1, b = 2, c = – 3

a+b

  

3 1 = 4 4

2

a +

a2 +



a2

a2

– 3 × – 3 = 9

1

a =

– 2 = 16



2 b = 2 c = – 1

168. (c)

= 18

1 1  + 3 a – a    a2



3x  2 – 3

 4x – 3  











3x  2 = 0

3x  2 = 0

=1 , b +

1

= 1 b c Values of a, b, c assume

4 3 x 2 + 8x – 3x – 2 3 = 0

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1

a +

abc =



1

× 2× – 1 = – 1 2 a + b + c = 2s let a = 2 b = 1 c = 1 s = 2

s–a 

2

2 2 2 + s–b  +  s–c  +s 2 2 2 a +b +c

2

2 2 2 +  2–1 + 2–1 +2 2 2 2 2 +1 +1

4 3x 2  5x – 2 3 = 0

163. (d)

b c   a  b+c  c+a  a+b   

(– 1 – 1 – 1) (– 1 – 1 – 1)

167. (a)

= 18 + 3 × 4 = 18 + 12 = 30

2abc x = bc  ac – ab

c+a

 1  2 2 – 3 –3  1   1  2  –3   –3  1  2   2–3 –3+1 1+2     



1

b+c

    c  a  b 



1

2

x3  y3  z 3 zxy



a –

a +

4x

5 × 3

y3 x3 z3 + + zxy xyz yxz

1 = 4 a Squaring both sides

1 1 1 1 1 1 1 a + b – c = y + x + z +x

2 bc  ac – ab = x abc



5

z2 x 2 y2 yz + zx + xy = ?

a 2 – 4a – 1= 0 a 2 – 1 = 4a

1 1 1 1 1 1 1 y + x = a, z + x = b , z

1 1 1 2 + – = b c x a

3

3xyz

3 3 q =  =  4 2

1 1 y = c



2

q2 = 1 –

=

162. (b)

1 1 y – z

x = 3 + 5 – 2

 zxy = 3

1 q2 + = 1 4

Now we have to find the value of x

–

5

x – 8 = – 2 15 (Squaring both sides) x² + 64 – 16 x = 60 x² + 4 – 16x = 0 x² + 6 – 16x = 2 165. (d) x + y + z = 0

+ q2 = 0

Comparing constant term of equation (i) and (ii)

x y 1 xz 1 yz  xy = a , xz = b , yz =

3 –

x = 8 – 2

2

1 1 + 2 × × x + q2 2 4

x2 + x +  q

wwM wa. th Les B aryn

Now

 =

x =

3

R Enak

1

– 3  x  y  z  = – 12  

164. (c)

r

1

=

geisnh eeYa ridna gv.i Sni

1–x

 2–2

0 11 4 4 11



6 6

1

150

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a b  

x 2 = (3 + 2 2 ) 2 (Squaring both sides) = 9 + 8 +12 2 = 17 + 12 2

2

2

a b b  + 2 × b × a = 25 a 

+ 

a2

b2

b

a2

2 +

173. (d)

 3x  3x

–

x

3

–

x –

3 x

171. (a)



2

=

1

x2 +

= 3

1 2 + 2 = 9

1 x + 2 = 7 x

= 172. (c)

x2 +

 

1

4 –1 2  –8

a b 15 4 3 × = × = b 5 c 16 4



a 3 = c 4

 9+9+9  27 175. (d) a 2 + b 2 + 4c 2 = 2(a + b – 2c) – 3  a2+b2 + 4c2 – 2a – 2b + 4c + 3=0  a 2 – 2a + 1 + b² – 2b + 1 + 4c 2 +

4c + 1 = 0 (a – 1) 2 + (b – 1) 2 + (2c + 1) 2 = 0 a = 1  a–1=0 b – 1 = 0 b = 1

 a2 + b2 + c2 = 1 + 1 + 1 4

=

9 4

= 2

2

4

63 225  4 1 16 = = = 45 16  225 4 45  4 18 –

1 1 1 1 1 1 2 + y 2 + 2 = xy + yz + zx x z Go through options 'd' take x = y = z

178. (d)

1 1 1 1 1 1 2 + 2 + 2 = 2 + 2 + 2 x x x x x x

 179. (b)

4

Option d is right x = 3t, y =

= 2 +

1

2 x = 2y

1

176. (b) 4x – y = 2 2x – 8y + 4 = 0

Rakesh Yadav Readers Publication Pvt. Ltd.

–1

1

45c 2 +20a 2

2

For minimum value a = b = c = 3 a 2 + b 2 +c 2 = 3² + 3² +3²

c =

18c 2 –7a 2

 a2  c2 18 – 7 2  c   =  a2  c 2  45  20 2  c  

a 9 18 – 7   18 – 7  c  16 = 2 = 9 a  45  20  45  20   16 c  

9 = 3 3

2c + 1 = 0

a 4 b 15 = and = b 5 c 16





3a = 9 a =

a b + = 5 ab ab

inf init e

So, the equations have only one solution

174. (c) a + b + c = 9 For minimum value a = b = c



t hen

a1 b1 c1   then no solution. a2 b2 c2

1

1 x y z × 0 = 0

2

a b + = 5 b a

1



1 1 2 + x + x x

7 + 3 = 10 a 2 + b 2 = 5ab 2

(iii)

177. (d)

1  zy  xz  xy    x + y + z   xyz 

x

2



a1 b c  1  1 , a2 b2 c2 soultion



y   xy + yx 



x + 1 = 3x

Squaring both sides



1

2

     x + y + z   x y z 

2

x

+



x 2 – 3x + 1 = 0

x +

z – xy

+ z x +y+ z  

3

1 4 4 + 2 = 2 2 = x 9 9

1

2

1 1 x x + y + z  + y x + y + z 



= 2

1 4 x2 + 2 – 2 = 9 x

x2 +

y – zx

+

z 2   yz + zx 

wwM wa. th Les aBryn



x   xy+zx +

Squaring both sides



2

a1 b  1 then there will be unique a2 b2 solution.

1

 x

x

1



x

1

1

1

2

1

= 2

x –

=

x – yz

+

Puting values of –yz, –zx , –xy from abov e

3

– 2 =

  3x 



  

2

1

1

2

ERna

3  

(ii)

yz + zx = –xy



1 2 2 +x = 17+12 2 +17–12 2 = 34 x

(i)

xy + yz = –zx

= 17 – 12 2

170. (b)x

= 25 – 2 = 23

xy + yz + zx = 0 xy + zx = –yz

  

1 1 17 – 12 2 × 2 = x 17  12 2 17 – 12 2



Note:For two linear equaltions a 1x + b 1y + c 1 = 0 a 2x + b 2y + c 2 = 0 Where x and y are variable.

Squaring both sides

x = 3 + 2 2

kgei snhe eYari dnag v.iSn ir

169. (d)



x = 2×

(I)

( t + 1)

1

2 x = t + 1

(t + 1) (II)

151

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3t = t + 1(from equation (i) and (ii) 2t = 1 1

t =

180. (c)

x +

1

x + a2

5

184. (c)

1

×x+a²=  x 

10

 

1



1

1 3 3 – x³ – y³ = 3 x y

4ac–b²

4a



x² +

  x 

2 4×–4×2–  –3 



4×–4

182. (a)

  

–16

=

41

16

x³ –

x 4 – 2x² + k

A² – 2 × A × B +

k = k = 1

2x

2

x³ –

2

 

x³ –

1

186. (d)

5x

183. (d)

2

 k

k

 5x  1

=

1

1 x

1 x

1

2x ²

5x

1 =   3 x x

x

2x 

 2x  



x

1 x

 

 5  = 15



x³ +



x³ +

= 81

3 1



 

x

1

x³

= (5)³

+ 3 × 5 = 125

 a  

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x²

x  

–

1 x

  

1

x² +

= 729

2 1 a

 

a   1

a³

= 3

=

a

3

1

x²

a³

+ 2 = 16

= 14

1 4 = 196 – 2 =194

x

x + y + z = 6 x² + y² + z² = 20



a

3

 

+ 3a ×

1

= 4

x

(x + y + z) 2 = (6) 2 x² + y² + z² + 2 (xy + yz + zx) = 36 20 + 2 (xy + yz + zx) = 36 2(xy + yz + zx) = 16 xy + yz + zx = 8

1



1

= 110

Squaring again

   

+ 3

=

 3

1 

a

a 



3 = 3

x³ + y³ + z³ – 3xyz

= (x + y +z) (x² + y² + z² – xy – zx – yz)

Take cube on both sides

= a³ +

1

x² +

= 729 + 27 = 756

x³

1

x³

Squaring both sides

189. (c)

1

= a³ +

x +

– 3 × 9 = 729

x³

1

5 = 3

1 = 3 + 2 = 5 x



x4 +

1

5

1

 

– 3

x³

a +

1

x +

x  

= 9

3

5

1 – 2 = 3 x

Taking cube on both sides

188. (b)

2

Take cube on both sides

(x²)² – 2 × x² × 1 +

x +

– 2 = 83 – 2

x²

–

x –

wwM wa. th Les B aryn

=

1

R Enak

In –4x²–3x + 2 a < 0 Maximum value Maximum value

–32 – 9

= 83

x²



3

Subtracting 2 from both sides

4ac–b²

Maximum value =

1

x2 +

185. (c)

(ii) When a < 0

=

1 3 3 x y

3

1

– 2

x

1

x

=

1

x 

 1   x  y  xy   

1 x³ + y³ + 3 = 3 3 x y

4a



x

=

1

1



1 (x + y) 3 = 3 3 x y

10

2 – 3x – 4x 2 = 0 – 4x² –3x+2= 0 ax² + bx +c = 0 In quadratic equation (i) When a > 0

181. (d)

2x

–

x

x³ + y³ + 3xy (x + y)=

Minimum value =

x²

xy

10

3

1



Cubing both sides

10 

1

=

x ² – 2x  1

2x 2 xy (x + y) = 1

x + y =

2

1

B = a =



= 0

a³

x

187. (b)

= 5

1

A = x



1

= a³ +

geisnh eeYa ridna gv.i Sni

 

1

x²+2×

10

=

A 2 + 2 × AB +B 2 = (A + B )²

B =

= 10

x

Divide by 2 both sides

2

x2 +

1

2x +

r





3

x³ + y³ + z³ – 3xyz = 6 (20 – 8) = 6 × 12 = 72

190. (b)

1 a

3

  

1

=3 3

x

=

x = 1 –

2

1 =

1– 2

1 2 1–2

=

×

1 2 1 2

1 2 –1

= 

k i fj es; d j . k)





2 1

152

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= 1 –

x

2 +

Take cube

–

3 1

 x



= (2)

3

  x 

–

3 1

 x

= 8

x = a – b y = b – c z = c – a x+y+z=a– b+b– c+c– a=0  x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx) = 0

3 – 2

x =

x =



2 (ab + bc + ca) = 225 – 83 = 142 ab + bc + ca = 71



a³ + b³ + c³ – 3abc =15 (83 – 71) = 15 × 12 = 180 1 =1 x 1 adding (1) both sides

195. (b) x+

y 1

 xy

x

= 1

3 – 2 3 2 2

 

2





(x+1)+ 

1 3

 10

+ 3xy (x + y)

x³ + y³ = 1000 – 30 = 970



1

x³ 

x  

1 x



1

x 1

1

1

 

3

 x

=

3 2  

x³ +

x³ +

1

x³ 1

x³ 1 x³

=

 

+ 3x 

+ 3 ×

=

27 8

3 2

–

1 x

= 9 2

 

=

27 8

200. (b)

+

1

b

a (b – a) = b² ab – a² = b² a² + b² – ab = 0  a³ + b³ = (a + b) (a² + b² – ab)  a³ + b³ = 0 201. (b) p – 2q = 4 Take cube on both sides (p – 2q)³ = (4)³ p³ – 8q³ – 3p × 2q (p – 2q) = 64 p³ – 8q³ – 6pq × 4 = 64 p³ – 8q³ – 24pq = 64 p³ – 8q³ – 24pq – 64 = 0 202. (c) x = –1

x 99

=

a b ab

1

1

a b

=



2

1

2

8



 –1

1

1

x 97

1

+

1 1 + 94 + x 95 x

+

x 96

1

 –1

+

98

95

+

1 97

 –1

+

1

 –196

+

1 94

 –1

+

–1

 1 1 –1

–1 = – 2

1 203. (a) 3 =a 3 4 +b 3 2 +c 4  32  1



1 3 4  32  1 =

1 2 1 1   2 23   23  1

 

 (  A³– B³=(A–B)



 

(A²+AB+B²)

1 Put, A = 3 , B = 1 2



(333 + 333 + 334)[ (333 + 333) 2 +

(1000) (0 + 1 + 1) = 1000

+

= – 1+1–1+1–1+1+

=

2 199. (a) (a² – b²)³ + (b² – c²)³ + (c² – a²)³ Let a² – b² = A b² – c² = B

Rakesh Yadav Readers Publication Pvt. Ltd.

+

1

1

 –1

(x+y+z)((x–y)² + (y – z)²) + (z – x)²

1

x 98

99

 1   2 3 – 1     2  1  1   1  2  3 3  2 – 1   2   2 3  1            

 1   2 3 – 1    

(333 – 334)² + (334 – 333)²]

27

1

+

1

a b

(a + b)² = ab

1

b–a



= 1

=

b²

a =

1 –1 x

a² + b² + 2ab = ab a² + b² + ab = 0 a³ – b³ = (a – b) (a² – ab + b²) a³ – b³ = 0 197. (d) If a + b +c = 0 then, a³ + b³ + c³ – 3abc = 0 a³ + b³ + c³ = 3abc 198. (c) x = y = 333, z = 334  x³ + y³ + z³ – 3xyz

 a ³ – 3a

x³ +

a

 

a

x³

196. (a)

3

3 1

(C – A)²] A³ + B³ + C³ – 3ABC = 0 A³ + B³ + C³ = 3ABC Where A = a² – b² etc.  A³ + B³ + C³ = 3 × (a² – b²) (b² – c²) (c² – a²) Hence (a + b) (a – b) is a factor

1

=2

= 1+1 = 2



= x 2 Taking cube on both sides

x 

x +

(A + B + C) [(A – B)² + (B – C)² +

2

 (x + 1) 5 +  x  15

= 3

x

= 1 + 1

 1    x  1

and

(10) 3 = x 3 + y 3 + 3 × 1(10)

2

x 1

Put x + 1 = 1

2

32–2 6 322 6 (x + y) 3 = x 3 + y

x+1+

3 – 2

3

1

 3 2

+

1

83 + 2 (ab + bc + ca) = 225

3 2

3 – 2

193. (c) 2x +

=

a² + b² + c² + 2ab + 2bc +2ca = 225

wwM wa. th Les aBryn

  



  

1



a + b + c = 15 a² + b² + c² = 83 (a + b + c )² = (15)²

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)

x+y =



194. (b)

3 – 2

y =



= x³ +

–9  16 7 + 2 = = 8 8



1



1 x³

3 2

y =

c² – a² = C A + B + C = a² – b² + b² – c² + c² – a² = 0  A³ + B³ + C³ – 3ABC =

191. (b)

192. (d)

1 –9 + 2 = + 2 x³ 8

x³ +

ERna

x  

27 – 36 –9 = 8 8

=

2 + 1 = 2

kgei snhe eYari dnag v.iSn ir

1

 x–

3  1 3 3  2  – 1

 

  

1

  

  2 3 – 1

 

153

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1 b  2 3 + c

(Comparing the terms ) a = 0 b = 1 c = –1  a+b+c=0+1–=0

1 x

3

1

=

2– 3 2– 3

×

2 3

1

 x³ + x ³ = 2 + 205. (d)



x =

3

3 + 2 –

3 = 4

210. (b)

5 + 2

x – 2 =

3

5

(x – 2)³ =

  

x³ – 8 – 3 × 2 × x [x – 2] = 5 x³ – 6x² + 12x – 13 = 0

p²–p p²–1 p² 2p³+p² + p²+3p + p+1 a

=





a² =

=

4 + 3 + 4

=

7 + 4



1 1 = a² 74 3

=

7  4 3  7 – 4 3 

=

1 49 – 48 = 64 24

211. (c)

3 x³ + = 4 (a³ + b³) x

and 3x +



x³ +

1 1 = 2



2 = 2 (satisfy)

Rakesh Yadav Readers Publication Pvt. Ltd.





3

1 = 7+ 4 3 + 7– 4 3 a²

a² +

= 14

1 + 4

p +



p +



 p



A 2 + 2 × A × B + B²



A =

1  p +  k²  4    2

B =



+2×

1  p + k²  4  

1 × 2

1  p B² =  k²  4    1 2 1 4 1 4

2

=

=

1 2   1 4

k² = 0 k = 0 215. (a)

1 = 3 x²

k² +

k² +

1  1    x – x   x ²  x ²  1   

x² + 1 +

p + k²

214. (a)

3 1 = 3x + x x³

1  = 3 x – x   

3

7–4 3 1

=

1 = 4 (a³ + b³) x³

1 3 = 3x – x x³

2

7–4 3

1  1    x³ – x ³ = 3 x – x     

1   1   x²  x²  x³  x³     

= (1 + 1) (1 + 1) = 2 × 2 = 4 208. (a) a, b, c, are +ve integers So, minimum value is a = b = c = 1  Putting the value of x in equation a³ + b³ + c³ – 3abc = 1 + 1 + 1 – 3 × 1× 1× 1 =0 Hence minimum value is 0.

x³ –

3

2  3 

2

Put x = 1

1 +

a = 2 +

12

7 49  2  – 26 – 12   = = 4 6 6

Hence option (b) is Answer.

1 x + = 2 x

213. (b)

    ² – 2

wwM wa. th Les B aryn

1–1 1–1 1 + + 2+1 1+3 1+1

1 1 2p² = 2



c –b ,  +  = a a

  ²  ²  + =   



Now check options (b)



+

7

1 1 = 0 + 0 + = 2 2

207. (b)

  ,

    = + 2 ,  = 2 = 6

x³ – 8 – 6x² + 12x = 5

In such type of question ssume values of p.  Let p = 1



 =



3



206. (b)

2x² – 7x + 12 = 0 roots are

Take cube on both sides

3 5 

8 52 10 – + +7 3 27 9

27 –252  279 = = 1 27 27

=

a = 1 b = 0 a² – b² = 1 – 0 = 1 or you can directly put the value of x also, 212. (a) 121a² + 64b² = (11a)² + (8b)² + 2 × 11a × 8b = (11a + 8b)²  So term added t o make perfect square = 176 ab



–96 – 156  90  189 = 27

2– 3 = 2– 3 4–3

a³ + b³ = 1

Let

2 3  –2   –2   –2  = 12  –13      3   3   3 

= – 12×

gj d k i fj es; d j . k



2 3

 –2  – 5  3  + 7  

3

x =1

4(a³ + b³) = 1 + 3



f



1 = 2 x²

x² +

(3x + 2) then 3x + 2 = 0 x = –

x = 32  3 x3 = 2 +



R Enak

204. (d)

209. (d) f(x) = 12x³ – 13x² – 5x + 7 If we divide f(x) by

r

 2  23  +    

geisnh eeYa ridna gv.i Sni

1 – 1 = a 23



1

Reciprocal of  x  x   

1 x  =  x  1  = x ² 1 x 

154

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F or

m in i m um

v al ue

1 1 1 + + a b c a = b = c a + b + c = 1



of

(given)

1

a = b = c =

3

1 1 1 = = = 3 a b c

220. (d) (x – 2) is a factor of x²+3Qx–2Q for (x – 2) = 0 x² + 3Qx – 2Q = 0

 

1

1

1

 Minimum value of a + b + c



= b 2– 3

b =

2 3



= 1



2– 3

1 b

a =

=

1 1 + a²+1 b²+1



1



1 1 b²

222. (b) x= 3 –

1

1





b² 1 + b²+1 b²+1



xy =

1 = a



2 – 3  b = 1



b

1



a

219. (a) a+

+

1

b

=

1

b



2– 3

=2– 3 +2+ 3 = 4

c

= c+

a



Let a = 2, b = – 1 & c = 1

–1

1

= – 1+

1/2

=

1

2



+

1



2

at abc = 2 × – 1 ×

3

1 2

= – 1



Rakesh Yadav Readers Publication Pvt. Ltd.

(b) x

–

1

2

– 3 xy



2



– 3

3x –

 b



1

4

–



3 x

1 3 x

5x x

–

3 x

1

x

3

–

x³ 5



 

3x –

1 x³

1 x

5 

1

= 1 x Take cube on both sides



2  3 3  4 =3 3 8

 b

2

1 a, B 2 2

 x 



–

3 1 x

x³ –



x³ –

x

be perfect square

1

 

x³ –

2

b = 2 a b= 4 a² a² = 4b

–

x –

, B = b

1   x  2a  

 5x – 3

x 2



A² + 2 × A × B + B² = (A + B) 2

2

3

8 3

1 2 x

–

x

=

1 a × x + 2

A = x, B =

2

3x

x

3

x² + 2 ×

3x

4

divide and multiply by x



8

8

x



= 1

x

xy

3 x² + ax + b

=

= 1 = 1 = 1



  

2 3 12 – 8 

1

2

–1 + 1 + 1 – 1= 0

x

xy

1

To save your time assume values of a, b, c according to equation.

2+

223. (b)



1

3

3

  3

1

1

1–2  1–0 1–0 



8

=



2  3

= b+

3 –

2 3 2 3

By rationals 1

    3 

1





=

225.

 x  y   x ²  y²+2xy–2xy – xy 

wwM wa. th Les aBryn 2  3  a

218. (d)

  

+

1  1  1 1 + + + –1  1  –1  –1 –1



1 1 3 – 3 + 3 + 3 =2 3

x+y=

x  y  x  y 

b²+1 = 1 b²+1



1

1–2 1–2 1–0 

xy

= 3 –



1

1

1–0  1–0  1–2 + 1–0  1–2 1–2

x ³  y³ x² y² y + x = xy

1

1 1 + 1+b² b²+1 b²

1–b 1–c 1–d 

Put a = 0, b = 0 and c = 2 and d = 2 a + b + c + d = 0 + 0 + 2 + 2 = 4 = 4 (satisify)

3

 x  y  x ² – xy  y² 



1 + b²+1

& y= 3 +

3

ERna

a =



1 +

1–c  1–d  1–a 

+

1

1

2  3 

a

1 +

a² + b² + 2ab = 144 a² + b² + 2 × 22 = 144 a² + b² = 144 – 44 = 100

= 3 + 3 +3 = 9 217. (b)

a + b + c + d = 4

1 1 + 1–a 1–b 1–c 1–a 1–b 1–c        

4 + 3 × Q × 2 – 2 × Q = 0

4 + 6Q – 2Q = 0 4Q = – 4 Q = – 1 221. (d) a + b = 12 ....... (I) ab = 22 ...... (II) Squaring both sides of equation (I)



224. (a)

kgei snhe eYari dnag v.iSn ir

216. (a)

=

3

–

= (1) 3

1 x³ 1 x³ 1 x³

 

– 3x –

1 x

 

= 1

– 3 = 1

= 4

1

4 1 4 x³ = = = 1 3  1  5 2 8   3x –   5 x  

226. (d)



x + y = 15 x – 10 = 5 – y

155

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x – 10 = – (y – 5) Take cube on both sides



(x – 10)³ = (– (y – 5))³ (x – 10)³ + (y – 5)³ = 0

227. (b)



= 66

x² 1

x²+

 For x = y –2k = 1 – 3k k = 1

–2 = 66 – 2

x²

231. (b)

–

1 x

 x ²  y²  z  x  y – 3z

+

 

= 64

3

xy³ z ²

(given)

x

1

x –

x² x ² – 1  2x



x

x

–

=

1 x x



2x



x



–

Then

x  

x

–



228. (c)

= + 8

+ 2 = 8 + 2 = 10

= – 8 x – 8 + 2 = – 6 (10, – 6) a² + a + 1 = 0

233. (a) x +

  x 

=1 x x² + 2 = x x² – x = –2 x – x² = 2

x²



x x² x



x



2 x

1 – x 

x³ 1

x³

1

x  

2

1 1 + + x x x³

18 + 3 =21

x

 

= 8



 x  13 –  x – 13  x  12 –  x – 12

= 2

– B³ = (A – B) (A² + AB + B²)

 x  1 – x  1   x  12   x –1  x  1   x –12   x  1 – x  1 x  1  x –1 =2

+ 6 = 8

x ²  1  2x  x ² – 1  x ²  1 – 2x   2 2x 

= 2



= 2× 2 = 4



= 5

x



3x ²  1 = 2 2x

3x² + 1 = 4x 3x² – 4x + 1 = 0 3x² – 3x – x + 1 = 0 3x (x –1) –1 (x – 1) = 0 (3x – 1) (x – 1) = 0 3x – 1 = 0

x³+



x³+





1 x

x x 1 – x 

234. (c)

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

 

= (5)³



1 3

x – 1 = 0

For x = 1 = 1 x³ 1 x³

+3×5 = 125

= 110

Squaring both sides

x6 +

x 6+

1

x =

3

x³   

= divide & mulitply by x

x





2

x ² 1 – x 

x² 1 x4 + + x³ x³ x³



 Take cube on both sides

x²  x  2



1

1

1  1    x ²  x ²   x ³  x ³ 

a³ – 1 = 0 a³ = 1 (a³)³ = 1³ a9 = 1 x+

+ 3x 

x³

x³ +

a ³  1³  a  1 a ² – a  1 a ³ – 1  a – 1 a ²  1  a     (a³ – 1) = (a – 1) × 0 

229. (a)

 

1

x³ +

wwM wa. th Les B aryn



+

x³ +

 A³



1

When x –

 x²  1



= + 2

x²

x³ +

 

6

x³

235. (b)

1

x² +

2 

1 x

x²

R Enak

When x –

4

A² – B² = (A – B) (A + B)

1 1

= 18

+ 2 = 4

Cubing equation (I)

1 x

1

x² +

x

 x 

x

........(I)

Squaring both sides



x



1 232. (b) x + = 2 x

= 8

x

1 x³

x³

 1 9 –1 1– 3  3 + 3 1   –33 1

= (8) 2

 

6

+ 3 × 3 = 27

x³

geisnh eeYa ridna gv.i Sni



–

2 1

x



= 3 + (– 3)= 0

x  

1

x³ +

x³ +

x = 1 y = – 3 z = –1

2

x  



= 3

x

(giv en)

1

x²+

230. (b)

1

x +

1 11 x = = 1 2 x – x² y = 1 – 3k and x = –2k

r

 

2

x 

By adding numinator and denominat or 1 +1 = 2 No option is satisfied



= (110)²

236. (c)

 1 6 + 2 = 12100 x 1 6 = 12100–2 = 12098

x

x² – 3x + 1 = 0 x² + 1 = 3x

x =

1 3

1 + 3 = 4

2

  x³  1

1 1

x = 1 x

5 =

+ 2

1 5 2

gj d k i fj es; d j . k ×

5 –2 5 –2

5 –2 = 5 –2 5–4 1  x– x = 5 +2– 5 + 2 = 4





2x ² – 3x – 2 3x ² – 4x – 3

156

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x

– –

x 3 x

– –

3x

2x –

x 4x



3x –

x

2 x 3 x

–3 –4

1  –3 x  2 4 – 3  1   3x –  – 4 3 4 – 4 x  2x –

8–3

=

12 – 4

237. (b)

 238. (d)

If

=

5 8

 239. (a)

=

a 4  a ²x ²  x 4

a² – ax  x² – a² – ax – x2 2ax + (a²  x²  ax )(a²  x² – ax ) a 4  a 2x ²  x 4 2

a 2  x 2 

=

–2ax 4

– ax 



 9a² = 9a² x  x – 3   –1 (x – 3) =

2

–1 x

  



x³ – 27 – 9x (x – 3) =



–1 x³ – 27 – 9 × – 1 = x³



x³ – 27 + 9 =



+

a 4  x 4  a 2x ² = 0

4

a  x 4  a2x ²

p³ + 3p² + 3p + 1 – 1

–1 x³

249. (c)

x+

=

a +

2

1 = a

3

3

 3

1 + 3 3 = 3 3 a³



a³ +



1 a³ + = 0 a³

247. (a)

Rakesh Yadav Readers Publication Pvt. Ltd.

3

=

a +

1 = a

1

1 x



a³+

2

1

x +

x



= 6

3 1 x

 

= (6) 3

x3 +

1 3 + 3 × 6 = 216 x

x3 +

1 3 = 216 – 18 = 198 x x

6

x

4 x

1 x

x 3

+

2

1

1 3

x

3

1 

1 +3× 3 a³

a3 +

5

= 3 + 2 2 + 3 – 2 2

= x3 + x +

1

 a³+ a ³ +3a× a a  a  

=

= 3 – 2 2

x



Take cube on both sides

1

3 –1

x = 3 + 2 2

x +

x  

= 3

9–4

=

3 –1

1000000–1 = 99999

1  a  a   

=3

x

33 – 4



RHS

1

2  1 3x 3 3x   – 4  –4 2 3x  3 – 4x  x x x   2 2 x  1  –1 x 1 x x  1– x  –  x   x x x

250. (d)

(100)³ – 1

1  a  a   

–1 x³

x³(x³ – 18) = –1 241. (a) a² + 4b² + 4b – 4ab – 2a – 8 = a² – 4ab + 4b² – 2a + 4b – 8 = (a –2b)² – 2 (a – 2b) – 8 Put t = a – 2b = t² – 2t – 8 = t² – 4t + 2t – 8 = t(t – 4) + 2(t – 4) = (t + 2) (t – 4) = (a – 2b – 4) (a – 2b + 2) (Put the value of assume t )

1 (1+1+4) = 3 2

=

Cube on both sides

–1 x³

1 [(x 2

1 [(997 – 998) 2 + (998 – 999) 2 2

(99 + 1)³ – 1

3



x 2 + y 2 + z 2 – xy – xz – yz =

248. (d)

2ax

245. (c) From option (c) LHS (x + 2)² = x² + 4x + 4 RHS = x² + 2x + 4

 –1    x 

(x – 3) 3 = 

a 18 + a 12 + a 6 + 1 (–1) 3 + (–1) 2 – 1 + 1 –1 + 1 –1 + 1 = 0

a 6 = –1

=

243. (b) x = 11 x 5 – 12x 4 + 12x 3 – 12x 2 + 12x – 1 = x5 – 11x4 – x4 + 11x3 + x3 – 11x2 –x² + 11x + x – 1 = (11) 5 – 11× (11) 4 – (11) 4 + 11 × (11) 3 + 11³– 11 × (11) 2 – (11 × 11) + (11 × 11) + (11) – 1 = 0 – 0 + 0 + 0 + 11 – 1 = 10 244. (c) p = 99 p(p² + 3p +3)

246. (a)



x³ – 18 =

a4  x4  a²x²

2ax

+

 LHS 

Taking cube on both sides



+

2

(p + 1)³ – 1 + 0 = 1 c²) = (a + b + c)² a = b = c a²) = 9a²

= = =

+ (999 – 997) 2]

a  x  x a²



a6 + 1 = 0

2ax

a4  x4  2x2a2 – a²x² 4

 

– y) 2 + (y – z) 2 + (z – x) 2]

–2ax

wwM wa. th Les aBryn

240. (a)

=

=

a = 2.234 b = 3.121 c = –5.355 a + b + c= 0 a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca) = 0 x² + y² + 1 = 2x x² – 2x + 1 + y² = 0 (x – 1)² + y² = 0 A² + B² = 0

x – 1 = 0 x = 1 y = 0 x³ + y 5 = 1 3(a² + b² + by options 3(a² + a² +

2ax

+

–2ax

=0.625

As powers are even it can possible only when A = 0 & B = 0



1 1 – a ²  ax  x ² a ² – ax  x ²

242. (d)

kgei snhe eYari dnag v.iSn ir

x 3x ²



2

ERna

2x ²

1 = 0 a³

=

3

 3

3 3

= = 251. (c)

  

x3 +

1 1 3 + x + x x

198 + 6 = 204 (a + b – c) 2 + (b + c – a) 2 + (c + a – b) 2 = ? a + b + c = 0

(given)

a + b = –c b + c = –a

157

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 

a + c = –b (a + b – c) 2 + (b + c – a) 2 + (c + a – b) 2

x2 +

   

(–c –c) 2 + (–a –a) 2 + (–b –b) 2

2 1

  253. (a)

2

2

2

4(a + b + c ) 3

p + 3p + 3p – 7 = 0 p 3 + 3p 2 + 3p + 1 = 7 + 1 (p + 1) 3 = (2) 3 p + 1 = 2 p = 1 p 2 + 2p = 1 + 2 = 3 x = 2015 y = 2014 z = 2013 1 2

[(x 175

256. (c)

337



144

3

2b 4



1  b ²   1 3 

3



2

x

= 2

1

12

=

x = 3

25

1 675 x2 + 2 + 2 = 144 x

1 625 x2 + 2 = – 2 x 144

1 625 – 288 x2 + 2 = x 144

a  a²  b³



1

b = 1

3

a  a ²  b³  3 a – a ³  b³



a ² – a ²  b ³ 



–b 3



2 3

2 3

 x  2

x 24  1 x12 x 24  1 12

12  x 



= 7



x 24 x12



1 x12

1 x12 = 7

Cubing both sides

 12  x  

36  x 

3

1  3  x12  = 7 1

x

36



3  x12  1 x12  1    x12   x12

= 343 36  x 

1 x 36 + 3 × 7 = 343

a  3b  3c Take cube both sides

36  x 

1 x 36 = 343 – 21

36  x 

1 x 72  1 x 36 = x 36

259. (d)

1 337 x + 2 = x 144

3

3

 3 a  3 b   3 c  1

2

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t3



3

1

a+b+3 a 3 b 3



x 

= 52

t +

1 3

x³ = 2a – 3bx x³ + 3bx = 2a 262. (d) Given

x

 t³ + = 64 – 12 = 52 t³ 3



 3 3 2 x³ = 2a + 3  b  x  

  1 1 + 3t t  t  t  = 64 t  

1

3

x³ = 2a + 3

 = 4 t

t +



a – a ²  b³



t 2 – 4t + 1 = 0 t 2 + 1 = 4t

12

a – a³  b³

a²  b³ ) +

x³ = 2a + 3

4t t 1 = t t

t+

2

1



x 4x – 1

–x

[take cube both sides]

x +



2

3b

3

255. (a)

x³ = (a +

(a – a ²  b ³ ) + 3

20736

3

Take cube on both sides

3 1 1 1 + = +1 = a b 2 2



258. (c)

2b ³

 b²

261. (c)x = 3 a  a ²  b ³ +

R Enak

3 2 b +3a b 2 2 a +b



wwM wa. th Les B aryn



144

9999 = 3333 3 3 3 257. (b) a + b = 9 a + b = 3 Assume values, a = 2,  (2) 3 + 1 = 9 2 + 1 = 3







175

=

58975

=

  3  4 x 2 – 1

=

2 2 2 2 a  b  2ab  a  b – 2ab

b²

  

x 4x 2 – 1

3 3 3 3 a  b  3ab a  b – a – b – 3ab a – b

b³  b ³

4x + 5y = 83 3x : 2y = 21 : 22 x: y = 7 : 11 let x = 7 and y = 11 y –x = 11 – 7 = 4

2x  1 6x – 3 = 3 2x  1 2x – 1

3



(a + b – c) 3 + 27abc = 0

260. (b)

1  2 1  2 1  x  x – 2  2 = 4 =  x  x  x

4x

(1 + 1 + 4) = 3 2 2 3a = b 2 (given)

3 2 2b +6a b 2 2 2a +2b

(a + b – c) 3 = –27 abc

x = 9999

1



 

12

1 2 x

–

x4 –



144

2

3



144

7

=

x

 2 x 



a+b –  a–b  a+b 2   a–b 2 

=

 1   1 25 7 175 =  x  x  x – x  = ×    12 12 144

[(2015 – 2014) + (2014 – 2 2013) 2 + (2013 – 2015) 2 ]

254. (a)

144

1

x –

1

=

=

2

– y) 2 + (y – z) 2 + (z – x) 2] =

 

1 3

1

 a + b – c = –3 a 3 b 3 c Again take cube both sides

49

4c 2 + 4a 2 + 4b 2

x 2 + y 2 + z 2 – xy – yz – zx =

=

x

1

337 – 288

r

= =

–

1

1

3 a + b + 3 a 3b 3 c = c



geisnh eeYa ridna gv.i Sni

252. (b)

x  

(–2c) 2 + (–2a) 2 + (–2b) 2

1

1 337 –2 2 – 2 = x 144

3a  3b  = c

= 322

158

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263. (d) Given P = 99 find P(P 2 + 3P + 3) = ? to put value in equation

    

=

2

2

2



5  3

2



x y 



x + y = 8 ..................(i)

99 × 10101

And, xy 

8 + 108 + 486 + 631

268. (c)

4

30

a

ERna 4a 

1 = 5 m 1 then m3 + = 125 – 15 m3 1 m3 + = 110 m3 m+

1

(x+3)3 +

1 3  a 4

3   –3 3 1 a –   =      a  4



a3 

1 1 1  27  3a   a    a3 a a  64



a3 

1  3  27  3   a3  4  64



a3 

269. (b)



2

?

 As

 3 

5 3

5 3

5

5 3

 

1 27 9   a 3 64 4

1 27 9 3  3 a3 64 4 192  171 1 21 3  a  3  3  a 64 64 According to the question, x = z = 225 y = 226 x 3+ y 3+ z 3– 3 x y z = ? we know,

 (– 4)3 – 3 (– 4)² + 3(– 4) + 3×(– 2) + 3 (– 2)² + (– 2)³

272.

 – 64 – 48  – 64 – 60  – 126 (b) 2 x – ky +

Rakesh Yadav Readers Publication Pvt. Ltd.

[(x –y ) 2+(y –z) 2+(z–x) 2]

– 12 – 6 + 12 – 8 – 2

7 = 0 6 x – 12y + 15 = 0 There has no solution for

a1

b1

=

a2

2

=

6 1

=

3

......(i) .....(ii)

c1



b2



c2

–k – 12

k 12



K= 4

273. (a) Here, x = 332 , y = 333 , z = 335 Find x³ + y³ +z³ – 3 xyz

a3 

1 x 3+y 3+z 3–3xyz =  x  y  z  2

= 110

3

x + 3

271. (d) Given, m = – 4, n = – 2 Find m³–3m²+3m+3n+3n²+n³ Putting value of m and n

4  3 a





x 2  y 2  2xy  xy x 2  y 2  2xy  xy

 

 

Here m = x + 3 then

a



x  y  xy ? Find: x 2  y 2  xy

Now, x  y 

5 3



5 3

2

 x  y   xy 2  x  y   xy

5

Given, 4a 

wwM wa. th Les aBryn 2

5 3

[Cubing both sides]

5 3

y

 3 

5 3



Given,

,

 

–1 m

m– 5=

1 Find: a  3  3  ? a

2 x 2 = 2

5 3

4 5  3 2

3



3

x + 3

Let x + 3 = m x =m – 3 then (m – 3)2 + (m –3) = 5 m2 + 9 – 6m + m – 3 = 5 m2 – 5m + 6 = 5 m2 – 5m = –1

 x  y   xy 2  x  y   xy



x 2 +1 = x (x 2 –x) = – 1 Putting value in,

x

     5  3

82  1 63  2 15 2  1  61

1 x  1 x

5 3

5 3

5 3

2

2 ? Find: x2  x  2

2

5 3

1

xy = 1 Substitutes values in the question. 

676  1  1  0  = 676 2

270. (b) x2 + x = 5 then (x + 3)3 +

(x  y )  2 15 .........(ii)



2 3 + 27×2 2 + (243 × 2)+631

x

 



to put value x = 2



2

Again, x y 

(100 – 1) (10101)

+

2

(+226–225) +(225–225) ]

(100 – 1) [10000 + 1 – 200+ 300 – 3 + 3] (100 – 1) (10000 + 100 + 1)

1 [225+225+226][(225226) 2 2 2



1 266. (b) Given, x   1 x



2

2



(100 – 1) [(100 – 1) 2 + [3×(100 – 1] +3)

1233 Given, x 2 + y 2 + z 2 = 2 (x + z – 1 ) Find- x 3+ y 3+ z 3 = ?  x 2 + y 2 + z 2 = 2 (x + z – 1 )  x 2 + y 2 + z 2 = 2x + 2z – 2  x 2 + y 2 + z 2 = 2x + 2z – 1– 1  (x 2+1–2x) + y 2 + (z2+1–2z) = 0  (x–1) 2 + y 2 + (z–1) 2 = 0  (x–1) 2 = 0  x = 1  y2 = 0  y = 0  (z – 1) 2 = 0  z = 1 Value substituted in question,  x 3+ y 3+ z 3  13 + 0 + 13  2



5 3

 

99 ((99) 2 + (3 × 99) + 3)

265. (b)

267. (d)

2

5 3 

5  3

99 99 99 264. (b) Given, x = 2 Find x 3 + 27x 2 + 243x + 631

   



kgei snhe eYari dnag v.iSn ir

 

x y 



=

1 2

a  b  c 

a – b  ²  b – c  ²  c – a  332  333  335 2     333 – 332 ²  335 – 333 2  





 335 – 332 ²  



1000 2

14 

1000 2

1²  2²  3² 

= 7000

159

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From equation (i)

1

Find x = ? 2

3 x 

2

 2

3

2 –

3 × x 

1



 2x 3 2– 3  x = -1 275.

(c)

    

276.

Given m–5n = 2 find m³ – 125 n³ – 30mn

x6 =



 x6 = a3b x from equation (i)  On dividing above eq. by x we get x



6

x x



x

a³b

=

5 a ³b

(d) Given: x 

The value of x

1887

if x = 1



1 x

= 2 ...........(i)

 x 

12

–

x

1 =? 12

x

Then, x

–

x

1 1 7  1  5 1 x5

3

a  3

1

x

2 3

x

=2

y

1 12

1 2 3

y

 

2 3

3

3

a  3 286.

278.

 

1 – 1=0

(a) Given:

x



2 3



1 =1 ........(i) x

x ²  3x  1

=?

x ²  7x  1

Rakesh Yadav Readers Publication Pvt. Ltd.

Put x = 2

1



3 [ 2 ] 8

3

 8  6  14 Ans.

a  3

2

2  3y  

3y

 3 27    3 27  

2

  2  3  

283. (d) According to the question,

x 2  x 2 x 2 x 2

3

a  3

3x  2y 5  2x  3y 6

(d)

3 x   3 x   

 8  1 7  2 3  7  2 3  = 112

a Find



1



x 27  y 8

,

2 3

8 2– 3 2 3  2– 3 

1

a  3

8 x  27y

8xy (x ²+y²)

1 12 1 – 12 1

 3. a  3 

18x  12y  10x  15y

x  2 3 , y 2 3



3

a  3

a  3

1 2 3

2 3

1



 1   a3   a  3   8    

( x  y ) = 1887

1 , 2 3

1

1+1=2

12

3  1  a 3   a  3   23    

x+y = 1 282. (a) According to the question,





Cubing both sides

888x + 999y = 555

=

1 53 a3 1

281. (b) 999x + 888y = 1332

x

a3

a  3  a  3  2

 1+1 = 2 280. (d) Given expression,  4x 2 + 8x Let P should be added,  4x 2 + 8x + p  (2x) 2 + 2 × (2x) × 2 [(a+b) 2 = a 2+b 2+2ab] Term that should be added = 2 2 = 4 P = 4

wwM wa. th Les B aryn



5

3 a bx

=

1 5 (a  3)

 2  Let x = 1



7

(b) a 

1 2 5



x

1

 x 

3 a b a 3 b .....

a 3b

7

R Enak



285.

1 1 To, put value in question,

on squaring both side

On cubing both sides

 0  1  0  (0  1)2

2

x

3 a b a 3 b ..... ......(i)

x² = a 3 b a 3 b ....

=

2

 x 

m³ – 125n³ – 30mn = 8 (d) Given



8x

 a 3  b 3  ab  (a 2  b 2 )2

1

x

(m–5n)³ = 2³ (cubing both sides)

m³ – 125n³ – 15mn×2 = 8

4x

a 2 – ax = 1 2 –1×2 =1 – 2 = –1 (a) Let a = 0 b = 1

11 = 0

1

Find x

m³–125n³–3m×5n (m–5n) = 8

284.



x  7x

279. (c) x 

m – 5n = 2

x =

277.

x  3x

4 1 4

a

x² + 1 = x

 x ²  1  3x   x ²  1  7 x

3

22  22

r





1

22  22

a

1  x  =1 x

2 3

geisnh eeYa ridna gv.i Sni

274. (d) If 2  x 3 

 287.

2

3

8  3 8 

 3  2 2     3  1

2  (5)  25

(d) According to the Question x =

3 +

2

y=

3 –

2

160

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292. (a)

(x 3 – 20 2 ) – (y 3 + 2 2 )

3 + 2 ) 3 – 20 2 – ( 3 –

3 2)

– 2 2] =3

Let

3 + 2 2

+ 9 2 + 6

20 2 – 3 3 + 2 – 2

So this is answer

a = b = c



According to the question,

1

1

x =

a +



x 4 – x²y² – 1 + y 4 – x²y² + 1 = x 4 – 2x²y² + y 4 = [x² – y²]²

  

  a

1

a+

& y= a –

a

2

 – 

  a

1

a–

a

2

2

 

1 1     2 – a –  2 a a  



2p

295. (d)

p²–2p+1 2

= a

=

1

=

Let m =

5 5

5

p +

......(i)

Let n =

or

or

or

or

5

296. (b)

Factor = (a)×(a+1) Here n = a From (i) & (ii) m – 1 = n m – n– 1 = 0

or

291. (b)

5– 5–

3 – 5x 2x

3

2x

–

5

2

3 – 5y

+

3

+

+

2y

2y

–

5

2

+

......(ii)

3 – 5z 2z

3

2z

–

5

2





= 0

297. (a)

=0

3 3 3 35 + + = 2x 2y 2z 2

1

4

p

302.

1 1 1 352 + + = x y z 2 3

2 2 2 3522 + + = x y z 2×3

 299. (a)

=10

Rakesh Yadav Readers Publication Pvt. Ltd.

y –a 2

+

1 – 2 

2 2  0 1 1 a² + b² + c² = ab + bc + ca Let a = b = c = 1

p

a² + b² + c² = ab + bc + ca



to find

1²+1²+1² = 1×1+1×1+1×1 3 = 3

1+1

a+c b

=?

=2

1

p³ – q³ = (p – q) {(p – q)² – xpq}



p³ – q³ = (p – q) [p² + q² – 2pq – xpq]



p³ – q³ = (p – q) [p² + q² – 2pq – (–3)pq]

  

p³ – q³ = (p – q) (p² + q² + pq) So, x = –3

because a³ – b³ = (a – b) (a² + b² + ab) (a) Given

 

– 2= 8

x + y+ z = 6 xy + yz + zx = 10

To find x³ + y³ + z³ – 3xyz = ?

 Using formula.  (x + y + z)² = x² +

= 10

According to the question, x = 2, y = 1, z = – 3 x³ + y³ + y³ – 3xyz = ? As we know that a + b + c = 0 then a³ + b³ + c³ – 3abc= 0 2 + 1 – 3 = 0 x³ + y³ + 3³ – 3xyz = 0 According to the question  (x 3 + y 6) (x 3 – y 6)

+ yz + zx)

 

= 6 [16 – 10] = 6 × 6 = 36 (d) Given:

   

1

x y

1



a

x

– a

36

=

a

+

y – a  =?

x² + y² + z² + 20

x² + y² + z² = 16 x 3 + y 3 + z 3 – 3xyz =

= 6 16 –  xy  yz  zx  

303.

x 1

x –yx y 2x  2 2 2 2 x –y x –y Given, x + y = 2a to Find

6² = x² + y² + z² + 2 × 10

 x –1

x –y

y² + z² + 2 (xy

(x + y + z ) x²  y²  z ² – xy – yz – zx 

x 6 + x 3y 6 – x 3y 6 – y 12

x 6 – y 12 298. (b) According to the question,



+

  

301. (d)

1

p +

3 – 2



p

wwM wa. th Les aBryn

or



300. (a)

1

Factor = (a)×(a+1) Here m = a + 1 m – 1 = a

a

x – a 



4

(Divide p both in nu. & de.)

290. (d)

2



1

ERna

p–2+

= [4] 2 = 16

1

2

S(S – C) + (S – a) (S – b) 15(15 – 10) + (15 – 10) (15 – 10) = 75 + 25 = 100 Now check from option. Option (a) ab = 10×10 = 100 (Satisfied) 293. (d) p + m = 6 .....(i) p³ + m³ = 72 (p + m) (p² + m² – pm) = 72 (p + m) [(p + m)² – 3pm] = 72 [  p² + m² = (p + m)² – 2pm] 6[(6)² – 3pm] = 72 from (i) 36 – 3pm = 12 pm = 8 294. (d) xm × xn = 1 x m+n = xº (  xº = 1) m + n = 0 m = –n

2

Let x = 3, y = 1, a = 2

a

10+10+10 2

S=

2

(d) SHORTCUT METHOD Always do these types of question with the help of Put a = b = c = 1 3(1 2 + 1 2 + 1 2) = (a + b + c) 2 3 = 3 satified

289. (d)



2 +9 2 –6 3

2 –9 3 +9 2 = 0



a = 10, b = 10, c = 10

30 S = = 15 2

= 9 3 –9 288.

3

–

x + y = 2a 3

a+b+c 2

S=



kgei snhe eYari dnag v.iSn ir

=[(

According to the question If 2S = a + b + c

=

a b

x a b = 1 a –b (using componendo & dividendo)





x=

a b ........(i) a –b

161

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so value of 2(x + y + z)



1 y a = 1– y b

309.



1 a b y = a –b



y =



a –b a b

........(i)

From question,

x –y 1  xy

x – y =

310.

x

 311.

3

3

= 4

–

2



   

2

3– 2

 a + 2 ² – a – 2 ² 8a x ² – y² 4a = =  x ² + y² a + 2 ² +  a – 2 ² 2a² + 8 a² + 4

 

307. (c)



312.

Rakesh Yadav Readers Publication Pvt. Ltd.

3 2

,b=

3 2

4c + 5 = 0

–5 4

c =

1  a + b + c  a – b  ²   b – c  ²   c – a  ²  2 a² + b² + c²

2 2 2 1  1 3 5   1 3  3 5  –5 1   –     –   –        2  2 4 4   2 4  4 4 4 2      2

2

1 3  –5        2 4  4 

2

2 2 2 1  2  3 – 5   2 – 3   8   –5 – 2            2 4 4 4 4         1 9 25   4 16 16

b

3– 2

a³ + b³ =? ab

a + b

3

a+

– 3ab  a + b  ab 3– 2

a +b =



3– 2

3 2 2

 

3 2

2

3 – 2

2



2

3  2

a × b =

 

2

+



3

cube both sides

3 2

a3 +

1 3

a

3– 2

1

a3 +

a3

2

 3a 

1 1  a = a  a 

3

 3

3 3

= 3 3



1

a3 + a + b = 10

 2 

3– 2 3

= 3

= ?

 2

3 2 3–



= 0

a + b 3 – 3ab a + b 

a 30 + a 24 + a 18 + a 12 +a 6 +1 24

= a (a 6+ 1)+a 12 (a 6+1)+a 6+1 = a 24 (0)+a 12 (0)+0 = 0

ab = 1

a³ + b³ + c³ – 3abc a² + b² + c²

a3

a6 + 1 = 0

ab 10³ – 3  1  10  1  1000 – 30 = 970 (c) (2a–1)²+(4b–3)²+(4c+5)² = 0 to find



1 = a

2

(a + b + c = 0)

2

1



313. (a)  a   a 

=?

a

1 0 =0 2



b²



1  2 × (5)  Again

Applying Componendo & Dividendo



1 2 = 18

3

a b + =2 b a a² + b² = 2ab (a – b)² = 0 a – b = 0 308. (a) Put (x + y + z) = 10 x = 2 y = 3 z = 5 x(x + y + z) = 20 2(10) = 20 Similarly other will satisfied



(b) Given



a + 2  x2 = 2 y a – 22

2

3 4

b =

2 1

a²

x a+2 = y a–2

1

4b – 3 = 0





x

to find

 x y z      0   = 3x yz  a b c abc   306. (b)





x

a =

 x  y   z   x  y  z   a    b    c    3  a               b  c 

2a – 1 = 0

a =

 = 4² x (squaring both sides)

wwM wa. th Les B aryn

x y z =b–c; =c–a; =a–b 305. (c) a b c

–

then,



3

R Enak

 a² + b² + c² = ab + bc + ca  Let a = b = c = 1  1²+1²+1²=1×1+(1×1)+(1×1)  3=3  So ratio of a : b : c = 1 : 1 : 1

1

  x 

(d) Given a² + b² + c² – ab – bc – ca = 0 to find a : b : c = ? According to the question, a² + b² + c² – ab – bc – ca = 0



x = 2 + 3 x² – 4x – 1 = 0

x

2ab 4ab   a ² – b² 2 a ² – b ² 



(d)

1  a + b + c  a – b  ²   b – c  ²   c – a  ²  2 a² + b² + c²

12

y =2 –

a  b a – b  1    a – b a  b 

ab 2 – a – b 2  a ² – b ² 1 1



x – y = 2 3 ...........(2) solve equation (1) and (2)

a + b a – b  – a – b a  b 



304.

2(10) = 20 (c) x² + y² = 14 x + y = 4 ...... (i) squaring both sides x² + y² + 2xy = 16 14 + 2xy = 16 2xy = 2 xy = 1 x² + y² = 14 subtrance (2xy)from both sides. x² + y² – 2xy = 14 – 2 xy (x – y)² = 14 – 2 × 1



geisnh eeYa ridna gv.i Sni



1– y b = 1 y a

r

Again,

314. (d)

1 1 1 = + b a b a

1 b a = a b ab ab = a² + b² +2ab a² + b² + ab = 0 a³ – b³ = (a–b) (a² + b²+ ab) = (a–b) (0) = 0

162

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CHAPTER

12

TRIGONOMETRY C h

Relations between Trigonometric Ratios :-

cosec 

(i)

p q

A

b

AC AB AC AB BC BC and . If BC ' BC ' AB ' AC ' AB ' AB

(ii)

(iii)

cos  =

or sec θ × cos θ =1

tan  =

tan =

AC AB

=

p b

=

t an  

(v)

cos  cot   sin 

b 5 p 12  , cot  = p  12 b 5

15

Sol.

8 b cot  = 15  p

Ex.1 Write all the six t-ratios value in the given figure: A

5

90°

12

B

A 12 =p

90°

q

C

h = 13

Let AC = 12 = p and AB = 5 = b Then from Pythagoras theorem, BC =

Clearly sin  and cosec  are reciprocals to each other. Similarly cos  and sec  are reciprocals to each other while tan  and cot  are reciprocals to each other.

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2

2

2

2

AB  AC  5  12

=

289k ²  17k

C

Sol. In  ABC is, a right angle triangle with  A = 90°,

BC h Hypotenuse cosec = = = AC p Perpendicular

Let b = 8k p = 15 k From pythagoras theorem, h2 = p2 + b2 = (15k)2 + (8k)2 2 2 or, h = 225 k + 64k2 = 289 k 2 or, h =

q

B

q

8

TYPE - 1

Base

AB b Base cot = = = AC p Perpendicular

b 5 h 13   , sec  = h 13 b 5

Ex.2 If 15 cot  = 8 then calculate the remaining trigonometric ratio.

sin  cos 

(iv)

Perpendicular

BC h Hypotenuse sec = = = AB b Base

h 13 p 12 = , cosec  = p  12 h 13

or cot θ ×tan θ =1

AC p Perpendicular = = BC h Hypotenuse

AB b Base cos = = = BC h Hypotenuse

1 t an 

cot  

b= 5

sin =

 sin  =

1 sec   cos 

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 B =  the n thes e ratio ar e respectively called sin  , cos  , tan  , cot  sec  and cosec  . Clearly for the given angle  , AC (p) is perpendicular, AB (b) is base and BC (h) is hypotenuse. Hence six different trigonometric ratios are follows (see the given figure) Trigonometric Ratios:-

1 sin 

or cosec θ × sin θ =1

B

To study different trigonometric ratio functions we will consider a right angled triangle. Suppose ABC is a right angled triangle with  A = 90°. We can obtain six different trigonometric ratio from the sides of these triangle. They are respectively

Side opposite to right angle is BC, which is hypotenuse h.

kgei snhe eYari dnag v.iSn ir

Trigonometric Ratio:

ERna

1.

25  144 =

169 = 13

Hence, sin  =

p 15k 15   h 17k 17

cos  =

b 8k 8   h 17k 17

tan  =

p 15k 15   b 8k 8

sec  =

h 17k 17   b 8k 8

h 17k 17 cosec  = p = = 15k 15

Ex.3 If t an  

4 , then cos   ? 3

Here side opposite to  is AC which is p.

(a) 4 5

(b)

3

Side adjacent to  is AB, which is b.

(c) 3 4

(d)

1

5

5

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BC 4  AB 3 2

2

 AC  (4)  (3)  5 C

Let XOX' and YOY' be two mutually perpendicular lines. These lines divide the plane into four parts and e ac h o n e o f t h e m i s c al l e d a quadrant.

1.

sin( )   sin 

2.

cos( )  cos 

3.

tan( )   tan 

Complementary Angle.

4.

cot( )   cot 

5.

sec( )  sec 

6.

cosec ( )  cosec 

(B)

T-ratios of ( 90 0   ) in terms of those of  :-

For a given angle  its complementary angle is (90°–  ).

4

A

B

cos(90 0   )  sin 

3.

t a n (9 0 0  ) c o t 

4.

cot (90 0   )  t an 

5.

sec(90 0   )  cosec

6.

cos ec(90 0   )  sec 

(C)

T-ratios of ( 90 0   ) in terms of those of  :-

1.

sin(90 0   )  cos 

 sin   cos(90 –  )

2.

cos(90 0   )   sin 

Similarly, we can prove that

3.

t an(90 0   )   cot 

 cos   sin(90 – )

4.

cot (90 0   )   tan 

5.

sec(90 0   )  cosec

6.

cosec (90 0   )  sec 

(D)

T-ratios of ( 180 0   ) in terms of those of  :-

1.

sin(180 0   )  sin 

2.

cos(180 0   )   cos 

3.

t an(180 0   )   t an 

4.

cot (180 0   )   cot 

I

5.

sec(180 0   )   sec 

All +ve

6.

cosec (180 0   )  cosec 

(E)

T-ratios of ( 180 o   ) in terms of those of  :-

1.

sin(180 0   )   sin 

2.

cos(180 0   )   cos 

3.

t an(180 0   )  tan 

4.

cot (180 0   )  cot 

5.

sec(180 0   )   sec 

6.

cosec (180 0   )  cosec 

B

, the value of



(a) 1 2

(b) 1 3

(c) 1 9

(d)

1 13

4 BC Sol.(c) t an    3 AB

side opposite angle θ AB sin  = = hypotenuse AC

and cos (90° –  ) =

side along with angle (90° – θ) AB = hypotenuse AC

R Enak

and AC  (3)2  (4)2  5

BC 4  AC 5

 90,270.....(odd   multiple of 90°) will be changed

 0,180,360...........(    multiple of 180°) will not be changed Change will be in following manner: sin  cos & cos   sin  tan  cot & cot   tan sec  cosec  & cosec   sec 

wwM wa. th Les B aryn

C

4

B

3

4 1 1  sin  5 1   1  sin  1  4 9 5 Quadrants:90° Y

Signs of Trignometric Ratios:Y

II

sin  cosec 

X’

Changing line

 ve

I Quadrant

0   9 0

0

Non-Changing line

180° X'

X 0°, z360°

O

IV Quadrant

III Quadrant

1 8 0 0   2 7 0 0

2 7 00   3 6 0 0

Y' 270°

t an  cot 

X

O

III

II Quadrant o 0 9 0  < 1 8 0

C

From definition,

1  sin  is:1  sin 

A

geisnh eeYa ridna gv.i Sni 2.

AB 3  AC 5

4 Ex.4 If tan  = 3

 sin  

sin(90 0   )  cos 



 cos  

1.

°–

3

90

A

r

Sol.(b)

t an  

 ve

IV cos  sec 

 ve

Y’ Trigonometric Ratios of Allied Angles (A) T-ratios of (  ) in terms of those of  :-

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sin(270 0   )   cos 

2.

cos(270 0   )   sin 

3.

t an(270 0   )  cot 

4.

cot (270 0   )  t an 

5.

sec(270 0   )   cos ec

6.

cos ec(270 0   )  sec

(J)



0

1 2

1 2

3 2

1

1.

sin(270 0   )   cos 

cos

1

3 2

1 2

1 2

0

2.

cos(270 0   )  sin  t an(270 0   )   cot 

0

1 3

1

3



3.

tan

4.

cot (270 0   )   t an 

cot



3

1

1 3

sec

1

2 3

2

2 

2

2 3

0

sec(270   )  cos ec

6.

0

cosec (270   )  sec

T-ratios of ( 360 o   ) in terms of those of  :-

cosec

0

1.

sin(360   )   sin 

2.

cos(360 0   )  cos 

3.

tan(3600   )   tan 

4.

cot (360 0   )   cot 

5.

sec(360 0   )  sec

 2

sin(360 0   )  sin 

2.

cos(360 0   )  cos 

3.

t an(360 0   )  t an 

4.

cot (360 0   )  cot 

5.

sec(360 0   )  sec

6.

cosec (360 0   )  cosec  o

T-ratios of ( n  360   ) in terms of those of  :1.

sin(n  360 0   )  sin 

2.

0

cos(n  360   )  cos 

3.

t an(n  360 0   )  t an 

4.

cot (n  360 0   )  cot 

0

=

– 3  sin 180       – sin   2

sin 90    cos  3 2

= cos 30° =

Sol.(ii) cos210° = cos (180 + 30)°

cos 180    – cos 

= – cos30° =

– 3 2

Sol.(iii) Tan 570° = Tan (540 + 30)° (540° multiple of 180°, Then no change Tan (540 +  ) = Tan  ) = Tan 30° =

Sol.(vi) cos(1020)° = cos (1080– 60)° 1080 multiple of 180°, so no change In Trignometry function. = cos (3 × 360 – 60)° 1 2 Sol.(vii) sec (1500°) = sec (1440 + 60)° = sec (4 × 360 + 60)°  sec n  360      sec  

1

Ex.5 find the value of following (i) sin 120° (ii) cos 210° (iii) Tan570° (iv) cot 780° (v) sin960° (vi) cos1020° (vii) sec 1500° Sol. (i) sin 120° = sin (90 + 30)°

1 3

Sol.(iv) cot 780° = cot (720 + 60)°  cot (n × 360 +  ) = cot 

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Sol.(v) sin 960° = sin(900 + 60)°  900° multiple of 180°, so no change of Trignometry function. = sin(2 × 360 + 180 + 60) = sin (180 + 60) = –sin60°

= cos 60° =

TYPE - II

T-ratios of ( 360 o   ) in terms of those of  :1.

1 3

= cot 60° =

0° 30° 45° 60° 90°

sin

6. cosec (360 0   )   cos ec (I)

6. cosec (n  360 0   )  cosec Value of some specific angle of trigonometrical (t)-ratio function. We must learn the following table to solve the question bas ed on trigonometrical (t)-ratio angle 0°, 30°, 45°, 60°, 90°

T-ratios of ( 270 o   ) in terms of those of  :-

5.

= cot (2 × 360 + 60)°

sec(n  360 0   )  sec

kgei snhe eYari dnag v.iSn ir

(H)

1.

5.

ERna

(G)

T-ratios of ( 270 o   ) in terms of those of  :-

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(F)

= sec 60° = 2 Ex.6

cos(90 0  A ). sec(360 0  A ). tan(180 0  A ) =? sec(A  720). sin(A  540 0 ).cot(A  90 0 )

(a) 0 (b) 1 (d) None of these Sol. (b)

(c) -1

cos  90 0  A  . sec  360 0  A  . tan 180 0  A  sec  A  720 0  . sin  A  540 0  . cot  A  90 0  

si n A . sec A t an A 

sec 2360  A .sin 3180 0  A . cot 90 0  A   

    

  

 

 sec     sec  and cot      cot  



sin A. sec A. tan A sec A   sin A    t an A 

sin A. sec A. t an A 1 sin A. sec A. t an A Ex.7 sin720° – cot 270° – sin 150° cos120° is equal to:– (a) 1 2 (b) 1 3 (c) 1 5 (d) 1 4 Sol.(d). sin720 º – cot270º – sin150º.cos120º = sin (2×360º+0º) –cot(360º–90º) – sin 

(90 0  60 0 ). cos(90 0  30 0 )

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 s i n 0 0  co t 9 0 0  c o s 6 0 0 . s i n 3 0 0

1 1  2 2

0 0 



Sol.(b) sec  

1 4

Sol.(d) sec 17° – sin 73° = sec 17° – sin (90° – 17°) = sec17°- cos17º

13 AC  5 AB

and BC  13 2  5 2  12

sin 37 0 Ex.8. Find the value of :cos 53 0 (a) 1 (b) -1 (c) 0 (d) 0 Sol. (a)

 sin  

=

BC 12  AC 13

2

C



3 We know that sin 60  2

r

Ex .12. If

3 cos 30  , cot 45 0  1 2 sec 60 = 2 0

x cos ec 2 30 0 . sec 2 45 0 8 cos 2 45 0. sin 2 60 0 = tan2 60º - tan2 30º, then the value of x is :(a) -1 (b) 0 (c) 1 (d) 2 Sol. (c)

53 (a)  10

 2

2

2

2



2

  3

 1     3

Sol. (c) 3 tan   4  0  t an   

13 Ex.11. If sec   and  lies in the 5 fourth quadrant, then the value of sin  is :12 (a) 13 (c) 

5 13

 sin  

5

4

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1

cosec 2 51 

1

(b) (d)

–

+

is:

1– x2 1 – x2

+ sin239° + tan251°

1

2

sin 51 . sec 2 39 = s in 2 5 1° + + tan2(90°– 39°)

–

sin 2 39 °

1 2

sin 90 – 39.sec 2 39

4 3 3 , cos    , cot    5 5 4

x , then the y

value of sec 17°– sin 73° is:

y2 – x2 (a) xy x2

1 2

cos 39.sec 2 39

[  sin (90°– θ )= c os θ ),

3

Ex.13 If sin 17° =

5 13

sin 239 °

= cos2 39° + sin2 39° + cot239°

[ sin  is positive and cos  is negative in II quadrant].

12 (b)  13 (d)

4 3

2

 1   3 8     2  2  1 3 8  8x    8     x  1 2 4 3

x2 –1

–

37 (d) 10

+

sin 2 51 sec 2 39

(c) x 2 – 1

Sol.(c)

7 (b) 10

23 (c) 10

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Ex.10 If

(a)

2 cot   5 cos   sin  is :-

0

 3 2  3 2  7 2 2 2 2        1  2  2  1   2 2   2     

cosec 2 51

tan251° –

3 tan   4  0 , whe re

R Enak

 cos ec 30  cos 0

2

1

12 13

     , then the value of 2

 sin 2 60 0  cos 2 30 0  cot 2 45 0  sec 2 60 0

x  2  

Ex. 14. If cosec 39° = x, the value of

B

 sin   –

0

2



since  lies in the fourth quadrant

(d) 2

0

x2 y2

x² x² y = y² ² – x² = y y² – x ² y²

geisnh eeYa ridna gv.i Sni

A

(b) 5 2

2

x2 y2 1–

Ex.9. Evaluate :- sin2 60º+cos2 30° + cot245º + sec2 60 º - cosec230º + cos2 0º:-

Sol. (c)

2

1 – cos 17 sin 17   cos 17 cos 17

sin 37 0 sin 37 0 sin 37 0   1 0 0 0 cos 53 cos(90  37 ) sin 37 0

(a) 3 2 (c) 7 2

1 - cos17º cos 17

x2 (b)

y2 – x2 x2

(c) y y 2  x 2 (d) y y2 – x 2

tan(90°– θ )= cot θ ] = 1 + cot2 39° – 1 = cosec2 39°– 1 = x2 – 1 Ex. 15 Find the value of cos (180° + A) + cos (180° + B) + cos (180° + C) + cos (180° + D) Where A, B, C and D are the vertices of a cyclic quadrilateral ? (a) 0 (b) 1 (c) 2 (d) 2 cos A Sol.(a) cos 180 0  A   cos 180 0  B   cos

180

0

 C   cos 180 0  D 

  cos A  cos B  cos C  cos D

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Proof (i)

 cos C  cos D [  A  C  B  D  180 0 c y c l i c quadrilateral]

 cos C  cos D  cos C  cos D  0

Some Useful formula (i)

2

2

sin  + cos  = 1 or sin2  = 1– cos2  or cos2  = 1 – sin2 

1 + tan2 = sec2  or sec2  – 1 = tan2  or sec2  – tan2  = 1 (iii) 1 + cot2  = cosec2  or cosec2  – 1 = cot2  or cosec2  – cot2  = 1 Proof we know, (ii)

Then, Now, 

0

b p sin  = h

b cos  = h

Now, sin2  + cos2  2

2

p2  b 2 h2

 In right angle  ABC p2 + b2 = h2 then sin2  + cos2 

h2 =1 h2 Same as we can proof all remaining results same this process

=

TYPE - III

If A + B = 90°, Results (i) sin A. secB = 1 or sinA = cosB (ii) cos A. cosecB = 1 or secA = cosecB (iii) tanA. tanB = 1 or tanA = cotB (iv)cotA. cotB = 1 2 2 (v)sin A +sin B = 1 2 2 (vi)cos A +cos B = 1

Sol.

2  + 3  = 90° 5  = 90°

 If A + B = 90°   tanA. tanB = 1

Now,

90 90 . cos 2 2 = sin45°. cos45°

= sin

=

1 1 1   2 2 2

Ex.23 If sin (x + 4)° sec (x – 4)° = 1

 If A + B = 90°   tanA. tanB = 1

So, t an 23 0. t an 67 0 = 1 Ex.18 The value of tan10°. tan25°. tan 65°. tan 80° is

5 5 . cos 2 2

sin

put the value of 5 

of

(tan 23  t an 67 ) is equal to :23° + 67° = 90°

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 p b  p2 b2 =   +   = 2 + 2 = h h h h

Sol.

5 5 . cos 2 2

sin

0

ERna

C

B

Ex.22 If cot2  cot3  = 1 find the value of

 If A + B = 90°   sinA. secB = 1

So, sin25°. sec65° = 1 Ex.17 The value

h

 If A + B = 90°   tanA. tanB = 1

1 sinA × =1  sin A Same as we can proof all remaining results same this process And their vice-versa are also true. when sinA. secB = 1, then we can say A + B = 90° Ex. 16 The value of (sin25°. sec65°) is equal to:Sol. 25° + 65° = 90°

A p

5  = 90°  = 18°

sin A. secB = 1 A + B = 90° (given) B = 90 – A sinA.sec.(90 – A) sinA.cosec.A

kgei snhe eYari dnag v.iSn ir

  cos 180 0  C   cos 180 0  D 

find the value of tan

Sol.

sin(x + 4)° sec (x – 4)° = 1 x + 4 + x – 4 = 90° 2x = 90° x = 45 Now,

Sol.

tan10° tan25° tan65° tan80°

=1 Ex.19 If sin(3x– 6) = cos (6x – 3) find the value of x. Sol. 3x– 6 + 6x – 3 = 90° 9x = 99° x = 11  If A+ B = 90,   then sinA = cosB  Ex.20 The value of cos40°.cosec50° Sol. 40° + 50° = 90°  If A + B = 90°   cosA. cosecB = 1

tan

2x 3

put of value of x = tan

2  45 90 = tan 3 3

= tan 30° =

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1 3

Ex.24 If cos(90 –  ) = sin (3  – 5) find the value of  Sol.

cos(90 –  ) = sin(3  – 50) cos(90 –  ) cosec (3  – 50) = 1 90 –  + 3  – 50 = 90°

So, cos40°. cosec50° = 1

Ex.21 If tan 2  tan3  = 1 find the value of  Sol. 2  + 3  = 90°

2x 3

 if cosA.cosecB = 1  A + B = 90° 

2  = 50°  = 25°

167

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cos–sin cos+sin Ex.26 Find the value of the following (i) sin75° (ii) cos75° (iii) tan15° (iv) tan75° Sol.(i) sin75° sin(45° + 30°) = sin45° cos30° + cos45°sin30°

sin75° = (ii)

(vii) (viii) (ix) (x)

1 1 3 –1 3 1 = × – × = 2 2 2 2 2 2

cos75° = (iii)

(ii) (iii) (iv)

(v)

tan 45 – tan30 1  tan45.tan30

1 3 1 = = 1 1  3 1–

tan15° =

(iv)

=

cos+sin cos–sin

(iv)

si n 3 6 0 

(vii)

sin 22

Sol.

tan 45  tan30 1 – tan45 tan30 1

3 1 3 –1

(i)

 5 1  sin 18 0     4 

(ii)

cos1 8 0 

10  2 5 4

2 2 2

cos 22

cos15 – sin15 cos15  sin15 = tan (45 – 15)

= tan 30° =

1 3

Ex.28 The value of tan40° + 2tan10° is equal to (a) tan40° (b) cot40° (c) sin40° (d) cos40° Sol. We know, 40° + 10° = 50° both sides take tan tan(40° + 10°) = tan50°



tan40  tan10 = tan50° 1 – tan40.tan10

tan40° + tan10° = tan50° – tan50°.tan40°.tan10 û 1 \ tanA.tanB = 1 if A + B = 90

tan40° + tan10° = tan50° – tan10° tan40° + 2tan10° = tan50°  Now, tan50° = tan.(90° – 40°) = cot40° Ex.29 The value of 

tan57  cot 37 is equal to tan33  cot 53 (a) tan33°.cot53° (b) tan53°.cot37° (c) tan33°.cot57° (d) tan57°.cot37°

3 1 tan75° = = cot15° 3 –1 Trignometric Ratios of Specific Angles

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10  2

cos15 – sin15 is cos15  sin15

 3 –1 3 1

10 – 2 5 4

10 2 2  2 2 Ex.27 The value of (viii)

3 –1 = cot75° 3 1

1 3 1 = = 1 –1  3

tan A  t an B 1  t an A. tan B tan A  t an B t an(A  B )  1  t an A. tan B cotA·cotB – 1 cot(A + B) = cotA + cotB cotB·cotA +1 cot(A – B) = cotB – cotA

1+ tan 1 – tan

 5 1  cos 36 0     4 

tan75° tan(45 + 30) =

t an(A  B) 

tan(45 +) =

3 –1 = sin15° 2 2

tan15° tan(45 – 30) =

Tangent Formulae

(i)

3 1 = cos15° 2 2

R Enak

(vi)

2 sin A. cos B = sin (A+B)+sin (A-B) 2 cos A. sin B = sin (A+B)sin (A-B) 2 sin A. sin B = cos (A-B)-cos (A+B) 2 cos A. cos B = cos (A+B)+cos (A-B) sin2 A-sin2 B = sin (A+B) . sin (A-B) cos2 A-cos2 B = cos (A+B) . cos (A-B)

(iii)

cos75° cos(45° + 30°) = cos45°cos30° –sin45°sin30°

wwM wa. th Les B aryn

(v)

1 1 1 3 1 3 × + × = 2 2 2 2 2 2

=

TYPE–IV Sum and Difference Formula (i) sin (A+B) = sin A. cos B+cos A sin B (ii) sin (A – B) = sin A. cos B – cos A sin B (iii) cos (A+B) = cos A. cos B – sin A sin B (iv) cos (A – B) = cos A. cos B+sin A sin B

1 – tan 1+ tan

=

 tan(– )  – tan    sin(– )  – sin   = – (tan60° + sin60°)

 3 –3 3 =–  3 2  =   2

tan(45 –) =

r

Now,

 if cotA.cotB = 1   then A + B = 90°  x – 50 + 80 – 2x = 90 – x + 30 = 90 x = – 60° tanx + sinx = tan(–60)° + sin (–60)° = – tan60° – sin60°

(vi)

geisnh eeYa ridna gv.i Sni

Ex.25 If cot (x – 50) = tan (80 – 2x) find the value of tanx + sinx Sol. cot (x – 50) = tan (80 – 2x) cot (x – 50) cot (80 – 2x) = 1

Sol.



tan57  cot 37 tan33  cot 53 tan 57  cot 37 tan(90 – 57)  cot 53

168

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1 tan37 cot 57  cot(90 – 37)

 

1 tan 57  tan 37 1  tan 37 tan 57

tan57 tan37  1 

tan37 tan57 tan37  1 tan57



1 × tan57° tan37



tan57°. cot37°

Then

B

x a = y b

Then,

Add 1 in both side.



a b ..........(i) b subtract 1 in both side.



x a –1 = –1 y b



.........(ii)

sin   cos  = 9 find the sin  – cos  value tan  and cos 

Ex.30 If

Sol.



C

3sin  2  2cos  1 Apply C and D

4 b = h 41

sec   tan  5 = sec  – tan  3 Apply C & D

3sin   2cos  2  1  3sin  – 2cos  2 – 1 3sin   2cos  3 3sin  – 2cos 

Alternate:-

sec   tan   sec  – tan  sec   tan  – sec  – tan 

sin   cos  9 = sin  – cos  1 Apply C & D

3sin   2cos  3sin  – 2cos 

divide all terms by cos 

3 tan   2 3 tan  – 2

2sec  8 = 2tan  2

4  tan  = (given) 3

1 cos   4 1 =4 sin   sin cos 

1 So, sin  = 4

x y a b x –y = a –b

sin   cos  5 Ex. 32 If = , find the sin  – cos  4

value of

Sol.

sin   cos   sin  – cos  sin   cos  – sin  – cos 

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4 3

sin  2  2  cos  3

53 = 5–3

=

tan  =

sin  4  cos  3

sec   tan  5 = , then find sec  – tan  3

wwM wa. th Les aBryn

x a 1 y +1 = b

x –y a –b = y b (i) /(ii)

2

  4

ERna

x a  y b

x  y y

2

5

The value of sin  Sol.

4 , then the value of 3

3sin   2cos  is 3sin  – 2cos 

Sol.

4

cos  =

Ex.31 If

x+y a+b = x–y a–b

Proof

Perpendicular Base

41

5

82 41 = 80 40

Ex. 33 If tan  =

41

A

Use of componendo and dividendox a = y b,

tan  =

=

=

5 4

Hypotenuse =

TYPE-V

If

Now,

2sin  10 = 2cos  8 tan =

Now,

2

9  1 tan2   1 = 2 92 – 1 tan   1

kgei snhe eYari dnag v.iSn ir



9 1 9 –1

=

tan57 

4 2 3 42 6  = = =3 4 4 – 2 2 3 –2 3 3

Ex.34 If 2cot  = 3, Then find the value of Sol.

2cot  = 3

2

tan   1 tan2   1

2cos  – sin  2cos   sin 

cot  =

3 2

sin   cos  5 = , sin  – cos  4

cos  3  sin  2

Apply C and D tan  = 9

2cos  3  sin  1

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Apply C and D 2cos   sin  3 1 = 2cos  – sin  3 –1 2cos   sin  4 = =2 2cos  – sin  2 2cos  – sin  1  2cos   sin  2

2sin  cos  =

289 – 169 169



2sin  cos  =

120 169



sin  cos  =

5 or 12

TYPE - VI

r

2 6 



2 6 

2

+ a  = (a + 2)2

2

= (a + 2)2 – (a)2

    

24 = (a + 2 – a) (a + 2 + a) 24 = 2 (2a + 2)



secx + tanx =

24 = 4(a + 1) 6=a+1

a=5 Hence, xz = (5 + 2) = 7 yz = 5 7 5 + 2 6 2 6

Sol.

Sol.

3 Ex.36 If sin  + cos  = , find the 17 value of sin  .cos  15 or 8

=

sin2  + cos2  + 2sin  cos 

289 169

Now, Check

A

a  2

2 5

8 or 15



B

C a

C

Let the triangle of side BC = a Then AB = BC + 2 = a + 2 Now,

8 15 23   17 17 17

Hence, sin  .cos  =

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A

17

p b p b + = h h h so, h  17 Apply pythagorean sides here hypotenuse is 17, then other sides 8 and 15

2

12 6 = = 6 2 6 6

AB – BC = 2, and AC = 2 5 find the value of cos2A – cos2C = ?

3 sin  + cos  = 17



289 –1 169



Ex. 38 In a  ABC,  B = 90°

5 12 60   Hence, sin  cos  = 13 13 169

17 13 squaring of both side

2sin  cos  =

b h

5 12 17   13 13 13 But we cannot find exact value of base and perpendicular, here no affect of value of sin  and cos  . This question because both are product.

sin  + cos  =

289 169

p b h

App ly p ythagore an here hypotenuse is 13, Then other sides of right angle triangle will be 5 and 12. Now, C h e c k

17 Ex.35 If sin  + cos  = find the 13 value of sin  .cos 

 17  (sin  + cos  )2 =   13

z

a

2

sinq + cosq = 17

wwM wa. th Les B aryn

Som e p y thagore an natural number will help in solving the pr oblem on trigonom etr ic ratio angle. pythagorean theorem (Bas e) 2 + (P erp end icular ) 2 = (Hypotenuse)2 32 + 42 = 52, 62 + 82 = 102, 2 2 2 5 + 12 = 13 , 102 + 242 =262, 82 + 15² = 172, 72 + 242 = 252, 202 + 212= 292,92 + 402 = 412, etc.



a  2

2 6

R Enak

3 –1 2 3 –1 2 1 = = = 3 2  1 3 1 4 2 2

1 + 2sin  cos  =

x

12 or 5

p h

2



x

Let yz = a, Then xz = 2 + yz = 2 + a Apply pythagorean theorem

13

3 (given)  cot  = 2

=

60 169

13

2cot  – 1 2cot   1



Sol.

y

2cos  – sin  2cos   sin  Divide all terms (in numerator and in denominator) by sin. 



xy = 2 6 and xz – yz = 2 Find the value of secx + tanx = ?

Alternate:-

Alternate:-

Sol.

Ex. 37 In a  xyz,  y = 90°

geisnh eeYa ridna gv.i Sni

So,



8 15 120   17 17 289



2





(a + 2)2 + (a)2 = 2 5

     

a2 + 4 + 4a + a2 = 20 2a2 + 4a = 16 a2 + 2a = 8 a2 + 2a – 8 = 0 a2 + 4a – 2a – 8 = 0 (a + 4) (a – 2) = 0 a = –4, a = 2 side of  is always positive hence, We take a = 2

170

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 4   2  =  –   2 5   2 5 

2

So, cot  =

16 4 12 3 –  = 20 20 20 5 Ex. 39 If 2sin  + 15cos2  = 7, (0°<  < 90°) find the value of cot 



3 5 1 1 (a) (b) (c) (d) 4 4 2 4 Sol. 2sin  + 15cos2  = 7, 2sin  + 15 (1 – sin2  ) = 7  2sin  + 15 – 15 sin2  = 7  15sin2  – 2sin  – 8 = 0  15sin2  – 12 sin  + 10sin  –  8=0 (3sin  + 2) (5sin  – 4) = 0  3sin  + 2 = 0 or 5sin  – 4 = 0 



–2 4 sin  = 3 5 Value of  between 0° and 90° so sin  is positive. Then we

Ex. 40 If 2 – cos2  = 3sin  cos  , find the value of tan  (a) Sol.

sin  = b=

2

2

– 4 

(b) 0

1 or 1 2 So, option (a) is correct

=3

A

wwM wa. th Les aBryn 3

5

1

3 4



h=

2

  4

=

value of tan Sol.

1 p = 2 b 2

2

SinA 1 A B

2

 1

5

Now,

2–cos2  = 3sin  cos 

=

25 = 5

Now, check 2sin  + 15cos2  = 7 2×

C

2

then, h =

b 3 cot  = = p 4 2

1 2

Alternate:Take options (a)

So, tan  =

3

1 3

tan  =

5

h=

(d)



B

We take option

2 3

2–cos2  = 3sin  cos  divide by cos2  both sides

tan  =

b 3 Hence, cot  = p = 4 Alternate:-

4

(c)

     

4 5

4 p = 5 h

5

1 2

3sin  cos  2 – cos 2  = 2 cos 2  cos  2 2sec  – 1 = 3tan  2(1 + tan2  ) – 1 = 3tan  2 + 2tan2  – 1 = 3tan  2tan2  – 3tan  + 1 = 0 (2tan  – 1) (tan  – 1) = 0 2tan  –1= 0, tan  – 1 = 0



sin  =

take sin  =

3 4

Functio n and Invers e function  (a) If sin θ + cosec θ = 2 then sin θ = cosec θ =1 sinn θ + cosecn θ =2  n  natural no. Ex.41 If sin θ + cosec θ = 2 find the value of sin100 θ +cosec100 θ Sol. sin θ + cosec θ =2 Then, sin θ = cosec θ = 1 so, sin100 θ + cosec100 θ = 1 = (1)100 + (1)100 = 2 (b) If cos θ + sec θ = 2 then cos θ = sec θ = 1 cosn θ + secn θ = 2  Ex.42 If cos θ +sec θ = 2, find the value of cos10 θ + sec10 θ = ? Sol. cos θ +sec θ = 2 cos θ = sec θ = 1 Then, cos10 θ + sec10 θ = (1)10 + (1)10 = 1+1 = 2 (c) If tan θ + cot θ = 2 so tan θ = cot θ =1 tann θ + cotn θ = 2 Ex.43 If tan θ + cot θ = 2 find the value of tan50 θ + cot60 θ Sol. tan θ + cot θ = 2 tan θ = cot θ =1 50 tan θ + cot60 θ = (1)50+(1)60=1+1=2 (d) If sinA + cosB = 2 Then A = 90º B = 0º Ex.44 If sinA + cosB = 2, then find the

kgei snhe eYari dnag v.iSn ir

2



TYPE -VII

8 27 35 + = =7 5 5 5 L.H.S = R.H.S

=

ERna

 

Now, AB = 4, BC = 2 cos2A – cos2C

4 3 + 15 ×   5 5

2

8 9 + 15 × 5 25

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Now,

tan

= =

6 6 = 5 5 L.H.S = R.H.S  tan  =

1 2

2 cosB=2 1 90º 0º

A+B

2

1 2  2  2–  =3× ×  5  5 5

+

A+B

= tan

2

90+0

= tan45º = 1 2 (e) If sinA + cosB =0 then A = 0°, B = 90º Ex.45 If sinA + cosB + sinC = 3, then find the value of cot

A+B+C 3

171

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1 1 A = 90º B = 0º C = 90º



1

tan  +cot  =2,

 2  2β    = 3     (a) sin (b) cos 3 2  2 (c) sin (d) cos 3 3 sin  + cos β = 2 sin   1; cos β  1

then tan7  +cot9  is equal to:

  = 90°; β = 0°

1 3

90+0+90 = cot 60º = cot 3

Ex .46

If

(a)



is

3 (b) 3

acute

(c) 2

and

Sol.

(d) 4

 sin

Sol. t an cot  2 tan  = cot  = 1

Ex .47

If

tan(x  y )  3

and

cos

cot (x  y )  3 , then what are the smallest positive value of x and y respectively? (a) 45 0 , 30 0

R Enak

 1  2  3 

then x + y = 60°

.....(i)

wwM wa. th Les B aryn Then x – y = 30° from (i) and (ii) x = 45° & y = 15°

.....(ii)

2 sin 22  3 , then what

Sol.

 2

(c) 1

(d)

 1 – sin2  + 1 – sin2  = 2

(a) 300, 200 (c) 200, 300

 sin2  + sin2  = 0

(b) 600, 400 (d) 450, 450

1 = 2

cos60°

3 2 sin 2 2  3  sin 22   sin 60 0 2

1 3

  =  =0

 cos160 0  cos 180 0  20 0    cos 20 0



similarly cos140° = – cos40°, cos120° = – cos60° cos100° = – cos80°

 cos 20

TYPE - VIII

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value

of

0

 cos160 0   0

cos 40  cos140   cos 60  cos120  0

0

  cos 80 0  cos100 0   cos180 0

= cos 20 0

cos 60

0

 cos 2 0 0  cos 40 0  cos 4 0 0 

 cos 60

0

     cos 80



0

 cos 80 0 

 cos180 0  cos180 0  1

(a) 39

(c) 40

Series Base Ex. 52 The

(d) 1 2

cos 1 80 0      cos 

Sol.

 tan3  + sin5  = 0

 2 2  60 0   2  30 0

1

Ex. 55 sin25° + sin26° + ...........sin284° + sin285° = ?

 sin  = sin  = 0

 3 θ 1 = 60°  θ 1 = 20°

(c) 3 (d)

[  tan (90° – θ ) = cot θ ,tan θ .cot θ = 1] Ex.54 The value of : cos20º + cos40º + cosº + ...... + cos160º + cos180º is:(a) 1 (b) -1

0

cos2  + cos2  = 2

Sol.

(b) 0

3 Sol. tan 1°.tan 2°. tan3°......tan 45° .....tan88°tan89° = (tan 1°.tan 89°) (tan2°. tan88°) .... tan 45° = (tan 1°.cot 1°). (tan 2°. cot2°) .... tan 45° = 1



will be the value of 1 and 2

2cos3 θ 1 = 1  cos3 θ 1 =

(a) 1

value of tan3  + sin5  is :

(a) – 1 (b) 0

(d) 0

Now,

Ex.51 If cos2  + cos2  = 2, then the

Ex.48 If 2 cos 31  1 and

(c) 2

 cos10 . cos 2 0................. cos179 0  0 Ex. 53 The value of tan1° tan2°tan3° .......tan 89° is :

(c) 0

 cos 1  cos 2  cos 3  0

3

(b) -1

 cos 90 0  0

 3  cos 30  3 2

 sin 1  sin 2  sin 3

3

Sol.

sin 1  sin 2  sin 3  3

Sol.

(d) 30 0 , 45 0

cot (x – y) =

3 2

cos 1  cos 2  cos 3  ? (b) 1 (c) 2 (d) 3

(a) 0

(c) 45 0 , 150

tan (x + y) =

3

 180   sin   3 

Ex.50 If sin 1  sin 2  sin 3  3 , then

(b) 150 , 60 0

Sol.

2  β 

= sin 60°  sin 60 

 tan7  + cot9  = 1 + 1 = 2

So,

(a) 1

r

Ex. 49 If sin  + cos β = 2(0°  β    90°), then sin

sinA + cosB + sinC=3

geisnh eeYa ridna gv.i Sni

Sol.

cos1 º,

cos2º,cos3º...... cos179 0 is:-

1 2

(b) 40 (d) 39

1 2

1 2

Sol. Let the number of terms be n, then By tn = a + (n – 1)d Here,  a = 5, d = 1

172

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Here,

 85 = 5 + (n – 1) 1  n – 1 = 85 – 5 = 80  n = 81  sin25°+ sin26° + .... + sin245°

 19 20  + = = 40 40 40 2  So, sum of 2 terms = sin² =1 2 n = 19

+ .... + sin284° + sin2 85° = (sin25° + sin285°) + (sin26° + sin2 84°) + .....+ to 40 terms + sin2 45° = (sin25° + cos2 5°) + (sin26° + cos²6)+.... + to 40 terms + sin245°

19 21 +1= 2 2 Ex.59 If 1+sinx + sin 2 x + sin³x 

Sol.

+.....  = 4+2 3 , find the value of x It is a G.P Series a 1– r a = 1, r = sinx

then, S  =

1 1– sin x



4–2 3







sin² 40 +sin² 40

...... Sin²

   +sin² + sin² 40 40 40

 19 + sin² ) + 40 40 2 18 (sin² + sin² ). .... + 40 40 20 sin² 40

4 +2 3

×



(a) 0

1 1 3 = 1– sin x 1 – 2

3 2 sin x = sin60º x = 60º

TYPE- IX

sin 2 θ + c o s 2 θ = 1 or

(b) 1

(c)

2 3 (d) 3 2

2 3  (cos 2  + sin 2  ) (cos 2  – 2 sin2  ) = 3 2  cos2  – sin2  = 3 2  cos2  – (1– cos2  ) = 3 2  2 cos2  – 1 = 3 Ex.64 sin  + sin2  = 1 Find the value of cos2  + cos4  Sol. sin  + sin2  = 1 sin  = 1 – sin2 

2

sin θ = 1 – c o s θ or

Now,

c o s 2 θ = 1 – s in 2 θ

sin  = cos2  cos2  + cos4  Put the value cos2 

2

Ex. 60 What is the value of sin 1000° + cos2 1000°? (a) 1000 (b) 100 (c) 10 (d) 1

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3

Sol.(c) cos4  – sin4  =

sin x =

(sin²

3 2

2 , then the 3 value of 2 cos2  – 1 is :

4 4 1 = 4 2 3 1– sin x – 4 4

2

(c)

Ex.63 If cos4  – sin4  =

comparing both equation

(A)

5

 4 sin2  = 3  sin  = tan  = tan60° =

(divide by 4 all terms)



(b)

(d) 6 3 Sol.(a) 3 sin2  + 7(1– sin2  ) = 4  3 sin2  + 7 – 7 sin2  = 4  7– 4 sin2  = 4



ERna



2

So,  = 60°

4 1 = 1– sin x 4–2 3

wwM wa. th Les aBryn

Ex.58 The value of  2 3 sin² + sin² + sin² + 40 40 40 20 ..... sin² 40  2 3 Sol. sin² +sin² +sin² +..... 40 40 40 19 20 + sin² + sin² 40 40

=

(a)

4–2 3

0



19 + sin² 90º 2

=

= 40 +

kgei snhe eYari dnag v.iSn ir

1 1 = 40 2 2 sin 90 – θ  cos θ   sin2 θ  cos 2θ= 1    Ex.56 The value of sin10º + sin20º + ... + sin340º + sin350º Sol. sin10º+sin20º+...+ sin340º+sin350º sin(360º – 350º) + sin(360º–  340º) +...sin180º ....sin340º + sin350º – sin350º–sin340º....+sin180º  +...sin340º+ sin350º = 0 [sin{360º– θ }= – sin θ ,sin180º = 0]) Ex.57 The value of cos²1º + cos² 3º + ...... + cos² 89º + cos² 90º Sol. cos²1º + cos² 3º +cos² 5º ...... + cos² 89º + cos² 90º 89  1 n= +1 = 45 2 45 sum = 2 45 cos² 90º 45 + =  2 2

19 20 + sin² 2 2

sum =

Sol.(d) sin21000° + cos21000° = 1 for every value of  in sin2  + cos2  will be 1 Ex. 61 If sin260° + cos2(3x – 9°) = 1 Then value of x is Sol. sin260° + cos2(3x – 9)° = 1 This is similar to sin2  + cos2  So,  60° = 3x – 9 69° = 3x x = 23° Ex.62If 3sin2  + 7 cos2  = 4, then the value of tan  is (where 0 <  cos 55° (b) cos61°>

1 2

(d)

218. What is the expression

wwM wa. th Les B aryn

cot² θ –



p 2 – q 2 tan  ?

(a) p

(c)

R Enak

1 is the value x2 +  2  ? x



geisnh eeYa ridna gv.i Sni

(c) a2 + b2 – 2 (d)  2  207. If  is the angle of first quadrant such that cosec4  = 17 + cot4  , then what is the value of sin  ? (a) 1/3 (b) 1/4 (c) 1/9 (d) 1/16

223. If cosec  = p/q and  is acute, then what is the v alue of

r

205. What is the value of

1 2 (d) tan 44° > 1 227. If sin  + cosec  = 2, then what is the value of sin4  + cos4  ? (a) 2 (b) 22 (c) 23 (d) 1

(c) sin 32° >

228. If   R be such that sec  >0 and 2sec2  + sec  – 6 = 0 Then, what is the value of cosec  ? (a)

5

(c) 3/ 5

(b)

3 /2

(d) 2/ 3

229. Under which one of the following conditions is the trigonometrical identity. sinx/(1 + cosx) = (1 – cosx)/sinx true? (a) x is not a multiple of 360° (b) x is not an odd multiple of 180° (c) x is not a multiple of 180° (d) None of the above 230. If 3sin  + 4cos  = 5, then what is 3 cos  – 4sin  equal to? (a) 0 (b) 3 (c) 4 (d) 5 231. If sec  = 13/5, then what is value 2sin  – 3cos  ? 4sin  – 9cos  (a) 1 (b) 2 (c) 3

of

(d) 4

192

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 1 233.If  = and sin θ= , then 2 2 the value of sin is (a) 1

(b)

1 2

1 2

(c)

(d)

241. If

90°), then the value of sin is 2 1 3 (b) (c) (d) 1 3 2 2 242. If sin x + cos x = c, then sin6 x + cos6 x is equal to.

(a)

3 2

(a)

234.If sec q + tan q = P, (P  0) the sec q is equal to:

(b)

1 1 (a)  P   , P 0 3 P 

(c)

1  1 (b)  0 P    ,P 2 P  

(d)

 1 P    ,P  0 (c) 2  P 

1  6c2  3c 4 4

   0    2  (a) 2 (b) 1 (c) 3 (d) 0 238.The value of following is cos2 4 ° + cos55° + cos 125° + cos 204° + cos300° (a) – 1/2 (b) 1/2 (c) 2 (d) 1 239. If sinA + cosecA = 3, then find the sin4 A  1 value of . sin2 A (a) 1 (b) 0 (c) 7 (d) 0 240. cos7° cos23° cos45° cosec83° cosec67° = ?

(a) 0

(b) 1

(c)

1 2

(d)

1 2

sin²A – cos²A

. sin³A  cos³A = ?

cos = n, then find the value of m² – n². (a)

mn

(c) 3 mn

(b) 2 mn (d) 4 mn

245. If

tan + sin = m and tan – sin = n, then find the

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39 35 91 65 (b) (c) (d) 72 72 144 144 237.If 1 + cos²q = 3 sinq cosq, then the integral value of cot q is

1  6c2  3c 4 16

(a) sinA (b) cosA (c) tanA (d) cosecA 244. If cot +cos = m and cot –

value of

(a)

mn .

(a)

1 (m²–n²) 2

(c)

1 1 (m²+n²) (d) (m²–n²) 4 4

(b) 2(m²–n²)

246. If cot +tan = x and sec– cos = y, then (x²y)2/3 –(xy²)2/3 =? (a) 4 (b) 3 (c) 2 (d) 1 sin8– cos8 =? cos 2(1  cos ²2)

247.

(a) 1

(b) –

1 (c) – 1 (d) 2 2

(sec + tan) = (sec – tan ) (sec – tan) (sec – tan), then each of the side is equal to

(a)  1 (b) – 1 (c) +1

(d) 4

 , then 4cos2 is equal to 2

(a) –

23 8

(b) –

31 8

(c) –

31 32

(d) –

33 32

252. If cos( –A) = a, cos( – B) = b, then sin²(A–B)+ 2ab cos(A – B) is equal to (a) a² – b² (b) a² + b² (c) b² – a² (d) 2ab 253. sin 4 sin4

 + 8

s in 4

3 + 8

sin 4

5 + 8

7 =? 8

3 1 (b) 1 (c) (d) 0 2 2 254. If sin + cos = a and sec + cosec = b, then the value of b (a² – 1) is equal to (a) 2a (b) 3a (c) 0 (d) 2ab (a)

255. cos15° cos7 (a)

1 2

(b)

1 1 °. cos82 ° = ? 2 2

1 8

(c)

1 4

(d)

1 16

cos(–) 256. 3tantan=1, then cos() =?

(a)

248. If (sec  +tan  ) (sec +tan )

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249. If asec + btan +c = 0 and psec + qtan +r = 0, then (br – qc)² – (pc – ar)² = ? (a) (aq – bp)² (b) (ap – bq)² (c) (aq + bp)² (d) (aq – bp)³ 250. If P = acos³x + 3a cosx. sin²x and Q = asin³x + 3acos²x.sinx, then (P +Q)2/3 + (P – Q)2/3 = ? (a) 2a2/3 (b) a1/3 (c) 2a1/3 (d) a1/3 251. If 8cos² + 8sec² = 65 and 0° 1 2 

12 tan θ = 5

= – 1  cosx  1 so value of x is none of the above



211

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=1 + tan²63° – sec²27° + cosec² 63° – cosec²27° = 1 + cot²27° – sec²27° + sec²27°– cosec²27° =1 + cot²27° – cosec²27° =1–1=0



=

= =

cos 1 – sin

5

1 2 5  2 2 5

 ×1 = 1² –2×1 + 2 2 sin90° = 1 – 2 + 2 1 = 1 (satisfied) sin

139. (a)

sin 43 cos19  – 8 cos² 60° cos 47 sin 71

cos1  sin

(If A + B = 90°, then sinA = cosB)

1 – sin ²

SinA cos B = 1 or =1 CosB sin A =2–2=0

cos ²

144. (a) Given,  +  = 90° By given condition,



R Enak

1 1 + cos² 7 2 2 = 1 (  sin² q + cos² q = 1) 141. (c) Given that, 3sinx + 5cosx = 5 On squaring both sides, we get 2 9sin x + 25cos2 x + 30sinxcosx = 25 9(1 – cos2x) + 25(1 – sin2x) +  30sinxcosx = 25 9 + 25 – (9cos2x + 25sin2x – 30sinx  cosx) = 25 9 = (3cosx – 5sinx)2 

wwM wa. th Les B aryn cos 0 1 x= = 1– 0 1 – sin 0

 x=1

cos cos 0 1   1  sin 1  sin 0 1  0

= 1 n o w c h e ck o p ti o n b y putting x = 1 only option (b) satisfying. 137. (b)



2 2 (90° –  ) = 3 3

146. (b)Given, sin3  = cos(  – 2°)

 sin3  = sin[90° – (  – 2°)]  3  = 90° –  + 2° 92

 4  = 92°   = 4 147. (b)

=

sin

2

= –1 2





–1

–

2 × (2)

1 2

3

3

sin  – cos  2

sin2  – cos2 

 – cos2 sin4   cos 4   sin2  cos 2  sin2  – cos 2 

= sin4  + cos4  + 2sin2  .cos2  – sin2  cos2 

sin cos   coseccosec ×  sin  cos  1 1   2 2   2 2   1 1     2 2 

= 23°

sin6  – cos 6  sin2  – cos 2  2

5

sin  cos   2 cos  sin 

 2  = 90º   = 45°

3cosx – 5sinx = 3

= 45º

2    = 36° 3

 sin2  = 1 = sin90°

Put  = 45º

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 =

sin2  + cos2  = 2sin  cos 



142. (a) If  +  = 90º

AC = 2²  1²  5 sinA + cotC BC BC  AC AB

2  3

145. (b)Given,

=sin² 7

put q = 0°

 =

 = 60° –



0  1  1 90º –7 7 = sin² + sin²  2   2

cos x = 1 – sin

....(i)

{from Eq. (i)}

1 1 140. (a) Sin² 7 + sin² 82 2 2

1  sin cos

143. (d) We know that, in a cyclic quadrilateral sum of opposite angle is 180° A + C = 180° .......(i)  and B + D = 180° .......(ii) cosA + cosB + cosC + cosD  = cosA + cosB + cos(180° – A) + cos(180° – B) From Eqs. (i) and (ii), = cosA + cosB – cosA – cosB = 0

2  1  8        =1+1–   2    

cos1  sin

1 =1 2

2 ×



x 138.(b) sin = x² – 2x +2 2 put value of x from options x=1

cos1  sin 1 – sin1  sin

cos 1  1  sin x Alternate:-

=



r

1 + – cosec²27° sin ²63

136. (b) x 

1

1 – sec²27° cot ²63

geisnh eeYa ridna gv.i Sni

135. (d) 1 +

= (sin2  + cos2  )2 – sin2  cos2  = 1 – sin2  cos2 

2

148.

(c) tanA =

1 – cos B sinB

Put A = 30º, B = 60º

212

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1 – cos 60º tan 30º = sin60º 1 1– 2 1 = 3 3 2



2tan A 2tan30º = 2 1 – tan A 1 – tan2 30º

= 1 + sec2  >1 + 1 > 2

cos  > 0 

( sec2  >1 for 0 <  0  C2 < 40 24

From Eqs. (i) and (ii), c 

B

40°

R

o

C

155. (d)

AB r cos40° =  cos40° = OB R

ERna

In  OAB,

So, the radius of the circle of latitude 40° S is R cos 40°. 150. (a) We know that, If value of cos  increases, then the value of  decreases.

2

3

q

A

 r = R cos40°

1

B

1  cos  > 2



sin  = +

  cos  > cos60º   < 3

For, 0 <  <

wwM wa. th Les aBryn   

Given, 7 cos2  + 3sin2  = 4 7(1 – sin2  ) + 3(sin2  ) = 4 7 – 4 sin2  = 4 4sin2  = 3

151. (b) Given, sin  + cos  = 1 On squaring both sides, we get

 sin  .cos  = 0

=

1 – sin  2

1  sin  cos2 

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60°

5cm

90°

AB 1 5 =  AC 2 AC  AC = 10cm

159. (a) Given, cos  + 3 sin  = 2



1 3 cos  + sin  = 1 2 2



sin30°cos  + cos30°sin  = 1

 

sin(30°+  ) = sin90°



 = 60° =

30°+  = 90°

= [cos2  .sec2  + tan2  ](cos2  + 1) ( cos²  .sec²  =1) = (1 + tan2  ) (cos2 + 1)

 3

160. (d) In  ABC, A

v

w

B

u

tanA =

3 = 3 1

( sin2  + cos2  = 1)

C

cos60° =

3 ,  = 60º 2

(cos²  + 1)

30°

B

3 2

156. (b) [(1 – sin2  ) sec2  + tan2  ]

2

=

A

C

 tan  =

1  sin 1  sin  1  sin  = 1 – sin 1  sin  1 – sin 

1  sin 2

158. (a) In  ABC,

 2

sin  =

(sin2  + cos2  ) + 2sin  cos  = 1  1 + 2 sin  cos  = 1

152. (b)

  0, 2  , sec2  is increasing from 1 to   p > 1

4,2 10 

154. (d) We know that, the value of cos  is decreasing from 0º to 90°. cos1° > cos89°  p>q  Also, cos 1° is close to 1 and cos 89° is close to 0. Hence, option (d) is correct.

149. (a) r

a2  b2 – C 2 153. (b) cos  = 2ab (By cosine rule)

(since, C cannot be negative) .......(i) Also, b + c > ac > 6 – 2  c > 4

3

3 A

= sec2  .cos2  + sec2 

 0 < c < 2 10

Now check the option, Option (c):- tanB tan60º =

= sec  + tan 

62  22 – C 2 40 – c 2 = 262 24 For acute angle,



2 3 2 = 3

= sec2  (cos2  +1)

=

1 1 = (Satisfy) 3 3

1 2 3 = =  1 2  1–     3

( sec2  – tan2  = 1)

1  sin  1 sin  = + cos  cos  cos 

kgei snhe eYari dnag v.iSn ir



=

BC u = AC v

andtanB =

v u

Also, u2 + v2 = w2 .....(i) (by Pythagoras theorem)



tanA + tanB = =

w2 uv

u v u2  v 2 + = v u uv [from Eq. (i)]

213

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161. (b) Given, C

3

1  ×   × 1 2– 2

=

A

B

3k

166.

k 3k

tan B =





3k

2



BC2 = 4k2  BC = 2k 162. (a) Given,



B 2

2x = AB + (x – 1)



AB2 = x + 1  AB = x  1



tan  =

BC x –1 = = AB x 1

x –1 x 1

163. (c) Given,

cos  cos  2 – 1 – sin  1  sin 



cos   sin  cos  – cos   cos  sin  2 1 – sin2 

( 1–sin²  =cos²  )

   

p –q q = p p  q q

=

p2 – q 2 p2  q 2

2sin  cos  = 2cos2  sin  = cos  tan  = 1  = 4

164. (b) As we know that, sinx is increasing from 0º to 90°. siny > sinx.  165. (b) sin360°. cot 30° – 2sec2 45° + 3cos60°. tan²45°– tan²60°

1 3 (  (a² – b²)(a²+b²) = a4 – b4)

cos4q – sin4q =

172. (c) tanq =

1 11

cos ec ² – sec ² cos ec ² sec ²

1 1 – sin ² cos ²  = 1 1  sin ² cos ²

cos ² – sin ² sin ².cos ² cos ² sin ² sin ².cos ²

cos ² – sin ²   cos ² sin ²

 sin ² cos ²  1 –  cos ²   sin ² cos ² 1     cos ²

1 1– 1 – tan ² 11  5   1  tan ² 1 6 1 11

173. (b) sinq=

3 5

(b) cosec 2  – 2 + s in 2 

= (sin  –cosec  )2 Hence, it is always non-negative. 168. (b) 1 – 2sin²q + sin4q  (1 – sin²q)²

wwM wa. th Les B aryn



p sec θ – q cosecθ p sec θ  q cosecθ

1 1 3

R Enak

A

9 3 –35 – 4 + –3 = 8 2 8

p

167.



 3

p tan– q = p tan q

x –1 sin   2x In ABC, using Pythagoras theorem, AC2 = AB2 +BC2 C

x–1

2

 p sec θ    cosecθ  – q  cosecθ  =  p sec θ  cosecθ  q    cosec 

+ (1k)2 = BC2

2x

+ 3

p (c) tan θ = q

=

In  ABC, AB2 + AC2 = BC2 ......(i) (by Pythagoras theorem)

 2

=

geisnh eeYa ridna gv.i Sni

1k

2

r

 3 =  2  × 3– 2  

 ( cos²q)²  cos4q 169. (c) Sinq = 0.7  sin²q + cos²q = 1  (0.7)² + cos²q = 1  0.49 + cos²q = 1  cos²q = 1 – 0.49  cosq = 0.51 170. (c) S i n ² 6 5° + si n ² 25° + cos²35° + cos²55° = sin²65°+ sin²(90° – 65°) + [cos²35° +cos²(90°– 35°)] = ( sin² 65° + co s²65°) + (cos²35° + sin²35°) = 1+1=2 1 (given) 3 cos²q + sin²q = 1 (property) (cos²q – sin²q) ( cos²q + sin²q)

171. (a)cos²q – sin²q =

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3 P = 5 H So, B = 4 P=3 H=5

sinq=

P B  tan cos B H   cot cos ec B H  P P

31 3 4 15  16   20 4 5  20  4 5 45 9  3 3 3 3 31 60 174. (b) (sina + coseca )2 + (cosa + seca )² = k +tan²a + cot²a put a = 45°,

=

214

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

(Sin45° +cosec45°)² + (cos45° + sec45°)² = K+ tan²45° + cot²45° 2



2



177. (d) 7 sina = 24 cosa

 1   1     2  2      2   2 

sin 24 = cos 7

=k+1+1



In ABD ,

3 sina = = 0.6 5

 tana =

  1 1  1 1  2 2   2 2   2  2  2  2 2 2    

BD AD  sin BAD  3 2

24  P 7B

 AD =

4

1 1 4 =k +2  k= 7 2 2

from ADC

CD AD   AD sin DAC sin45



175. (a)

14 tana – 75cosa – 7seca

7 24 25 – 75× –7× 25 7 7 = 48 – 21 – 25 = 2

= 14 ×

y² – x²

x In ABC , sin21°= y

178. (c)

AB = x AC = y

=

BC  y ² – x ²

= (tan²a + 1)sin²b =(tan²45°+ 1)sin²45°

AC BC – BC AC

2

2

2

y² –

=



y² – x²

y y² – x ²

179. (c)

y² – y ²  x ²



y y² – x²

x²

=

2

y y² – x²

1 sec– 1 = cos – 1 sec 1 1 1 cos

CD ....(ii) 2 sinDAC .

3 BD 1 CD .  . 2 sin BAD 2 sin DAC

=

 1   =2× 1 =1 = (1 + 1)   2  2

wwM wa. th Les aBryn

AC  – BC   BC AC 

ERna

 sec21° – sin69°

1

from equation (i) and (ii)





3 BD . ........(i) 2 sinBAD

kgei snhe eYari dnag v.iSn ir

=k+2



BD AD  (By sin rule) sin BAD sin60

sin BAD 3 1   2 sin DAC 2 3 1

3 2

=

...........(ii)

On adding equation (i) and (ii) 2seca = 2 

1 2

5 H seca = 4 B

1 – cos = 1  cos

2

1 – cos

182. (a) sec²q –

= cosecq – cotq 180. (c) A



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= sec²q –

sin ²

1 – cos 1 cos  – = sin sin sin

1

6

181. (a) sin3A = cos(A – 26°)  sin3A = sin[90º – (A–26º)]  3A = 90º – A + 26º  4A = 116°  A = 29°

176. (c) seca + tana = 2 ............(i) 1 seca – tana = 2



sin ²– 2sin4 2cos 4– cos ²

sin ²1 – 2sin ² cos ²2cos ² – 1

cos²– sin² 2cos²–1  1– 2sin² = sec²q – tan²q = 1 Alternate:sec² θ –

sin2– 2sin4 2cos 4– cos 2

Put θ = 0º sec² 0º –

60° 45° D 1 3 C   B  = 60°, C  = 45° 3 4 B

=1–

sin2  – 2sin4  2cos4  – cos2 

0–0 2 –1

=1–0=1

215

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183. (c) x =a(sinq + cosq) y = b(sinq – cosq) x = (sinq + cosq) a

x2  a2 = ( sinq + cosq)²

....(i) =

1 2 1 1 – 2 1 

=

x² y²  a ² b² = (sinq + cosq)2 + (sinq – cosq)²

=

x² y²  a ² b² = sin²q + cos²q +2sinq cosq + sin²q + cos²q – 2sinq. cosq

......(i) 3 r sinq = 1 ......(ii) squaring and adding equation (i) and (ii) r²cos²q + r² sin²q = 3 + 1 r²(cos²q + sin²q )= 4 r² = 4 r=2

2 2 3



4

3

1 = cos60° 2

R Enak

cos(2a + b) =

4a – b = 90° 2a + b = 60° adding 6a = 150° a = 25°  b = 10°  sin(a +2b)  sin(25°+ 2× 10°)

 sin45° =

 1  sin   1 – sin 1 – sin ²

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r 3 secq = 3 r r r  r tan sec 3  3 1 r sec tan r ²  3 3  2   r    3   2r  2  2  4 2 = r 1 r ²  1 2²  1 5 3 Alternate:r=2 cosq =

r sin 1  tan = r cos 3

 = 30° 2 tan 30 sec 30 2sec 30 tan 30

2 1 2  4 3 3 = 2 1 = 5 2  3 3 2

190. (c) cosecq =



2 3 – 1 1



2– 3

sin = b  sina = b sinb sin

3

1 P sinq = 3 H  cotq – cosecq

2

3

=

1

188. (c)

1  sin 1 – sin  1 – sin 1  sin

r sin 1  and rcosq r cos 3

tanq =

=

187. (d) sin(4a – b) = 1= sin90°

1  = cos60° or cos 2 3

2

=

2 sec30° =

 q= 3

186. (d)



4 3 1  = 3 1 3

wwM wa. th Les B aryn



1 2 1 1  2

1 –

Now check with option by putting q = 30°,



2 3 3   cos sin ² 1 – cos ²  2 – 2 cos²q = 3 cosq  2cos²q + 3cosq – 2 = 0 2cos² q + 4cosq – cosq – 2 = 0   2cosq (cosq+ 2) – 1( cosq + 2) = 0 (2cosq – 1) ( cosq + 2) = 0  2cosq – 1= 0 or cosq + 2  0 

3 )² = 0, ( rsinq – 1)² =

0r cosq =

1  sin 30 1 – sin 30 + 1 – sin 30 1  sin 30



x² y² + = 2(sin²+ cos²)  a ² b² =2×1=2 184. (d) sin5q = cos20°  5q + 20° = 90° ( If sinA = cosB then A + B = 90°)  5q = 70°  q = 14° 185. (c)2 secq = 3 cosec²q

+

r

y2 = (sinq – cosq)² ....(ii) b2 On adding equation (i) and (ii)



cosq =

(r cosq –

2 = 2sec cos Alternate:put q = 30°,

=

y = (sinq – cosq)  b



189. (a) ( r co s q – 3 )² ( rsinq – 1)² = 0

1  sin 1 – sin cos

geisnh eeYa ridna gv.i Sni



=

cos cos ² =a  = a² cos cos²



1 – sin ²  a² 1 – sin ²

 1 –sin²a = a² (1 – sin²b)  1 – b² sin²b = a² – a² sin²b [ value put in sina]

216

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193. (c) tan(A + B) = tan(A – B) =

1 – a² a² – 1   sin²b = b² – a ² a ² – b²

3

sin 3 cos = 3sinq

A=

3 =3 cos

= tan30°

90  2

then perpendicular =

194. (c) 6





(sin²q – cos²q) 2

2

 6  3       3  – 3      6 3 1 – = 9 9 3

192. (d)

AB = 20 cm BC : CA = ?





tan1  2cos ² – 2 2cos ²– 1

BC = cos30° CA

BC 3 =  CA 2

3 :2

sin1 – 2sin ²  cos 2 cos ²– 1   (1 – 2sin²q = 2 cos²q – 1 = cos2q = cos²q – sin²q)  tanq

2xy 195. (c) Given, cot  = x 2 – y 2 C

x 2 –y

2



B 2xy In  ABC, AC2 = (x2 – y2) + (2xy)2 2  AC = (x2 + y2)²  AC = x2 + y2 A

2xy AB  cos  = AC = x 2  y 2

1 + 2 = 2.5 2

 2.5 = 2.5

(Satisfy)

 θ = 30º 197. (d)Given, x cos60° + y cos0° = 3



x +y=3 2

 4x ×

sin – 2sin ³ 2 cos ³ – cos

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(  sin   2)

......(i)  x + 2y = 6  and 4x sin30° – ycot 45° = 2

2cos ²– 1

BC = cosC CA

( C  180 – 90 – 60  30 )





 tanq Alternate:-

wwM wa. th Les aBryn 



ERna

2



1 2

Put θ = 30º sin30º + cosec30º = 2.5

tan2cos ²– 1

2

 P  B     –   H  H 

 2sin2  – 5 sin  + 2 = 0  2sin2  – 4sin  – sin  + 2 = 0  2sin  (sin  – 2) –1 (sin  – 2) = 0  (2sin  – 1) (sin  –2) = 0

sin θ + cosec θ = 2.5

sin – 2sin ³ 2 cos ³ – cos

sin 1 – 2 1 – cos ²   cos 2 cos ² – 1

3

5 2

  = 30° Alternate:-.

sin1 – 2sin ²  cos 2 cos ²– 1  

6

1 

  sin   sin   =

 sin  =

A = 45° B = 15°

3 3

196. (a) Given, (sin  + cosec  ) = 2.5 

 (A + B) = 60° ......(i) (A – B) = 30° .....(ii) Adding both equation 2A = 90°

191. (c) 3 tanq = 3 sinq

cosq =

1

3 = tan60°

kgei snhe eYari dnag v.iSn ir

 1 –a² = b² sin²b– a² sin²b  1– a² = (b² – a²) sin²b

1 – y.1 = 2 2

......(ii)  2x – y = 2 On solving Eqs. (i) and (ii), we get x = y = 2 198. (a) log(tan1°) + log (tan2°) + .... ...+ log (tan89°) = log (tan 1° tan2° ...tan 45° ..tan 88° tan 89°) [  tan89° = tan(90° – 1) = cot1°] = log[tan1°cot1°) (tan2°cot2°) ....tan ....tan45°] = log (1°.1°...1°) = 0 199. (c) Now, (sinx – cosx)2 = (sin2x + cos2x) –2sinx cosx 1  = 1 – 2   2 1    sin x cos x  ,given  = 0 2

Alternate:sinx cosx =

1 2

Put x = 45º

217

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1 1 1 × = 2 2 2

Put x = y = 45º

1 1  = (Satisfy) 2 2  sinx – cosx sin45º – cos45º 1 1 – =0 2 2 200. (a) tan²y cosec²x –1 = tan²y Put x = y = 45º tan²45º cosec245º–1 = tan²45º 2

1×

 2 – 1 =

sin 45º cos 45º m  n and sin 45º cos 45º n = 1 m = 1  (m²–n²)sin²y = (1²–1²)sin²45º = 0 Now check from options Option (a):- 1 – n²  1 – (1)²  0 (Satisfy) 2 204. (b) Given, p = tan x + cot2x = (tanx + cotx)2 – 2

(1)²

2– 1 = 1 1 = 1  x=y

2

 sin2 x  cos ² x  = –2  sin x .cos x 

(Satisfy)

cos x cos x  2 1  cosecx cosecx –1

cos x cosec x 1  cot2 x

 pmin =

202. (c) Given, sinx : sin y = 3 : 1 3 1 : = sin60° : sin30° 2 2  x : y = 60 : 30 x : y = 2 : 1

=

cos x 203. (a) Given, cos y = n .....(i)

sin x sin y = m ...... (ii)

5sin75ºsin77º 2cos13º cos15º cos15ºsin77º

Now, (m2 – n2)sin2y

 sin2 x cos 2 x  2 =  sin2 y – cos 2 y  sin y

1 – cos x  cos 2

=

2

y – cos2 x 1 – cos 2 y 

cos2 y

cos2 y – cos2 x  1 – n 2 [from Eq. (i)] = cos2 y

2

 

–

= –

7sin81 cos 9

5 cos15 sin 77  2sin77 cos15 cos15 sin77

7cos9 cos 9

7cos15º.sin77º 7 cos 9º – cos15º.sin77º cos 9º =7–7=0 =

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2

b =

2

1 1  =1 2 2

Now check from option,

4 –2=2 1

 1 2  P =    3  3 P = 0.33 + 2.99 P = 3.32  P2 205. (b)

a =

 sinx siny + cosx cosy =

Option: (d)

p > 2 Hence, p > 2 Alternate:P = tan²x + cot²x Put x = 45º P = tan²45º + cot²45º P= 1+1 P= 2 Put x = 30º P = tan²30º + cot²30º

wwM wa. th Les B aryn

  tanx = 1  x = 4

Alternate:sinx + siny = a cosx + cosy = b Put x = y = 45º sin45º + sin45º = a

a 2  b2 – 2 2

2

R Enak

2cos x cosecx 2 cosec 2x – 1

a2  b2 – 2 2

2

4  2  =  –2  –2 = sin2x  sin ²2x Since, the maximum value of sin 2x is 1.

x–y=0 201. (c) Given,



=

geisnh eeYa ridna gv.i Sni



cos x sin x  n and m cos y sin y

206. (d) Given, sinx + sin y = a and cosx + cos y = b on squaring both sides, we get sin2x + sin2y + 2sinx sin y = a2...... (i) cos2x + cos2y + 2cos x cosy = b2...(ii) On adding Eqs. (i) and (ii), we get (sin2x + cos2x) + (sin2y + cos2y) + 2(sinx siny + cosx cosy) = a2 + b2 (sinx siny + cosx cosy) 

r

Alternate:-

1 sin45º cos45º = 2



2

 2    2

–2

2



2 2 – 2 2



2 = 1(Satisfy) 2

207. (a) Given, cosec4  – cot4  = 17  (cosec2  – cot2  ) (cosec2  + cot2  ) = 17  1  cos 2   = 17  sin2  

1

( cosec²  – cot²  =1) 2 2 – sin   = 17sin2 

 18sin2  =2  sin2  =

1 9

1 3 (since,  lie in first quadrant)

 sin  =

1 208. (c) Given, x +   = 2cos  x On squaring both sides, we get

x2 +

1 + 2 = 4cos2  x2

218

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1 = 2 (2cos2  – 1) x2

212. (c) Given, sin(x – y) =

= 2(2cos2  – sin2  – cos2  ) = 2cos2  – 2sin2  1 sin ²θ

209. (b) cot² θ –

– 1 – cos ²θ – sin ²θ  1 = = sin ²θ sin ²θ

( 1– cos² θ = sin² θ ) Alternate:-

cot  –1 =



 1 2      2

210. (c) Since, sinx = cosy As x and y are acute angles, then

215. ( a) B y us ing theorem A

wwM wa. th Les aBryn

m2 – n 2 211. (c) Given, sin  = 2 m  n2

1 unit =

C

q A In  ABC,

B

= =

2

 AC 

2

– BC 

m4 +n4 +2m2n2 – m4 +n4 –2m2n2  2

P ythagoras

5 units B

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A

tan x tan x – 1  sec x 1 – sec x

=

–2tan x sec x – tan2 x

2 cos x = 2cosecx = sin x cos x

tan x tan x – 1  sec x 1 – sec x Put x = 45º

1 1 – 1 2 1– 2 1 – 2 –1 – 2

3 P = 4 B

2



1–



–2 2 –1

 2

2 2 Now check from option. Option: (b) 2 cosec x = 2 × cosec45º

= 2×

4m n  2mn

m2 – n 2 2mn

4

tan x 1 – sec x – 1 – sec x  1 – sec 2 x



2

5 q

4

  + cos4 4 4

tan 45º tan 45º – 1 – sec 45º 1  sec 45º

P  B = 9  16 = 25 = 5 Let the length of hypotenuse = x cm C 2

2

 tan  =

   x= 4 4

=

15 3

216. (d) Since, tan  = H=

AB =

218. (b) =

2

m – n2

m

 tanx = tan

4

15 5 units = ×5 = 25 cm 3  AB = 25 cm

2

n

217. (c) Given, sinx – cosx = 0  sinx = cosx  tanx = 1

Alternate:-

C 15 cm AB²= AC²+BC² (5)²= (4)²+BC² BC = 3 units 3 units = 15 cm

 x+y= 2

+

2  5 10 = cm 3 3

 1   1  1 1 1 =  + = + =  2   2  4 4 2

1 2 1 × –1 2 –1 2 1

4 units

 x=y= 4

2

1 –1 2 –1

 2x = 36°  x = 18°

 1 – 2 = –1

2 3 = x 5

 sin4x + cos4x = sin4

2 –1

2 1 –1 = 2 2 –1 214. (d) Given, sin (x + 54°) = cosx  sin(x + 54°) = sin(90° – x) ( 0° < x < 90°)  x + 54° = 90° – x

=

1

 1–



 tan  =

=

1 cot² 45º – 2 sin 45º

sin  =

213. (b) Given, 1 + tan  = 2

1 s in 2 θ

Put θ = 45º





and cos(x + y) = cos60°  x – y = 30° and x + y = 60°  x = 45° and y = 15°

cos ²θ 1 cos ²θ –1 – = sin²θ sin ²θ sin²θ

cot² θ –

1 2

 sin(x – y) = sin30°

ERna

=

cos (x + y) =

1 and 2

kgei snhe eYari dnag v.iSn ir

 x2 +

3 B

4

2

(Satisfy)

4

219. (b) (sin x – cos x + 1) cosec2x = {(sin2x – cos2x) (sin2x + cos2x) + 1} cosec2x [ a² – b² = (a+b)(a–b)]

219

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p 2 = q

1 2 sin2 x

Alternate:(sin4x – cos4x + 1) cosec²x Put x = 90º (s in 49 0º – c os 4 9 0º + cosec²90º  (1 – 0 + 1) ×1

p = 2q

1)

cos x cosecy – cos x sin y

=

1 – cos 2 x =

=

1 3q 2  =q 3

( p = sin10x)

1 – 4(1)2 –2× 3 = 0 2

2

 x2 =

 3

 x2 =

2

2





 sin31° >

 sin30° =

B

 sin32°>

2

2

p2 – q 2

p – q . tan  = q



Alternate:p cosec θ = q

θ = 30º

3 1 1 = = 5 5 sin 3

sin x 1 – cos x = 1  cos x sin x

sin2x = (1 – cosx) (1 + cosx) (1 – cos2x) = (1 – cos2x) which is possible for all values of x except multiples of 180°. Since for x = 180°, sinx = 0 and 1 + cosx = 0

3cos  – 4 sin  = 0 231. (c)

226. (c) We know that, sin  is increasing in 0° to 90°.

q

In  ABC, tan  =

1 1 + 2 2

 sin31° + sin32° > 1

q

5

3cos  – 4sin  = x (Let) Using identity, 3² + 4² = 5² + x² x=0 So,

1 1 and sin 32° > 2 2

sin31° + sin32°>

p

2

4

230. (a) 3sin  + 4cos  = 5

On adding both sides, we get

p 223. (b)Given, cosec  = q

p 2– q

1 2

Value of sin increases 0° to 90°  sin31° > sin30° and sin32° > sin30°

( sin2  + cos2  = 1) = 1 + 2 – 2 + 15 = 16

C

229. (c)

 x = 3 +1

wwM wa. th Les B aryn

 3

2 3

 sin  = 1 – cos 2  = 1 – = 9 3

 

2

225. (d) We know that, sin 30° =

  – sec + 5 tan2 3 3

 cos  =

 cosec  =

+ (1)2 + 2 3 .1

3 1

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   2.   sin2  + 4cos2 222. (c)  cos 8 8 4

3 2

 x2 = 3 + 1 + 2 3

0  p  1

A

 sec  =

 x2 = 4 + 2 3

221. (b) We know that, 0  sin2x  1  0  sin10x  1



(sec   – 2, sec  > 0)

 x2 – 4 – 2 3 = 0

sin2 x = sinx

 1  = 1+4  –2+5  2 

4q 2 – q 2 × tan30º

 2x2 ×

= cos x cosec(90 – x ) – cos x.sin 90 – x  cos x .sec x – cos2 x

=

224. (d) Given, 2x2cos 60° – 4cot245° – 2 tan60° = 0

( x + y = 90°, given)

=

p 2 – q 2 ×tan θ



 2

220. (b)

 2sec2  + 4sec  – 3sec  – 6 = 0  2sec  (sec  + 2) – 3(sec  + 2) = 0  (2sec  – 3) (sec  + 2) = 0

geisnh eeYa ridna gv.i Sni

= 2sin2x.

sin 4  + cos 4  = sin 490° + cos490° =1+0=1  228. (c) Given, 2sec2  + sec  – 6 = 0



p cosec30º = q

r

( 1 – cos2x = sin2x) = {sin x – cos2x + 1) cosec2x ( 1 – cos2x =sin2x) 2

Given, sec  =

 sec2  =

1 2

169 25

 1 + tan2  =

1 2

13 5

169 25

 tan2  =

169 –1 25



1 2 sin  + sin 

 tan2  =

144 12  tan  = 25 5

  

sin2  – 2 sin  + 1 = 0 (sin  – 1)2 = 0  sin  = 1 sin  = sin90°   = 90°

sin  2 –3 2sin  – 3cos   = cos sin  4sin  – 9cos  4 –9 cos 

227. (d)

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Given, sin  + cosec  = 2



220

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235. (d) Given  sin222° + sin268° + cot2 30°  cos268° + sin268° + cot230°

12  2  – 3 2tan  – 3 5 = = 4 tan  – 9  12  4  – 9 5 

238. (b)Th e val u e o f co s 2 4 º +cos55º +cos125º +cos204º + cos300º We know that, cos ( 180º   ) = – cos co s24º + co s55º + co s (180º–55º) + cos(180º + 24º) + cos(360º – 60º) cos24º + cos55º – cos55º – cos24º + cos60º



sin(90 – )  cos  cos(90 – )  sin  

[From Eq. (i)]

2

 1 + cot 30°

24 –15 9 = =3 48 – 45 3

 1+

7 ......(i) 2

232. (d) r sinq =

[ cot30° =  4

7 3 .....(ii) 2 On squaring and adding both equation

r cosq =

49 = 7  2

1 sinq = 2

Therefore, sin  

30° + = 90°

= 60°

sin = sin60° =

3 2

234. (b) Given  Sec  + tan  = p.....(i) Then, 1

sec  – tan  = p .....(ii) From equation (i) + (ii) 1  2 sec  = p+ p

1 1 sec  = 2  p  p    Rakesh Yadav Readers Publication Pvt. Ltd.

1 +2=9 sin²A

sin4 A  1 sin2 A

240.

=9–2=7

(d) cos cosec = 1 if  + = 90° cos7° cos23° cos45° cosec83° cosec67° = (cos7° cosec83°) (cos23° cosec67°) cos45° =1×1×

130 288

65 144 237. (b)Given, 1+cos 2 = 3sin  cos [O Q (d) P = Q A cube of edge 6 cm is painted on all sides and then cut into unit cubes. The number of unit cubes with no sides painted is (a) 0 (b) 64 (c) 186 (d) 108 The length of diagonal of a square

R Enak

37.

38.

of the semicircle is ( take 

22 ) 7

(a) 126 (b) 14 (c) 10 (d) 7 31. A copper wire is bent in the form of square with an area of 121 cm2. If the same wire is bent in the form of a circle, the radius (in cm) of the circle is 22 ) 7 (a) 7 (b) 14 (c) 8 (d) 12 32. Water flows into a tank which is 200 m long and 150 m wide through a pipe of cross- section

( Take

39.

is 15 2 cm. Its area is (a) 112.5 cm2 (b) 450 cm2

255 2 cm2 (d) 225 cm2 2 40. A kite in the shape of a square with a diagonal 32 cm attached to an equilateral triangle of the base 8 cm. Approximately how much paper has been used to make it? (Use 3 = 1.732) (a) 539.712 cm2 (b) 538.721 cm2 (c) 540.712 cm2 (d) 539.217 cm2 41. A lawn is in the form of a rectangle having its breadth and length in the ratio 3 : 4. The area 1 of the lawn is hectare. The 12 breadth of the lawn is

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(a) 25 metres (b) 50 metres (c) 75 metres (d) 100 metres 42. The area of a rectangle is thrice that of a square. The length of the rectangle is 20 cm and the

3 2 times that of the side of the square. The side of the square, (in cm) is (a) 10 (b) 20 (c) 30 (d) 60 43. The length and breadth of a rectangular field are in the ratio 7 : 4. A path 4 m wide running all around outside has an area of 416 m2. The breadth (in m) of the field is (a) 28 (b) 14 (c) 15 (d) 16 44. How many tiles, each 4 decimeter square. will be required to cover the floor of a room 8 m long and 6 m broad?

r

breadth of the rectangle is

geisnh eeYa ridna gv.i Sni

0.3m × 0.2m at 20 km/hour. Then the time ( in hours ) for the water level in the tank to reach 8 m is (a) 50 (b) 120 (c) 150 (d) 200 33. A street of width 10 metres surrounds from outside a rectangular garden whose measurement is 200 m × 180 m. The area of the path (in square metres ) is (a) 8000 (b) 7000 (c) 7500 (d) 8200 34. The area of the square inscribed in a circle of radius 8 cm is (a) 256 sq. cm (b) 250 sq. cm (c) 128 sq. cm (d) 125 sq. cm 35. Area of square with diagonal

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24. The length of a rectangular hall is 5m more than its breadth. The area of the hall is 750m2. The length of the hall is : (a) 15 m (b) 22.5 m (c) 25 m (d) 30 m 25. A cistern 6 m long and 4 m wide contains water up to a depth of 1 m 25 cm. The total area of the wet surface is (a) 55 m2 (b) 53.5 m2 2 (c) 50 m (d) 49 m2 26. If the length and breadth of a rectangle are in the ratio 3 : 2 and its perimeter is 20 cm, then the area of the rectangle (in cm2) is (a) 24 cm2 (b) 36 cm2 (c) 48 cm2 (d) 12 cm2 27. The perimeter of a rectangle and a square are 160 m each. The area of the rectangle is less than that of the square by 100 sq m. The length of the rectangle is (a) 30m (b) 60m (c) 40m (d) 50m 28. A path of uniform width runs round the inside of a rectangular field 38 m long and 32 m wide, If the path occupies 600m 2 , then the width of the path is (a) 30 m (b) 5 m (c) 18.75 m (d) 10 m 29. The perimeter of the floor of a room is 18 m. What is the area of the walls of the room, If the height of the room is 3 m ? (a) 21 m2 (b) 42 m2 2 (c) 54 m (d) 108 m2 30. A copper wire is bent in the shape of a square of area 81 cm2. If the same wire is bent in the form of a semicircle, the radius (in cm)

(c)

(a) 200 (b) 260

(c) 280

(d) 300

45. A godown is 15 m long and 12 m broad. The sum of the area of the floor and the ceiling is equal to the sum of areas of the four walls. The volume (in m3) of the godown is: (a) 900 (b) 1200 (c) 1800 (d) 720 46. Length of a side of a square inscribed in a circle is a 2 units. The circumference of the circle is (a) 2a units

(b) a units

(c) 4a units

(d)

2a units 

47. The perimeter and length of a rectangle are 40 m and 12 m respectively. Its breadth will be (a) 10m (b) 8m (c) 6m(d) 3m 48. If each edge of a square be doubled. then the increase percentage in its area is (a) 200% (b) 250% (c) 280% (d) 300% 49. An elephant of length 4 m is at one corner of a rectangular cage of size (16 m × 30 m) and faces towards the diagonally opposite corner. If the elephant starts moving towards the diagonally opposite conrner it takes 15 seconds to reach this corner. Find the speed of the elephant (a) 1 m/sec (b) 2 m/sec (c) 1.87 m/sec (d) 1.5 m/sec

302

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51. If the side of a square is

1 (x  1) 2

58. The radius of a wheel is 21 cm, How many revolutions will it make in travelling 924 metres?

59.

3–x units and its diagonal is

2 units, then the length of the side of the square would be (a)

4 units 3

60.

(b) 1 unit

1 units (d) 2 units 2 52. A rectangular carpet has an area of 120 m2 and a perimeter of 46 metre. The length of its diagonal is: (a) 17 metres (b) 21 metres (c) 13 metres (d) 23 metres 53. If the length of a diagonal of a (c)

61.

62.

55.

56.

57.

63.

wwM wa. th Les aBryn

54.

(a) 24 2 cm 2 (b) 24 cm2 (c) 36 cm2 (d) 72 cm2 The length of a room is 3m more than its breadth. If the area of a floor of the room is 70 m2, then the perimeter of the floor will be(a) 14 metres (b) 28 metres (c) 34 metres (d) 17 metres The length of a rectangle is twice the breadth. If area of the rectangle be 417.605 sq. m., then length is(a) 29.08 metres (b) 29.80 metres (c) 29.09 metres (d) 28.90 metres The area of a sector of a circle of radius 5 cm, formed by an arc of length 3.5 cm is : (a) 8.5 cm2 (b) 8.75 cm2 (c) 7.75 cm2 (d) 7.50 cm2 The radius of a circular wheel is 1.75 m. The number of revolutions it will make in travelling 11

ERna

square is 6 2 cm, then its area will be

 22    use  =  7   (a) 7 (b) 11 (c) 200 (d) 700 The area (in sq. cm) of the largest circle that can be drawn inside a square of side 28 cm is : (a) 1724 (b) 784 (c) 8624 (d) 616 The area of the ring between two concentric circ les, whose circumference are 88 cm and 132 cm, is (a) 78 cm2 (b) 770 cm2 2 (c) 715 cm (d) 660 cm2 The diameter of a toy wheel is 14 cm, What is the distance travelled by it in 15 revolutions? (a) 880 cm (b) 660 cm (c) 600 cm (d) 560 cm A can go round a circular path 8 times in 40 minutes. If the diamete r of the cir cle is increased to 10 times the original diameter, the time required by A to go round the new path once travelling at the same speed as before is : (a) 25 min (b) 20 min (c) 50 min (d) 100 min The base of a triangle is 15 cm and height is 12 cm. The height of another triangle of double the area having the base 20 cm is (a) 9cm (b) 18 cm (c) 8 cm (d) 12.5 cm If a wire is bent into the shape of a square, the area of the square is 81 sq. cm, When the wire is bent into a semicircular shape, the are a of the sem icir cle



use  = km is  

(a) 800 (c) 1000

22 

:   (b) 900 (d) 1200 7

64.

22 ) is : 7 (a) 154 cm2 (b) 77 cm2 2 (c) 44 cm (d) 22 cm2 65. If the area of a triangle with base 12 cm is equal to the area of square with side 12 cm. the altitude of the triangle will be (a) 12 cm (b) 24 cm (c) 18 cm (d) 36 cm 66. The sides of a triangle are 3cm, 4 cm and 5 cm. The area (in cm2) of the traingle formed by joining the mid points of this triangle is: 3 3 (a) 6 (b) 3 (c) (d) 2 4

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67. Three circles of radius 3.5 cm each are placed in such a way that each touches the other two. The area of the portion enclosed by the circles is (a) 1.975 cm2 (b) 1.967 cm2 (c) 19.68 cm2 (d) 21.22 cm2 68. The area of a circular garden is 2464 sq. m. how much distance will have to be covered if you like to cross the garden along its 22 ) 7 (a) 56m (b) 48m (c) 28m (d) 24m 69. Four equal circles each of radius ‘a’ units touch one another. The area enclosed between them

kgei snhe eYari dnag v.iSn ir

50. A circle is inscribed in a square of side 35 cm. The area of the remaining portion of the square which is not enclosed by the circle is (a) 962.5 cm² (b) 262.5 cm² (c) 762.5 cm² (d) 562.4 cm²

(taking 

diameter? ( use 

( 

22 ). In square units, is 7

(a) 3a2

(b)

6a 2 7 2 a

41a 2 (d) 7 7 70. The area of the greatest circle inscribed inside a square of side

(c)

21 cm is ( Take  (a) 351

22 7

)

1 1 cm2 (b) 350 cm2 2 2

1 1 cm2 (d) 347 cm2 2 2 71. The are a of an equilate ral

(c) 346

triangle is 400 3 sq. m. Its perimeter is : (a) 120 m (b) 150 m (c) 90 m (d) 135 m 72. From a point in the interior of an equilateral triangle, the perpendicular distance of the sid es ar e

3 cm

2 3

cm

and 5 3 cm. The per imeter (in cm) of the triangle is (a) 64 (b) 32 (c) 48 (d) 24 73. The perimeter of a triangle is 30 cm and its area is 30 cm². If the largest side measures 13 cm, What is the length of the smallest side of the triangle? (a) 3cm (b) 4cm (c) 5cm (d) 6cm 74. Diameter of a wheel is 3 m. The wheel revolves 28 times in a minute. To cover 5.280 km distance, the wheel will take

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(a)

(a) 4

4 m 13

(b) 6

4 m 11

4 8 m (d) 12 m 11 11 76. The radius of a circular wheel is 1. 75m. The num ber of revolutions that it will make in travelling 11 km, is (a) 1000 (b) 10,000 (c) 100 (d) 10 77. The circumference of a circle is 100 cm. The side of a square inscribed in the circle is

(c) 12



 3  

cm2



(b)  3 – 2  cm2   (c)

   2 3 –   2 

(d)

   3 3 –   2 

cm2 cm2

82. The area of the largest triangle that can be ins crib ed in a semicircle of radius r cm, is (a) 2r cm2 (b) r2 cm2 1 2 r cm2 2 83. The area of the greatest circle, which can be inscribed in a square whose perimeter is 120 cm, is : 22 2 15 cm2 (a) 7 2 22  7     cm2  (b) 7 2 

(c) 2 cm2

(d)

2

22 15   (c)    cm2 7 2

100 2 50 2 (a) cm (b) cm  

2

(d)

22  9      cm2 7 2 

R Enak

84. The area of the incircle of an equilateral triangle of side 42 cm is ( Take   (a) 231 cm2

22 ) 7 (a) 22m (b) 20m (c) 21m (d) 4m 79. Four equal sized maximum circular plates are cut off from a square paper sheet of area 784 sq. cm. The circumference of each plate is

( Take 

( Take 

22 ) 7

(a) 22 cm (b) 44 cm (c) 66 cm (d) 88 cm 80. The circum -radius of an equilateral triangle is 8 cm. The in- radius of the triangle is (a) 3.25 cm (b) 3.50 cm (c) 4 cm (d) 4.25 cm 81. Three coins of the same size (radius 1 cm) are placed on a table such that each of them touches the other two. The area enclosed by the coins is

22 ): 7 (b) 462 cm2

(c) 22 3 cm2 (d) 924 cm2 85. The number of revolutions a wheel of diameter 40 cm makes in travelling a distance of 176 m, 22 is ( Take   ): 7 (a) 140 (b) 150 (c) 160 (d) 166 86. The length of the perpendiculars drawn from any point in the interior of an equailateral triangle to the respective sides are p1,p2 and p3. The length of each side of the triangle is

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100 (c) cm (b) 50 2 cm  78. A path of uniform wid th surrounds a circular park, The difference of internal and external circumference of this circular path is 132 metres. Its width is:

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(a)

2

3

p  p 1

2

 p3

1  p1  p2  p3 3 1 (c)  p1  p2  p3 3

(b)

(d)

87. A circle is inscribed in a square, An equilateral triangle of side 4 3 cm is inscribed in that circle. The length of the diagonal of the square (in cm) is (a) 4 2 (b) 8 (c) 8 2 (d) 16 88. The hypotenuse of a right angle isosceles triangle is 5 cm. Its area will be (a) 5 sq. cm (b) 6.25 sq. cm (c) 6.50 sq. cm (d) 12.5 sq. cm 89. From a point within an equilateral triangle, perpendiculars drawn to the three sides are 6 cm, 7 cm and 8 cm respectively, the length of the side of the triangle is : (a) 7 cm (b) 10. 5 cm

geisnh eeYa ridna gv.i Sni

22 ( Take  ) 7

   – 2

r

22 ): 7 (a) 10 minutes (b) 20 minutes (c) 30 minutes (d) 40 minutes 75. Find the diameter of a wheel that makes 113 revolutions to go 2 km 26 decameters.

( Take 

4 3

p  p 1

2

 p3

14 3 cm 3 90. In an isosceles triangle, the measure of each of equal sides is 10 cm and the angle between them is 45°, then area of the triangle is (c) 14 3 cm

(b)

25 2 cm2 2 (c) 25 2 cm2 (d) 2 3 cm2 91. The area of circle whose radius is 6 cm is trisected by two concentric circles. The radius of the smallest circle is

(a) 25 cm2

(b)

(a) 2 3 cm (b) 2 6 cm (c) 2 cm (d) 3 cm 92. The area of an equilateral triangle inscribed in a circle is 4 3 cm2. The area of the circle is 16 22  cm2 (b)  cm2 (a) 3 3 28 32  cm2 (d)  cm2 (c) 3 3 93. If the difference between the circumference and diameter of a circle is 30 cm, then the radius of the circle must be (a) 6cm (b) 7cm (c) 5cm (d) 8cm 94. The base and altitude of a right angled triangle are 12 cm and 5 cm res pectively. The perpendicular distance of its hypotenuse from the opposite vertex is 4 8 (a) 4 cm (b) 4 cm 13 13 (c) 5 (d) 7 cm

304

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(a) 48 cm2

(c) 16 3 cm2

(c) 192 3 cm2 (d) 192 cm2 96. The area of the shaded region in the figure given below is

102. A right triangle with sides 3 cm,4 cm and 5 cm is rotated about the side 3 cm to from a cone. The volume of the cone so formed is (a) 16 cm3 (b) 12  cm3 (c) 15 cm3 (d) 20  cm3 103. ABC is an equilateral triangle of side 2 cm. With A, B, C as centre and radius 1 cm three arcs are drawn. The area of the region within the triangle bounded by the three arcs is   (a) 3 3 –  cm2 2 

a c

 3   (b)  3 –  cm2 2  

(a)

 a 2   – 1  sq. units  2 2 

  (c)  3 –  cm2 2 

2 (b) a  – 1 sq. units

a2 (d) 2  – 1 sq. units b

104. The circumference of a circle is 11 cm and the angle of a sector of the circle is 60°. The area of the sector is ( use  (a) 1

29 cm2 48

isosceles triangle is 4 2 4 cm, the length of the hypotenuse is; (a) 4 cm (b) 6 cm (c) 8 cm (d) 10 cm

(b) 2

22 ) 7

29 cm2 48

27 27 cm2 (d) 2 cm2 48 48 105. If the difference between areas of the circumcircle and the incircle of an equilateral triangle is 44 cm2, then the area of the triangle

(c) 1

wwM wa. th Les aBryn

97. The area of a circle is increased by 22 c m, if its radius is increased by 1 cm. The original radius of the circle is (a) 6 cm (b) 3.2 cm (c) 3 cm (d) 3.5 cm 98. The area of the largest circle, that can be drawn inside a rectangle with sides 148 cm. by 14 cm is (a) 49 cm2 (b) 154 cm2 2 (c) 378 cm (d) 1078 cm2 99. A circ le is inscr ibed in an equilateral triangle of side 8 cm. The area of the portion between the triangle and the circle is (a) 11 cm2 (b) 10.95 cm2 2 (c) 10 cm (d) 10.50 cm2 100. In a triangular field having sides 30m, 72m and 78m, the length of the altitud e to the side measuring 72m is : (a) 25 m (b) 28 m (c) 30 m (d) 35 m 101. If the perimeter of a right-angled

22 is ( Take  ) 7

(a) 28 cm2

(b) 7 3 cm2

(c) 14 3 cm2 (d) 21 cm2 106. If the area of a circle inscribed in a square is 9 cm2, then the area of the square is (a) 24 cm2 (b) 30 cm2 2 (c) 36 cm (d) 81 cm2 107. The sides of a triangle are 6 cm, 8 cm and 10 cm. The area of the greatest square that can be inscribed in it, is (a) 18 cm2 (b) 15 cm2

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(c)

of the triangle is ( use (a) 50

1 cm2 7

(b) 50

22 ) 7

2 cm2 7

1

2 cm2 (d) 75 cm2 7 7 109. A wire, when bent in the form of a square, encloses a region having area 121 cm2 . If the same wire is bent into the form of a circle, then the area of the circle

(c) 75

is ( Take 

  (d)  – 3  cm2 2 

ERna

  – 1  sq. units 2 

2 (c) a  

108. The length of a sid e of an equilateral triangle is 8 cm. the area of the region lying between the circumcircle and the incircle

kgei snhe eYari dnag v.iSn ir

95. From a point in the interior of an equilateral triangle , the length of the perpendiculars to the three sides are 6 cm, 8 cm and 10 cm respectively. The area of the triangle is

2304 576 cm2 (d) cm2 49 49

22 7

)

(a) 144 cm2 (b) 180 cm2 2 (c) 154 cm (d) 176 cm2 110. If the perimeter of a semicircular field is 36 m. Find its radius ( use 

22 7

)

(a) 7 m (b) 8 m (c) 14 m (d) 16 m 111. The perimeter (in metres ) of a semicircle is numerically equal to its area ( in m²). The length of its diameter is (Take 

22 7

)

(a) 3

3 6 metres (b) 5 metres 11 11

(c) 6

6 2 metres (d) 6 metres 11 11

112. One acute angle of a right angled triangle is double the other. If the length of its hypotenuse is 10 cm, then its area is (a)

25 3 cm2 (b) 25 cm2 2

75 cm2 2 113.If a triangle with base 8 cm has the same area as a circle with rad ius 8 cm , then the corresponding altitude (in cm) of the triangle is (a) 12 (b) 20 (c) 16 (d) 32 (c) 25 3 cm2

(d)

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(b) 2 2 cm

(c) 2 3 cm (d) 4 cm 117. An equilateral triangle of side 6 cm has its corners cut off to form a regular hexagon. Area (in cm2) of this regular hexagon will be (a) 3 3

(b) 3 6

5 3 2 118. A 7 m wide road runs outside around a circular park, whose circumference is 176 m. the area

(c) 6 3

(d)

22 ) 7 (a) 1386 m2 (b) 1472 m2 2 (c) 1512 m (d) 1760 m2 119. The length (in cm) of a chord of a circle of radius 13 cm at a distance of 12 cm from its centre is (a) 5 (b) 8 (c) 10 (d) 12 120. The four equal circles of radius 4 cm drawn on the four corners of a s quar e touch each other externally. Then the area of the portion between the square and the four sectors is

22 ) 7 (a) 364.5 (b) 693.5 (c) 346.5 (d) 639.5 125. At each corner of a triangular field of sides 26 m, 28 m and 30 m, a cow is tethered by a rope of length 7m, the area (in m) ungrazed by the cows is (a) 336 (b) 259 (c) 154 (d) 77 126. In an equilateral triangle ABC, P&Q are mid point of sides AB & AC respectively such that PQ  BC. If PQ = 5 cm then find the length of BC. (a) 5 cm (b) 10 cm (c) 15 cm (d) 12 cm 127. ABC is an equilateral triangle, P and Q are two points on AB and

wire is ( take 

AC

of APQ is : (a)

(b) 9 4 – sq. cm (c) 5 6 – sq. cm (d) 6 5 – sq. cm

25 sq. cm 4

(b)

25 sq. cm 3

25 3 sq.cm (d) 25 3 sq. cm 4 128. The are a of a c ircle with circumference 22cm is (a) 38.5 cm² (b) 39 cm² (c) 36.5 cm² (d) 40 cm² 129. In ABC , O is the centroid and AD, BE, CF are three medians and the area of AOE = 15 cm2 then area of quadrilateral BDOF is (a) 20 cm2 (b) 30 cm2 2 (c) 40 cm (d) 25 cm2

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(c)

22 ) 7 (a) 70 (b) 35 (c) 17.5 (d) 140 135. Three circles of diameter 10 cm each are bound together by a rubber band as shown in the figure.

(in cm) is ( take 

r espe ctiv ely suc h that

PQ  BC . If PQ =5 cm, then area

(b)16 4 – sq. cm

(d) 169  – 4  sq. cm 121. If the four equal circles of radius 3 cm touch each other externally, then the area of the region bounded by the four circles is (a) 4 ( 9 – ) sq. cm

3 sq. cm

(c) 6 sq. cm (d) 3 sq. cm 124. A copper wire is bent in the form of an equilateral triangle and has area 121 3 cm2. If the same wire is bent into the form of a circle. the area (in cm2) enclosed by the

(a) 9( – 4 )sq. cm

(c) 99  – 4 sq. cm

(b)

r

 sq. cm 2

wwM wa. th Les B aryn

of the road is : (use

(a)

130. A straight line parallel to the base BC of the triangle ABC intersects AB and AC at the points D and E respectively. If the area of the  ABE be 36 sq. cm. then the area of the ACD is (a) 18 sq.cm (b) 36 sq. cm (c) 120 sq. cm (d) 54 sq. cm 131. The length of two sides of an isosceles triangle are 15 and 22 res pectively. What are the possible values of perimeter ? (a) 52 or 59 (b) 52 or 60 (c) 15 or 37 (d) 37 or 29 132. The diameter of a wheel is 98 cm. The number of revolutions in which it will have to cover a distance of 1540 m is (a) 500 (b) 600 (c) 700 (d) 800 133. The wheel of a motor car makes 1000 revolutions in moving 440 m. The diameter (in metre) of the wheel is (a) 0.44 (b) 0.14 (c) 0.24 (d) 0.34 134. A bicycle wheel makes 5000 revolutions in moving 11 km . Then the radius of the wheel

geisnh eeYa ridna gv.i Sni

(a) 3 cm

122. The length of each side of an equilateral triangle is 14 3 cm. The area of the incircle (in cm2) is (a) 450 (b) 308 (c) 154 (d) 77 123. Are a of the inc ircle of an equilateral triangle with side 6 cm is

R Enak

114. The measures (in cm) of sides of a right angled triangle are given by consecutive integers. its area (in cm2) is (a) 9 (b) 8 (c) 5 (d) 6 115. The are a of a r ight-ang led isosceles triangle having hypotenuse 16 2 cm is (a) 144 cm2 (b) 128 cm2 2 (c) 112 cm (d) 110 cm2 116. The area of an equilateral triangle is 4 3 cm2. The length of each side of the triangle is:

the length of the rubber band (in cm) if it is stretched is (a) 30 (b) 30 +10 (c) 10 (d) 60+20 136. If chord of length 16 cm is at a distance of 15 cm from the centre of the circle, then the length of the chord of the same circle which is at a distance of 8 cm from the centre is equal to (a) 10 cm (b) 20 cm (c) 30 cm (d) 40 cm 137. A semicircular shaped window has diameter of 63 cm, its 22 perimeter equals ( ) 7 (a) 126 cm (b) 162 cm (c) 198 cm (d) 251 cm

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(b) 7 3

10 7 7 10 (d) 3 3 139. The perimeter of a triangle is 40cm and its area is 60 cm2. If the largest side measures 17 cm, then the length ( in cm ) of the smallest side of the triangle is (a) 4 (b) 6 (c) 8 (d) 15 140. From four corners of a square sheet of side 4 cm four pieces each in the shape of arc of a circle with radius 2 cm are cut out. The area of the remaining portion is : (a) (8 –  ) sq. cm (b) (16 – 4 ) sq. cm (c) (16 – 8 ) sq. cm (d) (4 – 2 ) sq. cm 141. If the numerical value of the perim eter of an equilateral

(c)

triangle is 3 times the area of it, then the length of each side of the triangle is (a) 2 units (b) 3 units (c) 4 units (d) 6 units 142. Each side of an equilateral triangle is 6 cm. Find its area

(a) 4 units

(b) 2 3 units

(c) 17 units (d) 3 2 units 148. What is the area of a triangle having perimeter 32 cm, one side 11 cm and difference of other two sides 5 cm? (a) 8 30 cm2

(b) 5 35 cm2

(c) 6 30 cm2 (d) 8 2 cm2 149. Area of equilateral triangle having side 2cm is (a) 4 cm²

(b)

(c) 4 3 sq. cm (d) 8 3 sq. cm 143. The length of three medians of a triangle are 9 cm, 12 cm and 15 cm. The area ( in sq. cm) of the triangle is (a) 24 (b) 72 (c) 48 (d) 144 144. The area of the triangle formed by the straight line 3x +2y = 6 and the co-ordinate axes is (a) 3 square units (b) 6 square units (c) 4 square units (d) 8 square units 145.If the length of each side of an equilateral triangle is increased by 2 unit, the area is found to be increased by 3 + 3 square unit. The length of each side of the triangle is (a) 3 units (b) 3 units (c) 3 3 units

3 cm²

(c) 3 cm² (d) 6 cm² 150. The area of a circle is increased by 22 cm2 when its radius is increased by 1 cm. The original radius of the circle is (a) 3 cm (b) 5 cm (c) 7cm (d) 9 cm 151. The radii of two circles are 5 cm and 12 cm. The area of a third circle is equal to the sum of the area of the two circles. The radius of the third circle is : (a) 13 cm (b) 21 cm (c) 30 cm (d) 17 cm 152. The perimeter of a semicircular path is 36 m. Find the area of this semicircular path. (a) 42 sq. m (b) 54 sq. m (a) 63 sq. m (d) 77 sq. m 153. The area of a circle inscribed in a square of area 2m2 is   (a) m2 (b) m2 2 4 (c)  m2 (d) 2 m2 154. Three circles of radii 4 cm, 6 cm and 8 cm touch each other pair wise externally. The area of the triangle formed by the linesegments joining- the centres of the three circles is

wwM wa. th Les aBryn

(a) 9 3 sq. cm (b) 6 3 sq. cm

(c) 30 2 cm2 (d) 60 2 cm2 147. The area of an isosceles triangle is 4 square units, If the length of the unequal side is 2 unit, the length of each equal side is

(d) 3 2 units

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155. Two circles with centre A and B and radius 2 units touch each other externally at ‘C’, A third circle with centre ‘C’ and radius ‘2’ units meets other two at D and E. The n the area of the quadrilateral ABDE is (a) 2 2 sq. units (b) 3 3 sq. units (c) 3 2 sq. units

kgei snhe eYari dnag v.iSn ir

(a) 3 7

146. What is the area of the triangle whose sides are 9cm,10cm and 11 cm ? (a) 30 cm2 (b) 60 cm2

ERna

138. In an equilateral triangle ABC of side 10 cm, the side BC is trisected at D & E. Then the length (in cm) of AD is

(a) 144 13 sq. cm (b) 12 105 sq. cm (c) 4 3 sq. cm (d) 24 6 sq. cm

(d) 2 3 sq. units 156. If the perimeter of a right angled triangle is 56 cm and area of the triangle is 84 sq. cm, then the length of the hypotenuse is (in cm) (a) 25 (b) 50 (c) 7 (d) 24 157. If the length of each median of an equilateral triangle is 6 3 cm, the perimeter of the triangle is (a) 24 cm (b) 32 cm (c) 36 cm (d) 42 cm 158. The area of an equilateral triangle is 4 3 sq. cm. Its perimeter is (a) 12 cm (b) 6 cm

(c) 8 cm (d) 3 3 cm 159. A gear 12 cm in diameter is turning a gear 18 cm in diameter. When the smaller gear has 42 revolutions. how many has the larger one made? (a) 28 (b) 20 (c) 15 (d) 24 160. The perimeter of a semicircle is 18 cm, then the radius is: (using  (a) 5

1

22 ) 7 (b) 3 1 cm

cm

3

2

(c) 6 cm (d) 4 cm 161. A circle and a rectangle have the same perimeter. The sides of the rectangle are 18 cm and 26 cm. The area of the circle is 22 ) 7 (a) 125 cm2 (b) 230 cm2 2 (c) 550 cm (d) 616 cm2 162. The area of a circle is 38.5 sq. cm. Its circumference (in cm) is

(Take 

  use  = 

22  7

(a) 22 (b) 24

   (c) 26 (d) 32

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(a) 4 3 cm

(a)

(b) 8 3 cm

(c) 6 3 cm (d) 11 3 cm 164. The area (in sq. unit) of the triangle formed in the first quadrant by the line 3x +4y =12 is (a) 8 (b) 12 (c) 6 (d) 4 165. The height of an equilateral triangle is 15 cm. the area of the triangle is (a) 50 3 sq. cm (b) 70 3 sq. cm (c) 75 3 sq. cm

(b) a  b  c  ab  bc  ca (c) ab + bc + ca (d) None of the above 172. The radii of two circles are 10 cm and 24 cm. The radius of a circle whose area is the sum of the area of these two circles is (a) 36 cm (b) 17 cm (c) 34 cm (d) 26 cm 173. A circ le is inscr ibed in an equilateral triangle and a square is inscribed in that circle. The ratio of the areas of the triangle and the square is

(d) 150 3 sq. cm

(a)

166. The area of an equilateral triangle is 9 3 m 2 . The (in m) of the median is

(c) 3 3 : 2

(d) 3 3 :1

2

b is: a (a) 3

(b)

1 3

(c)

5 times the 6 base . What is the area (in cm2) of the triangle ? (a) 38172 (b) 18372 (c) 31872 (d) 13872 170. The altitude drawn to the base of an isosceles triangle is 8 cm and its perimeter is 64 cm. The area (in cm2) of the triangle is (a) 240 (b) 180 (c) 360 (d) 120 the equal sides is

3

(d)

1

3

175.If ABC is similar to DEF such that BC = 3 cm, EF = 4 cm and area of ABC = 54 cm 2, then the area of DEF is : (a) 66 cm2 (b) 78 cm2 2 (c) 96 cm (d) 54 cm2 176. The area of two similar triangles ABC and DEF are 20cm2 and 45 cm2 respectively. If AB =5 cm, then DE is equal to (a) 6.5 cm (b) 7.5 cm (c) 8.5 cm (d) 5.5 cm 177.C 1 and C 2 are two concentric circles with centre at O, Their radii are 12 cm and 3 cm, respectively. B and C are the point of contact of two tangents drawn to C2 from a point A lying on the circle C1. Then, the area of the quadrilateral ABOC is

wwM wa. th Les B aryn

1 1 cm (b) 6 cm 5 3 2 (c) 6 cm (d) 6 cm 3 169. The perimeter of an isosceles triangle is 544 cm and each of (a) 6

3 :8

174. If area of an equilateral triangle is a and height b, then value of

(b) 3 3

(c) 3 2 (d) 2 2 167. The sides of a triangle are 16 cm, 12 cm and 20 cm. Find the area, (a) 64 cm2 (b) 112 cm2 2 (c) 96 cm (d) 81 cm2 168. 360 sq. cm and 250 sq. cm are the area of two similar triangles. If the length of one of the sides of the first triangle be 8 cm , then the length of the corresponding side of the second triangle is

(b)

3:4

R Enak

(a) 2 3

leng th

a  b  c abc

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178. From a point P which is at a distance of 13 cm from centre O of a circle of radius 5 cm in the same plane, a pair of tangents PQ and PR are drawn to the circle Area of quadrilateral PQOR is (a) 65 cm2 (b) 60 cm2 2 (c) 30 cm (d) 90 cm2 179. A circular road runs around a circular ground. If the difference between the circumference of the outer circle and the inner circle is 66 meters, the width of the road is:

r

12 2 cm. An equilateral triangle is inscribed in that circle. The length of the side of the triangle is

171. Three circles of radius a, b, c touch each other externally. The area of the triangle formed by joining their centre is

geisnh eeYa ridna gv.i Sni

163. A circle is inscribed in a square whose length of the diagonal is

9 15 sq . cm 2 (b) 12 15 sq. cm (a)

(c) 9 15 sq. cm (d) 6 15 sq. cm

22 ) 7 (a) 10.5 metres (b) 7 metres (c) 5.25 metres (d) 21 metres 180. A p erson ob serv ed that he required 30 seconds less time to cross a circular ground along its diameter than to cover it once along the boundary. If his speed was 30 m/ minutes. then the radius of the circular ground is ( Take  

( Take  

22 ): 7

(a) 5.5 m (b) 7.5 m (c) 10.5 m (d) 3.5 m 181. The difference of perimeter and diameter of a circle is X unit. The diameter of the circle is

X X unit (b) unit  –1 1 X  X (c) unit (d)   – 1  unit    182. The area of the circumcircle of an equilateral triangle is 3 sq. cm .The perimeter of the triangle is (a)

(a) 3 3 cm (b) 9 cm (c) 18 cm (d) 3 cm 183. A horse is tied to a post by a rope. If the horse moves along a circular path always keeping the rope streched and describes 88 metres when it has traced out 72° at the centre, the length of the

22 ) 7 (a) 70 m (b) 75 m (c) 80 m (d) 65 m 184. Three circles of radii 3.5 cm, 4.5 cm and 5.5 cm touch each other externally. Then the perimeter of the triangle formed by joining the centres of the circles, in cm is rope is (Take  

308

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1 × area of ABC 8 191. A wire of length 44 cm is first bent to form a circle and then rebent to form a square. The difference of the two enclosed areas is (a) 44 cm2 (b) 33 cm2 2 (c) 55 cm (d) 66 cm2 192. ACB is an ang le in the semicircle of diameter AB = 5 cm and AC : BC = 3: 4. The area of the triangle ABC is (d)

(a) 6 2 sq. cm (b) 4 sq. cm (c) 12 sq. cm (d) 6 sq. cm 193. If the lengths of the sides AB, BC and CA of a triangle ABC are 10 cm, 8 cm and 6 cm respectively and If M is the mid-point of BC and MN  AB to cut AC at N. then area of the trapezium ABMN is equal to (a) 18 sq. cm (b) 20 sq. cm (c) 12 sq. cm (d) 16 sq. cm 194. In an equilateral triangle of side 24 cm, a circle is inscribed touching its sides. The area of the remaining portion of the triangle is

ERna

(a) 1 : 2 (b) 2 : 3 (c) 1 : 3 (d) 1 : 2 189. Three circles of equal radius ‘a’ cm touch each other. The area of the shaded region is :

1 (c) × area of ABC 4

( 3 =1.732) (a) 98.55 sq. cm (b) 100 sq. cm (c) 101 sq. cm (d) 95 sq. cm 195. Two sides of a plot measuring 32 m and 24 m and the angle between them is a perfect right angle. The other two sides measure 25 m each and the other three angles are not right angles. The area of the plot in m2 is (a) 768 (b) 534 (c) 696.5 (d) 684 196. A and b are two sides adjacent to the right angle of a right angled triangle and p is the perpendicular drawn to the hypotenuse from the opposite vertex. Then p2 is equal to

wwM wa. th Les aBryn

 3     (a)  2  a2sq. cm  

 6 3 –    2 (b)   a sq. cm 2   (c)



3 – a2sq. cm



 2 3 –    2 (d)   a sq. cm 2  

190. ABC is a right angled triangle. B being the right angle Mid- points of BC and AC are respectively B ' and A ' . Area of  A ' B ' C is

1 × area of ABC 2 2 (b) × area of ABC 3 (a)

externally, then the area of the circle with diamter AB is (a) 36 (b) 64 (c) 144 (d) 256 198. If the numerical value of the height and the area of an equilateral triangle be same. then the length of each side of the triangle is (a) 2 units (b) 4 units (c) 5 units (d) 8 units 199. If the length of a side of the square is equal to that of the diameter of a circle, then the ratio of the area of the square and that

kgei snhe eYari dnag v.iSn ir

(a) 27 (b)  [(3.5)2 +(4.5)2+ (5.5)2] (c) 27 (d) 13.5 185. Three sides of a triangular field are of length 15 m, 20m and 25 m long respectively. Find the cost of sowing seeds in the field at the rate of 5 rupees per sq. m (a) Rs.300 (b) Rs.600 (c) Rs.750 (d) Rs.150 186. A chord of length 30 cm is at a distance of 8 cm from the centre of a circle. The radius of the circle is: (a) 17 cm (b) 23 cm (c) 21 cm (d) 19 cm 187. The radius of the incircle of a triangle whose sides are 9 cm, 12 cm and 15 cm is (a) 9 cm (b) 13 cm (c) 3 cm (d) 6 cm 188. The ratio of inradius and circumradius of a square is :

(a) a2 +b2

1 1 (b) 2  2 a b

a 2b 2 (d) a2 – b2 a 2  b2 197. A is the centre of circle whose radius is 8 and B is the centre of a circle whose diameter is 8. If thes e tw o c i r c le s to uc h

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(c)

22 ) 7 (a) 14 : 11 (b) 7 : 11 (c) 11 : 14 (d) 11 : 7 200. The median of an equilateral triangle is 6 3 cm. The area ( in cm2) of the triangle is (a) 72 (b) 108 of the circle ( 

(c) 72 3 (d) 36 3 201. If the numerical value of the circumference and area of a circle is same, then the area is (a) 6 sq. units (b) 4 sq. units (c) 8 sq. units (d) 12 sq. units 202. The area of an equilateral triangle is 48 sq. cm. The length of the side is (a) 4 8 cm

(b) 4 3 cm

(c) 8 cm (d) 8 4 3 cm 203. The external fencing of a circular path around a circular plot of land is 33m more than its interior fencing. The width of the path around the plot is (a) 5.52 m (b) 5.25 m (c) 2.55 m (d) 2.25 m 204. The perimeter of a triangle is 54 m and its sides are in the ratio 5 : 6 : 7. The area of the triangle is (a) 18 m2

(b) 54 6 m2

(c) 27 2 m2 (d) 25 m2 205. A circular wire of diameter 112 cm is cut and bent in the form of a rectangle whose sides are in the ratio of 9 : 7. The smaller side of the rectangle is (a) 77 cm (b) 97 cm (c) 67 cm (d) 84 cm

309

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(c) 3 3 cm (d) 2 2 cm 207. Two equal maximum sized circular plates are cut off from a circular p ape r sheet of circumference 352 cm. Then the circumference of each circular plate is (a) 176 cm (b) 150 cm (c) 165 cm (d) 180 cm 208. The inradius of an equilateral triangle is 3 cm, then the perimeter of that triangle is (a) 18 cm (b) 15 cm (c) 12 cm (d) 6 cm 209. The dif ference between the circumference and diameter of a circle is 150 m. The radius of that

22 ) 7 (a) 25 metre (b) 35 metre (c) 30 metre (d) 40 metre 210. The perimeters of a circle, a square and an e quilater al triangle are same and their areas are C, S and T respectively. Which of the following statement is true ? (a) C = S = T (b) C > S > T (c) C < S < T (d) S < C < T

(a) 36 3 cm² (b) 144 3 cm² (c) 72 cm² (d) 12 cm² 215. Let C1 and C2 be the inscribed and circumscribed circles of a triangle with sides 3 cm, 4 cm and 5 cm, then

(a)

1 seconds to 2 complete a round around a circular field. If the speed of the horse was 66 m/sec, then the radius of the field is, [Given

211.A horse takes 2

22 ] 7 (a) 25.62 m (b) 26.52 m (c) 25.26 m (d) 26.25 m 212. The diameter of the front wheel of an engine is 2x cm and that of rear wheel is 2y cm to cover the same distance, find the number of times the rear wheel will revolve when the front wheel revolves ‘n’ times,



(a)

n times xy

(b)

yn times x

(c)

nx times y

(d)

xy times n

area of C1 is area of C2

9 25

(b)

16 25

9 4 (d) 16 25 216. A circular swimming pool is surrounded by a concrete wall 4m wide. If the area of the concrete wall surrounding the (c)

11 that of the pool, then 25 the radius(in m) of the pool : (a) 8 (b) 16 (c) 30 (d) 20 217. If the area of a circle is A, radius of the circ le is r and circumference of it is c, then pool is

wwM wa. th Les B aryn

circle is ( Take  

22 ) 7 (a) 625 (b) 1250 (c) 1875 (d) 2500 214. If the altitude of an equilateral triangle is12 3 cm, then its area would be; km is (Assume  

(a) rC = 2A

(b)

r C = 2 A

r² A (d) =C 4 r 218. The sides of a triangle having area 7776 sq. cm are in the ratio 3 : 4 : 5. The perimeter of the triangle is: (a) 400 cm (b) 412 cm (c) 424 cm (d) 432 cm 219. The perimeter of a sheet of paper in the shape of a quadrant of a circle is 75 cm. Its area would (c) AC =

22 ) 7 (a) 512.25 cm² (b) 346.5 cm² (c) 100 cm² (d) 693 cm² 220. A circ le is inscr ibed in an equilateral triangle of side 8m. The approximate area of the

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D 4m

O

be ( =

A

45°

r

(b) 2 3 cm

unoccupied space inside the triangle is (a) 21 m² (b) 11 m² (c) 20 m² (d) 22 m² 221. In the figure, OED and OBA are sectors of a circle with centre O. The area of the shaded portion.

geisnh eeYa ridna gv.i Sni

(a) 3 2 cm

213. A bicycle wheel has a diameter (including the tyre) of 56 cm. The number of times the wheel will rotate to cover a distance of 2.2

R Enak

206. If the perimeter of an equilateral triangle be 18 cm, then the length of each median is

3m

B E

(a)

11 m² 16

(b)

11 m² 8

(c)

11 m² 2

(d)

11 m² 4

222. If the circumference of a circle is

30 , 

then the diameter of the circle is (a) 30 (b)

15 30 (c) 60 (d)  ²

223. The outer and inner diameter of a circular path be 728 cm and 700 cm respectively. The breadth of the path is (a) 7 cm (b) 14 cm (c) 28 cm (d) 20 cm 224. A piece of wire when bent to form a circle will have a radius of 84 cm. If the wire is bent to form a square, the length of a side of the square is (a) 152 cm (b) 168 cm (c) 132 cm (d) 225 cm 225. The area of a circle is 324  sq.cm. The length of its longest chord (in cm.) is (a) 36 (b) 38 (c) 28 (d) 32 226. The circumference of a triangle is 24 cm and the circumference of its in-circle is 44 cm. Then the ar ea of the triangle is

22 ) 7 (a) 56 square cm (b) 48 square cm (c) 84 square cm (d) 68 square cm (taking  

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(b) 25 2 square cm (c) 12 2 square cm (d) 15 2 square cm 228. The inner-radius of a triangle is 6 cm, and the sum of the lengths of its sides is 50 cm. The area of the triangle (in sq. cm.) is (a) 150 (b) 300 (c) 50 (d) 56 229. One of the angles of a rightangled triangle is 15°, and the hypotenuse is 1 m. The area of the triangle (in sq. cm.) is (a) 1220 (b) 1250 (c) 1200 (d) 1215 230. If an isosceles triangle the length of each equal side is 'a' units and that of the third side is 'b' units, then its area will be

a 4a²–a² sq. units 4

(b)

b 4a²–b² sq. units 4

(c)

a 2a²–b² sq. units 2

(a) 12 cm

b a²–2b² sq. units 2 231. What is the position of the circumcentre of an obtuse-angles triangle? (a) It is the vertex opposite to the largest side. (b) It is the mid point of the largest side. (c) It lies outside the triangles. (d) It lies inside the triangles. 232. The ratio of cirumference and diameter of a circle is 22 : 7. If (d)

4 m, then 7 the radius of the circle is: the circumference be 1

1 m 3

(a)

1 m 4

(b)

(c)

1 m 2

(d) 1 m

(b) 136 cm,

(c) 16 cm (d) 44 cm 237. The perimeter of a rhombus is 40 m and its height is 5m its area is: (a) 60 m2 (b) 50 m2 (c) 45 m2 (d) 55 m2 238. The perimeter of a rhombus is 40 cm. If the length of one of its diagonals be 16 cm, the length of the other diagonal is (a) 14 cm (b) 15 cm (c) 16 cm (d) 12 cm 239. The area of a rhombus is 150 cm 2. The length of one of its diagonals is 10 cm. The length of the other diagonal is : (a) 25 cm (b) 30 cm (c) 35 cm (d) 40 cm 240. The area of a regular hexagon of side 2 3 cm is :

wwM wa. th Les aBryn

(a)

(c) 6 (d) 16 234. The diagonals of a rhombus are 32 cm and 24 cm respectively. The perimeter of the rhombus is: (a) 80 cm (b)72 cm (c) 68 cm (d) 64 cm 235. The diagonals of a rhombus are 24 cm and 10 cm. The perimeter of the rhombus (in cm) is : (a) 68 (b) 65 (c) 54 (d) 52 236. The perimeter of a rhombus is 40 cm, If one of the diagonals be 12 cm long, what is the length of the other diagonal ?

60° and the measure of one of its sides is 10 cm. The length of its smaller diagonal is : (a) 10cm

(b) 10 3 cm

(c) 10 2 cm

(d)

5 2 cm 2

244. The perimeter of a rhombus is 100 cm, If one of its diagonals is 14 cm.Then the area of the rhombus is (a) 144 cm2 (b) 225 cm2 (c) 336 cm2 (d) 400 cm2 245. The ratio of the length of the parallel sides of a trapezium is 3 : 2. The shortest distance between them is 15 cm. If the area of the trapezium is 450 cm2 the sum of the length of the parallel sides is (a) 15 cm (b) 36 cm (c) 42 cm (d) 60 cm 246. A parallelogram has sides 15 cm and 7 cm long. The length of one of the diagonals is 20 cm. The area of the parallelogram is (a) 42 cm2 (b) 60 cm2 (c) 84 cm2 (d) 96 cm2 247. Sides of a parallelogram are in the ratio 5 : 4. Its area is 1000 sq. units. Altitude on the greater side is 20 units. Altitude on the smaller side is (a) 20 units (b) 25 units (c) 10 units (d) 15 units 248. The perimeter of a rhombus is 40 cm and the measure of an angle is 60°, then the area of it is:

kgei snhe eYari dnag v.iSn ir

(a) 20 2 square cm

233. The area of a circle whose radius is the diagonal of a square whose area is 4 is: (a) 4 (b) 8

ERna

227. If the length of each of two equal sides of an isosceles triangle is 10 cm. and the adjacent angle is 45°, then the area of the triangle is

(a) 18 3 cm2

(b) 12 3 cm2

(c) 36 3 cm2 (d) 27 3 cm2 241. Each side of a regular hexagon is 1 cm. The area of the hexagon is (a)

3 3 cm2 2

(b)

3 3 cm2 4

(c) 4 3 cm2 (d) 3 2 cm2 242. The length of one side of a rhombus is 6.5 cm and its altitude is 10 cm. If the length of its diagonal be 26 cm, the length of the other diagonal will be: (a) 5 cm (b) 10 cm (c) 6.5 cm (d) 26 cm 243. The measure of each of two opposite angles of a rhombus is

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(a) 100 3 cm2 (b) 50 3 cm2 (c) 160 3 cm2 (d) 100 cm2 249. Two ad jace nt side s of a parallelogram are of length 15 cm and 18 cm, If the distance between two smaller sides is 12 cm, then the distance between two bigger sides is (a) 8 cm (b) 10 cm (c) 12 cm (d) 15 cm 250. A parallelogram ABCD has sides AB = 24 cm and AD = 16 cm. The distance between the sides AB and DC is 10 cm. Find the distance between the sides AD and BC. (a) 15 cm (b) 18 cm (c) 16 cm (d) 9 cm

311

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(b) 3 3 sq. units (c) 3 2 sq. units (d) 2 3 sq. units

256. In ABC , D and E are the points of sides AB and BC respectively

such that DE  AC and AD : BD = 3 : 2. The ratio of area of trap ezium ACED to that of BED is (a) 4 : 15 (b) 15 : 4 (c) 4 : 21 (d) 21 : 4 257. ABCD is a trapezium in which

AB  DC and AB = 2 CD. The

diagonals AC and BD meet at O. The ratio of area of triangles AOB and COD is

(b)

(c) area ( ΔABC ) < area ( ΔDCQ )

(d) area ( ΔABC )  area ( ΔDCQ )

260. ABCD is a parallelogram. P and Q are the mid- points of sides BC and CD respectively. If the area of ABC is 12 cm2, then the area of

APQ is

(a) 12 cm2 (b) 8 cm2 2 (c) 9 cm (d) 10 cm2 261. The area of a rhombus is 216 cm2 and the length of its one diagonal is 24 cm. The perimeter (in cm) of the rhombus is (a) 52 (b) 60 (c) 120 (d) 100 262. One of the four angles of a rhombus is 60°. If the length of each side of the rhombus is 8 cm, then the length of the longer diagonal is (a) 8 3 cm (c) 4 3 cm

(b) 8 cm (d)

8

cm 3 263. The diagonals of a rhombus are 12 cm and 16 cm respectively. The length of one side is (a) 8 cm (b) 6 cm (c) 10 cm (d) 12 cm 264. A parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. Its area is (a) 500 15 m2 (b) 600 15 m2

(c) 400 15 m2 (d) 450 15 m2 265.ABCD is a trapezium with AD and BC parallel sides. The ratio of the area of ABCD to that of  AED is

(a) 1 : 1

(b) 1 : 2 (c) 4 : 1 (d) 1 : 4 258. The length of each side of a rhombus is equal to the length of the side of a square whose

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B

A

E

N

C

BE EC

AD  BE AD  BC (d) AD  CE AD 266. perimeter of a rhombus is 2p unit and sum of length of diagonals is m unit, then area of the rhombus is (c)

(a)

1 2 m p sq unit 4

r

(b) area ( ΔABC ) > area ( ΔDCQ )

wwM wa. th Les B aryn

254. The perimeter of a non-square rhombus is 20 cm. One of its diagonal is 8 cm. The area of the rhombus is (a) 28 sq. cm (b) 20 sq. cm (c) 22 sq. cm (d) 24 sq. cm 255. The perimeter of a rhombus is 100 cm and one of its diagonals is 40 cm. Its area (in cm2) is (a) 1200 (b) 1000 (c) 600 (d) 500

(a) area ( ΔABC ) = area ( ΔDCQ )

(a) AD BC

geisnh eeYa ridna gv.i Sni

(a) 2 2 sq. units

diagonal is 40 2 cm. If the length of the diagonals of the rhombus are in the ratio 3 : 4, then its area (in cm2) is (a) 1550 (b) 1600 (c) 1535 (d) 1536 259. ABCD is a parallelogram BC is produced to Q such that BC = CQ. Then

R Enak

251. The ad jace nt side s of a paralleogram are 36 cm and 27 cm in length, If the distance between the shorter sides is 12 cm, then the distance between the longer sides is (a) 10 cm (b) 12 cm (c) 16 cm (d) 9 cm 252. If the diagonals of a rhombus are 8 cm and 6 cm, then the area of square having same side as that of rhombus is (a) 25 (b) 55 (c) 64 (d) 36 253. Two circles with centres A and B and radius 2 units touch each other externally at ‘C’ A third circle with centre ‘C’ and radius ‘2’ units meets other two at D and E. The n the area of the quadrilateral ABDE is

1 mp 2 sq unit 4 1 m 2 – p 2 sq unit (c) 4 1 m 2  p 2 sq unit (d) 4 267. Area of regular hexagon with side ‘a’ is (b)









3 3 2 a sq. unit 4 12 (b) a2 sq. unit 2 3 (a)

(c)

9

2 3

a2 sq. unit

6

a2 sq. unit 2 268. In ABC , D and E are two points on the sides AB and AC respectively so that DE  BC and (d)

2 . Then 3 the area of trapezium DECB is the area of ABC equal to 5 21 (a) (b) 9 25 4 1 (c) 1 (d) 5 5 4 269. The sides of a rhombus are 10 cm each and a d iagonal measures 16 cm. Area of the rhombus is (a) 96 sq. cm (b) 160 sq. cm (c) 100 sq. cm (d) 40 sq. cm 270. The lengths of two parallel sides of a trapezium are 6 cm and 8 cm. If the height of the trapezium be 4 cm, then its area is (a) 28 cm² (b) 56 cm² (c) 30 cm² (d) 36 cm²

AD = BD

D

312

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(b) 24

(c)

(d) 28 137 273. The perimeter of a rhombus is 60 cm and one of its diagonal is 24 cm. The area of the rhombus is (a) 432 sq.cm (b) 216 sq.cm (c) 108 sq.cm (d) 206 sq.cm 274. The area of the parallelogram whose length is 30 cm, width is 20 cm and one diagonal is 40 cm is (a) 200 15 cm² (b) 300 15 cm² (c) 100 15 cm²

wwM wa. th Les aBryn

(d) 150 15 cm² 275. The area of a rhombus is 256 sq.cm. and one of its diagonal is twice the other in length.Then length of its larger diagonal is (a) 32 cm (b) 48 cm (c) 36 cm (d) 24 cm 276. The length of two parallel sides of a trapezium are 15 cm and 20 cm. If its area is 175 sq.cm, then its height is: (a) 25 cm (b) 10 cm (c) 20 cm (d) 15 cm 277. The cost of carpenting a room is Rs. 120 . If the width had been 4 metres less, the cost of the Carpet would have been Rs. 20 less. The width of the room is : (a) 24 m (b) 20 m (c) 25 m (d) 18.4 m 278. The floor of a corridor is 100 m long and 3 m wide. Cost of covering the floor with carpet 50 cm wide at the ratio of Rs. 15 per m is (a) Rs. 4500 (b) Rs. 9000 (c) Rs. 7500 (d) Rs. 1900 279. A playground is in the shape of a rectangle. A sum of Rs. 1,000 was

(a) 2 : 5 (b) 2 : 5 (c) 4 : 25 (d) 4 : 5 288. The ratio of base of two triangles is x : y and that of their areas is a : b. Then the ratio of their corresponding altitudes will be: a b : (a) (b) 1 : 1 y x x b (d) a : y 289. The area of a field in the shape of a trapezium measures 1440m². The perpendicular distance between its parallel sides is 24m. If the ratio of the parallel sides is 5 : 3, the length of the longer parallel side is : (a) 75 m (b) 45 m (c) 120 m (d) 60 m 290. If the ratio of areas of two squares is 225 : 256, then the ratio of their perimeter is : (a) 225 : 256 (b) 256 : 225 (c) 15 : 16 (d) 16 : 15 291. The area of a triangle is 216 cm2 and its sides are in the ratio 3 : 4 : 5. The perimeter of the triangle is: (a) 6 cm (b) 12 cm (c) 36 cm (d) 72 cm 292. A circular wire of radius 42 cm is bent in the form of a rectangle whose sides are in the ratio of 6 : 5. The smaller side of the

(c) ay : bx

kgei snhe eYari dnag v.iSn ir

(a) 2 137

spent to make the ground usable at the rate of 25 paise per sq. m. The breadth of the ground is 50 m. If the length of the ground is increased by 20 m. what will be the expenditure (in rupees) at the same rate per sq. m.? (a) 1,250 (b) 1,000 (c) 1,500 (d) 2,250 280. A hall 25 metres long and 15 metres broad is surrounded by a verandah of uniform width of 3.5 metres. The cost of flooring the varandah, at Rs. 27.50 per square metre is (a) Rs. 9149.50 (b) Rs. 8146.50 (c) Rs. 9047.50 (d) Rs. 4186.50 281. The outer circumference of a circular race-track is 528 metre. The track is everywhere 14 metre wide. Cost of levelling the track at the rate of Rs.10 per sq. metre is : (a) Rs. 77660 (b) Rs. 67760 (c) Rs. 66760 (d) Rs. 76760 282. The length and breadth of a rectangular field are in the ratio of 3 : 2. If the perimeter of the field is 80m, its breadth (in metres) is : (a) 18 (b) 16 (c) 10 (d) 24 283. The sides of a rectangular plot are in the ratio 5 : 4 and its area is equal to 500 sq.m The perimeter of the plot is : (a) 80 m (b) 100 m (c) 90 m (d) 95 m 284. ABC is a triangle with base AB, D is a point on AB such that AB = 5 and DB =3. What is the ratio of the area of  ADC to the area of  ABC ? (a) 2/5 (b) 2/3 (c) 9/25 (d) 4/25 285. If the area of a triangle is 1176 cm2 and the ratio of base and corresponding altitude is 3 : 4, then the altitude of the triangle is: (a) 42 cm (b) 52 cm (c) 54 cm (d) 56 cm 286. The sides of a triangle are in the

ERna

271. If diagonals of a rhombus are 24 cm and 32 cm, then perimeter of that rhombus is (a) 80 cm (b) 84 cm (c) 76 cm (d) 72 cm 272. The ar ea of an isosceles trapezium is 176 cm² and the height is 2/11th of the sum of its parallel sides. If the ratio of the length of the parallel sides is 4 : 7, then the length of a diagonal (in cm) is

1 1 1 : : . If the perimeter of 2 3 4 the triangle is 52 cm, the length of the smallest side is : (a) 24 cm (b) 10 cm (c) 12 cm (d) 9 cm 287. If the diagonals of two squares are in the ratio of 2 : 5. Their area will be in the ratio of

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ratio

22 ): 7 (a) 60 cm (b) 30 cm (c) 25 cm (d) 36 cm 293. The ratio of the outer and the inner perimeter of a circular path is 23 : 22, If the path is 5 meters wide the diameter of the inner circle is: (a) 110 m (b) 55 m (c) 220 m (d) 230 m 294.The angles of a triangle are in the ratio 3 : 4 : 5. The measure of the largest angle of the triangle is (a) 60° (b) 75° (c) 120° (d) 150° 295.The ratio of the area of a square to that of the square drawn on its diagonal is: (a) 1 : 1 (b) 1 : 2 (c) 1 : 3 (d) 1 : 4 296.A square and an equilateral triangle are drawn on the same base. The ratio of their area is (a) 2 :1 (b) 1 : 1

rectangle is ( Take 

(c)

30 : 4

(d) 4 : 3

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(c) 5 : 6 (d) 5 : 6 299. If the length and the perimeter of a rectangle are in the ratio 5 : 16. then its length and breadth will be in the ratio (a) 5 : 11 (b) 5 : 8 (c) 5 : 4 (d) 5 :3 300. Through each vertex of a triangle, a line parallel to the opposite side is drawn. the r atio of the perimeter of the new triangle. thus formed, with that of the original triangle is (a) 3 : 2 (b) 1 : 2 (c) 2 : 1 (d) 2 : 3 301. The ratio of the number giving the measure of the circumference and the area of a circle of radius 3 cm is (a) 1 : 3 (b) 2 : 3 (c) 2 : 9 (d) 3 : 2 302. The height of an equilateral

(d) 3 15 42 309. The ratio of the areas of the circumcircle and the incirle of an equilateral triangle is (a) 2 : 1 (b) 4 : 1 (c) 8 : 1 (d) 3 :2 310. In ABC , the medians CD and BE intersect each other at O, then the ratio of the areas of ODE and ΔOBC is (a) 1 : 4 (b) 6 : 1 (c) 1 : 12 (d) 12 : 1 311. The ratio of the area of two isosceles triangles having the same vertical angle (i.e. angle between equal sides) is 1 : 4. The ratio of their heights is (a) 1 : 4 (b) 2 : 5 (c) 1 : 2 (d) 3 : 4 312. The ratio of length of each equal side and the third side of an isosceles triangle is 3 : 4. If the area is

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triangle is 4 3 cm. The ratio of the area of its circumcircle to that of its in-circle is (a) 2 : 1 (b) 4 : 1 (c) 4 : 3 (d) 3 : 2 303. The radius of circle A is twice that of circle B and the radius of circle B is twice that of circle C. Their area will be in the ratio (a) 16 : 4 : 1 (b) 4 : 2 : 1 (c) 1 : 2 : 4 (d) 1 : 4 : 16 304. A circle and a square have equal areas. the ratio of a side of the square and the radius of the circle is

(c)

(a) 1 : 

(b)  :1

(c) 1 : (d)  :1 305. The sides of a triangle are in the 1 1 1 ratio : : and its perimeter 3 4 5 is 94cm . The length of the smallest side of the triangle is: (a) 18 cm (b) 22.5 cm (c) 24 cm (d) 27 m 306. The sides of a quadrilateral are in the ratio 3 : 4 : 5 : 6 and its

8 5 units². the third side is (a) 3 units

(b) 2 5 square units units

(c) 8 2 units (d) 12 units 313. The ratio of sides of a triangle is 3 : 4 : 5. If area of the triangle is 72 square unit then the length of the smallest side is : (a) 4 3 unit

(b) 5 3 unit

(c) 6 3 unit (d) 3 3 unit 314. The ratio of sides of a triangle is 3 : 4 : 5 and area of the triangle is 72 square units. Then the area of an equilateral triangle whose perimeter is same as that of the previous triangle is (a) 32 3 square units

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(d) 60 3 square units 315. The parallel sides of a trapezium are in a ratio 2 : 3 and their shortest distance is 12 cm.If the area of the trapezium is 480 sq. cm., the longer of the parallel sides is of length : (a) 56 cm (b) 36 cm (c) 42 cm (d) 48 cm 316. An equilateral triangle is drawn on the diagonal of a square . The ratio of the area of the triangle to that of the square is

r

(d)  : 2 298.The area of two equilateral triangles are in the ratio 25 : 36. Their altitudes will be in the ratio: (a) 36 : 25 (b) 25 : 36

(c) 96 square units

geisnh eeYa ridna gv.i Sni

(c)  : 2

perimeter is 72 cm. The length of its greatest side (in cm) is (a) 24 (b) 27 (c) 30 (d) 36 307. The ratio of the radii of two wheels is 3: 4. The ratio of their circumference is (a) 4 : 3 (b) 3 : 4 (c) 2 : 2 (d) 3 : 2 308. The sides of a triangle are in the ratio 2 : 3 : 4. the perimeter of the triangle is 18 cm. The area (in cm2) of the triangle is (a) 9 (b) 36

R Enak

297. If the area of a circle and a square are equal, then the ratio of their perimeter is (a) 1 : 1 (b) 2 : 

(b) 48 3 square units

(a)

3 :2

(b) 1 : 3

(c) 2 : 3 (d) 4 : 3 317. Two triangles ABC and DEF are similar to each other in which AB = 10 cm, DE = 8 cm. Then the ratio of the area of triangles ABC and DEF is (a) 4 : 5 (b) 25 : 16 (c) 64 : 125 (d) 4 : 7 318. The ratio between the area of two circles is 4 : 7. What will be the ratio of their radii ? (a) 2 : 7

(b) 4 : 7

(c) 16 : 49

(d) 4 :

7

319. The area of a circle is proportional to the square of its radius. A small circle of radius 3 cm is drawn within a larger circle of radius 5 cm. Find the ratio of the area of the annular zone to the area of the larger circle (Area of the annular zone is the difference between the area of the larger circle and that of the smaller circle) (a) 9 : 16 (b) 9 : 25 (c) 16 : 25 (d) 16 : 27 320. The diameter of two circles are the side of a square and the diagonal of the square. The ratio of the area of the smaller circle and the larger circle is (a) 1 : 2 (b) 1 : 4 (c)

(d) 1 : 2 2: 3 321. The ratio of the area of an equilateral triangle and that of its circumcircle is (a) 2 3 : 2

(b) 4 :

(c) 3 3 : 4

(d) 7 2 : 2

314

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3 :1

(b) 1: 3

(c) 2 : 3 (d) 4 : 3 324. The radius of a circle is a side of a square. The ratio of the area of the circle and the square is (a) 1 :  (b)  : 1 (c)  : 2 (d) 2 :  325. ABC is an isosceles right angled triangle with B = 90°, On the sides AC and AB, two equilateral triangles ACD and ABE have been constructed. The ratio of area of ABE and ACD is (a) 1 : 3 (b) 2 : 3 (c) 1 : 2

(d) 1 : 2

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326. Two triangles ABC and DEF are similar to each other in which AB= 10 cm, DE=8 cm. Then the ratio of the area of triangles ABC and DEF is (a) 4 : 5 (b) 25 : 16 (c) 64 : 125 (d) 4 : 7 327. ABC is a right angled triangle, B being the right angle. Mid-points of BC and AC are respectively B' and A' .The ratio of the area of the quadrilateral AA'B'B to the area of the triangle ABC is (a) 1 : 2 (b) 2 : 3 (c) 3 : 4 (d) None of the above 328. The sides of a triangle are in the

1 1 1 ratio : : and its perimeter 4 6 8 is 91 cm. The difference of the length of longest side and that of shortest side is (a) 19 cm (b) 20 cm (c) 28 cm (d) 21 cm

breadth is increased by 10%, By what percent is its area changed? (a) 0% (b) 1% (c) 5% (d) 100% 337. The percentage increase in the area of a rectangle. If each of its sides is increased by 20%, is: (a) 40% (b) 42% (c) 44% (d) 46% 338. If the circumference of a circle is reduced by 50%, its area will be reduced by (a) 12.5% (b) 25% (c) 50% (d) 75% 339. If the side of a square is increased by 25%, then its area is increased by: (a) 25% (b) 55% (c) 40.5% (d) 56.25% 340. If the rad ius of a cir cle is increased by 50% . its area is increased by : (a) 125% (b) 100% (c) 75% (d) 50% 341. If the length of a rectangle is increased by 20% and its breadth is decreased by 20%, then its area (a) increases by 4% (b) decreases by 4% (c) decreases by 1% (d) None of these 342. If each side of a rectangle is increased by 50%, its area will be increased by (a) 50% (b) 125% (c) 100% (d) 250% 343. If the altitude of a triangle is increased by 10% while its area remains same, its corresponding base will have to be decreased by (a) 10% (b) 9%

kgei snhe eYari dnag v.iSn ir

(a)

329. If the arcs of unit length in two circles subtend angles of 60° and 75° at their centres, the ratio of their radii is (a) 3 : 4 (b) 4 : 5 (c) 5 : 4 (d) 3 : 5 330. ABCD is a parallelogram in which diagonals AC and BD intersect at O. If E, F, G and H are the midpoints of AO, DO, CO and BO respectively, then the ratio of the perimeter of the quadrilateral EFGH to the perimeter of parallelogram ABCD is (a) 1 : 4 (b) 2 : 3 (c) 1 : 2 (d) 1 : 3 331. If the circumference of a circle increases from 4 to 8 , what change occurs in its area ? (a) It doubles (b) It triples (c) It quadruples (d) It is halved 332. If the length of a rectangle is increased by 25% and the width is decreased by 20%, then the area of the rectangle : (a) Increases by 5% (b) decreases by 5% (c) remains unchanged (d) increases by 10% 333. The area of a circle of radius 5 is numerically what percent of its circumfernce ? (a) 200% (b) 255% (c) 240% (d) 250% 334. If the circumference and area of a circle are numerically equal, then the diameter is equal to : (a) area of the circle

ERna

322. If the perimeters of a rectangle and a square are equal and the ratio of two adjacent sides of the rectangle is 1 : 2 then the ratio of area of the rectangle and that of the square is (a) 1 : 1 (b) 1 : 2 (c) 2 : 3 (d) 8 : 9 323. The perimeter of a rectangle and an equilateral triangle are same. Also, one of the sides of the rectangle is equal to the side of the triangle. The ratio of the area of the rectangle and the triangle is

 (c) 2 (d) 4 2 335. If D and E are the mid-points of the side AB and AC respectively of the ABC in the given figure here, the shaded region of the triangle is what per cent of the whole triangular region ? (b)

B

A

D

E

C

(a) 50% (b) 25% (c) 75% (d) 60% 336. The length of a rectangle is de creased by 1 0% and its

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1 1 % (d) 11 % 9 11 344. If the circumference of a circle is increased by 50% then the area will be increased by (a) 50% (b) 75% (c) 100% (d) 125% 345. The length and breadth of a rectangle are increased by 12% and 15% respectively. Its area will be increased by :

(c) 9

1 % 5 (c) 27%

(a) 27

4 % 5 (d) 28%

(b) 28

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(a) 37

1 % 2

(b) 60%

12 1 % (b) 76 % 13 13 (c) 30% (d) 15% 359. The length and breadth of a rec tang le are d oubled. Percentage increase in area is (a) 150% (b) 200% (c) 300% (d) 400% 360. The length of a rectangle is increased by 10% and breadth decreased by 10%. The area of the new rectangle is (a) neither inc reas ed nor decreased (b) increased by 1% (c) decreased by 2% (d) decreased by 1% 361. How many circular plates of diameter d be taken out of a square plate of side 2d with minimum loss of material? (a) 8 (b) 6 (c) 4 (d) 2 362. What is the total area of three equilateral triangles inscribed in a semi-circle of radius 2 cm? (a) 23

wwM wa. th Les B aryn

(c) 75% (d) 120% 349. The length and breadth of rectangle are increased by 20% and 25% res pectively. The increase in the area of the resulting rectangle will be : (a) 60% (b) 50% (c) 40% (d) 30% 350. If each side of a square is increased by 10%. its area will be increased by (a) 10% (b) 21% (c) 44% (d) 100% 351. If the length of a rectangular plot of land is increased by 5% and the breadth is decreased by 10%. How much will its area increase or decrease? (a) 6.5% increase (b) 5.5% decrease (c) 5.5% increase (d) 6.5% decrease 352. The radius of circle is increased by 1%. How much does the area of the circle increase ? (a) 1% (b) 1.1% (c) 2% (d) 2.01% 353. The length of a room floor exceeds its breadth by 20m . The area of the floor remains unaltered when the length is decreased by 10 m but the breadth is increased by 5 m. The area of the floor (in square meters) is: (a) 280 (b) 325 (c) 300 (d) 420

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9 3 cm² 4

(d) 3 3 cm²

363. The area of a sector of a circle of radius 36 cm is 72 cm². The length of the corresponding arc of the sector is (a)  cm (b) 2 cm (c) 3 cm (d) 4 cm 364. A square is inscribed in a circle of diameter 2a and another square is circumscribing circle. The difference between the areas of outer and inner squares is (a) a² (b) 2a² (c) 3a²(d) 4a² 365. ABC is a triangle right angled at A. AB = 6 cm and AC = 8 cm. Semi-circles drawn (outside the triangle) on AB, AC and BC as diameters which enclose areas x, y and z square units, respectively. What is x +y–z equal to? (a) 48 cm² (b) 32 cm² (c) 0 (d) None of these 366. Consider an equilateral triangle of a side of one unit length. A new equilateral triangle is formed by joining the mid-points of one, then a third equilateral triangle is formed by joining the midpoints of second. The process is continued. The perimeter of all triangles, thus formed is (a) 4 (b) 5 (c) 6 (d) 7 367. What is the area of the larger segment of a circle formed by a chord of length 5 cm subtending an angle of 90º at the centre?

r

(b) 33

(c)

geisnh eeYa ridna gv.i Sni

 % 3 (c) 75% (d) 100% 347. Each side of a rectangular field is diminished by 40%. By how much percent is the area of the field diminished ? (a) 32% (b) 64% (c) 25% (d) 16% 348. The length of r ectangle is increased by 60%. By what percent would the breadth to be decreased to maintain the same area?

(a) 25%

354. In measuring the sides of a rectangle, there is an excess of 5% on one side and 2% deficeit on the other. Then the error percent in the area is (a) 3.3% (b) 3.0 % (c) 2.9% (d) 2.7% 355. The length and breadth of a rectangle are increased by 30% and 20% respectively. The area of the rectangle so formed exceeds the area of the square by (a) 46% (b) 66% (c) 42% (d) 56% 356. If side of a square is increased by 40%, the percentage increase in its surface area is (a) 40% (b) 60% (c) 80% (d) 96% 357. If the diameter of a circle is increased by 8%, then its area is increased by : (a) 16.64% (b) 6.64% (c) 165 (d) 16.46% 358. One sid e of a r ectangle is increased by 30%. To maintain the same area, the other side will have to be decreased by

R Enak

346. If the sides of an equilateral triangle are increased by 20%, 30% and 50% respectively to form a new triangle the increase in the perimeter of the equilateral triangle is

(a) 12 cm²

(b)

3 3 cm² 4

(a)

25      1  cm² 4 2 

25   (b) 4  2 – 1 cm²   25  3  (c) 4  2  1 cm²   (d) None of these 368. A rectangle of maximum area is drawn inside a circle of diameter 5 cm. What is the maximum area of such a rectangle? (a) 25 cm² (b) 12.5 cm² (c) 12 cm² (d) None of these 369. If AB and CD are two diameters of a cricle of radius r and they are mutually perpendicular, then

316

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(a)

8 m 93

16 (c) m 93

(b)

13 m 93

23 (d) m 93

D

O

B

Q

C

(a) 40 sq cm (b) 45 sq cm (c) 50 sq cm (d) 80 sq cm 380. If an isosceles right angled triangle has area 1sq unit, then what is its perimeter? (a) 3 units (b) 2 2 +1 units

2  1 units





(d) 2



2  1 units



381. A circular water fountain 6.6 m in diameter is surrounded by a path of width 1.5 m. The area of this path (in sq m) is (a) 13.62 (b) 13.15 (c) 12.15 (d) None of these 382. The area of a rectangular filed is 4500 sq m. If its length and breadth are in the ratio 9:5, then its perimeter is (a) 90 m (b) 150 m (c) 280 m (d) 360 m 383. The area of a square inscribed in a circle of radius 8 cm is (a) 32 sq cm (b) 64 sq cm (c) 128 sq cm (d) 256 sq cm 384. The short and long hands of a clock are 4 cm and 6 cm long, respectively. Then, the ratio of distances travelled by tips of short hand in 2 days and long hand in 3 days is (a) 4 : 9 (b) 2 : 9 (c) 2 : 3 (d) 1 : 27

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373. The area of an isosceles  ABC with AB = AC is 12 sq cm and altitude AD = 3 cm. What is its perimeter? (a) 18 cm (b) 16 cm (c) 14 cm (d) 12 cm 374. A hospital room is to accommodate 56 patients. It should be done in such a way that every patient gets 2.2 m² of floor and 8.8 m³ of space. If the length of the room is 14m, then breadth and the height of the room are respectively (a) 8.8 m,4m (b) 8.4 m,4.2m (c) 8 m, 4 m (d) 7.8 m,4.2m 375. How many 200 mm lengths can be cut from 10 m of ribbon? (a) 50 (b) 40 (c) 30 (d) 20 376. What is the area between a square of side 10 cm and two inverted semi-circular, crosssections each of radius 5 cm inscribed in the square? (a) 17.5 cm² (b) 18.5 cm² (c) 20.5 cm² (d) 21.5 cm²

A

(c)

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385. The arc AB of the circle with centre at O and radius 10 cm has length 16 cm. What is the area of the sector bounded by the radii OA, OB and the arc AB? (a) 40  sq cm(b) 40 sq cm (c) 80 sq cm (d) 20 sq cm 386.The length of a room floor exceeds its breadth by 20m. The area of the floor remains unaltered when the length is decreased by 10m bu t the breadth is increased by 5m. The area of the floor (in square metres) is: (a) 280 (b) 325 (c) 300 (d) 420 387.Find the perimeter of a square whi ch i s sy mmetrically inscribed in semicircle of radius 10 cm.

kgei snhe eYari dnag v.iSn ir

  (b)  (c) (d) 2 2 4 370. What is the area of a circle whose area is equal to that of a triangle with sides 7 cm, 24 cm and 25 cm? (a) 80 cm² (b) 84 cm² (c) 88 cm² (d) 90 cm² 371. If the area of an equilateral triangle is x and its perimeter is y, then which one of the following is correct? (a) y4 = 432x² (b) y4 = 216x² (c) y2 = 432x² (d) None of these 372. A rectangular field is 22 m long and 10 m wide. Two hemispherical pitholes of radius 2 m are dug from two places and the mud is spread over the remaining part of the field. The rise in the level of the field is (a)

377. The perimeter of a rectangle having area equal to 144 cm² and sides in the ratio 4:9 is (a) 52 cm (b) 56 cm (c) 60 cm (d) 64 cm 378. One side of a parallelogram is 8.06 cm and its perpendicular distance from opposite side is 2.08 cm. What is the approximate area of the parallelogram? (a) 12.56 cm² (b) 14.56 cm² (c) 16.76 cm² (d) 22.56 cm² 379. In the figure given below, the area of rectangle ABCD is 100 sq cm, O is any p oint on AB and CD=20cm. Then, the area of  COD is

ERna

what is the ratio of the area of the circle to the area of the ACD ?

(a) 80cm

(b) 80 cm

(c) 8 24 cm (d) 16 5 cm 388. Consider the following statement I. Area of a segment of a circle is les s than area of its corresponding sector. II. Distance trav elle d by a circular wheel of diameter 2d cm in one revolution is greater than 6d cm. Which of the above statements is/are correct? (a) Only I (b) Only II (c) Both I and II (d) Neither I nor II 389. The Perimeter of a rectangle is 82 m and its area is 400 sq m. What is the breadth of the rectangle? (a) 18 m (b) 16 m (c) 14 m (d) 12 m 390. The area enclosed between the circumference of two concentric circles is 16 sq cm and their radii are in the ratio 5:3. What is the area of the outer circle? (a) 9 sq cm (b) 16 sq cm (c) 25 sq cm (d) 36 sq cm 391. If the circumference of a circle is equal to the perimeter of square, then which one of the following is correct? (a) Area of circle = Area of square (b) Area of circle  Area of square (c) Area of circle > Area of square (d) Area of circle< Area of square

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(c) 21 3 cm (d) 42 cm 394. In the  ABC, the base BC is trisected at D and E. The line through D, Parallel to AB, meets AC at F and the line through E parallel to AC meets AB at G. If EG and DF intersect at H, then what is the ratio of the sum of the area of parallelogram AGHF and the area of the  DHE to the area of the  ABC? 1 1 1 1 (b) (c) (d) 2 3 4 6 395. If the area of a circle is equal to the area of a square with side

(a)

3 sq cm

triangle is 3 cm, then what is its perimeter? (a) 3 cm

(b) 3 3 cm

(c) 6 cm

(d) 6 3 cm

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(c) 1 :

4

(d) 1 : 4

409. If the four equal circles of radius

3cm to uch each oth er externally, then the area of the region bounded by the four circles is: (a) 49   sq.cm (b) 94   sq.cm

(b) 2 3 sq cm

(c) 3 3 sq cm (d) 4 3 sq cm 403. If a lawn 30 m long and 16 m wide is surrounded by a path 2 m wide, then what is the area of the path? (a) 200 m² (b) 280 m² (c) 300 m² (d) 320 m² 404. If a c ircle circumscr ibes a rectangle with side 16 cm and 12 cm, then what is the area of the circle? (a) 48 sq cm (b) 50 sq cm (c) 100 sq cm (d) 200 sq cm 405. The lengths of two sides of a right angled triangle which contain the right angle ar e a and b, respectively. Three squares are drawn on the three sides of the triangle on the outer side.What is the total area of the triangle and the three squares? (a) 2(a²+b²)+ab (b) 2(a²+b²)+2.5ab (c) 2(a²+b²)+0.5ab (d) 2.5(a²+b²) 406. A wall is of the form of a trapezium with height 4 m and parallel sides being 3 m and 5m. What is the cost of painting the wall, if the rate of painting is Rs.25 per sq m? (a) Rs. 240 (b) Rs. 400 (c) Rs. 480 (d) Rs. 800 407. A grassy field has the shape of an equilateral triangle of side 6

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2  units, then what is the diameter of the circle? (a) 1 unit (b) 2 units (c) 4 units (d) 8 units 396. A s quar e, a cricle and an equilateral triangle have same perimeter. Conside r the following statements I. The area of square is greater than the area of the triangle. II. The area of circle is less than the area of triangle. Which of the above statement is/ are correct? (a) Only I (b) Only II (c) Both I and II (d) Neither I nor II 397. If the area of a rectangle whose length is 5 more than twice its width is 75 sq units. What is the perimeter of the rectangle? (a) 40 units (b) 30 units (c) 24 units (d) 20 units 398. If the altitude of an equilateral

2 3 cm? (a)

r

(b) 42 3 cm

m. If a horse with 4.2 m long rope tied at a vertex. The percentage of the total area of the field which is available for grazing is best approximated by (a) 50% (b) 55% (c) 59% (d) 62% 408. The areas of two circles are in the ratio 1:2. If the two circles are bent in the form of squares, then what is the ratio of their areas? (a) 1 : 2 (b) 1 : 3

geisnh eeYa ridna gv.i Sni

(a) 21 cm

399. The area of a rectangle, whose one side is a is 2a². What is the area of a square having one of the diagonal of the rectangle as side? (a) 2a² (b) 3a² (c) 4a²(d) 5a² 400. If the outer and inner diameters of a stone parapet around a well ar e 11 2 cm and 70 cm respectively. Then, what is the area of the parapet? (a) 264 sq cm (b) 3003 sq cm (c) 6006 sq cm (d) 24024 sq cm 401. If the area of a  ABC is equal to area of square of side length 6 cm,then what is the length of the altitud e to AB , whe r e AB = 9 cm? (a) 18 cm (b) 14 cm (c) 12 cm (d) 8 cm 402. What is the area of an equilateral triangle having altitude equal to

R Enak

392. If the circumference of two circle are in the ratio 2:3, then what is the ratio of their areas? (a) 2:3 (b) 4:9 (c) 1:3 (d) 8:27 393. If the area of a circle inscribed in an equilateral triangle is 154 sq cm, then what is the perimeter of the triangle?

(c) 56   sq.cm (d) 65   sq.cm

410. If the diamete r of a circ le circumscribing a square is 15 2 cm, then what is the length of the side of the square? (a) 15 cm (b) 12 cm (c) 10 cm (d) 7.5 cm 411. Three congruent circles each of radius 4 cm touch one another. What is the area (in cm²) of the portion included between them? (a) 8

(b) 16 3 –8

(c) 16 3 – 4 (d) 16 3 –2 412. The two diagonals of a rhombus are of lengths 55 cm and 48 cm. If P is the perpendicular height of the rhombus, then which one of the following is correct? (a) 36 cm < p < 37 cm (b) 35 cm < p < 36 cm (c) 34 cm < p < 35 cm (d) 33 cm < p < 34 cm 413. The Perimeter of a triangular field is 240 m. If two of its sides are 78 m and 50 m, then what is the length of the perpendicular on the side of length 50 m from the opposite vertex? (a) 43 m (b) 52.2 m (c) 67.2 m (d) 70 m

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A

M

B

D

P

B

C

(a) 7cm² (b) 28 cm² (c) 14 cm² (d) 21 cm² 423. What is the maximum area of a rectangle, the perimeter of which is 18 cm? (a) 20.25 cm² (b) 20.00 cm² (c) 19.75 cm (d) 19.60 cm² 424. Three circular laminas of the same radius are cut out from a larger circular lamina. When the radius of each lamina cut out is the largest possible, then what is the ratio (approximate) of the area of the residual piece of the original lamina to its original total area? (a) 0.30 (b) 0.35 (c) 0.40 (d) 0.45 425. A wire is in the form of a radius 42 cm. If it is bent into a square, then what is the side of the square? (a) 66 cm (b) 42 cm (c) 36 cm (d) 33 cm 426. Seven semi- circluar areas are removed from the rectangle ABCD as shown in the figure below, in which AB = 2 cm and AD = 0.5 cm. The radius of each semi-circle, r,s,t, u and v is half of that of semi-circle p or q. What is the area of the remaining portion?

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2a a a a (a) (b) (c) (d) 3 2 3 4 417. A circle and a square have the same perimeter.Which one of the following is correct? (a) The area of the circle is equal to that of square (b) The area of the circle is larger than that of square (c) The area of the circle is less than that of square (d) No conclusion can be drawn 418. What is the radius of the circle inscribed in a triangle having side lengths 35 cm, 44 cm and 75 cm? (a) 3 cm (b) 4 cm (c) 5 cm (d) 6 cm 419. A rectangle area of 6 sq m is to be painted on a 3m × 4m board leaving a border of uniform width on all sides. What should be the width of the border? (a) 0.25 m (b) 0.5 m (c) 1 m (d) 3 m 420. A wheel of a bicycle has inner diameter 50 cm and thickness 10 cm. What is the speed of the bicycle, If it makes 10 revolutions is 5 s? (a) 5.5 m/s (b) 4.4 m/s (c) 3.3 m/s (d) 2.2 m/s

A

(c) (128 –15 )/128 cm² (d) None of these 427. PQRS is a diameter of a circle of radius 6 cm as shown in the figure above. The lengths PQ, QR and RS are equal. Semi-circles are drawn on PQ and QS as diamete rs. What is the perimeters of the shaded region?

kgei snhe eYari dnag v.iSn ir

. O

421. If a wire of length 36 cm is bent in the form of a semi-circle, then what is the radius of the semicircle? (a) 9 cm (b) 8 cm (c) 7 cm (d) 6 cm 422. In the given figure, the side of square ABCD is 7 cm. What is the area of the shaded portion, formed by the arcs BD or the circles with centre at C and A?

ERna

414. A piece of wire 78 cm long is bent in the form of and isosceles triangle. If the ratio of one of the equal sides to the base is 5:3, then what is the length of the base? (a) 16 cm (b) 18 cm (c) 20 cm (d) 30 cm 415. The length of a minute hand of a wall clock is 9 cm. What is the area swept (in cm²) by the minute hand in 20 min? (take  = 3.14) (a) 88.78 (b) 84.78 (c) 67.74 (d) 57.78 416. In the figure given below, AB is a line of length 2a, with M as midpoint. Semi-circles are drawn on one side with AM, and AB as diameters. A circle with centre O and radius r is drawn such that this circle touches all the three semicircles. What is the value of r?

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A

p

B r

s

t q

u

v

D (a) (128 –13 )/128 cm² (b) (125 –13 )/125 cm²

C

Q

R

S

(a) 12  cm (b) 14 cm (c) 16  cm (d) 18 cm 428. A person rides a bicycle round a circular path of radius 50 m. The radius of the wheel of the bicycle is 50 cm. The cycle comes to the starting point for the first time in 1 h. What is the number of revolutions of the wheel in 15 min? (a) 20 (b) 25 (c) 30 (d) 35 429. If a man walking at the rate 3 km/h crosses a square field diagonally in 1 min, then what is the area of the field? (a) 1000 m² (b) 1250 m² (c) 2500 m² (d) 5000 m² 430. The difference between the area of a square and that of an equilateral triangle on the same base is 1/4 cm². What is the length of side of triangle? (a) (4 –

1/2 3 ) cm

(b) (4 +

1/2 3 ) cm

(c) (4 –

–1/2 cm 3)

(d) (4 + 3 )–1/2 cm 431. A horse is tied to a pole fixed at one corner of a 50 m × 50 m square field of grass by means of a 20 m long rope. What is the area of that part of the field which the horse can graze? (a) 1256 m² (b) 942 m² (c) 628 m² (d) 314 m² 432. From a rectangular metal sheet of sides 25 cm and 20 cm; a circular sheet as large as possible is cut-off. What is the area of the remaining sheet? (a) 186 cm² (b) 144 cm² (c) 93 cm² (d) 72 cm²

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(a) 1 : 1

(b) 2 :

3

(c) 4 : 3 (d) 3 : 2 436. If the area of a circle, inscribed in an equilateral triangle is 4 cm², then what is the area of the triangle? (a) 12 3 cm² (b) 9 3 cm² (c) 8 3 cm² (d) 18 cm² 437. In the given figure,  ABC is a right angled triangle, right angled at A. Semi-circles are drawn on the sides AB, BC and AC. Then the area of shaded portion is equal to which one of the following?

travel in 15 min? (take =

B

A

B

(a) 425 cm² (c) 428 cm²

1 rd of 3 the first side of the rectangle. What is the area of the remaining portion?

C

(a) Area of  ABC (b) 2 times the area of  ABC (c) Area of semi-circle ABC (d) None of the above 438. In the given figure, ABC is a right angled triangle, right angled at B. BC = 21 cm and AB = 28 cm. Width AC as diameter of a semicircle and width BC as radius a quarter circle are drawn. What is the area of the shaded portion?

C

(b) 425.75 cm² (d) 428.75 cm²

27 ) 7

(a) 11 cm (b) 22 cm (c) 33 cm (d) 44 cm 443. A square of side x is taken. A rectangle is cut out from this square such that on side of the rectangle is half that of the

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A

445. A cycle wheel make s 10 00 revolution is moving 440 m. What is the diameter of the wheel? (a) 7 cm (b) 14 cm (c) 28 cm (d) 21 cm 446. A c ircle is ins crib ed in a equilateral triangle of side a. What is the area of any square inscribed in this circle? a² a² (b) 3 4

(c)

r

(a)

448.The ratio of the areas of the incircle and the circum-circle of a square is: (a) 1 : 2

(b)

7  x² (b)    8 

11  (c)    x² 12 

15  (d)    x² 16 

444. A rectangle cardboard is 18 cm×10 cm. From the four corners of the rectangle, quarter circles of radius 4 cm are cut. What is the perimeter (approximate) of the remaining portion? (a) 47.1 cm (b) 49.1 cm (c) 51.0 cm (d) 53.0 cm

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2 :1

(c) 1 : 2 (d) 2 : 1 449.The diagram represents the area swept by the wiper of a car. Wi th th e di men si on s given in the figure, calculate the shaded area swept by the wiper. O 30° 7cm A

D 14cm

square and the other is

3  x² (a)    4 

a² a² (d) 6 8

447. Consider a circle C of radius 6 cm with centre at O. What is the difference in the area of the circle C and the area of the sector of C subtending an angle of 80º at O? (a) 26 cm² (b) 16 cm² (c) 28 cm² (d) 30 cm²

geisnh eeYa ridna gv.i Sni

hypotenuse is 6 2 cm? (a) 12 cm² (b) 18 cm² (c) 24 cm² (d) 36 cm² 434. If A is the area of a triangle in cm², whose sides are 9 cm, 10 cm and 11 cm, then which one of the following is correct? (a) A < 40 cm² (b) 40 cm² < A < 45 cm² (c) 45 cm² < A < 50 cm² (d) A > 50 cm² 435. If x and y are respectively the areas of a square and a rhombus of sides of same length, then what is x : y?

439. The Perimeter of a square S1 is 12 m more than perimeter of the square S2. If the area of S1 equals three times, the area of S2 minus 11, then what is the perimeter of S1? (a) 24 m (b) 32 m (c) 36 m (d) 40 m 440. From a rectangular sheet of cardboard of size 5 cm×2 cm, the greatest possible circle is cut-off. What is the area of the remainting part? (a) (25 – ) cm² (b) (10 – ) cm² (c) (4 – ) cm² (d) (10 –2 ) cm² 441. A chord AB of a cricle of radius 20 cm makes a right angle at the centre of the circle. What is the area of the minor segment in cm²? (take = 3.14) (a) 31.4 cm² (b) 57 cm² (c) 62.8 cm² (d) 114 cm² 442. The minute hand of a clock is 14 cm long. How much distance does the end of the minute hand

R Enak

433. What is the area of a right angled is osce les triangle whose

B

C

(a) 102.67cm2 (b) 205.34cm2 (b) 51.33cm2 (d) 208.16cm2 450.If the length of a chord of a circle at a distance of 12cm from the centre is 10cm, then the diameter of the circle is : (a) 13 cm (b) 15 cm (c) 26 cm (d) 30 cm 451.Are a of the incircl e of an equilateral triangle with side 6cm is : (a)

 sq .cm 2

(c) 6 sq.cm

(b)

3 sq.cm

(d) 3 sq.cm

320

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of a circle with perimter equal to 24  : (a) 144 (b) 144  (c) 154  (d) none of these 456.A circle is inscribed in an equilateral triangle of side 8cm. The area of the portion between the triangle and the circle is : (a) 11cm2 (b) 10.95cm2 2 (c) 10cm (d) 10.50cm2 457.Find the ratio of the diameter of the circles inscribed in and circumscribed an equilateral triangle to its height. (a) 1 : 2 : 3 (b) 2 : 4 : 3 (c) 1 : 3 : 4 (d) 3 : 2 : 1 458.Find the area of the largest (or maximum sized) square that can be made inside a right angle triangle having sides 6cm, 8cm & 10cm when one of vertices of

the square coincide with the vertex of right angle of the triangle? 576 2 cm (a) (b) 24cm2 49 24 2 cm (c) (d) None of these 7 459. Area of the trapezium formed by x-axis; y-axis and the lines 3x + 4y =12 and 6x + 8y = 60 is: (a) 37.5sq.unit (b) 31.5sq.unit (c) 48sq.unit (d) 36.5sq.unit 460.A square having area 200sq.m, is formed in such a way that the

kgei snhe eYari dnag v.iSn ir

452.Th e adjace nt side s of a parellelogram are 36 cm and 27 cm in length. If the distance between the shorter sides is 12cm, then th e di stan ce between the longer sides is : (a) 10 cm (b) 12 cm (c) 16 cm (d) 9 cm 453.A circle and a rectangle have the same perimeter. The sides of the rectangle area 18cm and 26cm. The area of the circle is : (a) 125 cm2 (b) 230 cm2 2 (c) 550 cm (d) 616 cm2 454.The perimeter of a semicircular path is 36m. Find the area of this semicircular path. (a) 42sq.m (b) 54sq.m (c) 63sq.m (d) 77sq.m 455.Find the area of a rectangle whose area is equal to the area

length of its diagonal is 2 times of the diagonal of the given square. Then the area of the new square formed is: (a) 200 2 sq.m (b) 400 2 sq.m (c) 400sq.m (d) 800sq.m

ANSWER KEY 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. 80. 81. 82. 83. 84. 85. 86. 87. 88. 89. 90. 91. 92.

(b) (d) (b) (b) (b) (a) (c) (c) (d) (b) (c) (d) (d) (b) (b) (c) (b) (b) (b) (c) (b) (a) (b) (c) (a) (c) (c) (b) (b) (a) (b) (c) (b) (c) (b) (b) (a) (b) (a) (a) (c) (b) (c) (c) (a) (a)

93. (b) 94. (b) 95. (c) 96. (c) 97. (c) 98. (b) 99. (b) 100. (c) 101. (a) 102. (a) 103. (c) 104. (a) 105. (c) 106. (c) 107. (d) 108. (b) 109. (c) 110. (a) 111. (c) 112. (a) 113. (c) 114. (d) 115. (b) 116. (d) 117. (c) 118. (a) 119. (c) 120. (b) 121. (b) 122. (c) 123. (d) 124. (c) 125. (b) 126. (b) 127. (c) 128. (a) 129. (b) 130. (b) 131. (a) 132. (a) 133. (b) 134. (b) 135. (b) 136. (c) 137. (b) 138. (c)

139. (c) 140. (b) 141. (c) 142. (a) 143. (b) 144. (a) 145. (a) 146. (c) 147. (c) 148. (a) 149. (b) 150. (a) 151. (a) 152. (d) 153. (b) 154. (d) 155. (b) 156. (a) 157. (c) 158. (a) 159. (a) 160. (b) 161. (d) 162. (a) 163. (c) 164. (c) 165. (c) 166. (b) 167. (c) 168. (c) 169. (d) 170. (d) 171. (a) 172. (d) 173. (c) 174. (c) 175. (c) 176. (b) 177. (c) 178. (b) 179. (a) 180. (d) 181. (a) 182. (b) 183. (a) 184. (a)

185. (c) 186. (a) 187. (c) 188. (a) 189. (d) 190. (c) 191. (b) 192. (d) 193. (a) 194. (a) 195. (d) 196. (c) 197. (a) 198. (a) 199. (a) 200. (d) 201. (b) 202. (d) 203. (b) 204. (b) 205. (a) 206. (c) 207. (a) 208. (a) 209. (b) 210. (b) 211. (d) 212. (c) 213. (b) 214. (b) 215. (d) 216. (d) 217. (a) 218. (d) 219. (b) 220. (b) 221. (d) 222. (d) 223. (b) 224. (c) 225. (a) 226. (c) 227. (b) 228. (a) 229. (b) 230. (b)

ERna

(b) (c) (a) (c) (b) (b) (b) (c) (d) (d) (a) (d) (a) (b) (d) (a) (b) (d) (d) (b) (a) (c) (c) (d) (d) (a) (d) (b) (c) (d) (a) (d) (a) (c) (a) (a) (c) (b) (d) (a) (a) (a) (d) (d) (b) (a)

wwM wa. th Les aBryn

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46.

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231. (c) 232. (a) 233. (b) 234. (a) 235. (d) 236. (c) 237. (b) 238. (d) 239. (b) 240. (a) 241. (a) 242. (a) 243. (a) 244. (c) 245. (d) 246. (c) 247. (b) 248. (b) 249. (b) 250. (a) 251. (d) 252. (a) 253. (b) 254. (d) 255. (c) 256. (d) 257. (c) 258. (d) 259. (a) 260. (c) 261. (b) 262. (a) 263. (c) 264. (b) 265. (d) 266. (c) 267. (c) 268. (b) 269. (a) 270. (a) 271. (a) 272. (a) 273. (b) 274. (d) 275. (a) 276. (b)

277. (a) 278. (b) 279. (a) 280. (d) 281. (b) 282. (b) 283. (c) 284. (d) 285. (d) 286. (c) 287. (c) 288. (c) 289. (a) 290. (c) 291. (d) 292. (a) 293. (c) 294. (b) 295. (b) 296. (d) 297. (d) 298. (c) 299. (d) 300. (c) 301. (b) 302. (b) 303. (a) 304. (b) 305. (c) 306. (a) 307. (b) 308. (d) 309. (b) 310. (a) 311. (c) 312. (a) 313. (c) 314. (b) 315. (d) 316. (a) 317. (b) 318. (a) 319. (c) 320. (a) 321. (c) 322. (d)

323. (c) 324. (b) 325. (c) 326. (b) 327. (c) 328. (d) 329. (c) 330. (c) 331. (c) 332. (c) 333. (d) 334. (d) 335. (c) 336. (b) 337. (c) 338. (d) 339. (d) 340. (a) 341. (b) 342. (b) 343. (c) 344. (d) 345. (b) 346. (b) 347. (b) 348. (a) 349. (b) 350. (b) 351. (b) 352. (d) 353. (c) 354. (c) 355. (d) 356. (d) 357. (a) 358. (a) 359. (c) 360. (d) 361. (c) 362. (d) 363. (d) 364. (b) 365. (c) 366. (c) 367. (c) 368. (c)

369. (b) 370. (b) 371. (a) 372. (c) 373. (a) 374. (a) 375. (a) 376. (d) 377. (a) 378. (c) 379. (c) 380. (d) 381. (c) 382. (c) 383. (c) 384. (d) 385. (c) 386. (c) 387. (d) 388. (c) 389. (b) 390. (c) 391. (c) 392. (b) 393. (b) 394. (b) 395. (c) 396. (a) 397. (a) 398. (c) 399. (d) 400. (c) 401. (d) 402. (d) 403. (a) 404. (c) 405. (c) 406. (b) 407. (c) 408. (a) 409. (b) 410. (a) 411. (b) 412. (a) 413. (c) 414. (b)

415. (b) 416. (c) 417. (b) 418. (d) 419. (b) 420. (b) 421. (c) 422. (c) 423. (a) 424. (b) 425. (a) 426. (a) 427. (a) 428. (b) 429. (b) 430. (c) 431. (d) 432. (a) 433. (b) 434. (b) 435. (a) 436. (a) 437. (a) 438. (d) 439. (b) 440. (b) 441. (d) 442. (b) 443. (c) 444. (b) 445. (b) 446. (c) 447. (c) 448. (a) 449. (a) 450. (c) 451. (d) 452. (d) 453. (d) 454. (d) 455. (b) 456. (b) 457. (b) 458. (a) 459. (b) 460. (c)

321

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SOLUTION

Area of square = 2

= 2.

5.2

6.

3 x 4

7.

length

and

wwM wa.th Les B aryn

3 x 32 breadth = x– x = = =8m 4 4 4

(c) Since the room is in cuboid shape Length of largest rod = Diagonal of cuboid 32² 3²

8.

l ²b ²h ² =

=

1024 256144 9

3 x 4 According to question, Area = 192 m²

=

230412961024 9

3 x  x = 192 4

4 × side = 44 side

= 11 cm

area of square = (side)² = (11)² = 121cm² Circumference of circle = 44cm

a + b  ² 2

= (a + b)²

 a+b  ²

diagonal of square B = 11. (a)



2 a

+ b

5m 12 m

area of the rectangular garden = 12 × 5 = 60 m²  area of square = 60 (side)² = 60 side = 60 diagonal of the square

x = 644 = 16 cm d if fe re nc e of length

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2

= (a + b)

1924 = 64 × 4 3

breadth = x –

= 4 2 × 2 = 8 cm 10. (d) Diagonal of square A = (a + b) side of square Diagonal a+b = = 2 2

side of square B =

3 x ² = 192 4

x² =

32

= 4 2 Diagonal of new square

= 2×

=

2 68 4624 = = 22 m 3 9 3 (b) Perimeter of square = 44 cm

side of new square =

2 area of square B = 2 × area of square A

Diagonal = 10 m (c) Le t the le ng th of rectangular hall = x  Breadth of rectangular hall

=

=

4 2

=4 2 area of the square = 16 area of new square= 32 2

=

= 1010

=

16² 12² 

Diagonal

a + b area of square A =    2  a + b ²

3850  22 22 7 a² = 3025 m² (b) 2(l + b) = 28 l + b = 14 and l × b = 48 (l + b)² = l² + b² + 2lb (14)² = l² + b² + 48 × 2 196 - 96 = l2 + b2 l² + b² = 100 16a² 

256 4 = 32 m.

of

4a 4a × = 3850 2π 2π

R Enak geisnh

π×

3 x  x = 768 4 3 x² = 768 4 768 4 x² = = 256 × 4 3

5.

4a 2π

area of circle = 3850 m²

According to question, Area = 768 m²

4.

(b) Let the side of square = a and the radius of circle = r perimeter of square = circumference of circle 4a = 2 π r r=

Dif fe re nc e

=

Option (b) is the answer. (circle, 33 cm²)

a2   2 2 (a) Le t the le ng th of rectangular hall = x  Breadth of rectangular hall

x² =

9.

22 ×7×7 7 = 154 cm²

=

16 = 4 cm 4 (d) Side of the square =

area of circle =

5.2  5.2

2 2 = 2.6 × 5.2 = 13.52 cm² (c) Area of square

=

= 44

44 7 radius = 2 22 = 7 cm

 Diagonal 2     2  

Diagonal2

3.

2 π (radius)

Diagonal 2

r

(b) Side of a square =

eeYa ridn agv .iSn i

1.

3 x x = 4 4

=

2 side

=

2 × 60 =

120

= 2 30 m 12. (d) and

(a – 3) a

(a + 5)

322

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

 = 34 cm

x =

= 7

B

C

l

BD = length of diagonal = speed × time 52  15 = 13 m 60

ERna

= BD =

Again,

l²b²

4

68 60

 15 = 17

(l + b) = l² + b² + 2 l b 17² = 169 + 2 l b

= 8 cm

= 100 = 10 cm Hence perimeter of new square = 10 × 4= 40 cm 15. (d) (side)² = 484 cm² side = 22 cm perimeter of square = 4 × 22 = 88 cm According to question, 2 π r = 88 cm

l+b =

lb =

120 = 60 m² 2

19. (d) Let the breadth be = x m  length = (23 + x) m  2(x + 23 + x) = 206 4x = 206 – 46 x =

16 0 4

= 40 m

 length = 40 + 23 = 63 m  Required area = 63 × 40 = 2520 m² 20. (b) Length of rectangle = 48 m Breadth of rectangle =16 m According to question, Perimeter of square = Perimeter of rectangle = 2 (48 + 16) 4 × side = 2 × 64

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264 = 32 m 4

side =

 Area of the square = (side)² = (32)² =1024 21. (a)

a

b

4a = 40 4b = 32 a = 10 cm b = 8 cm area of third square = a² – b² = 10² – 8² = 100 – 64 = 36cm²

side of third square= 36 =6 cm perimeter of third square = 4 × 6 = 24 cm 22. (c) side of the square

perimeter 4 Sides of all five squares are =



b

wwM wa.th Les aBryn 32

So, area of this square = 8² = 64 cm² According to question, Area of new square = 64 + 36 = 100 cm² side of the new square

88 7 r = 2 22 = 14 cm

4

 l² + b² = 13² = 169

So, area of square = 6² = 36 cm² Again, side of square, whose perimeter is 32 cm =

28

The required length = x + 2 = 7 + 2 = 9 cm D 18. (d) A

=

24 = 6cm = 4

152 m

= 2 38 17. (b) Let the length of smaller line segment = x cm The le ng th of larg er line segment = (x + 2) cm According to question, (x + 2)² – x² = 32 x² + 4x + 4 – x² = 32

2

64  16 = 32 m 14. (b) Side of square , whos e perimeter is 24 cm



1003616 =

9

13. (a) According to question, 2(l + b) = 160 l+b = 80 ..... (i) l–b = 48 ....(ii) on solving (i) and (ii) l = 64, b = 16 are a of s quar e = ar ea of rectangle (side)² = 64 × 16 side

=

kgei snhe

 25 = 2  2

22  14  14 7 = 616 cm² 16. (a) l = 10 m, b = 6 m, h = 4m length of diagonal (longest rod)

area of circle=π r² =

v.iSn ir

to question, (a – 3) (a + 5) a² + 5a – 3a – 15 15 15 a = 2 15 25 Length = a + 5 = +5= 2 2 15 breadth = a – 3 = –3 2 15 – 6 9 = = 2 2 perimeter of the rectangle = 2 (l + b)

eYrai dnag

According a² = a² = 2a =

24 32 40 76 80 , , , , 4 4 4 4 4 = 6, 8, 10, 19, 20 ATQ area of another square = 6² + 8²+ 10²+19²+20² (side)² = 36 + 64 +100 +361 +400

=

side =

961 = 31

perimeter of square = 31×4 = 124 23. (c) Area of the tank = length × breadth = 180 × 120 = 21600 m² Total area of the circular plot = 40000 + 21600 = 61600 m²  area of circle = 61600 (radius)² = 61600 (radius)² = radius = =

61600 7 22 28 00  7

7  7  400 = 7 × 20

= 140m 24. (d) Let the breadth of rectangle =xm  length = (x +5) m  Area of hall = length × breadth 750 = (x + 5)x 750 = 30 × 25 (clearly 750 = 30 × 25)  x = 25, breadth = 25m length = 25 + 5 = 30 m

323

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27. (d) b l

a a

2(l +b) = 160m 4a = 160 l + b = 80m ......(i) a = 40 m ATQ a² – lb = 100 (40)² –lb = 100 1600 – lb = 100 lb = 1500 ....(ii) Clearly, 50 + 30 = 80 and 50 × 30 = 1500 length = 50 m

31. (a) a² = 121,a = 11  Perimeter of square = 11 × 4= 44 cm  Circum fere nce of circ le = 44 2r = 44 2×

22 7

× r = 44

32. (d) Let the no. of hours be x  (0.3 × 0.2 × 20000) × x = 200 × 150 × 8

x =

39. (d) Side of square =

=

 r = 7 cm

2

area of square = =

38 m

(70 – 2x)x=

600 2

= 300

70x – 2x² = 300 2x² – 70x + 300 = 0 x² – 35x + 150 = 0 x² – 30x – 5x + 150 = 0 x(x – 30) –5 (x – 30) = 0 (x – 30) (x – 5) =0 x = 30 not possible x = 5 (right) Alternate (l + b –2x) 2x = area of path = 600

Area of path = 200 × 220 – 200 × 180 = 44000 – 36000 = 8000 m² 34. (c) Diagonal of square = diameter of circle = 8 × 2 = 16 cm 16 = 8 2 cm  side of square = 2 2

 

 area of square = 8 2

35. (a) Side of square =

2

(32)² =

2

= 8 cm

1 2

1 (Diagonal)² 2

× 32 ×32 = 16 × 32

= 512 cm² area of triangle 3 1.73288 8 ² 4 4 = 1.732×2×8 =27.712 cm² Required are =(512+27.712)cm² = 539.712 cm² 41. (a) Area of the lawn

=

=

 Area of square = 8 × 8 = 64 cm² 36. (a) x² + 7x + 10 = x² + 5x + 2x + 10 = x(x + 5) + 2(x + 5) = (x+ 2) (x +5)

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1

= 128 cm² 8 2

2

15 2

2001508 32200

32m

area of path= 600 m² (l + b – 2x) 2x = 600 (38 + 32 – 2x)2x = 600 (70 – 2x)2x = 600

Diagonal

= 15 cm 2 area of square = (side)² = (15)² = 225 cm² 40. (a)

= 200 hrs. 33. (a)

x

wwM wa.th Les B aryn

28. (b)

36 = 7 cm 22 2 7

r =

r

Ratio of length and breadth =3:2 2(l + b) = 20 cm 2(3x + 2x) = 20 cm 2 ×5x = 20 cm 10x = 20 x=2  length = 3 × 2 = 6 cm, breadth = 2 × 2 = 4 cm area = length × breadth = 6 × 4 =24 cm²

 Two sides of rectangle = (x+ 2) (x +5)  Perimeter = 2(x + 2 + x + 5) = 2(2x + 7) = 4x + 14 37. (c) Let the sides of rectangle be 6 cm and 2 cm (or any other number)  Area of rectangle (Q) = 6 × 2 = 12 cm²  Side of square = 4 cm  Area of square(P) = 4 × 4 = 16 cm²  P>Q 38. (b) No. of cubes with no side is painted = (n–2)3 Where n is the side of the bigger cube Required number = (6–2)³ = 64

eeYa ridn agv .iSn i

2x 3x

take help from options to save your valuable time take option(b) x = 5m (38 + 32 – 2 × 5) 2 × 5 = (70 – 10) × 10= 60 × 10 = 600 29. (c) Area of walls = Perimeter of base × height = 18 × 3 = 54 m² 30. (d) a² = 81, a=9  Perimeter of square = 9 × 4 = 36 cm  2r +r = 36 r(2 + ) = 36

R Enak geisnh

25. (d) Required total area = Area of four walls + area of base = 2 × 1.25(6 + 4) + 6 × 4 = 49 m² 26. (a)

1 12

hectare

length × breadth = 4x × 3x =

1 12

× 10000m²

10000 m² 12

324

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2(15 × 12) = 2(15+12)× h 2 × 180 = 2 × 27 × h 180 20  m 27 3 Volume of the cuboid = l × b × h

20 = 15 × 12 × 3 = 60 × 20 = 1200 m³

20 ×

3 2

a = 3a²

46. (a) D

35 2 Required area of shaded portion

Radius of circle =

22 35 35   7 2 2 = 1225 – 962.5 = 262.5 m2

A

B

side of a square = AB =

2 a units

2 2 a

 Diameter = 2 a units

44 = 4, p=4 11  breadth = 4 × 4 = 16 m 44. (d) Area of the floor= 8 × 6 = 48 m² = 4800 dm² (1m =10 dm) Area of square tile = 4 × 4 = 16 dm²

kgei snhe

12 + b

4800 No. of tiles = = 300 16

=

in

100  100   = 100  100  % 100  

= 300% 49. (b) D

C 16 m

45. (b)

2

Shape of godown is cuboidcal length = 15 m , breadth = 12 m, height = h m Area of four walls = 2(l + b)× h area of floor = l × b area of ceiling = l × b ATQ l × b + l × b = 2(l + b)× h 2(l × b) = 2(l + b)× h

30 m

B

3–x 1 (x + 1) and d = 2 2

2a

3–x



2

=

2

=

30² + 16²

=

900256

1   2  x  1 

 x = 1 unit B

52. (a) A

l

C

Let ABCD is a rectangular carpet having length l metre and breadth b metre and BD is a diagonal  As we know ....(i)  Area = l × b = 120  Perimeter = 2(l + b) = 46 Using formula  (l + b)² = l 2 + b² + 2l b  (23)² = l 2 + b² + 2 × 120  529 = l 2 + b² + 240

 

AC = AB² + BC²

l 2 + b² = 529 – 240 l 2 + b² = 289

l 2 b 2 = 289 diagonal = 17 diagonal of carpet is17 metres 53. (c) Diagonal of a square



=

1156 = 34 metre. Distance travelled by elephant = 34 – 4 = 30 metre

speed of elephant =

Rakesh Yadav Readers Publication Pvt. Ltd.

 d=

D

Here, x = 100%, y = 100%

A

Here a =

2 side

b

xy   x  y  % 100  

area =

51. (b) Diagonal of square = of square

40 = 20 2

b = 20 – 12 = 8 m 48. (d) Pe rc entage inc re as e

wwM wa.th Les aBryn

p=

Circumference= π × diameter = π × 2a = 2 π a units. 47. (b) Perimeter of rectangle = 40 m Length = 12 metre  2(l + b) = 40 2 (12 + b) = 40

ERna



C

35

2

 AC = Diagonal =

Area of path = (l + b + 2x)2x where x = thickness of path Let l = 7p , b = 4p {7p + 4p + 2(4)}2(4)= 416 (11p + 8)8 = 416 11p + 8 = 52 11p = 44

D

  35  

a = 10 cm 43. (d)

35

According to the question,

C

.O

B

v.iSn ir

100 4 = 25 m 42. (a) Let the side of square = a cm ATQ l × b = 3a²

35

A 35

h=

100 Breadth = 3x = 3 × 12

=

50. (b)

eYrai dnag

10000 12 10000 x² = 12  12 100 x= 12

12x² =

30 15

= 2 m/s

= 6 2 cm Side of a square =

6 2 2

= 6 cm

Area of a square = 6 × 6 = 36 cm2

325

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=

x= x= Breadth =

4170.605 2

83521 400 289 cm 20

289 m 20

= 924 100 = 700 132

1  Area of sector = 2 lr 1  3.5  5 = 8.75 cm² 2 57. (c) Radius of circular wheel = 1.75 m Circ um fe re nc e of c ir cular wheel = 2 π r = 2 

22  1.75 m 7

No. of revolutions = =

r

Circumference = 2 π r Distance covered in 1 min

r 40

.

28 cm

14 cm

New circumference = 2 ×  × r × 10 Time taken =

Radius of the largest circle 1

  side of square  2 1  28 = 14 cm = 2 area of the circle = π (radius)²

=

=

22  14  14 = 616 cm² 7

.. r



2π r=

R 88

88 7 r = 2 22 = 14cm

2 π R = 132 cm 132 7 R = 2 22 = 21 cm

The area between two circles = π (21)² – π (14)² = π { 21² – 14² } = π (21+14)(21–14) =

22 35 7 = 770 cm² 7

61. (b)

.

7 cm

circumference of wheel = 2π r

11000 = = 1000 11

Rakesh Yadav Readers Publication Pvt. Ltd.

22 ×7 7 = 44cm = 2×

2r 10  40 2r  8

= 50 min

63. (b)

12 cm

15 cm

area of the triangle =

60. (b)

Distance to be covered Circumference of circle 11000m 22 2  1.75 m 7

.

= 2 × 8 × πr

59. (d)

wwM wa.th Les B aryn

289  2 = 28.90 m Length = 20 56. (b) Radius of circle = 5 cm Length of arc = 3.5 cm

22  21 cm = 132 cm 7 No. of revolutions Distance to be covered = Circumference of circle

= 2

 Total distance travelled by wheel in 15 revolutions = 15 × 44 cm = 660 cm 62. (c)

eeYa ridn agv .iSn i

x2

58. (d) Circumference of wheel = 2πr

R Enak geisnh

54. (c) Let the breadth of floor = x m The n the le ng th of f loor = (x+3) m A.T.Q, x × (x + 3) = 70 x2 + 3x – 70 = 0 x2 + 10x – 7x – 70 = 0 (x + 10) (x – 7) = 0 x = 7, x = – 10 Breadth = 7m Length = 10m Perimeter of floor = 2 (L + B) = 2 (10 + 7) = 34 m 55. (d) Let the breadth of rectangle =x m then the length of rectangle = 2x m A.T.Q, x × 2x = 417.605 2x2 = 417.605

1  base × height 2

1  15 × 12 2 = 90 cm² area of another triangle = 2 × 90 = 180 cm²

=

=

1  base × height = 180 2

=

1  20 × height = 180 2

height =

180  2 = 18 cm 20

64. (b)

area of the square = 81 cm² side of the square = 81 =9 cm perimeter of the square = 4 × 9 = 36 cm Now, According to question, π r + 2r = 36 r ( π + 2) = 36 r =

36 7 36 = 22 14 22 2 7

326

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22 72 r ² = × = 77 cm² 2 7 2 65. (b) Area of square = (12)² =144 cm² Area of triangle 1  base × height 2

=

1  12 × height 2

height = 66. (c)

144  2 = 24 cm 12

A

D

E

4

= 784 = 28m  diameter = 2r = 2 × 28 = 56 cm 69. (b) Re quire d are a =Ar ea of square – Area of circle (2a)² – π (a)² = 4a² –

=

1  12 × height = 144 2

3

2464  7 22

 r =

22 a² 7

28a² – 22a² 6a² = = 7 7 70. (c) Diameter of the circle = Side of square 2 r = 21 21 r = m 2

 3² + 4² = 5²  ABC is a right angled triangle 1  AB × BC 2

ar (ABC)=

1  3 × 4 = 6 cm² 2

22 21 21 693   = cm² 7 2 2 2

=

= 346

71. (a)

wwM wa.th Les aBryn

=

 21  Area = π r² = π   2

 Required Area of  DEF

B C  Are a of an equilate ral triangle = 400 3

1 ×6 4 3 = cm² 2

=

3

4

(side)² = 400 3

(side)² =

67. (b)

A

1 cm² 2 A

ERna

C

F

kgei snhe

2

B

3.5

3.5

3.5

C

3.5

3.5

3.5

7 cm

400 3 4 3

= 1600

side = 40 m perimeter = 3 × side = 3 × 40 = 120 m A 72. (c)

B

AB = BC = AC = 7 cm Area enclosed = Area of equilateral

 ABC –

1 2

 area of 1 circle 

1

B

D

F

1

1

1

2

2

2

x² =  x  3 +  x  2 3 +  x  5 3

1 ×x × 8 3 2 4 x = 16  perimeter of triangle = 3x = 3 × 16 = 48 cm Alternative:3

C

Let P be the point inside the equilateral  ABC

Rakesh Yadav Readers Publication Pvt. Ltd.

x² =

side of equilateral  =

PD = 3 , PE = 2 3 ,PF = 5 3

2

3 (sum of the altitudes draw from inte rnal p oint) 2

side =

=

3 2 3



3 2 35 3



 8 3 = 16 cm

perimeter =3 × sides = 3 × 16 = 48 cm

73. (c) Perimeter of  = 30 cm Area = 30 cm² Check the triplet

5, 12, 13 ,  3, 4, 5 whose largest side is 13. Also, 5² + 12² = 13² And perimeter = 5 + 12 + 13 = 30 cm Smallest side = 5 cm 74. (b) Diameter of the wheel = 3 m Circumference = π × diameter 22 66 3 = 7 7 Sinc e a whee l cove rs a distance e qual to its c ir cumf er ence in one revolution therefore distance covered in 28 revolutions

=

66 = 264 m 7 264 metres covered = 1 minutes

= 28 ×

1 metre covered =

P E

1

= 2 x  3 + 2 x 2 3 + 2  x 5 3

eYrai dnag

=

1

= 1.967 cm² 68. (a) π r² = 2464 cm²

=

=

and AB = BC = AC = x Ar (ABC)

 3 1 22 7 ² –  3.5 ²  7 4 2 

=

v.iSn ir

36  7 = = 7 36 area of the semi circle

1 minute 264

5280 metres covered =

5280 264

= 20 minutes 75. (b) distance covered = 2 km 26 decameters

327

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2260 = 20 m 113 Now, π × diameter = 20 =

20  7 22 70 4 = = 6 m 11 11 76. (a) Distance c ov er ed in 1 revolution = circumference of wheel

=2×

Circum radius

80. (c)

=

11  1000 = 1000 22 2  1.75 7

77. (b) Radius of circle circumference 100 = 2π 2π

=

=

Diagonal

100

=

=

50 2

2 2π π 78. (c) Let outer Radius = R and inner Radius = r 2 π R – 2 π r = 132 2 π (R – r) = 132

side 30 = = 15 cm 2 2

area of the circle 22 22 × (radius)² = ×(15)² 7 7

84. (b)

Circum radius of equilateral triangle =

side  3

=8

radius of in circle

side = 8 3 In radius of equilateral triangle side

=

8 3

60 3 (2)² – 3× × (1)² 360 4

1 3 ×4–3× × 6 4

=



=   3–



side 2 3

42

=

2 3

=

21 3

cm

area of incircle

= = 4 cm 2 3 2 3 81. (b) radius of each circle = 1 cm with all the three centres an equilateral triangle of side 1 cm is formed . area enclosed by coins = (area of equilateral triangle) – 3× (area of sector of angle 60°) =

=

   cm² 2

82. (b)

22

=

7

 21 2 22 2121     = 7  3  3

= 22 × 21 = 462 cm² 85. (a) r =20 cm

wheel of radius 20 cm no. of revolutions =

distance to cover circumference of wheel

=

176007 = 140 22220

86. (a)

132  7 2  22 = 21

R–r=

He nc e, wid th of path = 2 1 metres. 79. (b)

=

=

wwM wa.th Les B aryn

When a square is inscribed in the circ le , diag onal of the square is equal to diameter of the circle  Diagonal of square 100 100 =2× = 2π π  side of square

Perimeter 120 = = 30 cm 4 4 radius of the circle =

in radius

R Enak geisnh

22  1.75 m 7  Number of revolution

each

22  7 = 44 cm 7

diameter =

= 2×

side of the square

r

Total distance Number of revolutions

=

28 = = 7 cm 4  c ir cumf er ence of circular plate = 2 π r

eeYa ridn agv .iSn i

= (2 × 1000 + 26 ×10) (1 decameter = 10 meter) = 2260 m Distance covered in 1 revolution

.

M is the centre, BM = CM =r AM  BC,(AM = r) area of  ABC =

28 cm

.

83. (a)

1 r × 2r = r² 2

In an equilateral triangle side =

2 3

(P1 + P2 + P3)

87. (c)

side of square papersheet =

784 = 28 cm radius of each circle

Rakesh Yadav Readers Publication Pvt. Ltd.

328

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side of equilateral triangle

=

side 3

=

169 = 13 cm length of perpendicular,

100 1 50 2    2 2 2 2

AB×BC AC  length of perpendicular to hypotenuse to

= 25 2 cm²

4 3

=

=

=4

3

BD =

91. (a)

see the figure side of square = 2 × circum radius =2×4=8

Radius of circle = 6 cm Area of smallest circle

6²  = 12 3 Radius of smallest circle =

12  2 3 cm 

5 cm

x B

a

isosceles right triangle  x² + x² = 5² = 25 2x² = 25 x² =

B

C

3 a² = 4 3 4 a² = 4 × 4 a = 4 cm

25 2

Area of triangle

Circum radius =

a

ERna

1 1 25 = × x² = × 2 2 2 = 6.25 cm²



3 area of circle = r ²

wwM wa.th Les aBryn

89. (c)

4

3

3

2 3

P

1

× 24 =

3

48

3

×

3

3

Side = 16 3 cm area of triangle 3

3 (side)² = 4

=



 16 3

4

2



3 × 3× 16 × 16 4 = 192 3 cm² 96. (c)

=

2 16 2    3 cm 3

 =   

 P2 +P3 

(6 + 8 +10)

2

=

C

x

2

= =

A

92. (a)

125 60 8 = = 4 cm 13 13 13 95. (c) Side of equilateral triangle

eYrai dnag

A

kgei snhe

88. (b)

Hypotenuse

=

= Diagonal of square = 8 2 cm

perpendicular×Base

=

v.iSn ir

= 4 3 circumradius of triangle

a

4

2a

a

93. (b)

length of side =

=

=

2

3

42

2

3

(6 + 7 + 8) =

(P1 + P2 +P3)

2

3

2

Circumference – diameter = 30 cm 2r – 2r = 30 2r(– 1) = 30

× 21

r 

3

42 3   = 14 3 cm 3 3 3

30

 22 2  7

1 = (10)² × sin45° 2

 – 1  

=

30  7 2  15

= 7 cm

90. (c) Area of isosceles triangle

1 = a² sin( is angle between 2 equal sides)

area of shaded region = area of semicircle – area of triangle

r

=

a 1 – a2a 2 2

=

  a 2  – 1  sq units – a² = a²  2  2

97. (c) According to question (R + 1)² – R² = 22 {(R+1)2 – R²} = 22 (R +1 +R)(R +1– R) =

A

2R + 1 = 7 R = 3 cm

94. (b)

12 cm B

AC =

Rakesh Yadav Readers Publication Pvt. Ltd.

D 13 cm

5 cm

227 =7 22

98. (b)

C

12²5² = 14425

radius of largest circle

329

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=

2

4 2

= 4 cm

A

102. (a)

B

4

 a 2 a ² 2     3 – 12  = 44  

C

cone is rotated about 3 cm

8

2 3

4

=

2 3

side

3

 4 2   area of circle =     3 2216 22 44  = =16.76 21 7 3 Required area

=

3

=

4

2

8

22  16

–

21

3 × 64 – 16.76 4 = 16 3 –16.76 = 27.71 – 16.76 = 10.95 cm² 100. (c)

=

So, the triangle is right triangle

 a = 2 14

The cone so formed after rotating about Side AB. So, slant height of cone = 5 cm  Volume of cone =

1 × 3 r ²h

r = radius h = height  Volume of cone 1 22 = × × 4× 4 ×3 = 16cm³ 3 7 103. (c)

R Enak geisnh

=

Area of bounded region

wwM wa.th Les B aryn

3 1  22 – (1)² 4 2   2   3–  cm 2 

72

1 1 × 30 × 72 = × altitude × 72 2 2 altitude = 30 m 101. (a)

11  7 7  r = 22  2 = 4

=

=

 × r² 360

=

60 22 7 7 × × × 360 7 4 4

x

perimeter of triangle = 4 2 + 4 x + x + 2x = 4 2 + 4 2x + 2 x = 4 2 + 4

x=



= 4

 2  1

4

2 Hypotenuse =

P×b 86 24 = P+b = 86 = 7

2

77 29 =1 cm² 48 48 105. (c) Let the side of the triangle be ‘a’ cm

2x

Rakesh Yadav Readers Publication Pvt. Ltd.

8

cm 3 Radius of incircle 8 4 = = cm 2 3 3  Required area = (R² –r²)

=

 Circumradius =

 24 2  7    

576 cm² 49 108. (b) Radius of circumcircle

2x

2

= 14 3 cm² 106. (c) Side of square = diameter of the circle area of circle = r² = 9  r = 3 cm  Side of square = 3 × 2 = 6 cm  Area = 6 × 6 = 36 cm² 107. (d) The given triangle is a right angled triangle  side of the square

=

104. (a) 2r = 11

x



3 × 2 14 × 2 14 4

 Area of square =

Area of sector

x 2

 area=



78

30

4a ² – a ² 44  7 = = 14 12 22 3a ² = 14  12  a² = 56



eeYa ridn agv .iSn i

in-radius of circle (r) =

2 3

 a 2  a 2       = 44  –  2 3   3

5

3

a

and Inradius =

r

breadth 14 = = 7 cm 2 2 22 area = × 7 × 7 = 154 cm² 7 99. (b)

=

=

a 3

2



    22  8   4   =  3  –  3   7      

=

22 7

 64   3

–

16  3

  

22 2 × 16 = 50 cm² 7 7

330

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115. (b) 16 2

44  7

 r = 22  2 = 7 cm

Other sides =

22 ) = 36  r(2 + 7 36  7 =7m  r= 36

1  r(2 + ) = 2 r²  4 + 2 = r

3 a² = 4 3 4  a² = 16  a = 4 cm 117. (c) Side of hexagon







6 m 11 112. (a) The angles of the given triangle are 90°,30° and 60°

= 6

kgei snhe

4

  2   Diameter= 2 

176 2 = 28 m  Area of road = (28 + 7)² – (28)² = (35 + 28)(35 – 28)

118. (a) The radius of park =

22 × 7 × 63 = 1386 m² 7 119. (c) M A B 12 13 o

wwM wa.th Les aBryn

.

In  OMB MB = 13² – 12² = 5  AB = 5 × 2 = 10 cm 120. (b)

1  Area = 2 × 5 3 ×5

25 3 cm² 2 113. (c) Let the altitude = x cm

2 3

= 7 cm

22 × 7× 7 7 = 154 cm ² 123. (d) Radius of incircle

 Area = r² =

124. (c)

3 a² = 121 3 4  a = 22 cm

 3a = 66 cm

Circumference of circle = 66 cm 2r = 66 66  7 21 r = 2  22 = 2

21 21 22 × × 2 2 7 = 346.5 cm² 125. (b) Area grazed by the cow

Area = r² =

1 1 22 ×7×7 r² = × 2 2 7 = 77 m² =

=

10 P = =5 2 B = 5 3

14 3

=

3 3 3 3 = a² = ×4 2 2 = 6 3 cm²

4 +2 

=

6 = 3 cm 2 3 Area = r² = 3 cm²

116. (d)

ERna

 r =

1  Area = 2 × 16 × 16 = 128 cm²

Side of equilateral triangle = 3 = 2 cm Area of hexagon

r²

2

= 16 cm (as

eYrai dnag

22 = × 7 × 7 = 154 cm² 7 110. (a) 2r + r = 36  r(2 +) = 36

1

16 2

2 the isosceles  )

 Area = r²

111. (c) 2r + r =

Area of the shaded portion = 6 × 6 – (3)² = 36 – 9 = 9(4 – ) cm2 122. (c) Radius of incircle

v.iSn ir

109. (c) Side of square = 121 =11 cm Perimeter of square = Circumference of circle = 44 cm  2r = 44

26 + 30 + 28 = 42 2 Area of field S=

=

s s–a

=

42  16  14  12 = 336 m²

 s–bs–c

 Remaining area = 336 – 77 = 259 m² 126. (b)

=

1 × x × 8 = × 8² 2  64  x= 4  x =16 114. (d) The sides of the given triangle are 3,4 and 5 cm 1 area = × 3 × 4 = 6 cm² 2



Area of shaded portion = 8 × 8 – × 4² = 64 – 16 = 16(4 – ) cm² 121. (b)

Rakesh Yadav Readers Publication Pvt. Ltd.

As P and Q are mid-point and PQ||BC   APQ   ABC AP PQ 1 = = AB BC 2 BC  PQ = 2  BC = 2PQ = 2 × 5 = 10 cm



331

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127. (c)

r =

11  100  7 5  2  22 = 35 cm

135. (b) Length of rubber band = 3d + 2r = 30 + 10

3 25 3 (5)² = cm² 4 4

136. (c)

22  7

In  OMB OB =

15²8² = 17 cm OB = OD = radius In  OND

7

7 7 77 22 × × =  r² = 2 2 2 7 = 38.5 cm² 129. (b)

ND = 17² – 8² = 15 cm CD = 15 × 2 = 30 cm 137. (b) Perimeter = 2r +r

A

B

D F

3a =

E

C

AF = 5 3 In  ADF AD² = AF² + DF²

wwM wa.th Les B aryn

ar (  ABE) = ar(  ACD) = 36 cm² 131. (a) The third side will be either 15 or 22  Possible perimeter = 15 × 2 +22 = 52 and 22 × 2 + 15 = 59 132. (a) No. of revolutions =

=

Distance Circumference

1540100 = 500 22 98 2  7 2

133. (b) 2r =

440 1000

 10 2 5 –   AD² = 75 +  3  

10 7 AD = 3 139. (c) Let sides of triangle are a,b and c respectively  largest side given =17 cm Perimeter = a + b + c = 40 cm (given) area = 60 cm² (given) In such questions take the help of triplets which form right angle triangle

227  r = 50222 = .07

142. (a) Area of  =

143. (b) Area of

4 1

=

    9  12  3 2 

( 9,12,15 from triplet)

4 × 54 = 72 cm² 3 144. (a) 3x + 2y = 6 =

x y  =1 2 3

(Make R. H.S. equal to one)  Coordinates of  = (0,3),(2,0),(0,0)

 Area of

 =

1 ×3×2 2

= 3 square units 145. (a) Let each side of the triangle be a units 3

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4 (Areaof  3

 =

formed by median as side)

 So, here we have a side 17 cm

3 a² 4

3 × 36 = 9 3 cm2 4

=

 Diameter = .14 m 11000  100 134. (b) 2r = 5000

 3   a ² 3 4   

3 a 4 a = 4 units

In  AFB AF  BC AF² = AB² – FB² = 100 – 25

130. (b)

× 8 × 15  60

3=

R Enak geisnh

ar  BDOF = 2 × ar  AOE = 30 cm²

2

Area of shaded region = (4)² – (2)² = (16 – 4) cm² 141. (c) Let the side of the triangle be a  Perimeter = 3a

22 63 × 2 7 = 63 + 99 = 162 cm = 63 +

138. (c) ar (  AOE) = 15 cm²

1

Hence sides are 15,8. smaller side = 8 cm. 140. (b)

128. (a) 2r = 22

 r = 22  2 = 2

area =

r

  APQ =

= 8+15+ 17 = 40

eeYa ridn agv .iSn i

As PQ||BC   APQ   ABC   APQ is also an equilateral 

 by triplet we get sides 8 and 15  check the sides perimeter

4

((a+ 2)²– a²) = 3  3

1 (a² + 4 + 4a – a²) = 1 3 4

332

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 r² = 169

1 + a = 1 3 3 units

9  10  11 146. (c) S = = 15 2

=

s s–a

 s–bs – c 

2 4a ² – 4 = 4 4

 Area = =

4a ² – 4 = 8 4a² – 4 = 64 a² – 1 = 16 a² = 17

3 a= 6 3 2 a = 12 cm  Perimeter = 12 × 3 = 36 cm 158. (a) Area of equilateral 

=

 72 2

22  7  7 = 77 m² 72

153. (b) side of square= =

area

2 m = Diameter of circle  Radius of circle

a =

215 = 13 cm 2

1 2 = m 2 2

2

22  1  ×   Area =    7  2

=

16 16 – 13

=

16  3  8  5 = 8 30 cm²

 m² 2

6

area = s s  a  . s  bs  c   18  8  4 6  24 6cm 2

155. (b)

=

22  7 =7 22

2r + 1 = 7 r = 3 cm 151. (a) Area of two circles = (5² + 12²) = 169 cm²

12 = 504 2 No. of revolution by big gear

= 84×

504 2 9 = 28 160. (b) Perimeter of semi-circle = 2r + r = r(2 +)  r(2 +) = 18

8

10  14  12 S  18 2

3 a² 4

r² + 1 + 2r – r² =

3 a² = 4 3 4 a² = 16 a=4  Perimeter = 4 × 3 = 12 cm 159. (a) Distance covered by small gear = 2r × 42

Area (  ABDE) = 3 × ar(  AEC) 3 (2)² (ADC is equilateral 4 triangle)

=3×

= 3 3 square units

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187 18 = 22 36 2 7

r =

Side of  ABC = 10, 14, 12

16 – 816 – 11

3 × (2)² = 3 cm² 4 150. (a) Let the original radius = r cm  ((r+ 1)² – r²) = 22

8

=

6

4

Sides of the  = 11,8,13 cm 13  8 11 S= = 16 2  area

4

154. (d)

wwM wa.th Les aBryn

21 – 5 b= = 8 cm 2

=

ERna

17 units 148. (a) Sum of other two sides (a + b) = 32 – 11 = 21 and a – b = 5

149. (b) Area of  =

  = 36 

r=7m

1800 = 30 2 cm² 147. (c) Let the length of each equal side be a unit



7

 36   = 36  7

15  6 5 4

 a =



 r

=



22 

1 × 24 × 7 = 84 cm² 2 Perimeter = 7 + 24 + 25 = 56 cm  Hypotenuse = 25 cm 157.(c) Length of median

area =

eYrai dnag

Area =



r 2 

using hero’s formula

kgei snhe

a =

r² = 169 r = 13 cm  Radius of third circle = 13 cm 152. (d) Let the radius of the semicircle be = r  2r + r = 36 r(2 +) = 36

156. (a) Check tripletes 3,4,5 6,8,10 7,24,25  7, 24, 25 fulfill the given conditions

v.iSn ir

1 (4 + 4a) = 1 3 4

=

7 2

cm = 3

1 2

cm

161. (d) Perimeter of circle = 2r = 2(18 + 26) = 88 cm r = 44 cm r = 14 cm 22 × 14 × 14  Area of circle = 7 = 616 cm² 162. (a) Area of a circle = 38.5 cm² π r² = 38.5 38.5  7 r² = 22 7 r= cm 2 Circumference of a circle = 2 π r = 2

22 7  = 22cm 7 2

333

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163. (c) Diameter of circle =

= 12cm

2

12 Radius of circle = = 6 cm 2 Rad ius of c ircumcir cle of a equilateral  = 3  a = Radius × 3 = 6 3 cm 164. (c)

tor, altitude is equal to

3 2

8 ² x ²

168. (c)

360 36 = 250 25

=

3x + 4y = 12 8  x

3x 4y  =1 12 12

1 ×4×3 2 = 6 sq units 165. (c) Height of equilateral  = 15 cm

40 20 2   6 cm 6 3 3 Note: The ratio of areas of two similar triangles is equal to the ratio of square of their corresponding sides 169. (d)

x=

wwM wa.th Les B aryn

area of  OAB =

AB = AC =

3 (side) = 15 2 152 side = 3

area =

3 (side)² 4

3

    4 3 

225  4 3

= 75 3 cm²

166. (b)

4

side ²  9

3

(side)² = 9 × 4 = 36 side =

36 = 6 cm

length of median =

6

3 2

side

Let AB = AC = a cm BD = DC = b cm  Altitude of isosceles triangle is also median In right ADC AD² = a² – b² 64 = a² – b² .........(i) Perimeter = 64 a + a + 2b = 64 2a+ 2b = 64 a + b = 32 .........(ii) a ² – b ² 64  On dividing =2 a b 32 [a² – b² = (a + b) (a–b)] a–b =2  a + b = 32 On solving a = 17 , b = 15

5BC 6BC 5BC = 544 6 16BC = 544 6 5446 BC = = 204 16 5  AB = AC = 204 = 170 cm 6 b 4a ² – b ² Area of  ABC = 4  where a = equal side b = base

3 = × 6 = 3 3 cm 2

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=

204 4

area of  ABC = =

1 × AD × BC 2

1 × 8 × 30 = 120 cm² 2

171. (a)

BC

5 5 BC + BC + BC = 544 6 6

2

3

5

AB + BC + AC = 544

3 15  2 

 =  4 

36 6  25 5

R Enak geisnh

x y  =1 4 3  Coordinates of point A = (0,3) point B = (4,0)

170. (d)

sides 167. (c) clearly, 12 cm , 16 cm and 20 cm from a triplet

They from a right triangle, 1 area of triangle = × 16 × 12 2 = 96 cm²

 Divide by 12 on both sides make R.H.S = 1

=51 × 73984 = 51 × 272 = 13872 cm²

r

2

12 2

eeYa ridn agv .iSn i

=

Diagonal

Note: In an equilateral triangle, length of median, angle bisec-

x = AB = b + c y = BC = a + b z = AC = a + c  semi-perimeter(s) AB BC  AC 2a 2b 2c  = 2 2 = a+b+c area of  ABC 

s s – x s – y s – z 

a + b + c abc 172. (d)  R ²  10 ² 24 ² =

R² = 10² + 24² = 100 + 576 R=

676 = 26 cm

173. (c)

  ² – 204 ²

4 170

= 51 11560 – 41616

334

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Let the side of an equilateral triangle = 'a' and the side of square = 'b' In-circle radius of equilateral

176. (b)

AB ² DE ²

DE² =

4525 225 = 20 4

a

a

= 3 2 3 Now, Diagonal a a   b= 2 3 6

DE =

225 4

2

Let radius of outer circle = R and radius of inner circle = r ATQ 2R – 2r = 66

2

=

66  7 2  22

=

21 2

width = 10.5 m 180. (d) Distance covered in 30 seconds

eYrai dnag

30 = 15 m 60 This is the difference of distance of the boundary and the diameter Let 'R' be the radius

= 30 m/min×

3 3  3 3 :2 2 174. (c) Let the side of equilateral triangle = s

=

s²

AB = AC tangents drawn from the same point equal OB = OC = 3 cm OA = 12 cm

kgei snhe

3 4

ABO  ACO = 90° In Right ABO

135

ERna

AB  12² – 3² = 3 2

15  9 = 3 15 ar (ABOC) = 2× ar(ABO) 1 = 2× × AB × OB 2

wwM wa.th Les aBryn

3 s 2

= 3 15 × 3

2

 3   3 s s² 2  b²  3     4  = a 3 3 3 s² s² 4 4

2R – 2R = 15

=

height of equilateral triangle =

2R  – 1 = 15



R=

3²  ar DEF  4²

54 9  ar DEF  16

(radius is  tangent) and PQ = PR (tangent drawn from same point are equal) PQ =

OP ² – OQ ²  13² – 5²

= 12 ar (PQOR) = 2 × ar(PQO)

1654 =96 cm² 9

Rakesh Yadav Readers Publication Pvt. Ltd.

= 3.5 m

2R  – 1 = X

X 2R = –1

OQP  ORP = 90°

ar ABC 

7 2

181. (a) Perimeter of the circle = circumference of circle Let 'R' be the radius ATQ 2R – 2R = X

3

 ABC   DEF



15 15 15  7 22 2R = = – 1 = 15 = 7 –1 7

= 9 15 cm² 178. (b)

175. (c)

ar DEF  =

66

R–r=

177. (c)

3 a² 3 6 4  a²  = 2 4 a² a       6

area of equilateral  =

179. (a)

2(R – r) = 66

15 = = 7.5 cm 2

Required ratio

= 12 × 5 = 60 cm²

v.iSn ir

= 2

ar DEF 



20 25  45 DE ²

a

 = 2 3  Diagonal of square

ar ABC 

=2×

X

Diameter= π – 1  2R = diameter of the circle 182. (b) a 3

1 × PQ × OQ 2

335

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a ² = 3 3 a² = 9, a=3 perimeter = 3 × a = 3 × 3 = 9 cm 183. (a) r

72° r

887360 r= = 70 m 72222

184. (a)

=

=

=

area of triangle semiperimeter of triangle

=

54 54  2  9  15  12 36 2

= 3 cm

Alternate: In a right triangle, with, P, B and H incircle radius =

P B – H 2

Hence, r =

186. (a) o

6 = 3 cm 2 Also Circumcircle radius

=

8 15 M15

B

= 1: 2

area of  ABC  =

3  4a ² = 4

H 15  = 7.5 cm 2 2

3 a²

=3×

60 360

× a ² =

=

Rakesh Yadav Readers Publication Pvt. Ltd.

Diagonal a 2 = 2 2

a ² 2

area of shaded region = area of  AB C – area of 3 sector a ² 2  2 3 –    =  2  a² cm²   190. (c) A

=

3a ² –

A'

B

B'

C

In  ABC and  A'B'C C  =  C (common)

 B' =  B ( AB||A'B')   ABC   A'B'C 

area of A'B'C  B'C  area of ABC =  BC 

2

2

1 1 =   = 4 2

Let the side of square = a a In circle radius of square = 2 Circumcircle radius of square

3 2a  ² 4

area of 3 sectors ( = 60°)

188. (a)

r

A

2

912 – 15 2

1  AB  BC 2

1 = 1520 = 150 2 Cost of sowing seeds = 150 × ` 5 = ` 750

1

189. (d)

=

 15,20,25 form a triplet Clearly, 25² = 15² + 20² ABC is a right triangle Area of Right ABC

a 2 a 2 2

1

area of  A B C = × 9 × 12 2 = 54 cm² In-circle radius of triangle

wwM wa.th Les B aryn

In ABC perimeter of  ABC = (AB + BC + AC) = 2(3.5 + 4.5 + 5.5) = (13.5)×2 = 27 185. (c)

Since, 9,12,15 forms a triplet

R Enak geisnh

72 22 2 r = 88 360 7

Incircle radius = Circumcircle radius

Hence ABC is an equilateral triangle AB = BC = AC = '2a' cm

88

 2r Length of arc = 360



r

 a 2  area of circumcircle =    3

AB = 30 cm OM  AB and OM = 8  AM = BM = 15 cm In Right OMA OA² = OM² + AM² OA² = 15² + 8² OA² = 289 OA = 28 9 OA = 17 cm Radius of circle = 17 cm 187. (c)

eeYa ridn agv .iSn i



Let the side of an equilateral triangle = 'a' a Circumcircle radius = 3

 ar  A'B'C =

1 (area  ABC) 4

191. (b) Perimeter of square = 44 cm  44  Area of square =    4  = 121 cm²

2

336

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194. (a)

Now, In  ADC,

.

C B

 AC B

=

90°

(angle

in

semi-circle) AC : BC = 3 : 4 AB² =

area of  ADC 24 cm

Area of equilateral triangle =

3  side  ² = 4

=

3 × 6 × 24

= 144 3 cm² = 144 × 1.732 = 249.408 cm² Inradius of an equilateral triangle =

AC 2 +BC 2 =

3²4²

= 5 units 5 units = 5 cm

24

=

= 4 3 cm 2 3 Now, Area of incircle

1

 ar  ABC = 2 × 3× 4 = 6 cm² 193. (a)

=

22   Inradius  ² 7

=

22 4 3 4 3 7

A

22  16  3 1056 = 7 7 = 150.86 cm² Area of remaining part = area

N B

M

of Δ – area of incircle = 249.408 – 150.86 = 98.548 cm² 195. (d) A 25

wwM wa.th Les aBryn

C

ERna

=

6

 24  24

4

 side    2 3 

8

1 Ar(  ABC)= × 6 × 8 = 24 cm² 2

D

 ABC   MCN  C = C

M = B

ar (CMN)

32

2

B

1

1

=   = 4  2

ar (  MNAB)

= ar(  ABC) – ar(  CMN) =4–1=3

24

C

 ABC = 90º

AC =

2

25

( MN||AB)

 CM   ar ( ABC) =  BC 

AB² + BC²

=

32² + 24²

=

1024576

=

1 600 = 40 m

Now, area of  ABC

24 ×3  ar (  MNAB) = 4 = 18 cm²

Rakesh Yadav Readers Publication Pvt. Ltd.

=

s(s–a)(s–b)(s–c)

= 45(4 5 – 25)(4 5 – 25 )(45 – 40 )

3

kgei snhe

A

252540 = 45 m 2

s =

=

45×20×20×5

= 20 × 3 × 5 = 300 m² Area of the plot = 384 + 300 = 684 m² 196. (c) A

eYrai dnag

22 × (7)² 7 = 154 cm² Required difference = 154 – 121= 33 cm²

 area of circle =  r² =

192. (d)

24 cm

24 cm

22  7 r= = 7 cm 22

v.iSn ir

Circumference of circle = 2  r = 44 cm

=

1 × AB × BC 2

=

1 × 32 × 24 = 384cm² 2

H

M

a

P

B

C

b

Length of perpendicular drawn f rom the right angle a×b hypotenuse, P = H a² b² P² = H² a² b² P² = a² + b² ( H² = a² + b²) 197. (a)

.

A

8

4

.B

Diameter of the circle AB = 8 + 4 = 12 units 12 = 6 units 2  Area of circle = π r² = π ×(6)² = 36 π sq. units

Radius =

3 side  = 3 side  ² 2 4 side = 2 units 199. (a) Let the length of side of square = a then the diameter of circle = d According to question, a= d

198. (a)



area of square = area of circle

=

a²  d²  π   4

a² × 4 a² × 4 = π d² π a²

337

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15 18  21 = 27 2  Area of  = s (s – a)(s – b)(s – c)

S=

3 side  2

equilateral triangle =

Length of median, altitude, and angle bisector is

3

=

3 4

 perimeter of rectangle = 352 2 (l + b) = 352

= 12 cm

 area of  ABC =

3 4

l+b=

a²

3  3  12 = 36 3 cm²

3 side  ² 4

7  176 16 = 77 cm 206. (c) p er im eter of equilate ral triangle = 18 cm 3 × side = 18 cm

side =

=

3 side 2

3  6 = 3 3 cm 2

wwM wa.th Les B aryn

207. (a)

1

. r. . .r .

1 2   1 side =  64 (3)2  = 8 (3)4   203. (b)

.r . R

18 = 6cm 3

length of median =

3  side  ² = 48 4 48  4 (side)² = = 64 3 3

.

352 = 176 2

 smaller side =

×12×12

201. (b) π r² = 2 π r r = 2 units  Area of circle = π (2)² = 4 π sq. units 202. (d) Area of equilateral triangle =

22  112 = 352 cm 7

R Enak geisnh

6 3 2

a=

=

Circumference of paper sheet = 352 2 π R = 352

.

33 33 33 7 (R – r) = = 2 22 2π 3 7 21 = 2 2 = 4 thickness = 5.25 m 204. (b) Ratio = 5 : 6 : 7 sum of sides = perimeter = 18 sides, 5 6 54 = 15 54 = 18 18 18

R=

2πR – 2πr =

r=



352 352  7 = 2  22 2 π

= 56 cm R 56 = = 28 cm 2 2

circumference of circular plate = 2πr = 2

22  28 7

= 176 cm 208. (a) Inradius of triangle =

Rakesh Yadav Readers Publication Pvt. Ltd.

 22  d – 1 = 150 7   15 = 150 7 150 7 d= = 70 15 d 70 = Radius = 2 2 = 35m 210. (b) Let radius of circle = r Side of square = a d×

eeYa ridn agv .iSn i

3 a = 6 3 2



27  12  9  6

= 54 6 m² 205. (a) circumference of circle = π × diameter

3 side  = 2

=

=

side 3 = 2 3 side = 6 cm perimeter of equilateral triangle = 3 × 6 = 18 cm 209. (b) Circumference of circle = π d  π d – d = 150 d( π – 1) = 150

r

7  54 = 21 metres 18

4×7 4 14 = = 22 π 11  14 : 11 200. (d) Length of median of an

=

Side of equilateral Δ = b According to question, 2 π R = 4a = 3b πR 2 b= πR 2 3 Ratio of their areas:

 a=

π R² : π R²

:

:

a² π R   2   

2

:

3 b² 4

3 2   π R  4 3

2

π 3 π : 9 4 C : S : T Here, we can see that C > S > T Quic ke r Ap pr oach : Whe n p er im eter of two or m or e figure s ar e same the n the figure who has more vertex is greater in the area. Since, circle has infinite vertex. Therefore, C>S>T 211. (d) Distance c ov er ed in 1 revolution = Circumference of circular field = 2 π r Distance = speed × time 1

:

5 s = 165 m 2  2 π r = 165

= 66 m/s ×

2

equilate ral

22  r = 165 7

165  7 2  22

side

r=

2 3

= 26.25 m.

338

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212. (c ) Circ um fe re nc e of f ront wheel × no. of its revolutions = circumference of rear wheel × no. of its revolutions 2 π x × n = 2 π y × m (let ‘m’ is the revolution of rear wheel)

r = 1cm

3

A

R=

=

12 5 2  

 

2

=

B



   

4 25

3 a = 12 3 (given) 2

a = 24 cm Then area of an equilateral triangle =

3 2 a 4

3  24  24 4 = 144 3 cm²

=

215. (d) Let a triangle ABC has sides of measurement 3 cm, 4cm and 5 cm using triplets (3, 4, 5)   ABC will be a right angled triangle  Inner radius of circle C1 =

Outer radius of Pool with concrete wall = (R + 4) According to question 11 =  (R + 4)2 –  R2 25 11 R2 × = R2 + 16 + 8R – R2 25 11 2 25 R = 16 + 8R R 2

11R2 – 200R – 400 = 0 By option (d), (In such type of equation go through the option to save your valuable time) R = 20 11 × (20)2 – 200 × 20 – 400 = 0 4400 – 4000 – 400 = 0 0 = 0 (satisfy) Radius of pool R = 20 cm 217. (a) Area of circle = A Radius of circle = r Circumference of circle = c

Rakesh Yadav Readers Publication Pvt. Ltd.

C

1 × 4x  3x  7776 2

6 x² = 7776 x² = 1296

l r

R

AB + BC – CA = 43–5 2 2

4x

x = 36 Perimeter of triangle = 3x + 4x + 5x = 12 x = 12 × 36 = 432 cm 219.(b)

R+4

ERna

3 a 2

5x

Area of right angled triangle = 7776

wwM wa.th Les aBryn

angle =



5 = 2.5 cm 2

Area of C1 πr2 = Area of C2 πR2



C

We know that in an equilateral triangle a median also be a altitude Altitude of an equilateral tri-

A

v.iSn ir

In a right angled triangle half of hypotenuse is circum radius

216. (d) Let the radius of Swimming Pool = R

a

rc = 2A 218. (d)

3x

G



C2 C

kgei snhe

220000 = = 1250 176

a

4

r A = 2 C

eYrai dnag

22  56 = 176 cm 7 Total distance = 2.2 km = (2.2 ×1000 × 100) cm = 22,0000 cm  Number of revolutions

B

5 C1

Hypotenuse R= 2

=

.

.

 Circum-radius of circle C2

213. (b) Distance to be covered in one revolution = Circumference of wheel = π × diameter

a

 r² A = 2 r C

B

nx m= y

214. (b)

From (i) ÷ (ii)

A

 r² = A

(i)

2 r = c

(ii)

O

90º r

According to the figure,  Perimeter = r + r + l 75 cm = 2r + length of arc

2r 4 22  r 75 cm = 2r + 7  2 r = 21 cm.  Its area 75 cm = 2r +

1  22   21  21  4 7  = 346.5 cm2 A 220. (b) =

8

8 O

B

D 8

C

According to the question, Here OD = radius, a 8 r = = 2 3 2 3 4 r  3

339

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Required area of shaded portion 

224. (c) According to the questions,

229. (b) According to the question,

 4 2 3 2  8    4  3

A 75º

r = 84cm 

3 4

 64  

1

16 3

15º B

22 16  7 3 = 10.95 m2 = 11 m2  3  16 

r

D

4

=

A O

a=?

45° 3

E According to the question,

Area of sector OED = r2 ×

 360

45 = 2m2 360 Area of the sector OAB

= ×4×4×

 360

= × 3 × 3 ×

45 360

9 m2 8 So, Area of shaded portion = Area of OED – Area of OAB

=

9 16 – 9 = 8 8

22  84 = 4a 7 132 cm = a 225. (a) Area of circle = 324  cm2  r2 = 324  r = 18cm Longest chord = diameter = 2r = 2×18 = 36 cm 226. (c) Circumference of a  = 24 cm a + b + c = 24 cm

2×

or S =

7 7 22 11 2  =  m = 4 8 8 7 222. (d) According to the question, Circumference of a circle = 2r 30 2r =  15 r 2  30 D  2r  2  223. (b) According to the question

=

1 227. (b) Area of  = ab sin θ 2

=

1 × 10 × 10 × sin45º 2

= 25 2 cm² 228. (a) According to the question r=

 S

50  25 2

inner radius

The breadth of the path = (R – r) = (364– 350)cm = 14 cm

Rakesh Yadav Readers Publication Pvt. Ltd.

a

a

B

C

b

Let AB = AC = a BC = b.

 S=

a a b 2

S = a+ Area =

b 2

S(S–a)(S–b)(S–c)

b  b b b      a+ 2   a+ 2 –a   a+ 2 –a   a+ 2 –b      

Area =

Area

= Semi-perimeter 6=

A

Area =

semiperimeter =

R = 364 cm r = 350cm

1

Area of ABC = × AB × BC 2 1 = × sin15º cos15º 2 1 = × sin2 × 15 4 [ sin2  = 2sin  cos  ] 1 = × sin30º 4 1 1 1 = × = m² 2 4 8 1 = × 100 × 100 8 = 1250 cm² 230. (b) According to the question,

a+b+c = 12 cm 2

Circumference of incircle 2  r (inner) = 44 cm r (inner) = 7 cm Area of  = S × r (inner) = 12 × 7 = 84 cm²

wwM wa.th Les B aryn

= 2–

Let the length of side of the square be a cm (circumference of circle = perimeter square) 2  r = 4a

R Enak geisnh

B

= r2×

P AB = H 1 AB = sin15º B BC Cos15º = = H 1 BC = Cos15º

Sin15º =

eeYa ridn agv .iSn i

221. (d)

C

Area 25

Area = 150 cm²

b  b  b  b   a+ 2   2   2   a – 2      

b² 4

Area =

b 2

a²–

Area =

b 4

4a²–b² sq. units.

340

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231. (c) As we know circumcentre always made by the intersection of half altitude  In obtuse angle it will always be out.

Then, Perimeter = 2 d d2



10 cm

11 7  2r



1 2r

=

OB =

10² – 6²

=

22 7

100 – 36

=

2 1

B

4 × side of rhombus = 40 m side of rhombus = 10 m Sinc e rhom bus is als o a parallelog ram ther ef or e its area = base × height = 10 × 5 = 50 m² C 238. (d) D

2 2 Area of circle =  r²

234. (a) We know that rhombus is parallelogram whose all four sides are equal and its diagonals bisect each other at 90º. B

A

D

 AB =

8

16

12

C

16²  12²

= 400 = 20cm = side of rhombus  perimeter of the rhombus = 20 × 4 = 80cm 235. (d) If d1 and d2 are the lengths of diagonals of rhombus.

O 10

OB =

36 = 6 cm Diagonal BD= 2 × OB = 2 × 6 = 12 cm Alternative Side of rhombus =

Rakesh Yadav Readers Publication Pvt. Ltd.

3 a² 4

= 6

2 3 × 2 3 4

1 2

1 2

2

d1

162 d 2

 d22

 

3  12 = 18 3 cm² 4 241. (a) area of hexagon

B

Therefore In right  OAB OB² = AB² – OA² =10² – 8² = 100 – 64 = 36

10 =

= 6

=6×

3 (side)² 4

=6×

3 (1)² 4

=6×

3 3 3 = cm² 2 4

10

Perimeter of Rhombus = 40 cm 4 × side = 40 side = 10 cm We k now that d iagonals of rhombus bisect each other at right angle,

3 (side)² 4

= 6

8

A

 ×  2 2  = 8  cm²

= 6

C

r=

2

A regular hexagon consists of 6 equilateral triangle ar ea of regular hexagon

5m

wwM wa.th Les aBryn

2 side = r

16

240. (a)

64 = 8cm Diagonal BD = 8 × 2 = 16 cm. 237. (b) A D

1 m  R = 4 233. (b) Given:  Area of square = 4 side² = 4 side = 2  Diagonal of square = radius of circle

12

B

10 cm

In  AOB,

22 7

=

=

A

10 cm

cm O

kgei snhe



6

C

cm

6

ERna



10 cm

D

2r 22  2r = 7

4 1 7 2r

40 = 10 cm 4

side =

144 = 12 cm

v.iSn ir

= 2  26 = 52 cm 236. (c) 4 × side = 40 cm(given)

232. (a) According to the question, 2r  circumference 2r  Diameter



d2 =

239. (b) Diagonal (d1) = 10 cm area of Rhombus = 150 cm² 1 ×d1 × d2 = 150 2 1 × 10 × d2 = 150 2 150 2 d2 = 10 = 30 cm

eYrai dnag

D  Circum center



256 + d22 = 400 d22 = 400 – 256 = 144

= 2 676 B



2

= 2 24²10²

A

C

256d 2 2

20 = 2 1

242. (a)

A

D 10

B

6.5

C

(  Rhombus is a ||gm  area of Rhombus = base × height ) area of Rhombus = base × height = 6.5 × 10 = 65 cm² Also area of Rhombus =

1 ×d1 × d2 2

341

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1 × 26 ×d2 = 65 2 13 × d2 = 65 d2 = 5cm A

B

60°

 d2 =

=

1

× d1 × d2

2 1

× 14 × 48 = 336 cm²

2

245. (d) Let the parallel sides be 3x and 2x 1 (3x + 2x)× 15 = 450 2

5x = 60 x = 12  Sum of length of parallel sides = (3 + 2)×12 = 60 cm 246. (c)

Using Hero’s formula 15720 = 21 cm 2

Area of  ABC 21 21 – 20

21 – 721 – 15



= 21114  6 = 42 cm²

 Area of  ABCD = 42 × 2 = 84 cm² 5 247. (b) 20

r

100 = 25 4

a =5  Side of square = 5 cm  Area of square = 25 cm²

as  ABCD is a rhombus   ABC is an equilateral 

 ar (  ABC) =

3 × 10 × 10 4

253. (b)

= 25 3 cm²

 ar (  ABCD) = 25 3 × 2

Area (  ABDE)= 3 × ar(  ADC) (ADC is an equilateral triangle)

= 50 3 cm²

249. (b)

= 3

AB = 24 cm AD = 16 cm AE = 10 cm (Given) Area of Parallelogram = AE × DC = 10 × 24 = 240 cm² also, area of Parallelogram = FC × AD = 240 FC × 16 = 240 FC = 15 Distance between AD and BC  = 15 cm

251. (d)

4

Area of parallelogram

Rakesh Yadav Readers Publication Pvt. Ltd.

3 2 2 = 3 3 unit2 4

254. (d) side of rhombus =

Area of parallelogram = AD × FC = 15 × 12 = 180 cm² Area of parallelogram = DC × AE = 180 18 × AE = 180 AE = 10 cm  Distance between bigger sides = 10 cm 250. (a)

wwM wa.th Les B aryn



=

4a² = 8² + 6² a² =

2304 = 48 cm

 Area =

S=

4a ²  d12 +d2 2

C

100 = = 25 cm 4 we know that in a rhombus 4a² = d1² + d2²  d2² = 4×(25)² – (14)² = 2500 – 196 = 2304



= AE × DC = CF × AD AE × 36 = 12 × 27 = AE = 9 cm  Distance between bigger sides = 9 cm 252. (a) In a rhombus

eeYa ridn agv .iSn i

D

In the above figure  ADC is an equilateral triangle (as AC is angle bisector)  AC = 10 cm (smaller diagonal) 244. (c) Side of rhombus

R Enak geisnh

243. (a)

Area of parallelogram =AB×AE 5x × 20= 1000 x = 10 Area of parallelogram =AD × AF 1000 = 4x ×AF 1000 = 4 × 10 × AF AF = 25 cm (smaller side altitude) 248. (b)

A

20 = 5 cm 4

D o 4

B

AC = 8 cm C

5

OC = 4 cm In Right  OBC OB² = BC² – OC² = 5² – 4² = 9 OB =

9 = 3 cm

BD = 2 × OB= 2 × 3 = 6 cm area of Rhombus 1 1 × AC × BD = × 8 × 6 = 24 cm² 2 2 Note: In the question do not get confused with the wor ds non–square its simply to clear that it is Rhombus 255. (c) A

=

D

o

B

C

side of Rhombus =

100 = 25 cm 4

BD= 40 cm OB = 20 cm In right  OBC OC² = BC² – OB²

342

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Alternate:-

25x² = 6400

OC =

25² – 20² = 15 cm  AC = 2 × OC = 2 × 15 = 30 cm 1 Area = × BD × AC 2 1 = × 40 × 30 2 = 600 cm² 256. (d)

x² =

6400 25

80 6400 = = 16 5 25  AC = 4x = 4 × 16 = 64 BD = 3x = 3 × 16 = 48

x=

1 × AC × BD 2

1 × 64 × 48 2 = 1536 cm² =

 DE || AC

A

D

2²

4



5²

25

In ΔABC & ΔDCQ

ABC =DCQ ACB =DQC

ar ACED

  21 = = 21 : 4 ar BDE 4

ar (  APQ) =

ABC  DCQ

1 × d1×d2 = 216 2

ar (ABC )  ar (DCQ ) O

 AB = 2 CD AB 2  CD 1



4

1

= 4:1

(  AOB   COD ) 40

D

C

40 2

o

3x 2

40

40

40

1 ar( POQC ) = × 24 = 6 4 1  area (  PQC) = × 6 = 3 2 ar (  PQC)= 3

ar (QMAD) =

1 × 24 = 12 2

ar (  QAD) =

1 × 12 = 6 2

40

A

40

Let AC = 4x

B

and BD = 3x

 OA = 2x and

d2 =

area of ABCD = 24 Draw QM and PN and intersect them at O

wwM wa.th Les aBryn

   2 ar COD 1  258. (d)

1 24d2 = 216 2

ERna

260. (c)

ar AOB

3  24 = 9 cm² 8

d1 = 24 cm area of Rhombus = 216

BC = CQ

257. (c)

2

A BC D 

Q

C

B

ar(trap.ACED) = ar(BAC) – ar(  BDE) = 25 – 4 = 21



8 3

 ar (ABC ) 8  ar ( ABC ) = 12  ABCD = 2 × 12 = 24

kgei snhe

ar (BAC )



3

261. (b)

  BDE   BAC ar (BDE )

ar (APQ )

eYrai dnag

259. (a)

ar ( APQ ) =

v.iSn ir

area =

In this question

OB =

3x 2

In Right  OAB

2

 3x  2x ²   2  = 40

9x ² = 40² = 1600 4 16x² + 9x² = 1600 × 4

ar (  ABP) = 6 ar(  PQC) + ar(  QAD) +ar(  ABP) = 15 ar(  APQ) = 24 – 15 = 9 cm² also

216 2 = 18 cm 24

1 1  d1   24 = 12 cm 2 2 Diagonals of Rhombus bisect each other at right angle

OA =



OD =

1 1  d2   18 = 9 cm 2 2

Now, In Right AOD AD² = AO² + OD² = 12² + 9² = 144 + 81 = 225 AD = 225 = 15 cm Perimeter of Rhombus  = 4 × AD = 4 × 15 = 60 cm 262. (a)

ar APQ

3 9 ar ABCD = 24 = 8

4x² +

 always it will be 3 : 8

Rakesh Yadav Readers Publication Pvt. Ltd.

Let ABC = 60° OBC = 30°

343

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Diagonals of Rhombus are the angle bisectors In right BOC

1 ar ABCD  2 AD  BC   EN  1 ar AED   EN  AD 2

OB = Cos 30° BC

266. (c)

AD BC AD

A

D

b B

a

60

B

80

40

D

40

C

60

608040 = 90 2

In Right  OBC,

=

= 90  10  30  50 = 300 15 m²

1

=

 d1 × d 2 =

2

265. (d)

B

A

E

N



1 2

(ar

DECB)

(ar

ABC)

D

Let EN  AD



21 25 B

 2a  2b

C

AB = BC = CD = DA = 10 cm BD = 16 cm

m² ......(ii)



In  ODC, OD = 8, CD = 10,  DOC = 90º OC =



CD² – OD² = 10² – 8² = 6 cm AC = 2 × OC = 2 × 6 = 12 cm Now, Area of Rhombus ABCD

=

1  d1 × d2 2

=

1  16 × 12 = 96 cm² 2

270. (a) Area of trapezium 1 2

60º 60º 60º

=  sum of parallel sides   height

60º

60º 60º

=

Re gular he xagon has 6 equilateral triangle Ar ea of Re gular hexag on = 6×area of equilateral triangle

1 area of AED  × EN × AD 2 area of trapezium ABCD

=6×

1 = AD  BC   EN 2

=

Rakesh Yadav Readers Publication Pvt. Ltd.

=

A

D

267. (c)

C

4 25

area (  DECB) = area (  ABC) – area (  ADE) = 25 – 4 = 21

269. (a)

.....(i) m

1 2ab= (m²–P²) 4

ar (  ABCD) = 2 × ar(  ABD) = 600 15 m²

 2 ²  5 ² =

O

4a² + 4b² = P² Also, 2a + 2b = on squaring, 4a² + 4b² + 8ab = 4a² + 4b² = m² – 8ab from (i) and (ii) m² – 8ab = P² 8ab = m² – P² 4× (2ab) = m² – P² 1 2ab = (m² – P²) 4 area of Rhombus

9 0 9 0 – 8 0  9 0 – 6 0   9 0 – 4 0 



P² 4

a² + b² =

ar(  ABD) =

 ADE =  ABC and  AED =  ACB  ADE   ABC ar  ABC  =

P perimeter 2P = = 2 4 4 Let, AC = 2a  OA = OC = a BD = 2b OB = OD = b

wwM wa.th Les B aryn

S(  ABD) =



ar  ADE 

R Enak geisnh

100 = 10 cm A

DE  BC



C

P 2

C



side of Rhombus

AC = 16, BD = 12 cm OA = 8 cm, OB = 6 cm Diagonals of rhombus bisect each other at 90° In Right OAB AB² = OA² + OB² = 8² + 6² = 100

264. (b)

E

B

O

=

AB =

5 D

r

=

= 2 × 4 3 = 8 3 cm 263. (c)

 

A 2

3

OB 3  8 2 OB = 4 3 BD = 2 × OB



268. (b)

eeYa ridn agv .iSn i



9 2 3

3 4 a²

a² =

3 3 a² 2

1 1 6  8   4 =  14  4 2 2

= 28 cm² A

271. (a)

D 12 O

16 12

16

B

C

AC = 24, BD = 32  OB = OD = 16 and OA = OC = 12

344

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In  OBC, BC² = OB² + OC² = 16² + 12² = 400 BC =

400 = 20 cm perimeter = 20 × 4 = 80 cm D 4x 272. (a) C

1  18  24 = 216 sq cm 2 274. (d) Let ABCD is a || gm area of  ABCD = 2 ×area of  ADC For area of (  ADC) 

30cm

A

2x A

N

4x

M

B

40cm 20cm D

B

7x

30cm

1 (s um of parallel 2 sides)× distance between them 1 (7 x 4 x )2x 176 2

a b c 20  30  40 = 2 2 = 45 cm area (  ADC)

11x 2  176  x 2  16

=

S=

=

AB = 7  4  28 cm

S(S – a)(S – b)(S – c)

45(45 – 20 )(45 – 30)(45 – 40 )

=

45  25  15  5

CD = 4  4  16 cm

AC 2  CM2  AM 2

AC= 64  484 

548  2 137

= 150 15 cm2 275. (a) Let the diagonal of rhombus d1 = x & d2 = 2x

256 =

B

P

1

D

1 (x)(2x) 2

16 = x Longer diagonal = 2x = 2 (16) = 32 cm

wwM wa.th Les aBryn

A

276. (b)

15cm

A

B

C

[Diagonals of rhombus bisect perpendicularly]

In  APB AB = 15,

=

BD  9  2  18

C

1  sum of parallel sides   height 2

1

 175 = 2  20  15   h

AP = 12

Area of rhombus 1 =  diagonal1  diagonal2 2

20cm

As we know  Area of trapezium

 BP = 9

(B y p yth agoras th eo rem )

h

D

 height = 10 cm

277. (a) let the rate of carpenting = Rs x/m²  length × breadth × x = Rs 120 ...(i) length × (breadth – 4) × x

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22

 2  7  R1 = 528  R1 = 84 cm  New Radius = R1 – 14 = R2  R2 = 84 – 14  R2 = 70 New Radius R2 = 84 – 14 = 70 Area of Road = π (R12 – R22)  = π × 14× 154  Total expenditure 22  14  154  10 7 = Rs. 67760 282. (b) Since the ratio of length and breadth = 3 : 2 Let length of rectangular field = 3x Breadth of rectangular field = 2x Perimeter of the field = 80 m 2 (l + b) = 80 2 (2x + 3x) = 80 2 × 5x = 80

=

ABCD is a rhombus

60 AB 15cm 4 (Perimeter = 60 cm) AC = 24, AP = 12

300100 = 600 cm 50 Cost of Carpet = ` 15 × 600 = 9000 279. (a) Old expenditure = `1000 increase in area = 50 × 20 m² Increase in expenditure = 50 × 20 × .25 = `250  New expenditure = 1000 + 250 = `1250 280. (d) Area of verandah = (25+3.5)×(15+3.5)–25×15 = 527.25–375 = 152.25 m² cost of flooring= 152.25×27.5 = Rs. 4186.50 (app.)

281. (b) 2π R1 = 528

ar (  ABCD) = 2 × 75 15

Area of rhombus = 2 d1 d2

AC 2  8 2  222

273. (b)

kgei snhe

12 = 6+16=22 (AN=BM= =6) 2

= 75 15 cm2

ERna

CM = 2  4  8 cm AM=AN+NM = AN+16

breadth = 24m 278. (b) Area of corridor = 100 × 3 = 300 m² Carpet length

eYrai dnag

x 4



breadth 120 6  breadth – 4 100 = 5

=

20cm C

a = 20cm, b = 30cm, c = 40cm

area =

= Rs 100 ...(ii)

v.iSn ir

(Diagonals of Rhombus bisect each other at 90º

80 x = 10 = 8 then breadth = 2x = 2× 8 = 16 cm 283. (c) The sides of a rectangular plot are in the ratio = 5 : 4 Let the length of rectangular

345

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2

.D

B 3

ar   ABC 

2

=

2   5



=

2

 (2x )2 2

=

 5x 

x=

 4 : 25 a

288. (c)

 AD     AB 

2

4

b

h1

1 1 1 = : : 2 3 4 (Take L.C.M of 2, 3, and 4 which is 12) = 6:4:3 Now, 6x + 4x + 3x = 52 13x = 52 x = 4 length of smallest side = 3x  = 3 × 4 = 12 cm

36 = 6  Perimeter =12 × 6 = 72 cm 292. (a) r 42cm

h2

x

y

1  h1  x 2 a = 1 b  h2  y 2 h1 ay h1 x a    , h2 y b h 2 bx ay : bx 289. (a) Ratio of parallel sides =5:3 Let sides are 5x and 3x

1

2

5x 6x

p e r im e t e r o f r e c t a n g l e = c ir c um f e r e nc e of c ir c ular wire

4x × 24= 1440 1440 x = 4  24 = 15m  length of longer side = 5x = 5 × 15 = 75 m 290. (c)

a1 a2 =

15 225 = 16 256

Ratio of their perimeters

4a1



4a 2

a1



R 23  r 22 Let R = 23x , r = 22x R –r =5

23x – 22x x

=5 =5

a2



16

291. (d) Clearly, 3,4 and 5 form a triplet therefore, consider the triangle, a right triangle Let the sides are 3x,4x,and 5x perimeter = 3x + 4x + 5x = 12x 1 area of triangle = ×3x ×4x 2

= 110

diameter of inner circle = 2r = 2 × 110 = 220 m 294. (b) Ratio of angles = 3 : 4 : 5 3 + 4+ 5 =180° 12 = 180° 1=

15

 15 : 16

Rakesh Yadav Readers Publication Pvt. Ltd.

2R 23  2r 22

 r = 22 × 5 a1 ² 225  a2 ² 256

ATQ,

=

22  42 7 22x = 2 × 22 × 6 x = 12 clearly, smaller side of rectangle = 5 × 12 = 60 cm 293. (c)

2(6x + 5x) = 2 ×

5x  3x × 24 = 1440

wwM wa.th Les B aryn

25 285. (d) Base : Corresponding altitude = 3 : 4 Let the base = 3x altitude = 4x  area of triangle = 1176 1  3x  4x = 1176 2 1176  2 x² = 3  4 = 196 x = 14 altitude = 4 × 14 = 56 cm  286. (c) According to question, Ratio of sides of triangle are

1 × 3x × 4x = 216 2 216  2 x² = 3  4 = 36

4 25

1 (sum of parallel sides)× per2 pendicular distance= 1440 m²

5

AB = 5 cm DB = 3 cm  AD = 2 cm ar   ADC 

A1 2 A2 = 1

R Enak geisnh

A

1

r

500 x² = 20 = 25 x=5 Length = 5x = 5 × 5 = 25 m Breadth = 4x = 4 × 5 = 20 m Perimeter of the rectangle = 2(25 + 20) = 2 × 45 = 90m 284. (d) C

287. (c) Let diagonals be 2x and 5x

eeYa ridn agv .iSn i

field = 5x and the breadth of rectangular fieled = 4x According to question, Area = 500 m² 5x × 4x = 500 m² 20x² = 500 m²

3

180 = 15° 12

: 4 : 5 ×15 ×15 ×15

45

60

295. (b)

largest angle

75

a

a

a 2

a

a 2

a a 2 a 2

346

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Diagonal = a 2



a²  a²  a 2

Area of square Area of square on diagonal a²



2

= a 2

 

a² 1  a²2 2 = 1 : 2

296. (d)

area of square area of equilateral triangle

=

a²

=

=

3 a² 4

4

= 4: 3

3

297. (d) r

a1 5 = a 6 = 5 : 6 2 299. (d) Let length = 5x l 5 2l  b = 16 8l = 5l + 5b 3l = 5b l 5  =5:3 b 3 300. (c)

When we draw such figures as mentioned in the question the vertex of the old triangle are the mid points of the sides of new triangle and the sides of the old triangle are half of the opposite side.  required ratio = 2 : 1

ERna

r² = a² a² r² =  a r= 

=

a  = 2a



r 2a



2

=

 :2

298. (c)

h1

h2

a1

4

3

4

a2

a  1

25

2

1 1 1 : : 3 4 5 = 20 : 15 : 12 20 + 15 + 12 = 47  47  94 1  2  Smallest side = 12 × 2 = 24 cm

=

 

36

a1 25 5  = a2 36 6

Rakesh Yadav Readers Publication Pvt. Ltd.



s– bs– c

9  5  3 1

   

a

:

3

a 2

2   3

=4:1

310. (a)

As D and E are mid-points of AB and AC  DE||BC   ODE   BOC and also

DE BC

=

1 2

(as D and E are mid-points) ar (ODE)  1 2 1  ar (OBC) =   = 2 4  

311. (c)

The given angle is same let vertical angle =

305. (c) Ratio of sides =

2

a2

=

2

= 4:1 303. (a) Ratio of area= (Ratio of radius)² A B C Radius 4 : 2 : 1 Area 16 : 4 : 1 304. (b) r² = a² a²  = r² 1 a = :1 r

=

s s– a

= 3 15 cm² 309. (b) Ratio of area = (Ratio of radius)²

 a    radius)² =  3   a      2 3 

wwM wa.th Les aBryn

2r Ratio of perimeter = 4a

3

 area =

Circumference 2r = Area r ² 2 2 = = r 3 302. (b) Ratio of area= (R atio of

301. (b)

=

306. (a) Let the sides be 3x,4x,5x and 6x x 4  18x  72,  Greatest side = 6 × 4 = 24 cm 307. (b) Ratio of circumference = Ratio of radius = 3 : 4 308. (d) Let the sides be 2x, 3x and 4x  9x = 18  x = 2  Sides are 4,6 and 8 cm respectively Using hero’s formula 468 S= = 9 cm 2

eYrai dnag



Ratio of altitudes =

kgei snhe



3 a1 2 3 a2 2

v.iSn ir

Let the side of square = a



(   ABC and  DEF are isoceles triangles) when two angles are equal then third angle is also equal   ABC   DEF  ABC is similar to  DEF



area of  ABC area of  DEF

347

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316. (a) Let the side of square = a

 Side of equilateral  =

side of  ABC 1 = side of DEF 4 1 side of  ABC = = 2 side of DEF 312. (a) Let the sides be 3x ,3x and 4x

3



Area =

4x  4

4 3x

2

2

  – 4x

= 4x ² 36x ² – 16x ² = 4x

2

20 x

1

 6x² = 72  x² = 12

6x² = 72 x² = 12



1 × 3x × 4x = 72 2

x = 2 3 Perimeter of equilateral  =12 × 2 3 = 24 3 units Side of  =

24 3 = 8 3 units 3

area of  =

3 × 4

2

8 3 

3 = × 64 × 3 = 48 3 unit² 4 315. (d) Let the parallel sides be 2x and 3x

1  area = (2x + 3x)× 12 = 480 2

5x= 80 x = 16  Longer parallel side = 16 × 3 = 48 cm

of

l  b 1 2 2  4 8 = 8: 9    a² 9 9  3 2     2

2

r12 4  318. (a) r22 7

r12 4  2 r2 7



323. (c) 2(l + b) = 3a  (a = side of equilateral triangle) Let (b = a)  2(l + a) = 3a 2(l + a) = 3a 2l + 2a = 3a 2l = a Required Ratio a a a² 4  2   2 3 3 3a ² a² a² 4 4

l b

=

5 ² – 3 ²  16 25 5 ²

2

=

3

320.(a) Let side of square = a

a radius of smaller circle = 2 2a 2

radius of larger circle =

321. (c)

B

324. (b) Required ratio=

r ²   r² 1

= π :1 325. (c) Let AB = 1, BC = 1

 AC = 1²1² = 2 (using pythagoras) 3 1 ² 1  4  = ar ACD  2 3 2 ² 4 ar ABE 

 

A a a/3

= 2: 3

2

a      2  Required ratio =  2a 2    2    a² 4 1 = 2a ² = 2  1 : 2 4

wwM wa.th Les B aryn



=

10   =    = 25 : 16 ar  DEF 8 ar  ABC

 16 : 25

= 6 3 314. (b) Let the sides be 3x, 4x and 5x =

3 ×2 = 3:2 4 317. (b) Ratio of area= (Ratio side)²

5 ²

Smallest side = 3 × 2 3

area

2

=

5 ² –3 ²

 x= 2 3



3 2 Required ratio

a=

r1 4 2   = 2: 7 r2 7 7 319. (c) Required ratio =

 2 × 3x × 4x = 72



a

2

= 8x 3 5 = 8 5 = x³ = 1 = x=1  3rd side = 3 × 1 = 3 units 313. (c) 3, 4 and 5 from triplet Let the sides be 3x,4x and 5x

 

R Enak geisnh



2

2a

eeYa ridn agv .iSn i

2

4

Required ratio =

2a

322. (d) 2(l + b) = 4a (a = side of square) 2(2 + 1) = 4a 2 × 3 = 4a

r

 side of  ABC 2 =      side of  DEF 

=1:2 C

Circumradius =

326. (b) side 3



a 3

2 2 arΔABC  AB  10  25   =    = =  arΔDEF  DE   8  16

327. (c)

Equilateral  3 a² 3 3 4 Required ratio =  2 4 a       3

Rakesh Yadav Readers Publication Pvt. Ltd.

= 3 3 : 4π

A'B' ||AB

348

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1  perimeter of ABCD = 2

ar A ' B 'C  ar ABC 

perimeter of EFGH 1 = perimeter of ABCD 2

331. (c) Old circumference = 4 π 2πr = 4π

 B ' C 2 1 2 1          4  BC   2 

ar(AA'B'B) ar(  A'B'C)

=

ar AA ' B ' B

  ar ABC 

=

ar(  ABC)

–

r=

1 1 1 : : 4 6 8

1 1 1  24 :  24 :  24 = 4 6 8 =6:4:3 (Take L.C.M = 24) ATQ perimeter = 91 6 + 4 + 3 = 91 13 units = 91 91 1 unit = =7 13 Diff. between long er and shorter side = 6 - 3 = 3 units  3 units = 7 × 3 = 21 cm 329. (c) By using result, R11 = R22

330. (c)

A

D

E

H

B

New area = 16 π cm² Option (c) is the answer ( area is quadruples) 332. (c) Length 4 5 Breadth 5

4

area

20

20

area remains unchanged 333. (d) Area of circle = π (5)² = 25 π Circumference of circle = 2 π (5) = 10 π 25π  100 = 250% = 10π 334. (d) According to question, Circumference of a circle = area of circle 2 π r = π r² r =2 diameter of circle = 2r  =2×2= 4 335. (c)

3 4

3 × 100 = 75% 4 336. (b) Increment in breadth = 10% =

10 = 100

Decrement in length = 10% =

10 = 100

% change =

–1 = 1% 100

Alternate:using x = 10% (breadth), y = –10% (length) % change = x + y = 10 – 10 +

xy 100

10  –10

 

100

= –1%

337. (c) % increase = x + y + = 20 + 20 +

20  20 100

xy 100

= 44%

338. (d) If circumference of circle is reduced by 50% then radius is reduced by 50%

F

O

G

C

In  OBC, H and G are the midpoints of OB and OC 1  HG = BC 2 1 similarly, FG = CD 2 1 and EF = AD, 2 1 HE = AB 2 on adding, HE + HG + FG + EF =

= 2cm

8π = 4cm 2π

R=

wwM wa.th Les aBryn

 R 1  2  75  5 R  60 4 = 5 : 4 2 1

2π

 = 

Percentage of ar (DECB)

=

Old area = π (2)² = 4 π cm² New circumference = 8 π 2πR = 8π

3 =3:4 4

328. (d) Ratio of sides =

4π

 

ar ABC

v.iSn ir

 A ' B ' C   ABC Let BB' = B'C = 1 BC = 2 ( B' is the mid-point of BC)

ar DECB



eYrai dnag



perimeter of EFGH.

kgei snhe



A' and B' are the mid-point. By mid point theorem

ERna



1  AB + BC + CD + AD  2

 

D and E are the mid points of sides AB and AC DE||BC (By mid point theorem) 1 BC 2 ADE ABC

( is constant) Reduction in area 3 = × 100 = 75% 4 339. (d) Increase in area

also DE =

= 25 + 25 +

ADE  ABC     AED  ACB 

 

 

ar ABC

=

 1 2 1 =     = 4 2 

Rakesh Yadav Readers Publication Pvt. Ltd.

use formula : (x + y + 2

ar ADE

25  25 100

 DE       BC 

xy ) 100

= 50 + 6.25 = 56.25% 340. (a) Increase in area 5050 100 = 100 + 25 = 125%

= 50 + 50 +

349

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% decrease =

20  –20

 

25 – 9 × 100= 64% 25

  

348. (a)

100

= – 4% (decrease by 4%) 342. (b) Increase in area

8–5 × 100  % Decrease = 8

50  50 = 50 +50 + 100 = 100 + 25 = 125% 343. (c) Increase in altitude = 10%

= 37 349. (b)

1 % 2

  

=

30 – 20 % Increase = × 100= 50% 20

350. (b)

Side 10 Area 100

Area no change decrease in base

 Increment  50%  1 2  Original

= 21%

352. (d) Radius 100  101 Area 10000  10201 201 ×100 = 2.01% 10000 353. (c) Let the breadth = x cm  length = (x + 20) cm According to the question, x(x + 20) = (x + 10)(x + 5)  x² + 20x = x² +15x + 50

%Increase =

wwM wa.th Les B aryn

xy 345. (b) use x + y + 100 percentage change

12  15 9 = 12 + 15 + = 27 + 100 5 4 4 = 27 + 1 = 28 % 5 5

 5x = 50  x = 10  Area = 10(10 + 20) = 300 m²

346. (b)

100

100

100

120

150

354. (c)

  

130

Pe rime ter of e quilater al triangle = 100 +100 +100 = 300 Perimeter of New triangle = 120 + 150 +130 = 400

%error = 355. (d)

100 1 % increase = ×100 = 33 % 300 3 347. (b)

121 – 100 × 100 100   

351. (b)

  

40%

4 2  10 5

Side 5

Surface area (5)² = 25 24 (7)² = 49

7

24  100 = 96% 25

% increase =

Alternate Per centage incr ease in surf ace area = 40 + 40 +

40  40

%

100

= 80 + 16 = 96%

xy ] 100 357. (a) per centage incr ease in area

[% effect using x + y +

=

 88   8  8   100  

= 16 + 0.64 = 16.64% 358. (a) Side of square is increased by 30% =

30

=

10 0

3 10

200 – 189 % Decrease = ×100=5.5% 200

5 ×100 = 125% 4

Increase % =

% Increase =

 11  121

R Enak geisnh

1 1 = × 100 = 9 % 11 11 344. (d) Increase in circumference = Increase in radius

356. (d) 40% =

eeYa ridn agv .iSn i

= 20 – 20 +

xy 100

r

341. (b) using x + y +

1029 – 1000 ×100 = 2.9% 1000   

% increase in area

Rakesh Yadav Readers Publication Pvt. Ltd.

156 – 100 = × 100 = 56% 100

Other s ide will hav e to be decreased by 3 1  100 = 23 % 13 13 359. (c) Percentage increase in area

=

= 100 +100 +

100 100 100

= 300%

Alternate L 2

Percentage increase 3 =  100 = 300% 1 xy 360. (d) x  y + 100 10   –10  = 10 – 10 + = –1% 100 (Negative sign shows decrease) 361. (c)

.d/2

.d/2

.d/2

.d/2

2d

350

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From the figure it is clear that, 4 circular plates of diameter d can be made of a. Square plate of side 2d with minimum loss of material. 362. (d)  AOB,  BOC and  COD

Area of circumscribed square = (2a)² = 4a² Difference between areas of  outer and inner squares = 4a² – 2a² = 2a² 365. (c) In  ABC, by Pythagoras

are equilateral  .  Side = 2 cm

A

2

2

A 2

O

2

3 ×4 = 3 3 cm² 4 363. (d) Area of sector = 72 cm²

=

cm



25 cm² 2 Now, value of x+y–z

72  360 = 20º  θ = 36  36

 9 16  25 =  2  2  – =0   2 366. (c) Perimeters of triangles,

ERna

r ²θ = 72 360º

Now, length of arc =

rθ 180º

wwM wa.th Les aBryn

×36×20 = = 4 cm 180 364. (b)

=

D

C

.

O

 



(3×1) + (3×0.5) + (3×.025) + (3×0.125)+ ... 3 + 1.5 + 0.75 + 0.375

B

For inscribed circle, Diameter of circle = Diagonal of square Sinc e, s id es of square are equal. Now, In  ABC by Pythagoras theorem AB² + BC² = AC² 2 AB² = 4a²

3 a 1 = 3  2 =6 units Sn = = 1  n 1  2 2 1

367. (c) In  AOB, AO = OB = r (radius of circle)

A

AB² = 2a²  AB = 2 a Area of inner square = AB² =

 2a ²

= 2a²

For circumscribed square, Diameter of circle = Side of square

100 – 25  50 75  50 = 8 8

=

25 25 (3 +2) = 8 4

90º 5

.

C

 3    1   cm² 2 

A

B

D

C

Diameter = Diagonal of rectangle Now, let x and y be the length and breadth of rectangle are respectively, Now In  ABD, AB² + AD² = (5)²  x² + y² = 25 Since, they form Pythagoras triplet,  x = 4 and y = 3

So, area of rectangle = 3×4 = 12cm² 369. (b) Required ratio

O

=

Area of circle Area of ΔACD C

B Minor Segment

Using Pythagoras theorem, AB² = OA² + OB²  (5)² = r² + r² 25 cm  r² = 2

Rakesh Yadav Readers Publication Pvt. Ltd.

=

368. (c) ABCD be the rectangle inscribed in the circle of diameter 5 cm.



2a

A

 25 – 50   =  r² –   8  

16 = cm² 2 Area of semi-circle = z =



B

6 x

3 ² 9  cm² 2 2 Area of semi-circle = y

=3×

36

=

BC² = AB² + AC² = 36 + 64 = 10 cm Now, area of semi-circle = x

D

25 r² 25 25 – = – 8 2 8 4  25 – 50    =  8   Area of major segment = Area of circle – Area of minor segment

10

eYrai dnag

2

y 8

C

2

90º 25 25 × × = cm² 360º 2 8 Now, area of minor segment = Area of sector – Area of triangle

v.iSn ir

3 (Side)² 4

θ × r² 360º

=

z

kgei snhe

2

=

C

Now, total area = 3 × B

How, area of sector AOB



r A

r

r

B

O r D

351

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 = 1  2r r 2

370. (b) Semi-perimeter of triangle

a+b+c 2

1364 704 ×h = 7 21



7+24+25 56 = = 28 cm 2 2 Area of circle = Area of triangle

=

= =

s s – a s – b s – c 

 h=

373. (a) Area of the  ABC

28 28 – 728 – 2428 – 25

=

1 × b×h 2

=

28  21 4  3

A

= 7056 = 84 cm² 371. (a) Area of equilateral triangle 3a² =x 4 and perimeter = 3a = y

=

y ......(ii) 3 Now, puting the value of a from Eq. (ii) in Eq. (i), we get  y 2 3    3  y2 3  = x  x = 9 4 4

B

b/2



b=

10 m

2 2 22 = 2×  r³ = 2 × × ×2³ 3 3 7

2  2  22  8 704 = m³ 21 21 Area on which the muds is spread over = Ar ea of fieled – Are a of pitholes = 1×b – 2 ×  r²

=

22 = 22 × 10 – 2× ×2² 7 176 1540 –176 = 220 – = 7 7

10 cm

10 cm

D

b/2

C

12  2 = 8 cm 3

Here, BD = CD =

b 8 = 2 2

= 4cm In r ig ht ang le d  AB D, b y pythagoras theorem, AB =

BD2 +AD2

 a= 42  32 =

=

2m

a

b

22 m

2m

3 cm (h)

1 ×b×3  12 = 2

wwM wa.th Les B aryn

x=

10000 = 50 200 376. (d) Area between square and semi-circles = Area of square – 2 Area of semi-circle

=

22 × (5)² 7 = 100 – 78.5 = 21.5 cm² 377. (a) Let l = 4x and b = 9x  Area of rectangle = l×b 144 = 4x ×9x

= (10)² – 2×

a=

y2 y2  x= 3 3 4 12 3  12 3 x = y² On squaring both sides, we get y4 = 432 x² 372. (c) Volume of mud dug out in two hemispherical pitholes 

a

.....(i)

h=

5 cm

R Enak geisnh



704  7 16 = m 1364  21 93

8.8  56 =4m 14  8.8 375. (a) 1 m = 1000 mm 10 m = 10000 mm  Number of 200 mm lengths that can be cut from 10 m of ribbon 

eeYa ridn agv .iSn i

=

Then, 14×8.8 ×h = 8.8 × 56

r

1364 = m² 7 Now, let the rise in level by h m, then Area of remaining field × h = Volume of mud dugged out

r ²

16  9

25 = 5 cm

Now, perimeter of an isosceles triangle = 2a+b=2×5+8=10+8 = 18 cm 374. (a) Let the breadth and height of r oom be b and h m , respectively. Then, according to the question,  1×b = n Area occupied by one patient  14×b = 56×2.2 

b=

x² 

A

O

B

D

Q

C



 AD × CD = 100 cm²  AD ×20 = 100 AD = 5 cm [in rectangle AB = CD = 20 cm and AD = BC = OQ = 5 cm]



Area of  ODC =

562.2 = 8.8 m 14

Now, total volume of the room is equal to total patients multiplied by volume occupied by each patient.

Rakesh Yadav Readers Publication Pvt. Ltd.

144 36  x² = 4  x = 2 Now, l = 4×2 = 8 cm and b = 9×2 =18 cm Perimeter of rectangle  = 2 (l+b) = 2(8+18) = 2×26 = 52 cm 378. (c) Area of parallelogram = Base × Height = 8.06 × 2.08 = 16.76 cm² 379. (c) Given that, CD = 20 cm and area of rectangle ABCD = 100 cm² 

1 ×PQ × CD = 2

1 ×5×20 2 = 5×10 = 50 cm²

352

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 B

 D

C

In  ABC, a²+a² = 2a²  BC = a 2



a² =1 2



a=

2



Perimeter of  ABC = 2a +

2 a=2 2 + 2 . 2

B



Le ng th of a diam eter of a square = Diameter of a cricle



a 2 = 16

 

a = 8 2 cm Area of square ABCD = a² = 8 2

2

 

= 64×2

= 128 sq cm 384. (d) Given that, length of hour hand = 4 cm and length of minute hand = 6 cm Hour hand rotating in 1 day =  2 ×360º = 720º Hour hand rotating in 2 days = 

5m 1.

.

3.3 r=

m

Similarly, Minute hand rotating in 1 day = 24 ×360º Minute hand rotating in 3 days 

5m 1.

= r (3.3 + 1.5)² –  (3.3)² = [(4.8)² – (3.3)²] = (23.04 –10.89) = 12.15 m² 382. (c) Let length and breadth of a rectangular field are 9x and 5x A

 radius 180 Distanc e trave lle d by hour



 = 32 180º and d is tanc e tr av e lle d b y minute hand hand = 4×1440º ×

D

5x

9x

D

 Area of a rectangular field

= 4500 m²  9x × 5x = 4500  x² = 100 = (10)²  x = 10

10 B

16 = 2 r.

θ 360º

θ 16 = 360º 2r Now, area of sector OAB

=  r².

θ 360º

16 = 8r = 8×10 2r = 80 sq cm 386.(c) Let the breadth of floor be x metre.  Length = (x + 20)metre  Area of the floor = (x + 20)x sq.metre In case II, (x + 10)(x + 5) = x(x + 20)  x2 + 15x + 50 = x2 + 20x  20x = 15x + 50

=  r².

 5x = 50  x = 10 metre Area of the floor = x (x + 20) = 10(10 + 20) = 300 387.(d) B A x

= 72 ×360º ×

= 6×72º×360º ×

B

 radius 180

2 ×720º = 1440 ×

= 6×144 Required ratio =

32 6 144

1 = 27

Rakesh Yadav Readers Publication Pvt. Ltd.

 180

16



 C

a

wwM wa.th Les aBryn

= 2(1+ 2 ) units 381. (c) Area of path = Area of (fountain + path) – Area of fountain

a

kgei snhe



a

θ 360º

A O



D

ERna

 

90º + θ + θ = 180º (s ince , sum of all inter ior angles of any triangle is 180º) 2 θ =90º θ =45º Now, In  ABD, a AD sin 45º =  AD = a 2 1 Area of  ABC = ×AD×BC 2 a 1 = ×a 2 × 2 2 = 1sq unit (given)

a

A

385. (c) Arc of length = 2 r.

v.iSn ir

A 90º

So, the length and breadth of a rectangular field are 90 m and 50 m. Perimeter of rectangular field  = 2 (Length + Breadth) = 2 (90+50)=2×140 = 280 m 383. (c) Given that, radius of a circle = 8 cm and diameter of a circle = 16 cm

eYrai dnag

380. (d) Let AB = AC = a  BC² = AB² + AC² (by Pythagoras theorem)

D

O

C

Here ABCD is a square of side x.

x  OC = 2 , BC = x, and OB = radius of circle = 10cm In  OCB, OB2 = OC2 + BC2 2

 

x   + x2 = (10)2 2

5x2 2 4 = 100  x = 80

 x = 4 5cm Hence, perimeter of the square ABCD = 4x = 16 5

353

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S

O



P

Q

R

So, the area of a segment of a circle is always less than area of its corresponding sector. II. Distance travelled by a circular wheel of diameter 2 d cm in one revolution

2d = 2

a=



 r 2  1 = 4 r  9  2



r12 Area of 1st circle = 2 Area of IInd circle r2

r2

.

O



r1 5 = r2 3

equilateral triangle =

r1

B



 a 2     = 154 2 3 



 a 2 154  7   = (7)²   = 22 2 3 

.....(i)

Let r1 = 5x and r2 = 3x Also, given that, area enclosed between the circumferences of two concentric circles = 16 cm²  (r12 – r22) = 16

...(i)

 

Rakesh Yadav Readers Publication Pvt. Ltd.

F x/3

x/3

.

x/3

x/3

B x/3

D

x/3

H

x/3

x/3

x/3

E x/3 C

where, AGHF form a rhombus and is als o an  HDE equilateral triangle. Area of rhombus = (Ar ea of   AGF + Area of  GFH )  x 2 3     + 3  4

3 = 4

= 2.

 x 2     3 

2 x 3      4 3 

2 3 x    Now, area of  HDE =   4 3 

3 x² 4

and area of  ABC =

Area of rhombus AGHF + Area of ΔHDE Area of ΔABC 2

a

a = 14 3 cm Perimeter of an equilateral triangle = 3a = 3(14 3 ) = 42 3 cm

x/3

x/3

3 x     4  1 3  = = 3 2 3 x 4 395. (c) Given that, Area of the circle = Area of the square = (Side)² 3

2 3 where, a be the length of the side of an equilateral triangle. Given that, area of a circle inscribed in an equilateral triangle = 154 cm²

A

G

x/3

By given condition,

 r 2 4 1 =   = r 9  2 393. (b) We know that the radius of a circ le ins cr ib ed in a

wwM wa.th Les B aryn

= 2×3.14×d = 6.28d 2 which is greater than 6d cm. 389. (b) Given that, perimeter of a rectangle = 82 m 2 (Length + Breadth) = 82 m   Length + Breadth = 41 m  l+b = 41 m. . . .(i) Also, its area = 400 m²  l.b = 400m² Now, (l –b)² = (l+b)² – 4lb =(41)² – 4 (400) = 1681 – 1600 = 81 l–b=9 .....(iii)  From Eqs. (i) and (iii), 2l = 50  l = 25 m and b = 16 m Required breadth (b) = 16 m  390. (c) Given that, ratio of their radii = 5:3 i.e., r1 : r2 = 5 : 3

 r = 1.57r 2 Now, area of the circle (AC) =  r² = 3.14r² and area of the square (As) = a² = 2.4649r² Area of circle > Area of square  392. (b) Let the radii of two circles are r1 and r2. respectively. Given, Circumference of 1st circle 2 = Circumference of IInd circle 3 2r1 2 r1 2   2r2 = 3  r2 3 

A x/3

r

. . .

394. (b)  ABC forms an equilateral triangle.

eeYa ridn agv .iSn i

r ²θ 1 of  OPQ = – r² sin θ 360 2

(5x)² – (3x)² = 16 25x²–9x² = 16 16x² = 16 x² = 1  x =1 r1 = 5 and r2 = 3 Area of the outer circle =  r12 =  (5)² = 25 cm² 391. (c) Let the radius of a circle is r and a be the length of the side of a square. Give n, c ir cumf er ence of a circle = Perimeter of a square  2 r = 4a      

R Enak geisnh

388. (c) 1. We know that, Area of segment (PRQP) = Area of sector (OPRQO) – Area

r² = (2  )²  r² = 4  r² =

4 =4 

4 = 2 units Diameter of circle (d)  = 2.r = 2.2 = 4 units 396. (a) Let the radius of circle is r and the side of a square is a, then by given condition, 

r=

2 r =4a  a =

r 2 2



r  Area of square =   2

=

²r² 9.86r ² = = 2.46r² 4 4

354

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and area of circle = r² = 3.14r² and let the side of equilateral triangle is x. Then, by given condition,

Perimeter of an equilateral triangle = 3a = 3×2 = 6 cm 399. (d) Given that, 

E

 2a

F

 Area of equilateral triangle =

3 4

4²r ² 3 × 9 4

a A

(AD) =

3 × Side 2

3×side 2  2 3 = 3 ×side

3=



kgei snhe

= (a 5 )² = 5a² sq units 400. (c) Outer diameter = 112 cm and inner diameter = 70 cm 70 cm

.



a

2m

2m

16 m

A 2m

B

30m

E

F

Area of path = Area of EFGH Area of ABCD = 34 × 20 – 30 × 16 = 680 – 480 = 200 m² 404. (c) In right  ABC, 

cm

D

C

. O

A

12 cm

BC =

1 ×Base×Altitude = (Side)² 2

3 a² 4

C

But altitude =

Rakesh Yadav Readers Publication Pvt. Ltd.

B

16 cm

AC² = AB² +BC² AC² = (16)² +(12)² = 256 + 144 = 400  AC = 20 cm AO = 10 cm (radius)  and area of circumcircle = r² =  × (10)² = 100 cm² 405. (c) In  ABC, AB²+AC² =

a²+b²

C a²+b²

1 36  2 ×9×l=36  l= 2 9  l = 8 cm 402. (d) Area of equilateral triangle

a

D a

C

112

1 =  (112² – 70)² 4 1 = (12544– 4900) 4 1 22 = ×7644× 4 7 1 = ×24024 = 6006 cm² 4 401. (d) Let the length of altitude AB =l By given condition, Area of  ABC = Area of square

= B

2m

D

b



A

G

Required area



2 3 Side = = 2 cm 3

H

= a 5 unit Hence, area of square

wwM wa.th Les aBryn

–15  x =5 and 2 Since, width cannot be negative. Width = 5 units and  length = 2x +5 = 2×5+5 = 15 units Pe rime te r of the r ec tang le  = 2 (15+5) = 40 units 398. (c) Height of equilater al triangle,



B

Area of rectangle = 2a² = l×b l×b = 2a² = l×a  l=2a Now, In  ACD, AC² = AD² + CD² a² +4a² = 5a² Side of square, AC

3 (4)² = 4 3 cm² 4 403. (a) Required area of the path EF = 30+4 = 34m, GF = 16+4 =20 m

eYrai dnag



² r ² = 1.89r² = 3 3 Hence, Area of circle > Area of square > Area of equilateral triangle 397. (a) Let the wid th of the rectangle = x unit  Length = (2 x + 5) unit According to the question, Area = x (2x+5)  75 = 2x² +5x  2x² + 5x – 75 = 0  2x² + 15x – 10x – 75 = 0  x (2x +15) – 5 (2x +15) = 0  (2x +15) (x–5)=0



C

ERna

x² =

D

v.iSn ir

2r 3x = 2 r  x = 3

3 2 3 = a 2  a = 4 cm Area of equilateral triangle = 

A

B

a

 Required total area = a² + b²

3 a 2

+



a²  b²

2

1

 + 2 ab

= 2(a²+b²)+0.5ab

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D

D

A

C

Now, area of curve ADE =

r²θ 360º

=

22  (4.2)²  60º = 9.24 m² 7  360º

3 3 (side)² = × (6)² = 15.57 4 4

9.24 ×100 15.57 = 59.34% = 59% (approx) 408. (a) By given condition,

4 60º

wwM wa.th Les B aryn .....(i)

r1 r2 and a² = 2 2

r1 2     A1 a r 21 1  2   = =  2 = A2 a r 22 2 r2       2  [from Eq. (i)] 409. (b)

=

4 60º

1 1 ×d1×d2= ×55×48 2 2 = 1320 cm

=

A

B

O

P

D

C

60º 3 2 8 –3× × (4)²  360º 4

B

D

Area of triangle =

E

C

 Area of rhombus = Base × Height = DC × AE  DC×AE = 1320

Rakesh Yadav Readers Publication Pvt. Ltd.

1 × Base × 2

Height

A

B

B

4

= (16 3 –8 ) cm² 412. (a) Area of rhombus

2 1 2 2

A



Ar ea of Portion includ e between circles = Area of triangle – Area of 3 sectors

If circle or bent in the form of square, then 2 r1 = 4a1  a1 =

x



C

A1 r 21 1 = = A 2 r 22 2



cm

4 60º 4

=

 r 2 1  1 = r  2  2

1320 =36.16 36.5 36 cm < p < 37 cm 413. (c) Perimeter = a + b + c 240 = a + 78 + 50 a = 112

411. (b) Since, all sides of a  ABC are equal, so their all angles are equal to 60º

4

5329 = 1320 4

p=

 (15 2 )² = 2x²  2x² = 225 ×2  x² = 225  x = 15 cm Hence, length of the side of the square be 15 cm.

A

Required percentage

 p×

B

x

AC² = x² +x²

and area of equilateral  ABC



2

R Enak geisnh

6m

C

15

60º

B

=

 36  9  94   sq. cm.

E

60º

  1 1  OD  BD and OC  AC    2 2  

r

60º

32 4

410. (a) Let the sides of a square be x cm, In  ABC, AC² = AB² +BC²

 55 2  48 2         =1320 2  2 

 p×

78 m

4. 2

m

A

364

 p× OD²+OC² = 1320

b=

1 (3+5) × 4 = 16 m² 2  Total cost of painting Rs. 25 per sq m = 16×25 = Rs. 400 407. (c) Suppose a horse is tied at vertex A. Then, area available grazing field is ADE.

=

Area of the shaded region = Area of square of side 6cm - 4  a right angled sector

eeYa ridn agv .iSn i

406. (b) Area of trapezium



c =50 m

h

a =112 m

C

and also, Δ= s s–a s–bs–c Area of = 120 120 – 112120 – 78120 – 50 = 120  8  42  70 = 1680 m² Area of triangle =

1 × Base × 2

Height

1 ×50×l 2 2 1680 h= = 67.2 m  50 414. (b) Let the sides of isosceles triangle be 5x, 5x and 3x cm, respectively. By given condition, Pe rime te r of isosc eles triangle = Length of wire 5x + 5x +3x = 78  13x = 78  x = 6 cm  Length of base = 3×6 = 18 cm 415. (b) The angle made b y the minute hand in 20 min = 120º, (1 minute = 6º )  The are a swep t by the minute hand in 20 min  1680 =

θ 120º ×r2 = ×3.14×9×9 360º 360º = 84.78 cm²

=

356

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416. (c) Given, AB = 2a

= 7×11×2×3 = 462 cm² Hence, radius of incircle

R

.

O r

C

M

D

B

D

AM = a a 2 Now, OC = OP+PC = OP+CM

2



F

xm

B

4m

Given, area of EFGH = 6m²  (4–2x) (3–2x) =6  12–8x – 6x +4x² = 6  4x²–14x+12 =6  4x²–14x+6=0  2x²–7x+3=0  2x²–6x–x+3=0  2x(x–3)–1(x–3)=0  (x–3)(2x–1) = 0 

x = 3,

1 2

1 = 0.5 ( x  3) 2 420. (b) Inner radius, r1 = 25 cm and external radius, r2 = 25+10 = 35 cm Distance covered in 1 revolu

x=

ERna

r 2

F

x

22 7 = 220 cm. and distance covered in 10 revolutions = 2200 cm  Speed of bicycle

tion = 2 ×35 = 70×

wwM wa.th Les aBryn

a=

7 cm

G

A

 a 2  a 2   r       = (a–r)²+  2   2 a² +ar  r² + 4 a² = a² +r² – 2ar + 4 a r = 3 417. (b) Let r be the radius of circle and a be the side of square. By given condition, 2 r = 4a

r  Area of square =   2

2r 2 9.86r 2 = = 2.46 r² 4 4 and area of circle =  r² = 3.14r² Hence, area of the circle is larger than that of square. 418. (d) Let a = 35 cm, b = 44 cm and c = 75cm =

 s=

a+b+c 2

35+44+75 = 77 2 Now, Area of D =

=

s s–a s–bs–c

=

77×42×33×2

=

7×11×2×3×7×3×11×2

E

B E

a 2 So,  OCD is an isosceles triangle. ( OC = OD)   OMC = 90º In  OMC, OC² = OM² + CM²



xm

3m xm

a = r+ and 2

OD = OQ+QD = OQ+MD = r +

D

C H

and AC = CM = MD = BD =

A

Covered distance Time 2200 22 = cm/s = m/s 5 5 = 4.4 m/s 421. (c) Length of wire = 36 cm  Perimeter of semi-circle =  r+2r  22   2    36 = r  7  {  Perimeter of semi-circle = Length of wire}

=

36  7 = 7 cm 36 Hence, radius of semi-circle=7 cm 422. (c) Are a of c ur ve B CDE =

Rakesh Yadav Readers Publication Pvt. Ltd.

C

7 cm

v.iSn ir

A



1 49 ×7×7 = cm² 2 2



Re quir ed are a of s hade d region = 2 Area of curve BEDF

 77 49   28  = 2  2 – 2  =2  2  =28 cm²     423. (a) Let sides of a rectangle be l and b. Then, 2(l+b) = 18  l+b = 9 Area of rectangle = l×b  F or m axim um , ar ea of rectangle, l=b 2l = 9  l = 4.5  Maximum area of rectangle =  l×b = (4.5)² = 20.25 cm² 424. (b) Let r = Radius of 3 smaller laminas In  ADC, (2r)² = r² + DC²

eYrai dnag

P

 462 = =6 cm s 77 419. (b) Width of the border = x m

=

Q

kgei snhe

r

Area of  BCD =

2 3r  OC = 3 DC

 DC =

2r 2 × 3r = 3 3 Radius of larg er lamina = OE

=

2+ 3 r 3

Area of 3 laminas = 3 r² Area of larger lamina  2 3    r =  3    



= =

2



4  3  4 3  r² 3

7  4 3r² 3 D

A

r =

1 22 77 ×7×7= cm²  (7)² = 4 7 4 2

2r +r= 3

OC + CE =

c ir cular

r

r r B r

r

O

r

C E

Residual area 7  4 3   – 3 =  3 r2  

357

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3  Required ratio 4 3 – 2 2  r   3   = 74 3 r 2 3

= =

PQ = QR = RS = 4 cm Perimeter of shaded region = Perimeter of semi-circle PTS + Perimeter of semi-circle QUS + Perimeter of semi-circle PVQ =  (6) +  (4) +  (2) = 12 cm 428. (b) Circumference of circular path = 2 × 50 m = 10000 cm and circumference of wheel = 2 × 50 = 100 cm Distance covered in 60 min =  10000 cm Distance covered in 15 min 10000 =  ×15 = 2500 cm 60 Number of revolutions   

2

4 3 –2 7–4 3 × 74 3 7–4 3 28 3 – 48 – 14  8 3 49 – 48

= 36 3 – 62 = 36 × 1.732 – 62 = 62.352 – 62 = 0.35 425. (a) Circ um fe re nc e of c ir cle = 2 ×42 22 ×42 = 264 cm 7 Perimeter of square = 4x  264 = 4x  x = 66 cm 426. (a) Ar ea of 2 bigg er s em i-

= 2×

r² circles = 2× 2  0.5 2 1 0.25  × = = 2  cm²  2 4  2  and area of 5 smaller semi –

5r² 1 circles = =5× × × 2 2

 0.5 2      4 

0.25 1.25 = 1– – 4 32 = 1–

3 1000 ×1 = 50m 60 1 Area of field = d²  2 1 = × (50)² = 1250 m² 2 430. (c) Let the side of an square be a cm. By given condition, Area of square – Area of an

d=

equilateral triangle =

=

128 – 8 – 5 128

1 3 a² = 4 4   3 1   a² 1 – 4  =   4  a² –

T U

. .R

QO

D

50 m F 20 m C

50 m S

a

A

E

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B

a

B

 a 2 =6 2  a = 6 cm

1 Area of  ABC = × a² 2 1 = ×6×6=18 cm² 2 434. (b) a = 9cm, b = 10 cm and c = 11 cm 

s=

9  10  11 = 15 cm 2

A =

s s–a s–bs–c

15 15–915–1015–11

=

15×6×5×4 =30 2 = 42.3 cm² 435. (a) As we know that, if the length of square and rhombus are same, then the area should be same. 436. (a) Area of circle = 4 cm² (given)   r² = 4  r = 2 cm C

50 m

O

V

A

6 2 cm

a²+a² = 6 2

=

20 m

128 – 13 cm² 128 427. (a) Given, OS = 6 cm =

P

1 4

1  a² (4 – 3 )=1  a²= 4– 3 a = (4 – 3 )–1/2 cm  431. (d) Suppose a pole is fixed at a point C. Area of field in which the horse  can graze = Area of field in which the horse can graze = Area of curve CFE

 5 – 16 128

C

2500 = 25 100 429. (b) The distance covered by a man diagonally is

=

wwM wa.th Les B aryn

5 0.25 1.25 = × = cm² 2 16 32 Area of rectangle ABCD = 2×0.5 = 1 cm² Area of remaining portion

1 3.14  20  20 ( r²) = 4 4 = 314 m² 432. (a) Here width of sheet is 20 cm, which is the maximum diameter of the circular sheet.  Remaining area of sheet = Area of rectangle sheet – Area of circular sheet = 25×20 –  (10)² = 500 – 314 = 186 cm² 433. (b) Let the other sides of a right isosceles triangle be a cm. In  ABC,

=

r

 r

eeYa ridn agv .iSn i

3 –2

R Enak geisnh

4 = 

A

º 30 30º

2c

D

m

B

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E x

y

B

C

BC = 



x ²  y² Area of  ABC 1 1 = × x × y = xy 2 2 Area of semi-circle BACB

x ²  y² = 4 Area of shaded portion = Semi-circle ABDA + Area of semi-circle AECA(Area of semi-circle BACB – Area of  ABC)

C

of  ABC = Area of  ABC

A

438. (d) In  ABC, AC² =

28 cm

=

28²  21² =

784  441

1225  AC = 35 cm A

D

E

443. (c) Let the length of rectangle = and breadth of rectangle =

1 × OA × OB 2

x x x² × = 2 6 12 and area of square = x² Hence , ar ea of re maining x² 11x ² = 12 12 444. (b) Remaining perimeter

portion = x² – 4 cm

2 cm

cm

cm 20

4 cm

440 m 1000

and circumference =  ×d B

C

1 = ×20×20 = 200 cm² 2 and area of sector OACBO

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18 cm

 2r  =  4  4+10+2+10+2   = 2 × 3.14 × 4 + 24 = 25.12 +24 = 49.12 cm = 49.1 cm (approx) 445. (b) Distance travel in 1 revolu-

O

A

10 cm

4 cm

tion = 20

r² 1  2 + ×BC×BA – ×r 1 2 2 4 22 1 35 35 1 = × × × + × 21 7 2 2 2 2

=

4 cm 10 cm

B

90º

x 6

=

B

441. (d) Area of  AOB =

x 2

 Area of rectangle

5 cm

Area of rectangular sheet = 5×2 = 10 cm² and area of circle =  (1)² =  cm²  R equired area = Area of sheet – Area of circle = (10 – ) cm²

C 21 cm Area of shaded portion = Area of Semi - circle ACE + Area of  ABC – Area of quadrant circle BCD

2(14)90º 22 14  2 = × 360º 7 4 = 22 cm.

=

2 cm 4 cm 4 cm

x ² y²   x ²  y²   –  4  + Area   4 4

D

wwM wa.th Les aBryn

=

v.iSn ir

D

360º ×15º = 90º 60º  Required distance

=

4 cm

3 3 (AB)² = (4 3 )² 4 4 = 12 3 cm² 437. (a) In  ABC, A

=

r²θ 3.14  20  20  90º = 360º 360º 3.14  400 = = 314 cm² 4  Area of minor segment = Area of sector OACBO – Area of  AOB = 314 – 200 = 114 cm² 442. (b) Angle made in 60 min by minute hand of a clock = 360º and angle made in 15 min by minute hand of a clock =

eYrai dnag

Area of equilateral  ABC

2 cm

Now, AB = 2 AD = 4 3 cm

kgei snhe

 AD = 2 3 cm



22 × 21 × 21 7 4 51135 1 = + (21×28–33×21) 4 2 1925 1 = + (–105) 4 2 = 481.25 – 52.50 = 428.75 cm² 439. (b) Let the sides of squares S1 and S2 are a and b, respectively. So, perimeters of square S1 and S2 are 4a and 4b, respectively. By given condition, 4a = 4b + 12 ....(i) a=b+3 and a² = 3 (b)²–11  (b+3)² = 3 b²–11  b²+6b+9 = 3b² – 11  2b² – 6b – 20 = 0  2b² – 10b + 4b – 20 = 0  2b (b–5)+4(b–5) = 0  (b–5)(2b+4) = 0  b = 5m ( b  –2) On putting the value of b in Eq. (i), we get a = 5+3 = 8  Perimeter of S1 = 4×8 = 32 m 440. (b) From a rectangular sheet of cardboard of size 5×2 cm², a circle of radius 1 cm, can be cut-off.

× 28 –

4 cm

OD AD

ERna

In  OAD, tan 30º =

44000 cm 1000 44000  7 = 14 cm  d= 1000  22 446. (c) Side of an e quilater al triangle is a =

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3 a 2

 Radius of incircle

=

a a 3 1 × = 3 2 3 2 C

aS

R

a

Q

x

A B D  Diameter of incircle A

 Diameter of circle

 2 13  26cm A 451. (d)

wwM wa.th Les B aryn B

AD 

AB 2  BD 2  6 2  3 2

2

= R

r 2 r 2   Ratio of area  R 2 R 2

B

a 2 3

3 88 16 3cm2 4 radius of incircle (r)



a =

6 2 3

 3



2 3

8



2 3

4 3

Area of inscribed circle =  r 2 2

 4   = 22 × 16   3 7  3

 

2

 Area = r 2 =  3 = 3  452.(d) Area of parallelogram = base  height = 27  12 = 324sq. cm. Again,

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3 2 a 4



   3  3  3 sq.cm Alternatively : Inradius(r)=

C

a=8cm

Area of triangle 

1  3 3  3cm. 3

Radius of incircle  r 

=

r 2 1 22   77 2 2 7

r

 Area of circle  r 2

a

Area 

C

D

OD = In-radius

r

a 2 and Radius of circum-circle

367  7 metre 36

A

 36  9  27  3 3cm.



=

B

r 

DB = DC = 3cm.

R

a

22   r   2  36  7 

O

r²θ ×6²  80º = = 8 cm² 360º 360º  Required difference = 36 – 8 = 28 cm²

A

22 1414 616sqcm . . 7 454. (d) r  2r  36 

 77sq .metre 455. (b) Let the radius of circle be ‘r’  2  r = 24   r = 12  Area of circle =  (12)2 = 144   Area of the rectangle = area of circle = 144  456. (b)

 169  13cm

C

22 r  442  r  14cm 7

 Area of circle  r2

OC 2  AC 2

Radius ‘OA’ =

R Enak geisnh

a² = 2x² 3

D

2

B

C

 122  52  144  25

a² = Area of square  x² = 6 447. (c) Radius of circle, r = 6 cm  Area of circle= r² =  ×6² = 36 cm² and area of sector subtending an angle of 80º at O

448.(a)

453.(d) 2r  2(18  26)

OC = 12cm AC = CB = 5cm



 a 2      = x² + x²  3

324  9cm 36

22 14  r    36  7 

O

 a  a  =2    = 2 3 3   Let side of a square be x.



h 

2 = R

O P 30º

   r 2 360 360 30 (21 21  7  7) =  360 22 1   28 14 = 102.67cm2 = 7 12 450. (c)

324 = 36  h

eeYa ridn agv .iSn i

triangle is

a2 /4 2 1   1: 2 a2 /2 4 2 449.(a) Larger Radius (R) = 14 + 7 = 21cm Smaller Radius (S) = 7cm  Area of shaded portion 

r

then the altitude of equilateral

Required area  22 16   = 16 3   21 



360

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=

16 14.372 10.95cm2 21

86 24 = 86 7

 area of square

24 576 2     cm  7  49

a 2 3 incirle’s diameter (d) inradius =

Alternatively, let side of largest square = x  AP = (8 - x)cm and MC = (6 - x)cm In  ABC and  DMC,

 B =  M = 90°  C =  C = (common)

2a a = = 2 3 3

 ABC ~  DMC,

circum-radius =

a

(D) =

2a

3  d:D:H =

a

:

2a

3

3

:



3

 circumcircle’s diameter

BC AB 6 8 =   MC DM 6x x

24 3x  24 4x  x  cm 7  area of square = x2 =

3 a  2:4:3 2

576 2 cm 49

459.(b) For 3x + 4y = 12 By putting x = 0, y = 3 By puttting, y = 0, x = 4 For 6x + 8y = 60,

ERna

458. (a)

 Area of OAB

kgei snhe

=

15 2 By putting y = 0, x = 10

wwM wa.th Les aBryn

By putting x = 0, y 

Side of maximum sized square

AB×BC AB+BC

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1 1  OB OA  43  6 2 2  Area of trapezium 



75 6 2

75 12 63   31.5sq.units 2 2

460. (c) Side of the first square



Area



200  10 2metre

Its diagonal 

2  side

 10 2  2  20metre

 Diagonal of new square  2 20  20 2 metre  Its area 

Y

M

1 × OD× OC 2

1 15 75  10  2 2 2

2

457.(b) height = h 3 a (a ? side of ? ) 2

 Area of Δ OCD =

v.iSn ir

16 (211.732  22) 21

eYrai dnag

=

1 2 diagonal  2

C (0,15 2 A (0,3) X’

O

Y’

X D (10,0) B (4,0)

1  20 2 20 2 2  400sq.metre

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CHAPTER

17

There are two types of geometric shapes:1. 2D 2. 3D

What is 3D....?

A three-dimensional shape is a solid shape that has height and depth. For example, a sphere and a cube are three-dimensional, but a circle and a square are not.

CUBOID

CUBE

TRIANGULAR PRISM

TRIANGULAR BASED

SQUARE BASED PYRAMID

CYLINDER

PYRAMID

CONE

surface area is the circumference of the base of the solid and the face parallel to it. The Total Surface area is the sum of both the curved surface area and the area of the base and top.

What is Volume.....?

Volume is the measure of the amount of space inside of a solid figure, like a cube, ball, cylinder or pyramid. It's units are always "cubic", that is the number of little element cubes that fit inside the figure. Difference between Curved surface area and Total surface area The area of all the curved surfaces of any solid. The Curved

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HEMISPHERE

SPHERE

Cuboid (Parallelopiped) A cuboid is a 3 dimensional shape. It is a solid figure which has 6-regular faces. 12 edges, 8 vertices and 4 diagonals.

Height

wwM wa. th Les B aryn

(1) Cube (2) Rectangular Prism (Cuboid) (3) Cylinder (4) Cone (5) Sphere and Hemisphere (6) Prism (7) Pyramid

Face

R Enak

3D shapes: They have surface area and volume.

geisnh eeYa ridna gv.i Sni

Mensuration is the branch of mathematics which deals with the study of different geometrical shapes, their areas and Volume. In the broadest sense, it is all about the process of measurement. It is based on the use of algebric equations and geometric calculations to provide measurement data regarding the width, depth and volume of a given object or group of objects. while the measurement results obtained by the use of mensuration are estimates rather than actual physical measurements, the caluclations are usually considered very accurate.

r

MENSURATION 3–D (THREE DIMENSIONAL)

Le ng th

ea br

dth

Look at this shape. There are 3 different measurements: Length, Breadth, Height

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Formula The volume is found using the formula: Volume = Length × Breadth × Height Which is usually shortened to: V=l×b×h Or more simply: V = lbh (a) V = l  b  h cubic units

=

l +b +h

=

162 +182 + 242

2.

Sol.

EXAMPLES

1.

l 2  b2  h 2

d=

900 + 576 + 324

=

The dimensions of a cuboid are 16 cm, 18 cm and 24 cm. Find: (a) Volume (b) Surface area (c) Diagonal Sol. (a) Volume = l × b × h = 16 ×18 × 24 = 6912 cm³ (b) Surface area = 2 (lb + bh + hl) = 2(16×18 + 18×24 + 24×16) = 2208 cm²

1800

d = 30 2 m A brick measures 20 cm × 10 cm × 7.5 cm. How many bricks will be required for a wall 20 m × 2 m × 0.75 m ? Sol. Number of bricks = 3.

=

20 × 2 × 0.75 ×100 ×100 ×100 20 ×10 × 7.5

= 20,000 A rectangular sheet of metal is 80 m by 30 m. Equal squares of side 8 m are cut off at the corners and the remainder is folded up to form an open rectangular box. Find: (i) Volume (ii) Total surface area (iii) Surface area of box. Sol. (i) When four square of 8 cm are removed from four corners of rectangular sheet. 4.

8m

64 m

8m

14 m

8m

14 m

64 m

8m

Length and Breadth of remaining rectangular sheet will 64 cm and 14 cm & height of sheet will be 8 cm. Volume of open rectangular box = Length × Breadth × Height = 64 × 14 × 8 = 7168 m³ (ii) Surface area = 2(Le ngth + Breadth) × Height = 2(64 + 14)×8 = 2 × 78 × 8 = 1248 m²

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(iii) Total surface area = Surface area + Base area = 1248 + Length × Breadth = 1248 + 64 × 14 = 1248 + 896 = 2144 m² 5. In swimming pool measuring 90 m by 40 m, how much water will be displaced by 150 men, if the displacement of water by one man is 8 cm³, what will be the rise in water level? Sol. Volume of water displaced by 150 men = Volume of water came out (Let the height raised in water = h)  8 × 150 = 90 × 40 × h  h=

1 m = 33.33 cm 3

6.

Total Volume of wall Volume of one brick

ERna

(v)

l 2  b 2  h 2 units To find total surface area of a cuboid if the sum of all three sides and diagonal are given. Total surface area = (Sum of all three sides)² – (Diagonal)² Note : For painting the surface area of a box or to know how much tin sheet is required for making a box, we use formulae (iii) i.e. Total surface Area. To find the length of the longest pole to be placed in a room, we use formulae (iv) i.e. Diagonal.

2

d=

wwM wa. th Les aBryn

=

2

= 1156 = 34 cm Find the length of the longest pole that can be placed in a room 30 m long, 24 m broad and 18 m high.

(b) V = A1×A2 ×A 3 cubic units Where, A1 = area of base or top = l b sq. units A2 = area of one side face =bh sq. units A3 = area of other side face = hl sq. units (ii) Lateral surface Area /Curved surface area/ Area of four walls = Perimeter of Base  height = 2 (l + b)  h sq.units (iii) Total surface Area = 2(lb + bh + hl) sq. units (iv) Diagonal of cuboid

2

kgei snhe eYari dnag v.iSn ir

(i)

(c) Diagonal

A rectangular water reservoir is 15 m × 12 m at the base. Water flows into it through a pipe whose cross section is 5 cm by 3 cm at the rate of 16 m/s. Find the height to which water will rise in the reservoir in 25 minutes. Sol. Volume of water comes out fro m pi pe in 1 sec =

5 3 × × 16 m³ = 0.0240 m³ 100 100

Volume of water comes out from pipe in 25 min = 0.0240 × 25 × 60 = 36 m³  Volume of water pour into tank = Volume of water comes out from pipe.  15 × 12 × h = 36 (  h = rise in level of water)  h = 0.2 m 7. The sum of length, breadth and height of a cuboid is 25 cm and its diagonal is 15 cm long. Find the total surface area of the cuboid. Sol. We have the total surface area, = (25)² – (15)² = 625 – 225 = 400 sq. cm. CUBE A cube whose length, breadth and height are all equal is called a cube. A cube has 6 equal faces, 12 equal edges, 8 vertices and 4 equal diagonals.

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a

Consider a cube of edge a units. It is a special type of cuboid in which l = b = h = a units i.e. each face is a square. (i) Volume = a3 cubic units (ii) Lateral surface Area = 4a2sq.units (iii) Total surface Area = 6a2sq.units (iv) Diagonal of cube (d) = (v)

3 a units Face diagonal of cube =

(vi)

2 a units Volume of cube

 su rface area 3  cubic units =   6  

EXAMPLES 1.

Diagonal = a 3 = 5 3 = 8.660 = 8.66 cm 2. Three cubes of volume 1 cm³, 216 cm³ and 512 cm³ are melted to form a new cube. What is the diagonal of the new cube ? Sol. Volume of new cube = 1 + 216 + 512 = 729 cm³  Edge of new cube =

3

729 = 9 cm

Surface area = 6a² = 6 × (9)² = 486 cm²  Diagonal of the new cube 

r 2 r

h

h

Lateral surface (unrolled) d d

r

Right Circlular Cylinder Ab = r² = 0.25 d² AL = 2 rh V = r²h

Base (circle)

34398 Total surface area = 13

= 2646 cm² 6a² = 2646 a² = 441 a = 21 cm  Volume of cube = a³ = (21)³ = 9261 cm³ 5. A solid cube with an edge of 10 cm is melted to form two equal cubes. The edge of smaller cube to the bigger cube is. Sol. Vo lume of larger cube = summati on o f vo lume of smaller cubes  Let the edge of smaller cubes be ‘a’  (10)³ = (a)³ + (a)³  (10)³ = 2(a³)

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Edge of a cube is 5 cm. Find: (a) Volume (b) Surface area (c) Diagonal Sol. Volume = a³ = (5)³ = 125 cm³ Surface area = 6a² = 6 × (5)² = 150 cm2

A right circular cylinder is a cylinder whose base is a circle and whose elements are perpendicular to its base.

geisnh eeYa ridna gv.i Sni

a

R Enak

a

The surface area of a cube is 864 cm². Find the volume. Sol. 6a² = 864  a² = 144  a = 12 cm a³ = (12)³ = 1728 cm³ 4. The Cost of painting the whole surface area of a cube at the rate of 13 paise per sq. cm is Rs. 343.98. Then the volume of the cube is: Sol. Cost of painting the whole surface are = Rs. 343.98 = 34398 paise

r

3.

 a=

10 1 3

2

10 edge of smaller cube  edge of bigger cube = 3 2 10 1 = 3 2

= a 3= 9 3 = 15.6 cm (approx) Rakesh Yadav Readers Publication Pvt. Ltd.

Lateral Surface Circular Base

3.

4.

A b  r ²  A b  d² 4 Lateral surface Area, AL A L = 2rh

Volume, V

A Right circular cylinder is a threedimensional object with two congruent circles as parallel bases and a lateral surface consisting of a rectangle. Volume and surface area of a Right Circular Cylinder. if ‘r’ is the radius of a circular base of the cylinder and 'h' is the height of the cylinder. Circular Base

2.

A L = dh

Right Circular Cylinder

Solid View

1.

Properties of a Right Circular Cylinder The axis of a right circular cylinder is the line joining the centers of the bases. For any oblique or non-oblique sections which do not pass any one base, the center of which is at the axis. A right circular cylinder can be formed by revolving a rectangle about one side as axis of revolution. Every section of a right circular cylinder made by a cutting plane containing two elements and parallel to the axis is a rectangle. Fromulae for Right Circular Cylinder Area of the base, Ab

Wire-Frame View h r

V = Ab h V = r²h  V = d2 h 4

Total surface Area, AT Total surface area (open both ends), AL = A Total surface Area (open one end), A = Ab + AL Total surface Area (closed both ends), A = 2Ab + AL

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1.

Find the volume of an iron rod which is 7 cm long and whose diameter is 1 cm. Sol. Diameter = 1 cm. Radius =

5.

Th e radius of a wire is decreased to one third. If volume remains same, length will increase: Sol. Let, radius = R and length = h, Volume = πR2h

1 cm 2

New radius =

Height = 7 cm Volume of Cylinder =  r²h =

Let, new length = H 1

=

2.

A cylindrical iron rod is 70 cm long, and the diameter of its end is 2 cm. What is its weight, reckoning a cubic cm of iron to weigh 10 grams ? Sol. Volume of the iron rod = π r² h =

22  1  1  70 = 220 cm³ 7

 weight of the cylinder

= 7.

220  10 = 2.2 kg. 1000

A cylinderical vessel, whose base is 14 dm in diameter holds 2310 litres of water. Taking a litre of water to occupy 1000 cubic cm, what is the height of the vessel in dm?

ERna

3.

22  70  70  h = 2310 × 1000 Sol. 7

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4.

πR 2H 9

6.

= 1540 cubic m

A powder tin has a square base with side 8 cm and height 13 cm. Another is cylindrical with radius of its base 7 cm and height 15 cm. Find the difference in their capacities. Sol. Difference in capacities

22  7  7  15 – 8  8  13 = 7 = 2310 – 832 = 1478 cubic m A metallic sphere of radius 21 cm is dropped into a cylinderical vessel, which is partially filled with water. The diameter of the vessel is 1.68 metres. If the sphere is completely submerged, find by how much the surface of water will rise.

4 Sol. Volume of sphere = πr ³ 3  4 22  =  3  7  21  21  21 = 38808 cubic cm. 

2

πR 2H or H = 9h  πR² h = 9

22 35 35 10000  =  7 100 100  60  24

 1540  Rise in level=  25  12  = 5.13 m



Volume = π   R  × H 3

22 1 1 11   7 7 2 2 2

= 5.5 cubic cm Water flows at 10 km per hour through a pipe with cross section a circle of radius 35 cm, into a cistern of dimensions 25 m by 12 m by 10 m. By how much will the water level rise in the cisten in 24 minutes ? Sol. Volume flown in 24 minutes

1 R 3

= 150 cm = 15 dm. Find how man y pieces of 3 money cm in diameter and 4

down to form a cube whose edge is 3 cm long ? Sol. Volume of one piece of money

 h = 1.75 cm

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22 24.5  24.5   32 7 22 = 15092 m³ Since 1 cubic metre = 1000 kg.  1 cubic metre = 1 metric ton

=

1000kg 1metric ton

 Volume of cylinder = 15092 metric tones

Right Circular Cone

Co ne i s a three dimen sion al geometric shape. If one end of a line is twisted about a second set line while keeping the lines other end fixed, we get a cone. The point about which the line is curved is known as the vertex and the base of the cone is a circle. The vertex is directly above the centre of the bottom. Vertex

Height

= π r² h =

22 3 3 1    7 8 8 8



22 3 3 1    n  3  3  3 7 8 8 8

n

=

n

= 488.72

3337888 22  3  3  1

θ

Semi Vertical Angle Slant Height

2310  1000  7 22  70  70

1 cm thick must be melted 8

22  84  84  h = 38808 7

The diameter of a cylindrical tank is 24.5 metres and depth 32 metres. How many metric tons of water will it hold ? (One cubic metre of water weighs 1000 kg.) Sol. Volume of the cylinder

1dm 10 cm

 h=

8.

9.

kgei snhe eYari dnag v.iSn ir

EXAMPLES

Center

Radius

Properties of a Cone There are number of properties of a cone. Some of them are as follows: • Volume of a Cone • Lateral surface Area of a Cone • Total Surface Area of a Cone

Definition of Right Circular Cone A right circular cone is one whose axis is perpendicular to the plane of the base. We can generate a right cone by revolving a right triangle about one of its legs.

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Where, Base area = π r²

Radius

 Curved surface

Volume of a Circular cone =

1 3

×  ×Radius² × Height EXAMPLES

Axis 1.

Vertex

Formulae of Right Circular Cone

Find the volume of a cone whose diameter of the base is 21 cm and the slant height is 37.5 cm. 2

37.5

– 10.5

 volume =

1 2 πr h 3

The surface area of a right circular cone is the sum of area of base and lateral surface area of a cone. The surface area is measured in terms of square units. Surface area of a cone = Base Area + Lateral surface Area of a cone

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 l=

Area of a right circular cone = r ( r +l ) Here, l= r2  h2 Where, r = Radius h = Height and l = Slant height of cone

Volume of a Right Circular Cone

The volume of a cone is one third of the product of the area of base and the height of the cone. the volume of a right circular cone is measured in terms of cubic units. Volume of a right circular cone can be calculated by the following formula. Volume of a right circular cone

3.

r 2  h2 =

Sol.  Volume =

1 =  Base area  height 3

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2

10

2

 24

1 2 1 22 2 πr h =   r  24 3 3 7 = 1232 = r² = 49 = r = 7

= 25 cm

1 2 πr h = 3

1 2 π2 h 3

1 22 h = 22 6 3

 h = 18 cm From a solid right circular with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and base is removed. Find the volume of the remaining solid. Sol. Volume of remaining solid

6.

= πr 2 h –

2

22 1 22 × 6 × 6 × 10 – × × 7 3 7 6 × 6 × 10

24

7. 2

 7 

1 2 πr  h 3

=

1  r2 h 3

Slant height =

A cylindrical piece of metal of radius 2 cm and height 6 cm is shaped into a cone of same radius. The height of cone is:

Sol. Volume of Cone =

1 2 πr h 3

= 26 cm  Curved surface area of cone =  r l = 3.14 × 10 × 26 = 31.4 × 26 = 816.4 cm² If a right circular cone of vertical height 24 cm has a volume of 1232 cm³, then the area of its curved surface in cm² is:



5.

=3

πr 2 h 3

Volume of cylinder = π × (2)² × 6  Volume of cone = Volume of cylinder

1  3.14  25x 2   12x  = 2512 3 x=2  Radius = 5x = 10 cm & Height = 12x = 24 cm

= πr 2 πrl = πr r l  Surface Area of a Right Circular Cone can be calculated by the following formula.

πr2 h

Number of cones = 1

1 22  10.5  10.5  36 =  3 7

 volume =

1 2 πr h 3

Volume of 1 cone =

Surface Area of a Right Circular Cone:-

Note:Area is measured in square units and volume is measured in cubic units.

Sol. Volume of 1 cylinder = πr 2 h

= 36 cm

= 4158 cm³ 2. The radius and height of a right circular cone are in the ratio 5 : 12 and its volume is 2512 cm³. Find the slant height, radius and curved surface area of the cone. (Take π = 3.14) Sol. Let radius = 5x, height = 12x

1  r²h 3

A right cylindrical vessel is full with water. How many right cones having same diameter and height as those of right cylinder will be needed to store that water ?

geisnh eeYa ridna gv.i Sni

Sol. h =

2

R Enak

For a right circular cone of radius r, height h and slant height l, we have Lateral surface area of a right circular cone =  rl Total surface area of a right circular cone =  (r + l)r Volume of a right circular cone=

4.

r

Slant

 22  =  7  7  25  = 550 cm²

=

22 2  6  6  10  7 3

=

5280 2 = 754 cm³ 7 7

The slant height of a conical tomb is 17

1 metres. If its 2

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diameter be 28 metres, find the cost of constructing it at Rs. 135 per cubic metre and also find the cost of white-washing its slant surface at Rs. 3.30 per square metre. Sol. Height of the cone 2

=

= 11.

If the heights and the curved surface areas of two circular cylinder are in the ratio 1 : 3 and 4 : 5 respectively. Find the ratio of their radii.

22 35  14  = 770 m³ 7 2  Cost of white washing = 770 × 3.30 = Rs. 2541 8. Radius of the base of a right circular cone is 3 cm and the height of the cone is 4 cm. Find the total surface area of the cone. Sol. Applying to the question, Total surface area= × r (l + r)

A frustum of a cone or truncated cone is the result of cutting a cone by a plane parallel to the base and removing the part containing the apex. Upper part containing the vertex



4 3 3

l  h2  r 2



16 1  4) = 25 4 = 4 : 25 [  ratio of diameters = ratio of radii] 10. If the volumes of the two cones are in the ratio 4 : 1 and their heights in the ratio 4 : 9, what is the ratio of their radii ?

Sol. Ratio of radii =

4 : 1  

Sol.

l

6 cm

The height is the line segment that joins the two bases perpendicularly.

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2

102  6 – 2

= 10.77 cm

AL = π (6 + 2)×10.77 = 270.78 cm² 1 3

h

l

2.

R

2

h2  R – r 

Unfold of a Truncated Cone

1 1 :  4 9

l =

2 2 V= π 10  6  2  6  2 =544.54 cm³

The radii are of their bases are ‘r’ and ‘R’. The slant height is the shortest possible distance between the edges of the two bases. The slant height of the truncated cone is obtained by applying the Pythagoras theorem for the shaded triangle: l ² = h² + (R – r)² l=

l ² = 10² + (6 – 2)²

AT = 270.78 + π×6² + π×2² =396.35 cm²

22  3  8 528 3 = = = 75 sq. cm 7 7 7

If the heights of two cones are in the ratio 1 : 4 and their diameters in the ratio 4 : 5 what is the ratio of their volumes ? Sol. We have, Ratio of Volumes = (4 : 5)² × (1 :

Calculate the lateral surface area, surface area and volume of a truncated cone of radii 2 and 6 cm and height of 10 cm.

Frustum of a right cone

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

1.

10 cm

ERna

2

1 .π.h R 2  r2  R.r  3

EXAMPLES

Frustum of a right circular cone

=

9.

V=

kgei snhe eYari dnag v.iSn ir

1 22 21   14  14  = 2156 m³ 3 7 2 Cost of constructing the conical tomb = 2156 × 135 = Rs. 291060 Curved surface area of the conical tomb = πr l

=



AT = π [ l (R + r)+ R² +r² ] Volume of a Truncated Cone

= (4 : 5) × (3 : 1) = 12 : 5

Volume of the cone

2

AL = π (R + r) l

Surface Area of a Truncated Cone

 1 Sol. Required ratio = (4 : 5) × 1 :   3

21  35  2 m   – 14 = 2  2

22 3 = 7

Lateral Area of a Truncated Cone

4 : 1  9 : 4 = 3 : 1

Calculate the lateral surface area, surface area and volume of a truncated cone of radii 10 and 12 cm and a slant height of 15 cm.

Sol.

h

15 cm

12 cm

AL = π (R + r) l =π(12 + 10)×15 = 1,036.73 cm² AT=1036.72 + π ×12² + π ×10² = 1803.27 cm²  l² = h² + (R – r)² 15² = h² + (12 – 10)²

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= 5,666.65 cm³

Surface area of a sphere = 4 times the area of its g r e a t circle = 4πr2 = πd²

(ii)

Volume of a sphere =

Sphere

=

4 πr 3 3

π 3 d 6

(iii) For a spherical shell if R and r are outer and inner radii respectively, then the volume of a shell is = =

1.

1 cm. = d1 4 Diameter of metal ball = 3cm = d2 Volume of leaden ball

=

4 3

3 Volume of sphere = πr =

Volume of metal ball =

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Two spheres each of 10m diameter are melted down and recast into a cone with a height equal to the radius of its base, Find the height of the cone. Sol. Here, d = 10m  Radius of cone = height of the cone (Given)  r=h Volume of sphere =

π 3 (d) 6

π π  103 =  1000 6 6

= 523.599 cu. m. Volume of two sphere = 1047.2 cu. m. Volume of the cone = Volume of two spheres

π 3 d2 6

π 3 3 = 14.137 cu. m. 6  Number of leaden ball =

=

π 3 d 6

π (13.5)3 = 1288.25 cu. m. 6

3 3 π 1  4  d1  π 3   d = =     1 6 4 3 2 6

= 0.0082 cu. m.

Surface area = 4πr2 = πd2 = π(13.5)² = 572.56 sq. m.

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= 1000 = 10m How many leaden ball of a

=

The diameter of a sphere is 13.5m. Find its surface area and volume. Sol. Here, d = 13.5 m

=

3  1047.2 π

1 cm. in diameter can be cast 4 out of metal of a ball 3 cm in diameter supposing no waste. Sol. Here, diameter of leaden ball

EXAMPLES

2.

Every section made by a plane passed through a sphere is a circle. If the plane passes through the centre of a sphere, the plane section is a great circle; otherwise, the section is a small circle (Fig. 2). Clearly any plane through the centre of the sphere contains a diameter. Hence, all great circles of a sphere are equals have for their common centre, the centre of the sphere and have for their radius, the radius of the sphere. Surface Area and Volume of a Sphere: If ‘r’ is the radius and ‘d’ is the diameter of a great circle, then

3.

π 3 (D – d3 ) 6

=

Great and Small Circles:

=

 h³  h

4 π(R 3 – r3 ) 3

R Enak

A sphere is a solid bounded by a closed surface every point of which is equidistant from a fixed point called the centre. Most familiar examples of a sphere are baseball, tennis ball, bowl, and so forth. Terms such as radius, diameter, chord, and so forth, as applied to the sphere are defined in the same sense as for the circle. Thus, a radius of a sphere is a straight line segment connecting its centre with any point on the sphere. Obviously, all radii of the same sphere are equal. Diameter of the sphere is a straight line drawn from the surface and after passing through the centre ending at the surface. The sphere may also be considered as generated by the complete rotation of a semicircle about a diameter.

 h³

r

1 π 14.866 122  102  12  10 3

(i)

geisnh eeYa ridna gv.i Sni

h = 152 – 22 = 14.866 cm V =

14.137 = 1728.00 0.0082

4.

A metal sphere of diameter 14 cm is dropped into a cylindrical vessel, which is partly filled with water. The diameter of the vessel is 1.68 metres. If the sphere is completely submerged, find by how much the surface of water will rise. Sol. Radius of the sphere = 7 cm Volume of sphere  4 22   7  7  7 =    3 7

1 cu. m 3 Volume of water displaced by 1 sphere = 1437 cu cm. 3 Let the water rise by h cm,

= 1437

Then,

22  28  28  h 7



1 2 πr h = 1047.2 3

= 1437

1 3

or h =



1 πh3 = 1047.2 3

4312  7 7  22  28  28  3 12

= 0.58 cm

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5.

Find the weight of an iron shell, the external and internal diameters of which are 13 cm and 10 cm respectively, if 1 cubic cm of iron weighs 8 gms. Sol. Volume of iron shell

=

4 22 1197   = 627 cu cm. 3 7 8

6.

Weight of iron shell = 627 × 8 = 5016 gms. = 5.016 kg. Find the surface area of a sph ere whose vo lume is 310464 cu cm.

Sol.

4 3 πr = 310464 3

or r3 =

310464  3  7 = 74088 4  22

 r = 42 cm

Volume of hemisphere =

= 19404 cm³ Find the curved surface area of a hemisphere of radius 21 cm. Sol. According to the question, We have Curved surface area = 2πr² 2.

22   =  2  7  21  21 cm²

= 2772 cm² A hemisphere of lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find the radius of the base. Sol. Volume of hemisphere 3.

=

2156 cubic cm 3

ERna

Let, the radius of the base of cone be ‘r’ cm.

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Then,

1 22 2 2156   r  49 = 3 7 3

or r² =

2.

Curved surface area or Surface

3.

area of hemisphere = 2πr2 Total surface area of solid hemisphere = 2πr2 πr2  3πr2 EXAMPLES

There are all Prism: Square CrossPrism Section

2156  7  3 = 14 3  22  49

 Radius (r) =

1.

For example, a cylinder is not prism, because it has curved sides. Try drawing a shape on a piece of paper (using straight lines) Then imagine it extending up from the sheet of paper ...... that's a prism !

1  4 22  = 2  3  7  7  7  7 

22  42  42 7 = 22176 sq. cm.

2 3 Volume of hemispher = πr 3

A prism is a polyhedron, which means all faces are flat.

 2 22  =  3  7  21  21  21

= 4

A plane through the center of the sphere cuts it into two equal parts. Each part is called hemisphere.

2 r 3 3

Volume of hemisphere

2  surface area = 4πr

Hemisphere

No Curves :

kgei snhe eYari dnag v.iSn ir

3 4  13   3 π = 3   2  – 5   

r = 21 cm

14 = 3.74 cm

Cube

Cross-Section

Right Prism

A prism is a solid object with: • Identical ends • Flat faces • Same cross section all along its length The cross section of this object is a triangle ..... it has the same cross section all along its length ... so it's a triangular prism.

(yes, a cube is a prism, because it is a square all along its length)

Triangular Prism

Pentagonal Prism

CrossSection

CrossSection

1.

Find the volume of a hemisphere of radius 21 cm. Sol. According to the question, We have Rakesh Yadav Readers Publication Pvt. Ltd.

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Cross-Section

It is "irregular" because the cross-section is not "regular" in shape.

1.

What is the volume of this prism ?

regular pentagon. If the area of the cross-section is 30 cm², then what is the volume of the prism ?

9 2 2 2 2 2 2

2 2 2 2 2

m 5c 12.

Sol. Volume = Base area × height There are 5 squares  Base area = 5 × 2 × 2 = 20 Volume = 20 × 9 = 180 units. 2. The diagram show a prism whose cross-section is a right triangle. What is the volume of the prism ?

Sol. Volume = Base area × Height = 30 × 12.5 = 375 cm³ 5. The diagram shows a prism whose corss-section is an equilateral triangle of lengths 10 cm. Given that its volume is 866 cm², what is the total surface area of the prism ?

r

2

3 in

Sol. Volume = base area × height.

Surface Area = 2 × Base Area + Base Perimeter × Length

3.

1  4  3 = 6 in² 2

Volume = 6 × 9 = 54 in³ The diagram shows a prism wh ose cross-se ctio n is a square. The length of the base of the prism is 19 cm and its volume is 1,539 cm³. What is the total surface area of the prism ?

wwM wa. th Les B aryn

Volume of Prism

R Enak

Base area =

cm

n 9i

4 in

10

10 cm

Surface Area of Prism

EXAMPLES

geisnh eeYa ridna gv.i Sni

Regular and Irregular Prisms All the previous examples are of Regular Prisms, because the cross section is regular (in other words it is a shape with equal edge lengths, and equal angles.) Here is an example of an Irregular Prism: Irregular Pentagonal Prism:

10 cm

Sol. Volume = Base area × Height Base area =

3 × (10)2 = 43.3 4

cm2 Height = Volume  Base area

866 = 20 43.3 C.S.A = Base perimeter × height C.S.A = 3 × 10 × 20 = 600 cm² T.S.A = C.S.A + 2 base area T.S.A = 600 + 2 × 43.3 = 600 + 86.6 = 686.6 cm² 6. There is a 10 cm long prism whose cross-section is an isosceles trapezoid: =

19 cm

Sol. Volume = Base area × Height 1539 = Base area × 19 Base area =

The Volume of a prism is the product of its base area and length. Volume = Base Area × Length Volume = Base Area × Length Curved Surface Area = Base perimeter × Height Total Surface Area = CSA + 2 Base Area

1539 = 81 19

Base area = a² = 81 a =9 C.S.A = Base perimeter × Height = 4 × 9 × 19 = 684 cm² T.S.A = C.S.A + 2 Base area = 684 + 2 × 81 = 684 + 162 = 846 cm² 4. The diagram shows a prism with a cross section that is a

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6 cm 5 cm

10 cm

4 cm 12 cm

What is the total surface area of the prism ? Sol. Base area =

1 × Sum of  2

sides × height

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1 = × (6 + 12) × 4 = 36 cm² 2

= 1350 + 150×2.45= 1717.5 ft³ The diagram shows a prism whose cross-section is an isosceles trapezoid.

8.

Base perimeter = 5 + 6 + 5 + 12 = 28 cm C.S.A. = Base perimeter × h C.S.A. = 28 × 10 = 280 cm² T.S.A. = C.S.A. + 2 Base area T.S.A. = 280 + 2 × 36 = 352 cm² 7.

6 cm

10 cm

What is the volume of the prism ? Sol. Volume = Base area × Height

15 ft

D

E

1 (S um of || 2 sides) × height

The diagram shows a barn. What is the volume of the barn? (The length of the hypotenuse in the right triangle is rounded to the nearest foot.) Sol. Volume = Base area × height Total base area = Area of

Base area =

 BCED + Area of  ABC Area of BCDE = 10 × 9 = 90 ft²

=

1  10  6  5 2

=

1  16  5 = 40 cm² 2

ERna

C N 10

When we think of pyramids we think of the Great Pyramids of Egypt. They are actually Square Pyramids, because their base is a Square.

In  ABC, AB² = AN² + BN² AN² = 49 – 25 AN² = 24

Base

Pyramids

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B

7

Pyramid

Volume = 40 × 8 = 320 cm³

A

7

Base

Notice these interesting things: • It has 4 Faces • The 3 Side Faces are Triangles • The Base is also a Triangle • It has 4 Vertices (corner points) • It has 6 Edges • It is also a Tetrahedron (if all triangles are equilateral triangles) 2. Square Pyramid

kgei snhe eYari dnag v.iSn ir

7 ft 7 ft B 10 ft 9 ft

Pyramid

8 cm

5 cm

A

and they are named after the shape of their base. 1. Triangular Pyramid

Notice these interesting things: • It has 5 Faces • The 4 Side Faces are Triangles • The Base is a Square • It has 5 Vertices (corner points) • It has 8 Edges 3. Pentagonal Pyramid Pyramid

Base

AN = 2 6 ft

 Area of  ABC

=

1  BC  AN 2

=

1  10  2 6 2

Parts of Pyramid

= 10 6 ft²  Total base area

Apex

Notice these interesting things: • It has 6 Faces • Th e 5 Side Faces are Triangles • The Base is a Pentagon • It has 6 Vertices (corner points) • It has 10 Edges Right vs Oblique Pyramid:-

= 10 6  90  ft²





Volume = 10 6  90  15 = 1350 + 150 6

A pyramid is made by connecting a base to an apex.

h

Types of Pyramid

There are many types of Pyramids,

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Sol. Volume =

1 × Area of base × 3

height Volume =

Volume = 98 cm³ The diagram shows a squarebased pyramid with base lengths 6 in and height 8 in. What is the volume of the pyramid?

8 in

2.

AB = 10 ft BC = 18 ft VO = 12 ft VM = 15 ft VN = 13 ft What is the total surface area of the pyramid ? Sol. Total surface area = 2 × area  ABV + 2 × area  VBC + Area of ABCD

geisnh eeYa ridna gv.i Sni

This tells us where the top (apex) of the pyramid is. When the apex is directly above the center of the base it is a Right Pyramid, otherwise it is an Oblique Pyramid. Regular vs Irregular Pyramid:This tells us about the shape of the base. When the base is a regular polygon it is a Regular Pyramid, otherwise it is an Irregular Pyramid.

1  42  7 3

r

h

1 2

1 2

= 2   10  15  2   18  13  10  18

Regular Pyramid

Irregular Pyramid

5.

Sol. Volume

1 × Area of base × 3

height

FORMULAE Volu me

of

pyramid

(ii)

1 × (area of base) × height 3 Curved surface area

3.

=

=

1 × (perimeter of base) × 2

slant height

(iii) Total surface area = curved surface area + area of the base. EXAMPLES 1. The diagram shows a pyramid whose base is a regular pentagon of area 42 cm² and whose height is 7 cm. What is the volume of the pyramid?

1  6  6  8 = 96 in³ 3

Th e di agram shows a rectangular-based pyramid with base length 15 cm and width 8 cm. The height of the pyramid is 20 cm. What is the volume of pyramid ?

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(i)

Volume =

15 cm

1 × Area of base × 3

1 height =  15  8  20 3

4.

7 cm 42 cm²

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3 cm

5 cm

8 cm

Sol. Volume =

8 cm

4 cm

20 cm

Base is Irregular

R Enak

Base is Regular

= 150 +234 +180 = 564 ft² The diagram shows a pyramid with a triangular base ABC. The point D is vertically above the point C. What is the volume of the pyramid ?

Volume = 800 cm³. The diagram shows a pyramid with vertex V and a rectangular base ABCD. M is the midpoint of AB, N is the midpoint of BC and O is the point at the center of the base.

Sol. Volume 1 =  Area of base  height 3 =

1 1  348 3 2

Volume = 16 cm³

Tetrahedron Tetrahedron Facts

372

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2

3 × EdgeLength

Surface Area = Volume =

Area of 4 triangles = 4

= 140.30 The length of one edge of a regular tetrahedron is 9 units. What is its volume?

2.

Sol. Volume = =

EXAMPLES

=

2

2 2 3  9 =  729 12 12

=

3  Edge length

=

3  4 = 16 × 1.732

Volume =

=

=

2 3  Edge length 12

2 2 3  6 =  216 = 25.46 cm³ 12 12

Hollow Cylinder

(iv)

81 3 4

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Thickness (t) = (r2 – r1) EXAMPLES

1.

A hollow garden roller 63cm wide with a girth of 440cm is made of iron 4cm thick. The volume of iron is: Sol. Circumference = 440 cm  2πr = 440

 440  r =  2  22  7  = 70 cm

Inner radius = 70 – 4 = 66 cm Volume of Iron 2 2 = π70  – 66   63

=

22  136  4  63 = 58752 cm³ 7

2.

A hollow cylindrical tube open at both ends is made of iron 2 cm thick. If the e xternal diameter be 50 cm and the length of tube be 140 cm, find the volume of iron in it. Sol. External diameter = 50 cm 50  External radius = 2 = 25 cm

1 bh 2

1 3 = 99 2 2

(r1 + r2)h Total surface area = inner curved surface area + Outer curved surface area + area of Base an d To p = 2πr1  r2   h  (r2  r1 )

2

3 9 3 9 = 2 2

Area of triangle =

(iii)

= 27.71 cm² 4. The total length of the edges of a tetrahedron is 36cm. What is its volume? Sol. A tetrahedron has 6 edges So one Edge Length = 36cm ÷ 6 = 6cm

3  9 = 81 × 1.732

= 140.30 Alternate:Calculate the area of a side, which is an equilateral triangle. The base is 9, the height is

Cu r ved su r face ar ea = 2 

2

=

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=

2

3  Edge length

(ii)

ERna

The length of one edge of a regular tetrahedron is 9 units. What is its surface area? Sol. Surface area

Volume (V) = π(r22  r12 )h

2 3  Edge length 12

= 85.91 units³ 3. The total length of the edges of a tetrahedron is 24 cm. What is its surface area ? Sol. Note:- In tetrahedron has 6 edges. One edge length = 24 ÷ 6 = 4 cm Surface area

2 3 × Edge Length  12

1.

81  3 = 81  3 4

(i)

kgei snhe eYari dnag v.iSn ir

Notice these interesting things: • It has 4 Faces • Each face is an Equilateral Triangle • It has 6 Edges • It has 4 Vertices (corner points) and at each vertex 3 edges meet • It is one of the Platonic Solids The tetrahedron also has a beautiful and unique property ... all four vertices are the same distance from each other! And it is the only Platonic Solid with no parallel faces. When we say "tetrahedron" we often mean "regular tetrahedron" (in other words all faces are the same size and shape)

r1

Volume =

h

22  140 [(25)2 – (23)2] 7

r2

=

22  140  48  2 = 42240 cm³ 7

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EXERCISE

3.

4.

5.

6.

7.

8.

9.

(c) 24

(a) 16 (b) 27 (c) 64 (d) 8 A cuboidal water tank has 216 1 litres of water. Its depth is 3 1 of its length and breadth is 2 1 of of the difference of length 3 and breadth. The length of the tank is (a) 72 dm (b) 18 dm (d) 6 dm (d) 2 dm The volume of cuboid is twice the volume of a cube. If the dimensions of the cuboid are 9 cm, 8 cm and 6 cm, the total surface area of the cube is: (a) 72 cm2 (b) 216 cm2 (c) 432 cm2 (d) 108 cm2 The length, breadth and height of a room is 5m, 4 m and 3m respectively. Find the length of the largest bamboo that can be kept inside the room. (a) 5 m (b) 60 m (c) 7 m (d) 5 2 m A wooden box measures 20 cm by 12 cm by 10 cm . Thickness of wood is 1 cm. Volume of wood to make the box ( in cubic cm) is (a) 960 (b) 519 (c) 2400 (d) 1120 A cuboidal block of 6 cm × 9 cm × 12 cm is cut up into exact number of equal cube. The least posible number of cubes will be (a) 6 (b) 9 (c) 24 (d) 30 A cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.1 m and its walls are 5 cm thick. The thickness of the bottom is: (a) 1 m (b) 13.5 m (c) 1 dm (d) 90 cm

(a) 15 3

(b) 15

(c) 10 2 (d) 5 3 13. A rectangular sheet of metal is 40 cm by 15 cm . equal squares of side 4cm are cut off at the corners and the remainder is folded up to form an op en rectangular box The volume of the box is (a) 896 cm3 (b) 986 cm3 3 (c) 600 cm (d) 916 cm3 14. The areas of three consecutive faces of a cuboid are 12 cm 2, then the volume (in cm3) of the cuboid is (a) 3600 (b) 100

wwM wa.th Les B aryn

2.

(b) 12

(c) 80 (d) 24 3 15. The length of the longest rod that can be placed in a room which is 12 m long, 9 m broad and 8 m high is (a) 27m (b) 19m (c) 17m (d) 13m 16. The floor of a room is of size 4 m × 3 m and its height is 3 m. The walls and celling of the room require painting. The area to be painted is (a) 66 m2 (b) 54 m2 2 (c) 42 m (d) 33 m2 17. If the sum of three dimensions and the total surface area of a rectangular box are 12 cm and 94 cm 2 respectively, then the maximum length of a stick that can be placed inside the box is

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(a) 5 2 cm

(b) 5 cm

(c) 6 cm (d) 2 5 cm 18. The area of the four walls of a room is 660 m2 and its length is twice of its breadth. If the height of the room is 11 m, then area of its floor (in m2) is (a) 120 (b) 150 (c) 200 (d) 330 19. If the length of the diagonal of

r

(d) 3 2 How m any c ub e s , e ac h of edge 3 cm, can be cut from a cube of edge 15 cm ? (a) 25 (b) 27 (c) 125 (d) 144 What is the volume of a cube (in cubic cm) whose diagonal measures 4 3 cm? (a) 8

10. The are a of thr ee adjac ent faces of a cuboid are x, y, z square units respectively. If the volume of the cuboid be v cube units. then the correct relation between v,x, y, z is (a) v2 = xyz (b) v3 =xyz 2 3 3 3 (c) v = x y z (d) v3 =x2y2z2 11. The largest sphere is carved out of a cube of side 7 cm. The volume of the sphere (in cm 3) will be (a) 718.66 (b) 543.72 (c) 481.34 (d) 179.67 12. The length (in meters) of the longest rod that can be put in a room of dimensions 10 m× 10 m × 5 m is

eeYa ridn agv .iSn i

If diagonal of a cube is 12 cm, then its volume in cm3 is :

R Enak geisnh

1.

a cube is 8 3 cm, then its total surface area is (a) 192 cm2 (b) 512 cm2 2 (c) 768 cm (d) 384 cm2 20. The maximum length of a pencil that can be kept in a rectangular box of dimensions 8cm × 6cm × 2cm is (a) 2 13 cm

21.

22.

23.

24.

25.

(b) 2 14 cm

(c) 2 26 cm (d) 10 2 cm The volume of a cubical box is 3.375 cubic metres. The length of edge of the box is (a) 75 m (b) 1.5 m (c) 1.125 m (d) 2.5 m Two cubes of sides 6 cm each are kept side by side to form a rectangular parallelopiped. The area (in sq. cm) of the whole surface of the rectangular parallelopiped is (a) 432 (b) 360 (c) 396 (d) 340 2 cm of rain has fallen on a square km of land. Assuming that 50% of the raindrops could have been collected and contained in a pool having a 100 m× 10 m base, by what level would the water level in the pool have increased ? (a) 1 km (b) 10 m (c) 10 cm (d) 1 m A parallelopiped whose sides are in ratio 2 : 4 :8 have the same volume as a cube. The ratio of their surface area is: (a) 7 : 5 (b) 4 : 3 (c) 8 : 5 (d) 7 :6 If two adjacent sid es of a rectangular parallelopiped are 1 cm and 2 cm and the total surfac e ar ea of the parallelopiped is 22 square cm,

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(b) 2 3 cm

(c)

14 cm

(d) 4 cm

26. If the sum of the length, Br eadth and height of a rectangular parallelopiped is 24 cm and the length of its diagonal is 15 cm, then its total surface area is (a) 256 cm2 (b) 265 cm2 2 (c) 315 cm (d) 351 cm2 27. If the total surface area of a cube is 96 cm2, its volume is (a) 56 cm3 (b) 16 cm3 3 (c) 64 cm (d) 36 cm3 28. The length of the large st possible rod that can be placed in a cubical room is 35 3 m. The surface area of the largest possible sphere that fit within the cub ical room (asuming

22 ) (in sq. m) is 7 (a) 3,500 (b) 3,850 (c) 2,450 (d) 4,250 The volume of air in a room is 204 m3. The height of the room is 6 m. What is the floor area of the room? (a) 32 m2 (b) 46 m2 2 (c) 44 m (d) 34 m2 A square of side 3 cm is cut off from each corner of a rectangular sheet of length 24 cm and breadth 18 cm and the remaining sheet is folded to form an open rectangular box. The surface area of the box is (a) 468 cm2 (b) 396 cm2 2 (c) 615 cm (d) 423 cm2 Three solid iron cubes of edges 4 cm, 5 cm and 6 cm are melted together to make a new cube. 62 cm3 of the melted material is lost due to improper handing. The area (in cm2) of the whole surface of the newly formed cube is (a) 294(b) 343 (c) 125 (d) 216 Area of the floor of a cubical room is 48 sq. m. The length of the longest rod that can be kept in that room is (a) 9 metre (b) 12 metre (c) 18 metre (d) 6 metre

29.

30.

31.

32.

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

41. A cuboidal shaped water tank, 2.1 m long and 1.5 m broad is half filled with water. If 630 litres more water is poured into tank, the water level will rise (a) 2 cm (b) 0.15 cm (c) 0.20 m (d) 0.18 cm 42. A solid cuboid of dimensions 8 cm × 4 cm × 2 cm is melted and cast into identical cubes of edge 2 cm. Number of such identical cubes is. (a) 16 (b) 4 (c) 10 (d) 8 43. A metallic hem isphere is melted and recast in the shape of cone with the same base radius (R) as that of the hemisphere. If H is the height of the cone, then:

v.iSn ir

10 cm

33. Three cubes of sides 6 cm, 8 cm and 1 cm are melted to form a new cube. The surface area of the new cube is (a) 486 cm2 (b) 496 cm2 2 (c) 586 cm (d) 658 cm2 34. Some bricks are arranged in an area measuring 20 cu.m. If the length, breadth and height of each brick is 25 cm, 12.5 cm and 8 cm respectively, then the number of bricks are (suppose there is no gap in between two bricks) (a) 6,000 (b) 8,000 (c) 4,000 (d) 10,000 35. The whole surface of a cube is 150 sq. cm. Then the volume of the cube is (a) 125 cm3 (b) 216 cm3 3 (c) 343 cm (d) 512 cm3 36. The ratio of the length and br eadth of a r ectangular parallelopiped is 5 : 3 and its height is 6 cm. If the total surface area of the parallelopiped be 558 sq. cm, then its length in dm is (a) 9 (b) 1.5 (c) 10 (d) 15 37. If the sum of the dimensions of a rectangular parallelopiped is 24 cm and the length of the diagonal is 15 cm, then the total surface area of it is (a) 420 cm2 (b) 275 cm2 2 (c) 351 cm (d) 378 cm2 38. The length, breadth and height of a cuboid are in the ratio 3 : 4 : 6 and its volume is 576 cm3. The whole surface area of the cuboid is (a) 216 cm2 (b) 324 cm2 2 (c) 432 cm (d) 460 cm2 39. If the number of vertices, edges and fac es of a rectangular parallelopiped are denoted by v, e and f respectively, the value of (v – e + f) is (a) 4 (b) 1 (c) 0 (d) 2 40. A low land, 48 m long and 31.5m broad is raised to 6.5 dm. For this, earth is removed from a cuboidal hole, 27 m long and 18.2 m broad, dug by the side of the land. The depth of the hole will be. (a) 3 m (b) 2 m (c) 2.2 m (d) 2.5 m

eYrai dnag

(a)

the

kgei snhe

of

ERna

then the d iagonal parallelopiped is

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(a) H = 2 R

(b) H =

2 R 3

(c) H = 3R (d) B = 3R 44. If the radius of a sphere is increased by 2 cm, its surface area increased by 352 cm2. The radius of sphere before change is : (a) 3 cm (b) 4 cm (c) 5 cm (d) 6 cm 45. The height of a conical tank is 60 cm and the diameter of its bas e is 64c m. The c ost of painting it from outside at the rate of Rs. 35 per sq. m. is : (a) Rs. 52.00 approx, (b) Rs. 39.20 approx, (c) Rs. 35.20 approx, (d) Rs. 23.94 approx, 46. A solid metallic cone of height 10 cm, radius of base 20 cm is melted to make spherical balls each of 4 cm diameter. How many such balls can be made? (a) 25 (b) 75 (c) 50 (d) 125 47. A cylindrical tank of diameter 35 cm is full of water. If 11 litres of water is drawn off, the water level in the tank will drop by : (a) 10

1 2

cm

(c) 14 cm

(b) 12 (d) 11

6 7 3 7

cm cm

48. The volume of a right circular cylinder whose height is 40cm, and circumference of its base is 66 cm is: (a) 55440 cm3 (b) 3465 cm3 (c) 7720 cm3 (d) 13860 cm3

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total S.A. is 156 3 cm2. (a) 60 3

(b) 180 3

meters) nearly. (a) 3017.1 (c) 301.7

(taking



22 7

(b) 3170 (d) 30.17

)

is

22 7

)

1 1 cm (b) cm 3 2 2 (c) cm (d) 2 cm 3 58. A solid metallic spherical ball of diameter 6 cm is melted and recasted into a cone with diameter of the base as 12 cm. The height of the cone is (a) 6 cm (b) 2 cm (c) 4 cm (d) 3 cm 59. The volume of a right circular cone is 1232 cm3 and its vertical height is 24 cm . Its curved surface area is (a) 154 cm2 (b) 550 cm2 2 (c) 604 cm (d) 704 cm2 60. The volume of a sphere is

(a)

88 3  14  cm3 The curved sur21 face area of the sphere is

( Take  

22 7

)

(a) 2424 cm (b) 2446 cm2 2 (c) 2484 cm (d) 2464 cm2 61. The surface area of a sphere is 2 6 4 cm Its diameter is equal to (a) 16 cm (b) 8 cm (c) 4 cm (d) 2 cm 62. The diameter of the base of a cylinderical drum is 35 dm. and the height is 24 dm. It is full of kerosene. How many tins each of size 25 cm × 22 cm × 35 cm can be filled with kerosene from the drum ? (use 

2



22 7

)

(a) 1200 (b) 1020 (c) 600 (d) 120 63. A hollow iron pipe is 21 cm long and its exterior diameter is 8 cm. If the thickness of the pipe

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is ( Take



22 7

):

(a) 3.696 kg (b) 3.6 kg (c) 36 kg (d) 36.9 kg 64. The volume of a right circular cylinder, 14 cm in height, is equal to that of a cube whose edge is 11 cm Take

 

22 7

the

radius of the base of the cylinder is (a) 5.2 cm (b) 5.5 cm (c) 11.0 cm (d) 22.0 cm 65. If the volume of a right circular cylinder is 9h m3, where h is its height (in metres) then the diameter of the base of the cylinder is equal to (a) 3 m (b) 6 m (c) 9 m (d) 12 m 66. Each of the measure of the radius of base of a cone and that of a sphere is 8 cm. Also, the volume of these two solids are equal. the slant height of the cone is

r



(a) 1760 cu.cm (b) 880 cu.cm. (c) 440 cu.cm (d) 220 cu.cm 57. A sphere of radius 2 cm is put into water contained in a cylinder of base- radius 4 cm. If the sphere is completely immersed in the water, the water level in the cylinder rise by

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(c) 120 3 (d) 240 3 53. Three solid spheres of a metal whose radii are 1 cm, 6 cm and 8 cm are melted to form an other solid sphere. The radius of this new sphere is (a) 10.5 cm (d) 9.5 cm (c) 10 cm (d) 9 cm 54. The slant height of a conical mountain is 2.5 km and the area of its base is 1.54 km 2. then find the height of conical mountain. (a) 2.2 km (b) 2.4 km (c) 3 km (d) 3.11 km 55. The base of a conical tent is 19.2 metres in diameter and the height is 2.8 metres . The area of the canvas required to put up such a tent ( in square

in making the tube is (

is 1 cm and iron weights 8 g/ cm3, then the weight of the pipe

eeYa ridn agv .iSn i

(c) 0.054 cc (d) 0.54 cc 50. The volume of a right circular cylinder is equal to the volume of that right cir cular cone whose height is 108 cm and diameter of base is 30 cm. If the height of the cylinder is 9 cm, the diameter of its base is (a) 30 cm (b) 60 cm (c) 50 cm (d) 40 cm 51. Three solid metallic spheres of diameter 6 cm, 8 cm and 10 cm are melted and recast into a new soild sphere. The diameter of the new sphere is : (a) 4 cm (b) 6 cm (c) 8 cm (d) 12 cm 52. Find the volume of a prism which is based on a regular Hexagon & of height 10cm. If

56. A hollow cylinderical tube 20 cm long. is made of iron and its external and internal diameters are 8 cm and 6 cm respectively. The volume of iron used

R Enak geisnh

49. The circumference of the base of a circular cylinder is 6 cm. The height of the cylinder is equal to the diameter of the base. How many litres of water can it hold ? (a) 54 cc (b) 36 cc

(a) 8 17 cm (b) 4 17 cm (c) 34 2 cm (d) 34 cm 67. A well 20 m in diameter is dug 14 m deep and the earth taken out is spread all around it to a width of 5 m to form an embankment. The height of the embankment is: (a) 10 m (b) 11 m (c) 11.2 m (d) 11.5 m 68. The diameter of the iron ball used for the shot-put game is 14 cm. It is melted and then a solid cylinder of height

2

1 3

cm

is made. What will be the diameter of the base of the cylinder? (a) 14 cm (b) 28 cm 14 28 cm (d) cm 3 3 69. The sum of radii of two spheres is 10 cm and the sum of their volume is 880 cm3. What will be the product of their radii?

(c)

(b) 26

(a) 21 (c) 33

1 3

1 3

(d) 70

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)

3

3

(a) 460 cm (b) 462 cm (c) 624 cm3 (d) 400 cm3 71. A copper rod of 1 cm diameter and 8 cm length is drawn into a wire of uniform diameter and 18 m length. The radius ( in cm) of the wire is (a)

1 15

1 (b) 30

2 (c) 15

(d) 15

72. 12 spheres of the same size are made by melting a solid cylinder of 16 cm diameter and 2 cm height. The diameter of each sphere is : (a) 2 cm (b) 4 cm (c) 3 cm (d) 3 cm 73. When the circumference of a toy ballon is increased from 20 cm to 25 cm its radius ( in cm) is increased by : 5  5 (c) 2 (d)  5 If the volume and surface area of a sphere are numerically the same, then its radius is (a) 1 unit (b) 2 units (c) 3 units (d) 4 units In a right circular cone, the radius of its base is 7 cm and its height 24 cm. A cross- section is made through the midpoint of the height parallel to the base. The volume of the upper portion is (a) 169 cm3 (b) 154 cm3 3 (c) 1078 cm (d) 800 cm3 Some solid metallic right circular cones. each with radius of the base 3 cm and height 4 cm, are melted to form a solid sphere of radius 6 cm. The number of right circular cones is (a) 12 (b) 24 (c) 48 (d) 6 A right circular cylinder of height 16 cm is covered by a rectangular tin foil of size 16 cm × 22 cm, The volume of the cylinder is (a) 352 cm3 (b) 308 cm3 3 (c) 616 cm (d) 176 cm3

74.

75.

76.

77.

(b)

(Use



22 7

)

(a) 4400 cm3 (b) 15400 cm3 (c) 35000 cm3 (d) 144 cm3 80. The radius of the base and height of a metallic soild cylinder are r cm and 6 cm respectively. It is melted and recast into a solid cone of the same r adius of bas e. The height of the cone is: (a) 54 cm (b) 27 cm (c) 18 cm (d) 9 cm 81. The total surface ar ea of a metallic hemisphere is 1848 cm2. The hemisphere is melted to form a solid right circular cone. If the radius of the base of the cone is the same as the radius of the hemisphere its height is (a) 42 cm (b) 26 cm (c) 28 cm (d) 30 cm 82. A right circular cylinder is formed by rolling a rectangular paper 12 cm long and 3 cm wide along its length. The radius of the base of the cylinder will be

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(a) 5

(c) 5 5 0 5 cm³ (d) 6 1 6 5 cm³ 79. The size of a rectangular piece of paper is 100 cm × 44 cm. A cylinder is formed by rolling the paper along its breadth. The volume of the c ylinder is



22 7

)

(a) 1 cm (b) 5.2 cm (c) 2.3 cm (d) 3.7 cm 85. Two iron sphere each of diameter 6 cm are immersed in the water contained in a cylinerical vessel of radius 6 cm. The level of the water in the vessel will be raised by (a) 1 cm (b) 2 cm (c) 3 cm (d) 6 cm 86. The height of the cone is 30 cm. A small cone is cut off at the top by a plane parallel to its

v.iSn ir

22 7

( Take 

base. If its volume is

1 27

of the

eYrai dnag



kgei snhe

(Take 

78. If the area of the base of a cone is 770 cm2 and the area of its curved surface is 814 cm². then find its volume. (a) 2 1 3 5 cm³ (b) 392 5 cm³

ERna

70. A rectangular paper sheet of dimensions 22 cm × 12 cm is folded in the form of a cylinder along its length. What will be the volume of this cylinder?

(a)

3 2

6 (b) cm 

cm

9 (c) cm 2

(d) 2 cm

83. What part of a ditch, 48 metres long. 16.5 metres broad and 4 metres deep can be filled by the ear th g ot b y digging a cylinderical tunnel of diameter 4 m etre s and le ngth 56 metres?

(a)

1 9

(b)

(Use  2 9

(c)

7 9



22 7

(d)

)

8 9

84. The volume of the metal of cylinderical pipe is 748 cm 3. The length of the pipe is 14 cm and its external radius is 9 cm. its thickness is

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volume of the cone. at what height above the base, is the section made ? (a) 6 cm (b) 8 cm (c) 10 cm (d) 20 cm 87. The total surface area of a solid hemisphere is 108 cm2. The volume of the hemisphere is (a) 72 cm3 (b) 144 cm3 3 (c) 108 6 cm (d) 54 6 cm3 88. A solid metallic sphere of radius 3 decimetres is melted to form a circular sheet of 1 milimetre thickness. The diameter of the sheet so formed is (a) 26 metres (b) 24 metres (c) 12 metres (d) 6 metres 89. Water flows through a cylinderical pipe. whose radius is 7 cm, at 5 metre per second. The time, it takes to fill an empty water tank with height 1.54 metres and area of the base (3 × 5) square metres, is ( take 



22 7

)

(a) 6 minutes (b) 5 minutes (c) 10 minutes (d) 9 minutes 90. If S denotes the ar ea of the c u r v e d s u r f ac e o f a r i g h t circular cone of height h a n d s e m i v e r ti c a l a n g le  then S e quals (a) h 2 tan2 1 3

h 2 tan2 (c) h sec tan (b)

2

(d)

1 2 sec tan 3 h

91. The height and the radius of the base of a right circular cone are 12 cm and 6 cm respectively. The rad ius of the c ircular

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2 3

S2

S2

(d) S2

(a)

(c)

4R 2 r12r22

4R 3  r22

r12

volume of the cone is 3

2 314 7

cm , the slant height (in cm) of the cone will be (a) 12 (b) 13 (c) 15 (d) 17 97. A solid metallic cone is melted and recast into a solid cylinder of the same base as that of the cone. If the height of the cylinder is 7 cm, the height of the cone was (a) 20 cm (b) 21 cm (c) 28 cm (d) 24 cm 98. A copper wire of length 36m and diameter 2mm is melted to form a sphere. The radius of the sphere (in cm) is (a) 2.5 (b) 3 (c) 3.5 (d) 4

(b)

R3 r12r22

(d)

R2 r12r22

103. The ratio of height and the diameter of a right circular cone is 3 : 2 and its volume is 1078

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93. The volume of a right circular cylinder and that of a sphere are equal and their radii are also equal. If the height of the cylinder be h and the diameter of the sphere d. then which of the following relation is correct? (a) h =d (b) 2h =d (c) 2h = 3d (d) 3h =2d 94. Water is being pumped out through a circular pipe whose internal diameter is 7cm. If the flow of water is 12 cm p er second, how many litres of water is being pumped out in one hour? (a) 1663.2 (b) 1500 (c) 1747.6 (d) 2000 95. The lateral surface area of a cylinder is 1056 cm 2 and its height is 16cm. Find its volume. (a) 4545 cm3 (b) 4455 cm3 (c) 5445 cm3 (d) 5544 cm3 96. The radius of the base and height of a right circular cone are in the ratio 5 : 12. If the

ing the value of  

r

(c)

(b)

1 2

106. A s olid cylinde r has total surface area of 462 sq. cm. Its curved surface area is one third of the total surface area. Then the radius of the cylinder is (a) 7 cm (b) 3.5 cm (c) 9 cm (d) 11 cm 107. The diameter of a cylinder is 7 cm and its height is 16 cm. Us22 7

, the lat-

eral surface area of the cylinder is (a) 352 cm2 (b) 350 cm2 2 (c) 355 cm (d) 348 cm2 108.The height of a solid right circular cylinder is 6 metres and three times the sum of the area of its two end faces is twice the area of its curved surface, The radius of its base (in meter) is (a) 4 (b) 2 (c) 8 (d) 10 109. A semi-circular sheet of metal of diameter 28 cm is bent into an open conical cup. The depth of the cup is approximately (a) 11 cm (b) 12 cm (c) 13 cm (d) 14 cm 110.A right angled sector of radius r cm is rolled up into a cone in such a way that the two binding radii are joined together . then the curved surface area of the cone is r 2 (a) r 2 cm 2 (b) cm 2 4

eeYa ridn agv .iSn i

3 (a) S2 4

99. The diameter of the base of a right circular cone is 4 cm and its height 2 3 cm. The slant height of the cone is (a) 5 cm (b) 4 cm (c) 2 3 (d) 3 cm 100. The rain water from a roof 22 m × 20 m dr ains into a cylinderical vessel having a diameter of 2 m and height 3.5 m, If the vessel is just full, then the rainfall (in cm) is : (a) 2 (b) 2.5 (c) 3 (d) 4.5 101. From a solid cylinder of height 10 cm and radius of the base 6 cm, a cone of same height and same base is removed. The volume of the remaining solid is : (a) 240 cu. cm (b) 5280 cu. cm (c) 620  cu. cm (d) 360 cu. cm 102. Two solid right cones of equal height and of radii r1 and r2 are melted and made to form a solid sphere of radius R. Then the height of the cone is

R Enak geisnh

cross-section of the cone cut by a plane parallel to its base at a distance of 3 cm from the base is (a) 4 cm (b) 5.5 cm (c) 4.5 cm (d) 3.5 cm 92. If S1 and S2 be the surface areas of a sphere and the curved surface area of the circumscribed cylinder respectively, then S1 is equal to

cc, then (taking

 

22 7

) its

height is : (a) 7 cm (b) 14 cm (c) 21 cm (d) 28 cm 104. From a right circular cylinder of radius 10 cm and height 21 cm. a right circular cone of same base radius is removed. If the volume of the remaining portion is 4400 cm 3, then the height of the removed cone ( take  

22 7

) is :

(a) 15 cm (b) 18 cm (c) 21 cm (d) 24 cm 105. A child reshapes a cone made up of clay of height 24 cm and radius 6 cm into a sphere. The radius (in cm) of the sphere is (a) 6 (b) 12 (c) 24 (d) 48

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r 2 2 cm 2 (d) 2r cm 2 2 111.The radius of the base of a conical tent is 16 metre. If

(c)

3 sq. metre canvas is re7 quired to construct the tent, then the slant height of the tent 427

is : ( take  

22 7

)

(a) 17 metre (b) 15 metre (c) 19 metre (d) 8.5 metre 112. A circus tent is cylinderical up to a height of 3 m and conical above it. If its diameter is 105m and the slant height of the conical part is 63 m, then the total area of the canvas required to make the tent is ( take   (a) 11385 m (c) 9900 m2

2

(b) 10395 m (d) 990 m2

22 7

)

2

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) is

(c)

2 times of the previous volume (d) double of the p ervious volume 120.The base of a right circular cone has the radius 'a' which is same as that of a sphere. Both the sphere and the cone have the same volume. Height of the cone is (a) 3a (b) 4a 7 (c) a 4

r (b) 2r (c) r 2 (d) 2r 2 117. From a solid cylinder whose height is 12 cm and diameter 10 cm. a conical cavity of same height and same diameter of the base is hollowed out. The volume of the remanining solid

(a)

is approximately ( 

22 7

)

(a) 942.86 cm3 (b) 314.29 cm3 (c) 628.57 cm3 (d) 450.76 cm3 118. The radius of a cylinder is 10 cm and height is 4 cm. The number of centimetres that may be added either to the radius or to the height to get the same increase in the volume of the cylinder is (a) 5 cm (b) 4 cm (c) 25 cm (d) 16 cm

7 (d) a 3

121. The circumference of the base of a 16 cm high solid cone is 33 cm What is the volume of the cone in cm3 ? (a) 1028 (b) 616 (c) 462 (d) 828 122. A s olid s p h e r e of 6 c m diameter is melted and recast into 8 solid spheres of equal volume. The radius (in cm) of each small sphere is (a) 1.5 (b) 3 (c) 2 (d) 2.5 123. I n a c y lind e r i c al v e s s e l of diameter 24 cm filled up with sufficient quantity of water, a s o l i d s p h e r i c a l b a ll o f r ad ius 6 c m is c om p le te ly immersed. Then the increase in he ight of water level is : (a) 1.5 cm (b) 2 cm (c) 3 cm (d) 4.2 cm 124. A solid wooden toy is in the shape of a right circular cone mounted on a hemisphere. If the radius of the hemisphere is 4.2 cm and the total height of the toy is 10.2 cm find the the volume of wood en toy (nearly). (a) 104 cm3 (b) 162 cm3 3 (c) 421 cm (d) 266 cm3 125. If a solid cone of volume 27 cm 3 is kept inside a hollow c ylinde r whos e radius and height are that of the cone,

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(a) 75.43 sq. cm, (b) 103.71 sq. cm, (c) 85.35 sq. cm, (d) 120.71 sq. cm, 114. Marbles of diameter 1.4 cm are dropped into a cylinderical beaker containing some water and fully submerged. The diameter of the beaker is 7 cm. Find how many marble s have b een dropped in it if the water rises by 5.6 cm ? (a) 50 (b) 150 (c) 250 (d) 350 115.A cylindrical rod of iron whose height is eight times its radius is melted and cast into spherical balls each of half the radius of the c ylinde r. The number of such spherical balls is (a) 12 (b) 16 (c) 24 (d) 48 116.A cylinder has ‘r’ as the radius of the bas e and ‘ h’ as the height. The radius of base of anothe r cy lind er, having double the volume b ut the same height as that of the first cylinder must be equal to

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(a) 3 cm3

(b) 18 cm3

(c) 54 cm3 (d) 81 cm3 126.A cylindrical can whose base is horizontal and is of internal r ad ius 3 . 5 c m c ontains sufficient water so that when a s olid s p he r e is p lac e d inside, water just covers the sphere. The sphere fits in the c an e xac tly . The d e p th of water in the can before the sphere was put, is

v.iSn ir

7

th e n t he v ol um e of w at e r needed to fill the empty space is

eYrai dnag

22

kgei snhe

(takeing  

119. The radius of the base of a right circular cone is doubled keeping its height fixed. The volume of the cone will be : (a) Three times of the previous volume (b) four times of the previous volume

ERna

113. A toy is in the form of a cone mounte d on a hemis pher e. The radius of the hemisphere and that of the cone is 3 cm and height of the cone is 4 cm. The total surface area of the toy

(a)

35 cm 3

(b)

17 cm 3

7 14 cm (d) cm 3 3 127.The radius and height of a cylinder are in the ratio 5 : 7 and its volume is 550 cm 3 . Calculate it curved surface area in sq. cm. (a) 110 (b) 444 (c) 220 (d) 616 128.The area of the curved surface and the area of the base of a right circular cylinder are a square cm and b square cm respectively. The height of the cylinder is

(c)

(a)

(c)

2a

b a 2 b

cm

cm

(b)

(d)

a b 2 a 2 b

cm

cm

129.The volume of a solid hemisphere is 19404 cm3. Its total surface area is (a) 4158 cm2 (b) 2858 cm2 (c) 1738 cm2 (d) 2038 cm2 130. A solid hemisphere is of radius 11 cm. the curved surface area in sq. cm is (a) 1140.85 (b) 1386.00 (c) 760.57 (d) 860.57 131. The base of a cone and a cylinder have the same radius 6 cm. They have also the same height 8 cm. The ratio of the curved surface area of the cylinder to that of the cone is (a) 8 : 5 (b) 8 : 3 (c) 4 : 3 (d) 5 : 3 132. A right cylindrical vessel is full with water. How many right

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139. What is the height of a cylinder that has the same volume and radius as a sphere of diameter 12 cm ? (a) 7 cm (b) 10 cm (c) 9 cm (d) 8 cm 140. The perimeter of the base of a right circular cone is 8 cm. If the height of the cone is 21 cm, then is volume is :

22 (Take   ) 7 (a) 8000 (b) 400 (c) 800 (d) 125 134. The number of spherical bullets that can be made out of solid cube of lead whose edge measures 44 cm each bullet being of 4 cm diamete r, is (Take

108 (c) 112 cm3 (d) cm3  141 .If the volume of two right cir cular cones are in the ratio 4 : 1 and their diameter are in the ratio 5 : 4, then the ratio of their height is : (a) 25 : 16 (b) 25 : 64 (c) 64 : 25 (d) 16 : 25 142. The volume of a conical tent is 1232 cu. m and the area of its base is 154 sq. m. Find the length of the canvas required to build the tent, if the canvas is 2m in width.

112 cm3 

( Take 

22 ) 7

(a) 270 m (b) 272 m (c) 276 m (d) 275 m 143. If the ratio of the diameters of two r ight circular cones of equal height be 3: 4, then the ratio of their volume will be (a) 3 : 4 (b) 9 : 16 (c) 16 : 9 (d) 27 : 64 144. The surface area of two spheres are in the ratio 4 : 9. Their volumes will be in the ratio (a) 2 : 3 (b) 4 : 9 (c) 8 : 27 (d) 64 : 729 145. The total surface ar ea of a sphere is 8 square unit. The volume of the sphere is

wwM wa.th Les B aryn

4a 2V 4a 2 unit (b) unit V  4V a 2V (c) unit (d) 2 unit

(a)

4

a

(a)

m 2h

(c)

2h m

eeYa ridn agv .iSn i

(b)

R Enak geisnh

22 ) 7 (a) 2541 (b) 2451 (c) 2514 (d) 2415 135. The radius of a metallic cylinder is 3 cm and its height is 5 cm. It is melted and moulded into sm all cone s, e ach of height 1 cm and base radius 1 mm. The num ber of s uch cones formed is (a) 450 (b) 1350 (c) 8500 (d) 13500 136.A sector is formed by opening out a cone of base radius 8 cm and height 6 cm. Then the radius of the sector is (in cm) (a) 4 (b) 8 (c) 10 (d) 6 137. A solid cone of height 9 cm with diameter of its base 18 cm is cut out from a wooden solid sphere of radius 9 cm. The p e r c e nt ag e of wood wasted is : (a) 25% (b) 30% (c) 50% (d) 75% 138. The perimeter of the base of a right circular cylinder is ‘a’ unit . If the volume of the cylinder is V cubic unit, then the height of the cylinder is



(a) 108  cm3

22 ) is 7 (a) 624.26 cm3 (b) 622.36 cm3 (c) 622.56 cm3 (d) 623.20 cm3 147. A conical flask is full of water. The flask has base rad i us r a nd he i g ht h. Th is water is poured into a cylindrical flask of base radius m, height of cylindrical flask is

ity of the cup ( Take 

r

cones having the same diameter and height as that of the right cylinder will be needed to store that water ? 22 (take   ) 7 (a) 4 (b) 2 (c) 3 (d) 5 133. A spherical lead ball of radius 10 cm is melted and small lead balls of radius 5mm are made The total number of possible sm all lead balls is

(a)

8 2  cubic unit 3

(b)

8  cubic unit 3

(c) 8 3 cubic unit

8 3  cubic unit 5 146. A semicircular sheet of metal of diameter 28 cm is bent into an open conical cup. The capac-

(d)

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(b)

h 2 m 2

(d)

r²h 3m2

148.A solid spherical copper ball whose diameter is 14 cm is melted and converted into a wire having diam eter equal to 14 cm. The length of the wire is 16 (a) 27 cm (b) cm 3

28 cm 3 149. A sphere of diameter 6 cm is droped in a rig ht c ircular cylinderical vessel partly filled with water. The diameter of the cylinderical vessel is 12 cm. If the sphere is just completely submerged in water, then the ris e of water level in the cylindrical vessel is (a) 2 cm (b) 1 cm (c) 3 cm (d) 4 cm 150. A copper sphere of diameter 18 cm is drawn into a wire of diameter 4 mm. The length of the wire in metre is : (a) 2.43 m (b) 243 m (c) 2430 m (d) 24.3 m 151. A rectangular block of metal has dimensions 21 cm, 77 cm and 24 cm. The block has been melted into a s pher e. The radius of the sphere is (Take (c) 15 cm

(d)

22 ) 7 (a) 21 cm (b) 7 cm (c) 14 cm (d) 28 cm 152. The radius of cross-section of a solid cylindrical rod of iron is 50 cm. the cylinder is melted down and formed into 6 solid spherical balls of the same



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3vh 3 – c 2h 2  9v 2 is

(a) 1078 cubic cm (b) 1708 cubic cm (c) 7108 cubic cm (d) 7180 cubic cm 166.A hollow sphere of internal and external diameter 6 cm and 10 cm respectively is melted into a right circular c one of diameter 8 cm. The height of the cone is (a) 22.5 cm (b) 23.5 cm (c) 24.5 cm (d) 25.5 cm 167.A flask in the shape of a right circular cone of height 24 cm is filed with water. The water is pour ed in right cirular cylindrical flask whose radius

eYrai dnag

v.iSn ir

S3 V2

is (a) 36 units (b) 9 units (c) 18 units (d) 27 units 161. Assume that a drop of water is spherical and its diameter is one-tenth of a cm. A conical glass has a height equal to the diameter of its rim. If 32,000 drops of water fill the glass completely. Then the height of the glass (in cm) is (a) 1 (b) 2 (c) 3 (d) 4 162. A tank 40 m long, 30 m broad and 12 m deep is dug in a field 1000 m long and 30 m wide. By how much will the level of the field rise if the earth dug out of the tank is evenly spread over the field ? (a) 2 metre (b) 1.2 metre (c) 0.5 metre (d) 5 metre 16 3.A sphe re is cut into two hemispheres. One of them is used as bowl. It takes 8 bowlfuls of this to fill a conical vessel of height 12 cm and radius 6 cm. The radius of the sphere(in centimetre) will be (a) 3 (b) 2 (c) 4 (d) 6 164.A ball of lead 4 cm in diameter is covered with gold. If the volume of the gold and lead are equal then the thickness of

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(a) 2 (b) – 1 (c) 1 (d) 0 156. The total number of spherical bullets, each of diameter 5 decimeter, that can be made by utilizing the maximum of a rectangular block of lead with 11 metr e le ngth, 10 metre breadth and 5 metre width is (assume that   3 ) (a) equal to 8800 (b) less than 8800 (c) equal to 8400 (d) greater than 9000 157. If a metallic cone of radius 30 cm and height 45 cm is melted and re cast into me tallic spheres of radius 5 cm, find the number of spheres , (a) 81 (b) 41 (c) 80 (d) 40 158. A metallic sphere of radius 10.5 cm is melted and then recast into small cones each of radius 3.5 cm and height 3 cm. The number of cone s thus formed is (a) 140(b) 132 (c) 112 (d) 126 159. A right circular cone is 3.6 cm high and radius of its base is 1.6 cm. It is melted and recast

respectively, then value of

kgei snhe

(c) 3 2 times of the preivous volume (d) 6 time s of the pr evious volume 155. If h, c, v are respecitvely the height, curved surface area and volume of a right circular cone then the v alue of

into a right circular cone with radius of its base as 1.2 cm. Then the height of the cone (in cm) is (a) 3.6 cm (b) 4.8 cm (c) 6.4 cm (d) 7.2 cm 160. If surface area and volume of a sphere are S and V

ERna

radius as that of the cylinder. The le ngth of the r od (in metres) is (a) 0.8 (b) 2 (c) 3 (d) 4 153. Two right circular cones of equal height and radii of there respective base 3 cm and 4 cm are melted together and made to a solid sphere of radius 5 cm. The height of a cone is (a) 10 cm (b) 20 cm (c) 30 cm (d) 40 cm 154. The radius of the base and the height of a right circular cone are doubled. The volume of the cone will be (a) 8 time s of the pr evious volume (b) three times of the previous volume

gold [g iven 3 2 = 1.25 9) is approximately (a) 5.038 cm (b) 5.190 cm (c) 1.038 cm (d) 0.518 cm 165.A conical cup is filled with icecream. The ice-cream forms a hemispherical shape on its open top. The height of the hemispherical part is 7 cm. the radius of the hemispherical part equals to the height of the cone. Then the volume of the

  

ice-cream is  

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22   7 

1 rd of radius of the base of 3 the circular cone. Then the height of the water in the cylindrical flask is (a) 32 cm (b) 24 cm (c) 48 cm (d) 72 cm 168.A solid metallic spherical ball of diameter 6 cm is melted and re cast into a cone with diameter of the base as 12 cm. The height of the cone is (a) 2 cm (b) 3 cm (c) 4 cm (d) 6 cm 169. A he m i s p he r ic al b owl of inte r nal r a d ius 15 cm contains a liquid. The liquid is to be filled into cylindrical shaped bottles of diameter 5 c m and he ig ht 6 c m . The number of bottles required to empty the bowl is (a) 30 (b) 40 (c) 50 (d) 60 170. If V1, V2 and V3 be the volumes of a r ight cir cular c one. a sphere and a right circular cy lind er having the same radius and same height then is

(a) V1=

V1 V2 V2 V 3  = (b) = V3 3 2 3 4

V1 V2 V1 V3  = V3 (d) = V2 = 3 2 3 2 171. If the surface area of a sphere is 346.5 cm2, then its radius [ (c)

taking   (a) 7 cm (c) 5.25 cm

22 ] 7 (b) 3.25 cm (d) 9 cm

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22    7   (a) 1100 (b) 5500 (c) 500 (d) 450 173. If the volume of a sphere is numerically equal to its surface area then its diameter is; (a) 4 cm (b) 6 cm (c) 3 cm (d) 2 cm 174. 5 persons live in a tent. If each person requires 16 m² of floor area and 100 m³ space for air then the height of the c o ne o f s m al le s t s iz e to ac c o m od ate t he s e p e r s ons would be? (a) 16 m (b) 18.75 m (c) 10.25 m (d) 20 m 175. The numerical values of the volume and the area of the lateral surface of a right circular cone are equal. If the height of the cone be h and r a d i us b e r , th e v a lu e of

tion of a tunnel, cylindrical in shape, of diameter 4 m and

22   length 56 m is  Take  = 7   (a)



1 1 + is 2 h r2

9 1

(b)

3 1

(c)

1 3

(d)

1 9

176.There is wooden sphere of ra-

dius 6 3 cm. The surface area of the largest possible cube cut out from the sphere will be (a) 464 3 cm² (b) 646 3 cm²

(c) 864 cm² (d) 462 cm² 177. I f a he m is p he r e i s m e l te d and f our s p he r e s of e qual volume are made, the radius of each sphere will be equal to (a) 1/4th of the hemisphere (b) radius of the hemisphere (c) 1/2 of the rad ius of the hemisphere (d) 1/6 th of the radius of the hemisphere 178. The portion of a ditch 48 m long. 16.5 m wide and 4 m deep that can be filled with stones and earth available during excava-

1 Part 2

1 2 Part (d) Part 4 9 179. From a solid right circular cylinder of length 4 cm and diameter 6 cm, a conical cavity of the same height and base is hollowed out. The whole surface area of the remaining solid (in square cm.) is (a) 48 (b) 63

are 'r' units and 4rh square units respectively. The height of the cylinder is:

eeYa ridn agv .iSn i

(c)

(c) 15 (d) 24 180. A spherical ball of radius 1 cm is dropped into a conical vessel of radius 3 cm and slant height 6 cm. The volume of water (in cm³), that can just immerse the ball, is

 4 5 (b) 3 (c) (d) 3 3 3 181. If the height of a cylinder is 4 times its circumference, the volume of the cylinder in terms of its circumference, c is (a)

(a)

2c³ 

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(a)

(b)

R Enak geisnh

  cm, is 

1 Part 9

184. The total surface area of a right circular cylinder with radius of the base 7 cm and heigtht 20 cm is: (a) 140 cm2 (b) 1000 cm2 2 (c) 900 cm (d) 1188 cm2 185. The radius of base and curved surface area of a right cylinder

r

172. Deepali makes a model of a cylindrical kaleidoscope for her science project. She uses a chart paper to make it. If the length of the kaleidoscope is 25 cm and radius 35 cm, the area of the paper she used, in sq.

(b)

c³ 

(c) 4 c³ (d) 2  c³ 182. The radii of a sphere and a right circular cylinder are 3 cm eac h. I f their volumes are equal, then curved surface area of the cylinder is

22    Assume π 7   

(a) 75

3 cm2 7

(b) 65

3 cm2 7

(c) 74

3 cm2 7

(d) 72

3 cm2 7

183. The radius of a hemispherical bowl is 6 cm. The capacity of



the bowl is: Take Π = 22 7



(a) 452.57 cm³ (b) 452 cm³ (c) 345.53 cm³ (d) 495.51 cm³

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(a) 4h units

(b)

h units 2

(c) h units (d) 2h units 186. A hemi-spherical bowl has 3.5 cm radius. It is to be painted inside as well as outside. The cost of painting it at the rate of Rs. 5 per 10sq. cm. will be: (a) Rs. 77 (b) Rs. 175 (c) Rs. 50 (d) Rs. 100 187. The volume of a right circular cone which is obtained from a wooden cube of edge 4.2 dm wasting minimum amount of wood is: (a) 194.04 cu. dm (b) 19.404 cu. dm (c) 1940.4 cu. dm (d) 1940.4 cu. dm 188. If the radius of a sphere is increased by 2 cm, then its surface area increases by 352cm². The radius of the sphere ini 22  tially was: u se 7   

(a) 3cm (b) 5cm (c) 4cm (d) 6cm 189. A right triangle with sides 9 cm, 12 cm and 15 cm is rotated about the side of 9 cm to form a cone. The volume of the cone so formed is: (a) 432 p cm³ (b) 327 p cm³ (c) 334 p cm³ (d) 324 p cm³ 190. The volume of the largest right circular cone that can be cut out of a cube of edge 7 cm ?  22    u se  7   (a) 13.6 cm³ (b) 147.68 cm³ (c) 89.9 cm³ (d) 121 cm³ 191. By melting two solid metallic spheres of radii 1 cm and 6 cm,

382

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( Taking

 

22 7

). find the ratio

of its diameter to its height . (a) 7 : 6 (b) 6 : 7 (c) 3 : 7 (d) 7 : 3 206. The ratio of the volume of two cones is 2 : 3 and the ratio of radii of their base is 1 : 2. The ratio of their height is (a) 3 : 8 (b) 8 : 3 (c) 4 : 3 (d) 3 : 4 207. If the volume of two cubes are in the ratio 27 : 64, then the ra-

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tio of their total surface area is: (a) 27 : 64 (b) 3 : 4 (c) 9 : 16 (d) 3 : 8 208. A h e m i s p h e r e a n d a c on e hav e e q ual b as e . I f th e ir heights are also equal, the ratio of their curved surface will be :

v.iSn ir

(a) 1 : 2 (b) 2 :1 (c) 1 : 2 (d) 2 : 1 209. If the height of a given cone be doubled and radius of the base remains the same the r atio of the v olum e of the given cone to that of the second cone will be (a) 2 : 1 (b) 1 : 8 (c) 1 : 2 (d) 8 : 1 210. Spheres A and B have their radii 40 cm and 10 cm respectively. Ratio of surface area of A to the surface area of B is : (a) 1 : 16 (b) 4 : 1 (c) 1 : 4 (d) 16 : 1 211. If the radius of the base of a cone be doubled and height is left unchanged, then ratio of the volume of new cone to that of the original cone will be: (a) 1 : 4 (b) 2 : 1 (c) 1 : 2 (d) 4 : 1 212. A cube of edge 5 cm is cut into cubes each of edge of 1 cm. The ratio of the total surface area of one of the small cubes to that of the large cube is equal to: (a) 1 : 125 (b) 1 : 5 (c) 1 : 625 (d) 1 : 25 213. The diameter of two hollow spheres made from the same metal sheet are 21 cm and 17.5 cm respectively. The ratio of the area of metal sheets required for making the two spheres is (a) 6 : 5 (b) 36 : 25 (c) 3 : 2 (d) 18 : 25 214. B y m e l t i n g a s o l i d l e a d sp he re of diame te r 12 c m, three small spheres are made whose d iame te rs are in the ratio 3 : 4 : 5. The radius (in cm) of the smallest sphere is (a) 3 (b) 6 (c) 1.5 (d) 4 215. A cone is cut at mid point of i t s h e i g h t b y a f r us t u m parallel to its base. The ratio between the volumes of two parts of cone would be (a) 1 : 1 (b) 1 : 8 (c) 1 : 4 (d) 1 : 7

eYrai dnag

(b) 2 : 1 2 :1 (c) 1 : 2 (d) 1 : 4 201. If the volume of two cubes are in the ratio 27 : 1, the ratio of their edge is : (a) 3 : 1 (b) 27 : 1 (c) 1 : 3 (d) 1 : 27 202. The edges of a cuboid are in the ratio 1 : 2 : 3 and its surface area is 88 cm 2. The volume of the cuboid is : (a) 48 cm³ (b) 64 cm³ (c) 16 cm³ (d) 100 cm³ 203. The volume of two spheres are in the ratio 8 : 27. The ratio of their surface area is: (a) 4 : 9 (b) 2 : 3 (c) 4 : 5 (d) 5 : 6 204. The base radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their volumes is : (a) 27 : 20 (b) 20 : 27 (c) 9 : 4 (d) 4 : 9 205. The curved surface area of a cylinderical pillar is 264 m 2 and its v olu m e is 9 2 4 m 3 (a)

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(b) 1

kgei snhe

1 1 (c) 3 (d) 2 2 2 193.The volume (in m3) of rain water that can be collected from 1.5 hectares of ground in a rainfall of 5 cm is (a) 75 (b) 750 (c) 7500 (d) 75000 194. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour, How much water ( in-litres) will fall into sea in a minute ? (a) 4,00,000 m³ (b) 40,00,000 m³ (c) 40,000 m³ (d) 4,000 m³ 195. Water is flowing at the rate of 3 km/hr through a circular pipe of 20 cm internal diameter into a circular cistern of diameter 10 m and depth 2m. In how much time will the cistern be filled ? (a) 1 hour (b) 1 hour 40 minutes (c) 1 hour 20 minutes (d) 2 hours 40 minutes 196. Water flows at the rate of 10 metres per minute from cylindrcial pipe 5 mm in diameter, How long it will take to f ill up a c on ic al v e s s e l whose diameter at the base is 30 cm and depth 24 cm? (a) 28 mintues 48 seconds (b) 51 minutes 12 seconds (c) 51 minutes 24 seconds (d) 28 mintues 36 seconds 197. The radius of the base of conical tent is 12 m. The tent is 9 m high. Find the cost of canv a s r e qu ir e d t o m a k e t he tent, if one square metre of canvas costs Rs.120 (Take = 3.14) (a) Rs. 67,830 (b) Rs. 67,800 (c) Rs. 67,820 (d) Rs. 67,824

(a) 2

198. A plate of square base made of brass is of length x cm and thic k ne s s 1 m m . The p late weighs 4725 gm. If 1 cubic cm of b r as s we ig hs 8 . 4 gr am , then the value of x is: (a) 76 (b) 72 (c) 74 (d) 75 199. The diameter of a 120 cm long roller is 84 cm. It takes 500 complete revolutions of the roller to level a ground. The cost of levelling the ground at Rs. 1.50 sq. m. is: (a) Rs. 5750 (b) Rs. 6000 (c) Rs. 3760 (d) Rs. 2376 200. Two right circular cylinders of e qual v olume hav e their heights in the ratio 1 : 2. The ratio of their radii is :

ERna

a hollow sphere of thickness 1 cm is made. The external radius of the hollow sphere will be (a) 8 cm (b) 9 cm (c) 6 cm (d) 7 cm 192. Water is flowing at the rate of 5 km/h through a pipe of diameter 14 cm into a rectangular tank which is 50 m long 44m wide, The time taken (in hours ) for the rise in the level of water in the tank to 7 cm is

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(a)

4 3

(b)

2 3

(c)

3 4

3 (d) 2

224.A solid metallic sphere of radius 8 cm is melted to form 64 equal small solid spheres. The ratio of the surface area of this sphere to that of a small sphere is (a) 4 : 1 (b) 1 :16 (c) 16 :1 (d) 1 : 4

22 ) 7

eeYa ridn agv .iSn i

cm) of its base is ( use

233. The radius and the height of a c o ne a r e i n th e r a tio 4 : 3. The ratio of the curved surface area and total surface area of the cone is (a) 5 : 9 (b) 3 : 7 (c) 5 : 4 (d) 16 : 9 234. A sphere and a cylinder have equal volume and equal radius. The ratio of the curved surface area of the cylinder to that of the sphere is (a) 4 : 3 (b) 2 : 3 (c) 3 : 2 (d) 3 : 4 235. A right circular cylinder and a cone have equal base radius and equal he ig ht. If their curved surfaces are in the ratio 8 : 5, then the radius of the base to the height are in the ratio: (a) 2 : 3 (b) 4 : 3 (c) 3 : 4 (d) 3 : 2 236. A right prism with trapezium base of parallel side 8 cm & 14 cm. Height of prism is 12 cm & its volume is 1056 cm 3 then. Find the distance two parallel lines. (a) 8 (b) 10 (c) 16 (d) 6 237. The radii of the base of cylinder and a cone are in the ratio

r

225. The diameter of two cylinders, whose volumes are equal, are in the r atio 3 : 2 , The ir heights will be in the ratio. (a) 4 : 9 (b) 5 : 6 (c) 5 : 8 (d) 8 : 9 226. The radius of base and slant height of a cone are in the ratio 4 : 7. If slant height is 1 4 c m the n the r adius (in

R Enak geisnh

(a) 8 (b) 12 (c) 14 (d) 16 227. A right circular cylinder just encloses a sphere of radius r. The ratio of the surface area of the sphere and the curved surface area of the cylinder is (a) 2 : 1 (b) 1 : 2 (c) 1 : 3 (d) 1 : 1 228.The ratio of radii of two cone is 3 : 4 and the ratio of their height is 4 : 3. Then the ratio of their volume will be (a) 3 : 4 (b) 4 : 3 (c) 9 : 16 (d) 16 : 9 229.If a right circular cone is separated into solids of volumes V1, V2, V3 by two planes parallel to the base which also trisect the altitude, then V1 : V2 : V3 is (a) 1 : 2 : 3 (b) 1 : 4 : 6 (c) 1 : 6 : 9 (d) 1 : 7 : 19 230. The total surf ace area of a solid right circular cylinder is twice that of a solid sphere. If they have the same radii, the ratio of the volume of the cylinder to that of the sphere is given by (a) 9 : 4 (b) 2 : 1 (c) 3 : 4 (d) 4 :9 231. The respective height and volume of a hemisphere and a right circ ular cy lind er are equal, then the ratio of their radii is

wwM wa.th Les B aryn

216.The ratio of the area of the incircle and the circum-circle of a square is (a) 1 : 2 (b) 2 : 1 (c) 1 : 2 (d) 2 : 1 (d) remains unchanged 217. The ratio of the surface area of a sphere and the curved surface area of the cylinder circumscribing the sphere is (a) 1 : 2 (b) 1 : 1 (c) 2 : 1 (d) 2 : 3 218. The radii of two spheres are in the ratio 3 : 2. Their volume will be in the ratio : (a) 9 : 4 (b) 3 : 2 (c) 8 :27 (d) 27 : 8 219. The volume of a sphere and a right circular cylinder having the same radius are equal. The ratio of the diameter of the sphere to the height of the cylinder is (a) 3 : 2 (b) 2 : 3 (c) 1 :2 (d) 2 : 1 220. A cone, a hemisphere and a cylinder stand on equal bases and have the s ame he ight. The ratio of their respective volume is (a) 1 : 2 : 3 (b) 2 : 1 : 3 (c) 1 : 3 : 2 (d) 3 : 1 : 2 221. The radii of the base of two cylinders are in the ratio 3 : 5 and their heights in the ratio 2 : 3. The ratio of their curved surface will be : (a) 2 : 5 (b) 2 : 3 (c) 3 : 5 (d) 5 :3 222. If the radii of two spheres are in the ratio 1 : 4, then their surface area are in the ratio : (a) 1 : 2 (b) 1 : 4 (c) 1 : 8 (d) 1 :16 223. The radii of the base of two cylinders A and B are in the ratio 3 : 2 and their height in the ratio x : 1. If the volume of cylinder A is 3 times that of cylinder B, the value of x is

(a)

2: 3

(b)

3 :1

(c) 3 : 2 (d) 2 : 3 232. The ratio of the volume of a cube and of a solid sphere is 363 : 49. The ratio of an edge of the cube and the radius of the sphere is ( take  (a) 7 : 11 (c) 11 :7

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(b) 22 : 7 (d) 7 : 22

22 ) 7

3 : 2 and their heights are

in the ratio 2 : 3 . Their volumes are in the ratio of (a)

3: 2

(b) 3 3 : 2

(c) 3 : 2 2 (d) 2 : 6 238. The heights of two cones are in the ratio 1 : 3 and the diameters of their base are in the ratio 3 : 5, The ratio of their volume is (a) 3 : 25 (b) 4 : 25 (c) 6 : 25 (d) 7 : 25 239. A sphere and a hemisphere have the same volume . The ratio of their radii is (a) 1 : 2 (b) 1 : 8 (c) 1 : 2 (d) 1 : 3 2 240. The diameter of the moon is assumed to be one fourth of the diameter of the earth. Then the ratio of the volume of the earth to that of the moon is (a) 64 : 1 (b) 1 : 64 (c) 60 : 7 (d) 7 : 60 241. If A denotes the volume of a right circular cylinder of same height as its diameter and B is the vol-

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(b)

3 2

(c)

2 3

(d)

3 4

242. Diagonal of a cube is 6 3 cm. Ratio of its total surface area and volume (numerically) is (a) 2 : 1 (b) 1 : 6 (c) 1 : 1 (d) 1 : 2 243. A sphere and a hemisphere have the same volume. The ratio of their curved surface area is : 3

(a) 2 2 :1 2

2

(b) 2 3 :1 1

4

3

(a) 1 : 3 3

(b) 1 : 2 2

2

4

(c) 1 : 23 (d) 1 : 2 3 251. A plane divides a right circular cone into two parts of equal volume. If the plane is parallel to the base, then the ratio, in which the height of the cone is divided, is (a) 1 : 2

(b) 1 : 3 2

(c) 1 : 3 2 – 1 (d) 1 : 3 2 1 252. A rectangle based pyramid, length and width of the base is 18cm and 10cm respectively. Find the total surface area, if its height is 12cm : (a) 267cm2 (b) 564cm2 2 (c) 516cm (d)None of these 253. A cone of height 7 cm and base radius 1 cm is carved from a cuboidal block of wood 10 cm × 5 cm × 2 cm .[As-

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(c) 4 3 :1 (d) 2 3 :1 244. The volume of a cylinder and a cone are in the ratio 3 : 1. Find their diameters and then compare them when their heights are equal. (a) Diameter of cylinder = 2 times of diameter of cone (b) Diameter of cylinder = Diameter of cone (c) Diameter of cylinder > Diameter of cone (d) Diamete r of cylinde r < Diaameter of cone 245. A solid sphere is melted and recast into a right circular cone with a base radius equal to the radius of sphere. What is the ratio of the height and radius of the cone so formed ? (a) 4 : 3 (b) 2 : 3 (c) 3 : 4 (d) 4 : 1 246. Find the total surface area of a prism which is based on  of perimeter 45 cm & incircle radius 9cm, if its volume is 810 cm3. (a) 405 (b) 585 (c) 616 (d) 468 247. The ratio of weights of two spheres of different materials is 8 : 17 and the ratio of weights per 1 cc of materials of each is 289 : 64. The ratio of radii of the two spheres is (a) 8 : 17 (b) 4 : 17 (c) 17 : 4 (d) 17 : 8 248. If the ratio of volumes of two cones is 2 : 3 and the ratio of the radii of their bases is 1 : 2, then the ratio of their heights

22 ] The percent7 age of wood wasted in the process is :

suming 

2 3

(a) 92 % 1

(c) 42 %

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3

v.iSn ir

4 3

254.If the radius of a cylinder is de cr eased b y 50 % and the height is increased by 50% to form a new cylinder, the volume will be decreased by (a) 0% (b) 25% (c) 62.5% (d) 75% 255. Each of the height and base radius of a cone is increased by 10 0%. The perc entage increase in the volume of the cone is (a) 700% (b) 400% (c) 300% (d) 100% 256. If both the radius and height of a right circular cone are increased by 20%, its volume will be increased by (a) 20% (b) 40% (c) 60% (d) 72.8% 257. A cone of height 15 cm and b a s e d iam e t e r 3 0 c m is carved out of a wooden sphere of radius 15 cm. The percentage of used wood is : (a) 75% (b) 50% (c) 40% (d) 25% 258. If the height of a right circular cone is increased by 200% and the radius of the base is reduced by 50%, the volume of the cone (a) increases by 25% (b) increases by 50% (c) remains unaltered (d) decreases by 25% 259. If the height and the radius of the base of a cone are each increased by 100%, then the volume of the cone becomes (a) double that of the original (b) three times that of the original (c) six times that of the original (d) eight times that of the original 260. If the height of a cylinder is increased by 15 per cent and the radius of its base is decreased by 10 percent then by what percent will its curved surface area change ? (a) 3.5 percent decrease (b) 3.5 percent increase (c) 5 percent increase (d) 5 percent decrease

eYrai dnag

(a)

kgei snhe

A is: B

then

will be (a) 8 : 3 (b) 3 : 8 (c) 4 : 3 (d) 3 : 4 249. The volumes of a right circular cylinde r and a sphe re are equal. The radius of the cylinder and the diameter of the sphere are equal. The ratio of height and radius of the cylinder is (a) 3 : 1 (b) 1 : 3 (c) 6 : 1 (d) 1 : 6 250. A large solid sphere is melted and moulded to form identical right circular cones with base radius and height same as the radius of the sphere. One of these cones is melted and moulde d to form a s maller solid sphere. Then the ratio of the surface area of the smaller to the s ur f ac e ar e a of the larger sphere is

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ume of a sphere of same radius

1 3

(b) 46 % 1

(d) 41 % 3

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(a) 144 2 cu. cm,

R Enak geisnh

(b) 72 2 cu. cm, (c) 8 2 cu. cm,

(d) 12 2 cu. cm, 273. If the radii of the circular ends of a truncated conical bucket which is 45cm high be 28 cm and 7 cm then the capacity of the bucket in cubic centimetre is (use 

1 3 2 (c) (d) 2 2 3 268. If the radius of a sphere be doubled. the area of its surface will become (a) Double (b) Three times (c) Four times (d) None of the mentioned

(a) 2

(b)

22 ) 7

(a) 48510 (b) 45810 (c) 48150 (d) 48051 274. There is a pyramid on a base which is a regular hexagon of side 2a cm. If every slant edge 5a 2 cm, then the volume of this pyramid is

of this pyramid is of length

(a) 3a3 cm3

(b) 3 2 a2 cm3

(c) 3 3 a3 cm3 (d) 6a3 cm3 275. The base of a right pyramid is a square of side 40 cm long. If the volume of the pyramid is 8000 cm3, then its height is : (a) 5 cm (b) 10 cm (c) 15 cm (d) 20 cm 276.The base of a right prism is a trapezium. The length of the parallel sides are 8 cm and 14 cm and the distance between

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the parallel sides is 8 cm, If the volume of the prism is 1056 cm 3 , then the height of the prism is (a) 44 cm (b) 16.5 cm (c) 12 cm (d) 10.56 cm 277. Each edge of a regular tetrahedron is 3 cm, then its volume is 9 2 c.c. (b) 27 3 c.c. 4 4 2 (c) c.c. (d) 9 3 c.c. 9 278. The perimeter of the triangular base of a right prism is 15 cm and radius of the incircle of the triangular base is 3 cm. If the volume of the prism be 270 cm 3 then the he ight of the prism is (a) 6 cm (b) 7.5 cm (c) 10 cm (d) 12 cm 279. The base of a solid right prism is a triangle whose sides are 9 cm, 12 cm and 15 cm, The height of the prism is 5 cm. The the total surface area of the prism is (a) 180 cm2 (b) 234 cm2 2 (c) 288 cm (d) 270 cm2 280.The base of a right prism is an e qu il ate r a l tr i an g l e of area 173 cm 2 and the volume of the prism is 10380 cm 3 . The area of the lateral surface of the prism is

r

(a)

eeYa ridn agv .iSn i

269. If each edge of a cube is increased by 50%, the percentage increase in its surface area is (a) 150% (b) 75% (c) 100% (d) 125% 270. If the radius of a sphere be doubled, then the percentage increase in volume is (a) 500% (b) 700% (c) 600% (d) 800% 271. Find the radius of maximum size sphere which can be inscribed or put in a cone whose base radius and height are 6cm and 8cm respectively? (a) 4cm (b) 5cm (c) 3cm (d) None of these 272. If the length of each side of a regular tetrahedron is 12 cm, then the volume of the tetrahedron is

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261.If the radius of a sphere is doubled, its volume becomes (a) double (b) four times (c) six times (d) eight times 262. If the radius of a right circular cylinder is decreased by 50% and its height is increased by 60% its volume will be decreased by (a) 10% (b) 60% (c) 40% (d) 20% 263. The le ng th, b r e ad th and height of a cuboid are in the ratio 1 : 2 : 3. If they are increased by 100%, 200% and 200% respectively. Then compared to the original volume the increase in the volume of the cuboid will be (a) 5 times (b) 18 times (c) 12 times (d) 17 times 264. Each of the radius of the base and the height of a right circular cylinder is increased by 10%. The volume of the cylinder is increased by (a) 3.31% (b) 14.5% (c) 33.1% (d) 19.5% 265. If the height of a cone is increase by 100% then its volume is increased by : (a) 100% (d) 200% (d) 300% (d) 400% 266. A hemispherical cup of radius 4 cm is filled to the brim with coffee. The coffee is then poured into a vertical cone of radius 8 cm and height 16 cm. The percentage of the volume of the cone that remains empty is : (a) 87.5% (b) 80.5% (c) 81.6% (d) 88.2% 267. The height of a circular cylinder is increased six times and the base area is decreased to one ninth of its value. The factor by which the lateral surface of the cylinder increases is

( use

3 =1.73) (a) 1200 cm2 (b) 2400 cm2 2 (c) 3600 cm (d) 4380 cm2 281. The base of a right pyamid is a square of side 16 cm long . If its height be 15 cm, then the area of the lateral surface in square cm is : (a) 136 (b) 544 (c) 800 (d) 1280 282. Area of the base of a pyramid is 57 sq. cm. and height is 10 cm, then its volume (in cm 3), is (a) 570 (b) 390 (c) 190 (d) 590 283. The height of a right prism with a square base is 15 cm. If the area of the total surface of the prism is 608 sq. cm, its volume is (a) 910 cm3 (b) 920 cm3 (c) 960 cm3 (d) 980 cm3

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(a) 288 3

(b)144 2

(a) 5  3 3





(c) 360

(b) 36 3

(d) 72 5  3



10 3 cm. If the total surface

are of the pyramid is 270 3 sq. cm. its height is (a) 12 3 cm (b) 10 cm (c) 10 3 cm

(d) 12 cm



294. A right pyramid stands on a square base of side 16 cm and its height is 15 cm. The area (in cm 2) of its slant surface is (a) 514(b) 544 (c) 344 (d) 444 295. The base of a right prism is a right angled triangle whose sides are 5 cm, 12 cm and 13 cm. If the total surface area of the prism is 360 cm2, then its height (in cm) is (a) 10 (b) 12 (c) 9 (d) 11 296.A right pyramid 6 m high has a square base of which the diagonal is 1152 m. Volume of the pyramid is (a) 144 m3 (b) 288 m3 (c) 576 m3 (d) 1152 m3 297. The height of the right pyramid whose area of the base is 30 m2 and volume is 500 m2 is (a) 50 m (b) 60 m (c) 40 m (d) 20 m 298.The base of a right. prism is an equilateral triangle. If the lateral surface area and volume is 120 cm2, 40 3 cm3 respectively then the side of base of the prism is (a) 4 cm (b) 5 cm (c) 7 cm (d) 40 cm 299. Each edge of a regular tetrahedron is 4 cm. its volume (in cubic cm) is

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(c) 108 3 (d)144 3 288. The base of right prism is a triangle whose perimeter is 28 cm and the inradius of the triangle is 4 cm. If the volume of the prism is 366 cc, then its height is (a) 6 cm (b) 8 cm (c) 4 cm (d) None of these 289.If the base of a right pyramid s triangle of sides 5 cm, 12 cm and 13 cm and its volume is 330 cm, then its height (in cm) will be (a) 33 (b) 32 (c) 11 (d) 22 290.The base of a right pyramid is equilateral tr iang le of side

(a)

16 3 3

(b) 16 3

(c)

16 2 3

(d) 16 2

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v.iSn ir

square base of diagonal 10 2 cm. If the height of the pyramid is 12 cm, the area (in cm2) of its slant surface is (a) 520(b) 420 (c) 360 (d) 260 293. If the altitude of a right prism is 10 cm and its base is an equilateral triangle of side 12 cm, then its total surface area (in cm2) is

300. The base of a prism is a right angled triangle with two sides meeting at right angle are 5 cm and 12 cm. The height of the prism is 10 cm. The total surface area of the prism is (a) 360 sq. cm (b) 300 sq. cm (c) 330 sq. cm (d) 325 sq. cm 301. The radius of a cylinder is 10 cm and height is 4 cm. The number of centimetres that may be added either to the radius or to the height to get the same increase in the volume of the cylinder is : (a) 25 (b) 4 (c) 5 (d) 16 302. If the area of the base, height and volume of a right prism be

eYrai dnag

(c) 150 3 cubic cm (d) 300 3 cubic cm 285. A prism has as the base a right angled triangle whose sides adjacent to the right angles are 10 cm and 12 cm long. The height of the prism is 20 cm . the density of the material of the prism is 6gm/cubic cm. the weight of the prism is (a) 6.4 kg (b) 7.2 kg (c) 3.4 kg (d) 4.8 kg 286. If the slant height of a right pyramid with square base is 4 metre and the total slant surfac e of the pyr amid is 12 square metre, then the ratio of total slant surface and area of the base is : (a) 16 : 3 (b) 24 : 5 (c) 32 : 9 (d) 12 : 3 287. The length of each edge of a regular tetrahedron is 12 cm. The area (in sq. cm) of the total surface of the tetrahedron is

81 3 cm 3. the height (in cm ) of the prism is (a) 9 (b) 10 (c) 12 (d) 15 292.A right pyramid stands on a

kgei snhe

(a) 320 3 cubic cm (b) 160 3 cubic cm

291.A right prism stands on a base of 6 cm side equilateral triangle and its volume is

ERna

284. The base of a right prism is an equilateral triangle of side 8 cm and height of the prism is 10 cm. Then the volume of the prism is

3 3     2  p² cm², 10 3 cm and   7200 cm³ respectively, then the value of P (in cm) will be?

2

3 (c) 3 (d) 2 3 303. If the base of right prism remains same and the lateral edges are halved, then its volume will be reduced by (a) 33.33% (b) 50% (c) 25% (d) 66% 304. The total surface ar ea of a reg ular triangular pyramid with each edges of length 1 cm is? 4 2 cm² (b) 3 cm² (a) 2 (c) 4 cm² (d) 4 3 cm² 305. Base of a right pyramid is a square of side 10 cm. If the height of the pyramid is 12 cm, then its total surface area is (a) 360 cm² (b)400 cm² (c)460 cm² (d) 260 cm² 306. A right prism has a triangular base whose sides are 13 cm, 20 cm and 21 cm, If the altitude of the prism is 9 cm, then its volume is (a) 1143 cm³ (b) 1314 cm³ (c) 1413 cm³ (d) 1134 cm³ 307. Base of a prism of height 10 cm is square. Total surface area of the prism is 192 sq. cm. The volume of the prism is (a) 120 cm3 (b) 640 cm3 (c) 90 cm3 (d) 160 cm3 (a) 4

(b)

387

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00 3

equal  of side 18 3 & the height of pyramid is 12 cm. (a) 124 3

(b) 624 3

(c) 648 3 (d) 405 3 313. The base of a right prism is a  ABCD. If the volume of prism is 2070. Then find the lateral surface area. AB = 9, BC = 14, CD = 13, DA = 12 DAB = 90°. (a) 720 (b) 540 (c) 920 (d) 960 314. Find the volue of pyramid which is base d on a e quilater al traingle of side 4 cm & height of pyramid is 20 3 cm. (a) 100 (b) 160 (c) 80 (d) 40 315. Find the total surface area of pyramid of 4 cm height which is based on a square of side 6cm. (a) 48 (b) 72 (c) 96 (d) 120

(b)

100 119 3

100 119 (d) 100 119 9 317. A rectangular water tank is open at the top. Its capacity is 24 m3. Its length and breadth are 4 m and 3 m respectively. Ignoring the thickness of the material used for building the tank, the total cost of painting the inner and outer surfaces of the tank at the rate of Rs. 10 per m2 is: (a) Rs. 400 (b) Rs. 500 (c) Rs. 600 (d) Rs. 800 318. If V be the volume and S the surface area of a cuboid of di-

eeYa ridn agv .iSn i

(c)

(a) 50 paise (b) 25 paise (c) 75 paise (d) 1 paise 322. The volume of a rectangular block of stone is 10368 dm2 , its dimensions are in the ratio of 3:2:1, If its entire surface is polished at 2 paise per dm2 , then what is the total cost? (a) Rs. 31.68 (b) Rs. 31.50 (c) Rs. 63 (d) Rs. 63.36 323. A rectangular water tank measure 15m × 6m at top and is 10m deep. It is full of water. If water is drawn out lowering the level by 1 meter how much of water has been drawn out? (a) 90,000 litres (b) 45,000 litres (c) 80,000 litres (d) 40,000 litre 324. A rectangular tank is 45 m long and 26 m broad. Water flows into it through a pipe whose cross section is 13 cm2, at the rate of 9 km/hour. How much will the level of the water rise in the tank in 15 min? (a) 0.0016m (b) 0.0020m (c) 0.0025m (d) 0.0018 325. The diagonals of the three faces of a cuboid are x, y and z respectiv ely. Find the volume of cuboid?

r

(a)

mensions a, b and c then

1 is V

equal to:

(a)

S (a+b+c) 2

(b)

2  1 1 1  + +  S a b c

2S a+b+c (d) 2S(a + b + c) 319. An open box is made of wood 3 cm thick. Its external length is 1.46 m, breadth 1.16 m and height 8.3 dm. The cost of painting the inner surface of the box at 50 paise per 100 cm2 is: (a) Rs. 138.50 (b) Rs. 277 (c) Rs. 415.50 (d) Rs. 554 320. The areas of three adjacent faces of a cuboid are x, y & z square units respectively. If the volume of the cuboid be v cubic units, then the correct relation between v, x, y, z is: (a) v2 = xyz (b) v3 = xyz (c) v2 = x3y3z3 (d) v3 = x2y2z2 321. 1 m3 piece of copper is melted and recast into a square cross section bar 36 m long. An exact cube is cut off from this bar. If 1 m3 of copper cost Rs. 108, then the cost of the cube is. (c)

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v e – f is 2 (a) 2 (b) 4 (c) 5 (d) 10 309. The base of a right prism is a trapezium whose lengths of two parallel sides are 10 cm and 6 cm and distance between them is 5 cm. If the height of the prism is 8 cm, its volume is: (a) 300 cm³ (b) 300.5 cm³ (c) 320 cm³ (d) 310 cm³ 310. Base of a right prism is a rectangle, the ratio of whose length and breadth is 3 : 2. If the height of the prism is 12 cm and total surface area is 288 sq. cm, the volume of the prism is: (a) 288 cm (b) 290 cm³ (c) 286 cm³ (d) 291 cm³ 311. Height of a prism-shaped part of a machine is 8 cm and its base is an isosceles triangle, whose each of the equal sides is 5 cm and remaining side is 6 cm. The volume of the part is (a) 90 cm³ (b) 96 cm³ (c) 120 cm³ (d) 86 cm³ 312. Find the total surface area of pyramid which is based on a

316. Find the volue of a pyramid which is based on a square of side 10cm & lateral edge of pyramid is 12 cm.

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308. A right prism has triangular base. If v be the number of vertices, e be the number of edges and f be the number of faces of the pr ism. The value of

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(a)

xyz 2 2

 y +z –x z +x 2

(b)

2

2

2

2

–y2 x 2 +y2 –z2 

2 2

 y +z z +x x 2

(c)

2

2

2

2

+y2



2 2

(d) None of these 326. The same string, when wound on the exterior four walls of a cube of side n cm, starting at point C and ending at point D, can given exactly one turn. The length of the string, (in cm) is. D

C

(a) 2n

(b) 17n

(c) n

(d) 13n

388

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

5 – 2 cm

(b) 5(2 – 3) cm (c) 5 cm (d) 5 2 cm 329. A well of radius 'r' is dug 20 m deep and the earth taken out is spread all around it to a width of 1 m to form an embankment. The height of the embankment is 5 m then find the value of 'r': (a)

1 5 2

(b)

1 5 4

5 –1 5 –1 (d) 2 4 330. A cylinder is filled to 4/5th of volume. If is then tilted so that level of water coincides with one edge of its bottom and top edge of the opposite side. In the process, 30 litre of the water is spilled. What is the value of the cylinder? (a) 75 litre (b) 96 litre (c) Data insufficient (d) 100 litre 331. A monument has 50 cylindrical pillars each of diameter 50 cm and height 4 m. what will be the labour charges for getting these pillars cleared at the rate of 50 paise per m2 (Use π = 3.14). (a) Rs. 237 (b) Rs. 157 (c) Rs. 257 (d) Rs. 353 332. A right circular cyindrical tank has the storage capacity 38808 ml. If the radius of the base of the cylinder is three fourth of the height what is the radius of base? (a) 28 cm (b) 56 cm (c) 21 cm (d) 42 cm 333. A rectangular piece of iron sheet measuring 50 cm and

(a)

2 3 cm 3 π

(b)

2 2π cm 3 3

π 5 3 cm 3 (d) cm 3 3 π 336. A sector of circle of radius 3cm has an angle of 120º. if it is modulated into a cone, find the volume of the cone. (c)

π cm 3 3

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(c)

(a)

(b)

2 2π cm 3 3

2 3 3 cm 3 (d) cm 3 π π 337. If the slant height and the radius of the base of a right circular cone are H and r respectively then the ratio of the areas of the lateral surface and the base is: (a) 2H :r (b) H : r (c) H : 2r (d) H2 : r2 338. A sector of a circle of radius 15cm has the angle 120.It is rolled up so that two bounding radii are joined toghether to form a cone.the volume of the cone is.

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(c)

  (b) 100 2 π cm (c) 250 2 π /3 cm (a) 250 2 π cm 3



3



3

(d)  100 2 π / 3 cm 3

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v.iSn ir



339. The base radius and height of a c one is 5 cm and 2 5cm respectively.if the cone is cut parallel to its base at a height of h from the base. If the volume of this frustrum is 110 cm3 find the radius of smaller cone? (a) (104)1/3 cm (b) (104)1/2 cm (c) 5 cm (d) None of these 340. A hemisphe reic al b owl is 176cm round the brim. supposing it to be half full, how many persons may be swerved from it in hemispherical glasses 4 cm in diameter at the top? (a) 1372 (b) 1272 (c) 1172 (d) 1472 341. A sphere of radius 3 cm is dropped into a cylindrical vessel partly filled with water. The radius of the vesasel is 6 cm.If the sphere is submerged completely, then the surface of the water is raised by

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(a) 5

100 cm is rolled into cylinder of height 50 cm. If the cost of painting the cylinder is Rs. 50 per square meter, then what will be the cost of painting the surface of the cylinder? (a) Rs. 25.00 (b) Rs. 37.50 (c) Rs 75.00 (d) Rs. 87.50 334. Sixteen cylindrical cans, each with a radius of 1 unit, are placed inside a cardboard box four in a row. If the cans touch the adjacent cans and or the walls of the box, then which of the following could be the interior area of the bottom of the box in square units? (a) 16 (b) 32 (c) 64 (d) 128 335. Find the volume of a rig ht c i r c u l a r c o ne f o r m e d b y joining the edges of a sector of a circle of radius 4cm where the angle of the sector is 90°.

kgei snhe

327. A reservoir is supplied water by a pipe 6 cm in diameter. How many pipes of 1.5 cm diameter would disc harg e the same quantity, supposing the velocity of water is same? (a) 8 (b) 12 (c) 16 (d) 20 328. Given a solid cylinder of radius 10 cm and length 1000 cm, a cylindrical hole is made into it to obtain a cylindrical shell of uniform thickness and having volume equal to one-fourth of the original cy lind er. The thickness of the cylindrical shell is:

1 1 cm (b) cm 4 2 (c) 1 cm (d) 2cm 342. Let A and B be two solid spheres area of B is 300% higher than surface area of A. The volume of A is found to be k% lower than the volume of B.The value of k must be (a) 85.5 (b) 92.5 (c) 90.5 (d) 87.5 343. The base of a prism is a regular hexagon. If every edge of the prism measures 1 metre and height is 1 metre, than volume of the prism is

(a)

(a)

3 2 cu m 2

(b)

3 3 cu m 2

6 2 5 3 cu m (d) cu m 5 2 344. The base of a right prism is a pentagon whose sides are in

(c)

the ratio 1: 2 : 2 :1:2 and its height is 10 cm. If the longest side of the base be 6 cm, the volume of the prism is (a) 270 cm3 (b) 360 cm3 (c) 540 cm3 (d) None of these

389

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(d) 2 : 3 3 :2 346. The base of a right prism is a trapezium. The lengths of the parallel sides are 8 cm and 14 cm and the distance between the parallel sides is 8 cm. If the volume of the prism is 1056 cm 3, then the height of the prism is (a) 44 cm (b) 16.5 cm (c) 12 cm (d) 10.56 cm 347. If the base of a right rectangular prism is left unchanged and the measure of the late ral edges are doubled, then its volume will be (a) unchanged (b) tripled (c) doubled (d) quadrupled 348. Prism has as the base a right angled triangle whose sides adjacent to the right angles are 10 cm and 12 cm long. The height of the prism is 20 cm. The density of the material of the prism is 6 gm. cubic cm. The weight of the prism is. (a) 6.4 kg (b) 7.2 kg (c) 3.4 kg (d) 4.8 kg 349. The perimeter of the triangular base of a right prism is 15 cm and radius of the incircle of the triangular base is 3 cm. If the volume of the prism be 270 cm 3, then the height of the prism is(a) 6 cm (b) 7.5 cm (c) 10 cm (d) 12 cm 350. The base of a right prism is an equilateral tirang le. If its height is one-fourth and each side of the base is tripled, then the ratio of the volumes of the old to the new prism is(a) 4 : 3 (b) 1 : 4 (c) 1 : 2 (d) 4 : 9 351. A right pyramid is on a regular hexagonal base. Each side of the base is 10 m and the height is 30 m. The volume of the pyramid is (a) 2500 m3 (b) 2550 m3 (c) 2598 m3 (d) 5196 m3

(a) 3a3

(b) 3a 2 2

(c) 3a 3 3 (d) 6a3 353. If the area of the base of a regular he xagonal pyramid is 2 96 3 m and the area of one of

its side faces is 32 3 m3, then the volume of the pyramid is: (a) 380 3 m 3

(b) 382 2 m 3

(c) 384 3 m 3 (d) 386 3 m 3 354. What part of a ditch, 48 metres long 16.5 metres broad and 4 metres deep can be filled by the sand got by digging a cylindrical tunnel of diameter 4 metres and length 56 metres? 22    useπ  7 

1 2 7 8 (b) (c) (d) 9 9 9 9 355. A cylindrical rod of iron whose height is eight times its radius is melted and cast into spherical balls each of half the radius of the cylinder. The number of such pherical balls is (a) 12 (b) 16 (c) 24 (d) 48 356. Water flows at the rate of 10 meters per minute from a cylindrical pipe 5 mm in diameter. How long it take to fill up a conical vessel whose diameter at the base is 30 cm and depth 24 cm? (a) 28 minutes 48 seconds (b) 51 minutes 12 seconds (c) 51 minutes 24 seconds (d) 28 minutes 36 second 357. A semi-circular sheet of metal of diameter 28 cm is bent into an open conical cup. The depth of the cup is aproximately : (a) 11 cm (b) 12 cm (c) 13 cm (d) 14 cm 358. The height of a right prism with a square base 15 cm. If the total S.A. of prism of 608 cm2. The find its volume. (a) 480 (b) 460 (c) 1500 (d) 960

(a)

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r

6 :1

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(c)

(b)

5a , 2 then the volume of this pyramid is.

this pyramid is of length

359. A slab of ice 8 inches in length 11 inches in breadth, and 2 inches thick was melted and resolidified in the form of a rod of 8 inches diamete r. The length of such a rod, in inches, is nearest to. (a) 3 (b) 3.5 (c) 4 (d) 4.5 360. A storage tank consists of a circular c ylinder with a hemisphere adjoined on either side. If the external diameter of the cylinder be 14 m and its length be 50 m, then what will be the cost of painting it at the rate of Rs. 10 per sq m? (a) Rs. 38160 (b) Rs. 28160 (c) Rs. 39160 (d) None of these 361. The diameter of the iron ball used for the shotput game is 14 cm. It is melted and then a solid

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(a) 1 : 6

352. There is a pyramid on a base which is a regular hexagon of side 2a. If every slant edge of

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345. There are two prism, one has equilateral triangle as a base and the other regular hexagon. If both of the prisms have equal heights and volumes, then find the ratio between the length of each side at their bases.

1 cm is 3 made. What will be the diameter of the base of the cylinder? (a) 14 cm (b) 28 cm

cylinder of height 2

14 28 cm (d) cm 3 3 362. If the area of the circular shell having inner and outer radii of 8 cm and 12 cm respectively is equal to the total surface area of cylinder of radius R 1 and height h, then h, in terms or R1 will be.

(c)

(a)

3R12 –30 7R1

(b)

R12 –40 R 21

(c)

30 – R1 R 21

(d)

40 – R12 R1

363. Two solid right cones of equal heights are of radii r1 and r2 are melted and made to form a solid sphere of radius R. Then the height of the cone is: (a)

4R 2 r12  r22

(b)

4R r1  r2

4R 3 R2 (d) 2 2 r  r2 r1  r22 364. A solid wooden toy is in the shape of a right circular cone mounted on a hemisphere. If the radius of the hemisphere is 4.2 cm and the total height of the toy is 10.2 cm, find the

(c)

2 1

390

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22cm 10cm 8cm

(a)

a 2h cm 3p2

(b)

3hp 2 cm a2

3a 2 p2 cm (d) hp 2 cm 3h 2 368. The perimeter of an equilateral

(c)

triangle is 72 3 cm. Find its height. (a) 63 metres (b) 24 metres (c) 18 metres (d) 36 metres 369. A pit 7.5 metre long, 6 metre wide and 1.5 metre deep is dug in a field. Find the volume of soil removed in cubic metres. (a) 135 m³ (b) 101.25 m³ (c) 50.625 m³ (d) 67.5 m³

v.iSn ir

are of the largest area possible? (a) 85.71 cm² (b) 257.14 cm² (c) 514.28 cm² (d) 331.33 cm² 378.The areas of three adjacent faces of a cuboid are x, y, z. If the volume is V, then V² will be equal to (a) xy/z (b) yz/x² (c) x²y²/z² (d) xyz 379. The dimensions of a field are 20 m by 9m. A pit 10 m long, 4.5 m wide and 3m deep is dug in one corner of the field and the soil re move d has be en evenly spread over the remaining area of the field. What will be the rise in the height of field as a result of this operation? (a) 1m (b) 2m (c) 3m (d) 4m 380. A vessel is in the form of a hollow cylinde r mounte d on a hemispherical bowl. The diameter of the sphere is 14 cm and the total height of the vessel is 13cm. Find the capacity of the vessel. (Take = 22/7) (a) 321.33 cm (b) 1642.67 cm³ (c) 1232 cm³ (d) 1632.33 cm³ 381. A circular tent is cylindrical to a height of 3 metres and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 6 m wide to make the required tent. (a) 3894 m (b) 973.5 m (c) 1947 m (d) 1800 m 382. A steel sphere of radius 4 cm is drawn into a wire of diameter 4 mm. Find the length of wire. (a) 10,665 mm (b) 42.660 mm (c) 21,333 mm (d) 14,220 mm 383. A cylinder and a cone having equal diameter of their bases are placed in the Qutab Minar one on the other, with the cylinder placed in the bottom. If their curved surface area are in the ratio of 8 : 5, find the ratio of their heights. Assume the height of the cylinder to be equal to the radius of Qutab Minar. (Assume Qutab Minar to

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(a) 728.57 cm3 (b) 782.57 cm3 (c) 872.57 cm3 (d) 827.57 cm3 367. A conical flask has radius a cm and height h cm. It was completely filled with milk. The milk is poured into a cylindrical therefore flask whose base radius is p cm. What will be the height of the soultion level in the flask?

kgei snhe

18cm

370. In a shower, 10 cm of rain falls. What will be the volume of water that falls on 1 hectare area of ground? (a) 500 m³ (b) 650 m³ (c) 1000 m³ (d) 750 m³ 371. Seven equal cubes each of side 5 cm are joined end to end. Find the surface area of the resulting cuboid. (a) 750 cm² (b) 1500 cm² (c) 2250 cm² (d) 700 cm² 372. In a swimming pool measuring 90 m by 40m, 150 men take a dip. If the average displacement of water by a man is 8 cubic metres, what will be rise in water level? (a) 30 cm (b) 50 cm (c) 20 cm (d) 33.33 cm 373.A conical tent is to accommodate 10 persons. Each person must have 6 m² space to sit and 30 m³ of air to breath. What will be the height of the cone ? (a) 37.5 m (b) 150 m (c) 75 m (d) None of these 374. A hollow spherical shell is made of a metal of density 4.9 g/cm³. If its internal and external radii are 10 cm and 12 cm respectively, find the weight of the shell. (Take = 3.1416) (a) 5016 gm (b) 1416.8 gm (c) 14942.28gm (d) 5667.1 gm 375. A spherical cannon ball, 28 cm in diameter, is melted and cast into a right circular conical mould the base of which is 35 cm in diam eter . Find the height of the cone correct up to two places of decimals. (a) 8.96 cm (b) 35.84 cm (c) 5.97 cm (d) 17.9 cm 376. A rope is wound round the outside of a circular drum whose diameter is 70 cm and a bucket is tied to the other end of the rope. Find the number of revolutions made by the drum if the bucket is raised by 11 m. (a) 10 (b) 2.5 (c) 5 (d) 5.5 377. A cube whose edge is 20 cm long has circle on each of its faces painted black. What is the total area of the unpainted surface of the cube if the circles

ERna

volume of the wooden by (nearly). (a) 104 cm3 (b) 162 cm3 (c) 427 cm3 (d) 266 cm3 365. The volume of a cylinder and a cone are in the ratio 3 : 1. Find their diameters and then compare them when their heights are equal. (a) Diameter of cylinder = 2 times diameter of cone (b) Diameter of cylinder = Diameter of cone (c) Diameter of cylinder > Diameter of cone (d) Diameter of cylinder < Diameter of cone 366. A oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attatched to 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make a funnel.

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R Enak geisnh

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r

mon diameter is 3.5 cm and the heights of conical and cylindrical portion are respectively 6 cm and 10 cm. Find the volume of the solid. (Use = 3.14) (a) 117 cm² (b) 234 cm² (c) 58.5 cm² (d) None of these 397. A right elliptical cylinder full of petrol has its widest elliptical side 2.4m and the shortest 1.6m. Its height is 7m. Find the time required to empty half the tank through a hose of diameter 4cm if the rate of flow of petrol is 120 m/min (a) 60 min (b) 90 min (c) 75 min (d) 70 min 398. The radius of a right circular cylinder is increased by 50%. Find the percentage increase in volume (a) 120% (b) 75% (c) 150% (d) 125% 399. Water flows out at the rate of 10m/min from a cylindrical pipe of diameter 5 mm. Find the time taken to fill a conical tank whose diameter at the surface is 40 cm and depth 24 cm. (a) 50 min (b) 102.4 min (c) 51.2 min (d) 25.6 min 400. The section of a solid right circular cone by a plane containing vertex and perpendicular to base is an equilateral triangle of side 12 cm. Find the volume of the cone. (a) 72 cm³ (b) 144 cm³

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sphere is 4.2 cm and the total height of the toy is 10.2 cm, find the volume of wood used in the toy. (a) 353.72 cm³(b) 266.11 cm³ (c) 532.22 cm³(d) 133.55 cm³ 391.A cylindrical container whose diameter is12 cm and height is 15 cm, is filled with ice cream. The whole ice-cream is distributed to 10 children in equal cones having hemispherical tops. If the height of the conical portion is twice the diameter of the cylindrical container its base, find the diameter of the ice-cream cone. (a) 6 cm (b) 13 cm (c) 3 cm (d) 18 cm 392.A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 19 cm and the diameter of the cylinder is 7 cm. Find the total surface area of the solid. (Use = 22/7). (a) 398.75 cm²(b) 418 cm² (c) 444 cm² (d) 412 cm² 393.A cone, a hemisphere and a cylinder stand on equal bases and hav e the s am e height. What is the ratio of their volumes? (a) 2 : 1 : 3 (b) 2.5 : 1 : 3 (c) 1 : 2 : 3 (d) 1.5 : 2 : 3 394.The internal and external diam eter s of a hollow he mispherical vessel are 24 cm and 25 cm respectively. The cost of painting 1 cm² of the surface is ` 0.05. Find the total cost of painting the vessel all over. (Take  = 22/7) (a) ` 97.65 (b) ` 86.4 (c) ` 184 (d) ` 96.28 395. A cyclindrical cane whose base is horizontal is of internal radius 3.5 cm contain sufficient water so that when a solid sphere of max. size is placed, water just immersed it. Calculate the de pth of water in the cane before the sphere was put. 5 7 4 8 (a) (b) (c) (d) 2 3 3 3 396. A solid is in the form of a right circular cylinder with a hemisphere at one end and a cone at the other end. Their com-

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be hav ing same r ad ius throughout). (a) 1 : 4 (b) 3 : 4 (c) 4 : 3 (d) 2 : 3 384. If the curved surface area of a cone is thrice that of another cone and slant height of the second cone is thrice that of the first, find the ratio of the area of their base. (a) 81 : 1 (b) 9 : 1 (c) 3 : 1 (d) 27 : 1 385. A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If external radius of the base of the cylinder is 5 cm and its height is 32 cm, find the uniform thickness of the cylinder. (a) 2 cm (b) 3 cm (c) 1 cm (d) 3.5 cm 386. A hollow sphere of external and internal radius 6 cm and 4 cm respectively is melted into a cone of base diameter 8cm. Find the height of the cone (a) 25 cm (b) 35 cm (c) 30 cm (d) 38 cm 387. Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the s urface areas of the three cubes. (a) 7 : 9 (b) 49 : 81 (c) 9 : 7 (d) 27 : 23 388. If V be the volume of a cuboid of dimension x, y, z and A is its surface, then A/V will be equal to (a) x²y²z² (b) 1/2(1/xy+1/xz+1/yz) (c) None of these (d) 1/xyz 389. A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The height and radius of the cylindrical part are 13 cm and 5cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Calculate the surface area of the toy if the height of conical part is 12 cm. (a) 1440 cm² (b) 385 cm² (c) 1580 cm² (d) 770 cm² 390. A solid wooden toy is in the form of a cone mounted on a hemisphere. If the radii of the hemi-

(c) 74 cm³

(d) 72 3 π cm³

401. Iron weight 8 times the weight of oak. Find the diameter of an iron ball whose weight is equal to that of a ball of oak 18 cm in diameter. (a) 4.5 cm (b) 9 cm (c) 12 cm (d) 15 cm 402. A piece of squared timber is 7 metres long and 0.1 metre both in width and thickness. What is its weight at the rate of 950 kg per cubic metres? (a) 66 kg (b) 67 kg (c) 66.5 kg (d) 68.5 kg 403. How many c ubic metres of masonry are there in a wall 81

392

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(b)

4 ph2 3

3 2 ph (d) ph 2 2 412. How many bullets can be made out of a cube of lead whose edge measures 22 cm, each bullet being 2 cm in diameter? (a) 5324 (b) 2662 (c) 1347 (d) 2541 413. A cylindrical vessel 60 cm in diameter is partially filled with water. A sp here , 30 cm in radius is gently dropped into the vessle. To what further height will water in the cylinder rise? (a) 15 cm (b) 30 cm (c) 40 cm (d) Can't be determined 414. Th e d i f f er en ce b et w een t h e ou tside and inside su r face of a cylindr ical m etallic pipe, 14cm long, is 44cm 2. If the pipe is made of 99cm 3 of m etal. Find the ou ter r adii of the pipe? (a) 2cm (b) 2.5cm (c) 4cm (d) 5cm 415. How many bullets can be made out of a cube of lead whose edge measures 22 cm, each bullet being 2 cm in deameter? (a) 2341 (b) 2641 (c) 2541 (d) 2451 416. A right cylindrical vessel is full with water. How many right cones having same diameter and height as those of right cylinder will be needed to store that water? (a) 2 (b) 3 (c) 4 (d) 5 417. An open rectgangular cistern when measured from out side is 1 m 35 cm long; 1 m 8 cm broad and 90 cm deep, and is made of iron 2.5 cm thick. Find (i) the capacity of the cistern, (ii) the volume of the iron used. (a) 1171625 cu cm, 140575 cu cm (b) 1711625 cu cm, 104575 cu cm (c) 1171625 cu cm, 145075 cu cm (d) None of these

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(c)

418. Find the weight of a lead pipe 3.5 metres long, if the external diameter of the pipe is 2.4 cm and the thickness of the lead is 2 mm and 1 cc of lead weight 11.4 gm. (a) 5.5 kg (b) 5 kg (c) 8 kg (d) 10 kg 419. A closed rectangular box has inner dimensions 24 cm by 12 cm by 10 cm. Calculate its capacity and the area of tin foil needed to line its inner surface. (a) 2680 cu cm, 1296 sq cm (b) 2880 cu cm, 1396 sq cm (c) 2880 cu cm, 1296 sq cm (d) 2860 cu cm, 1296 sq cm 420. The dimension of an open box are 52 cm, 40 cm and 29 cm. Its thickness is 2 cm. If 1 cm3 of metal used in the box weight 0.5 gm, the weight of the box is: (a) 8.56 kg (b) 7.76 kg (c) 7.756 kg (d) 6.832 kg 421. Half cubic metre of gold sheet is extended by hammering so as to cover an area of 1 hectare. Find the thickness of the gold. (a) 0.05 cm (b) 0.5 cm (c) 0.005 cm (d) 0.0005 cm 422. Two cubic metres of gold are extended by hammering so as to cove r an are a of twe lve hectares. Find the thickness of gold. (a) 0.017 cm (b) 0.0017 cm (c) 1.7 cm (d) 0.17 cm 423. A cub of silver is drawn into a

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(a) 2ph2

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1 km/hr. Find how 2 many cubic m of water run into the sea per second. (a) 2500 cub metres (b) 2000 cub metres (c) 2200 cub metres (d) None of these 405. A cistern is constructed to hold 200 litres, and the base of the cistern is a square metre. what is the depth of the cistern? A cubic metre is 1000 litres. (a) 50 cm (b) 20 cm (c) 25 cm (d) 40 cm 406. A field is 500 metres long and 30 metres broad and a tank 50 metres long, 20 metres broad and 14 metres deep is dug in the field, and the earth taken out of it is spread evenly over the field, How much is the level of the field raised? (a) 0.5 m (b) 1.5 m (c) 1 m (d) 2 m 407. Find the volume and surface area of a cube, whose each edge measures 25 cm. (a) 15265 cu cm, 3750 sq cm (b) 15625 cu cm, 2500 sq cm (c) 15625 cu cm, 3850 sq cm (d) Data inadequate 408. The three co-terminus edges of a rectangular solid are 36, 75 and 80 cm respectively. Find the edge of a cube which will be of the same capacity? (a) 70 cm (b) 36 cm (c) 60 cm (d) Data inadequate 409. A cube of metal each edge of which measures 5 cm, weighs 0.625 kg. What is the length of each edge of a cube of the same metal which weighs 40 kg? (a) 20 cm (b) 25 cm (c) 15 cm (d) 30 cm 410. The sum of the radius of the base and the height of a solid cylinder is 37 m. If the total surface area of the cylinder be 1628 sq m, find the volume.

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rate of 4

(a) 4620 cu m (b) 4630 cu m (c) 4520 cu m (d) 4830 cu m 411. If the diameter of the base of a closed right circular cylinder is equal to its height h, then its whole surface area is:

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metres long, 4 metres high and 0.2 metre thick. (a) 64.8 cub m (b) 69 cub m (c) 68 cub m (d) 68.9 cub m 404. A river 10 metres deep and 200 metres wide is flowing at the

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1 mm in diameter, find 10 the length of the wire. (p = 3.1416) (a) 128 metres (b) 127.3 metres (c) 129.3 metres (d) 128.3 metres 424. A hollow cylindrical tube open at both ends ins made of iron 4 cm thic k. I f the internal diameter be 40 cm and the length of the tube be 144 cm, find the volume of iron in it. (a) 25344p (b) 23544p (c) 26344p (d) None of these 425. A hollow cylindrical tube open at both ends is made of iron 2 cm thic k. I f the internal diameter be 33 cm and the length of the tube be 70 cm, find the volume of iron in it. (a) 12400 cu cm (b) 15400 cu cm (c) 13800 cu cm (d) 16400 cu cm

wire

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r

is increased by 1016 cm3. Find the side of the cube. If each side of it is decreased by 2 cm, by how much will the volume decrease? (a) 12 cm, 729 cm3 (b) 8 cm, 512 cm3 (c) 9 cm, 729 cm3 (d) 12 cm, 728 cm3 439. Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the resulting cuboid to that of the total surface areas of the three cubes: (a) 5 : 7 (b) 7 : 9 (c) 9 : 7 (d) None of these 440. A hollow square shaped tube open at both ends is made of iron. The internal square is of 5 cm side and the length of the tube is 8 cm. There are 192 cm3 of iron in the tube. Find the thickness: (a) 2 cm (b) 0.5 cm (c) 1 cm (d) can't be determined 441. A cube of 11 cm e dge is im mers ed c ompletely in a rectangular vessel containing water. If the dimesions of base are 15 cm and 12 cm. Find the ris e in water level in the vessel: (a) 6.85 cm (b) 7 cm (c) 7.31 cm (d) 7.39 cm 442. A rectanular tank 25 cm long and 20 cm wide contains water to a depth of 5 cm. A metal cube of side 10 cm is placed in the tank so that one face of the cube rests on the bottom of the tank. Find how many litres of water must be poured into the tank so as to just cover the cube? (a) 1 L (b) 1.5 L (c) 2 L (d) 2.5 L 443. A rectangular block has length 10 cm, breadth 8 cm and height 2 cm. From this block, a cubical hole of side 2 cm is drilled out. Find the v olume and the surface area of the remaining solid: (a) 152 cm2, 512 cm2 (b) 125 cm2, 215 cm2 (c) 152 cm2, 240cm2 (d) 125 cm2, 512 cm2

R Enak geisnh

eeYa ridn agv .iSn i

433. The external dimensions of a wooden box closed at both ends are 24 cm, 16 cm and 10 cm respectively and thickness of the wood is 5 mm. If the empty box weighs 7.35 kg, find the weight of 1 cubic cm of wood: (a) 10 g (b) 12.5 g (c) 27 g (d) 15 g 434. The internal dimensions of a tank are 12 dm, 8 dm and 5 dm. How many cubes each of edge 7 cm can be placed in the tank with faces parallel to the sides of the tank. Find also, how much space is left unoccupied? (a) 35 ; 113 dm3 (b) 1313 ; 31.13 dm3 (c) 35 ; 31.013 dm3 (d) 1309 ; 13.31 dm3 435. The length, breadth and height of box are 2 m, 1.5 m and 80 cm respectively. What would be the cost of canvas to cover it up fully, if one square metre of canvas costs Rs. 25.00? (a) Rs. 260 (b) Rs. 290 (c) Rs. 285 (d) None of these 436. Three cubes each of edge 3 cm long are placed together as shown in the adjoining digure. Find the surface area of the cuboid so formed:

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426. One cm of rain has fallen on 2 square km of land. Assuming that 25% of the raindrops could have b een collected and contained in a pool having a 50 m × 5 m base, by what level would the water level in the pool have in creased? (a) 20 m (b) 40 m (c) 25 m (d) Data inadequate 427. Two cm of rain has fallen on a square km of land. Assuming that 40% of the raindrops could have b een collected and contained in a pool having a 200 m × 20 m base, by what level would the water level in the pool have incrased? (a) 2 m (b) 1 m (c) 4 m (d) 1.5 m 428. The length of a tank is thrice that of breadth, which is 256 cm deep and holds 3000 L water. What is the base area of the tank? (1000 L = 1 cubic metre) (a) 111775 m2 (b) 1171.875 m2 2 (c) 1.171875 m (d) None of these 429. A lid of rectangular box of sides 39.5 cm by 9.35 cm by 9.35 cm is sealed all around with tape suc h that there is an ovrelapping of 3.75 cm of the tape. What is the length of the tape used? (a) 111.54 cm (b) 101.45 cm (c) 110.45 cm (d) None of these 430. A cistern from inside is 12.5 m long, 8.5 m broad and 4m high and is open at top. Find the cost of cementing the inside of a cistern at Rs. 24 per sq. m: (a) Rs. 6582 (b) Rs. 8256 (c) Rs. 7752 (d) Rs. 8752 431. 250 men took a dip in a water tank at a time, which is 80m × 50m. What is the rise in the water level if the average displacement of 1 man is 4 m3? (a) 22 cm (b) 25 cm (c) 18 cm (d) 30 cm 432. The edge of a cube is increased by 100% the surface area of the cube is increased by: (a) 100% (b) 200% (c) 300% (d) 400%

3cm

m 3c 3cm 3cm 3cm (a) 182 sq. cm (b) 162 sq. cm (c) 126 sq. cm (d) None of these 437. A room is 36 m long, 12 m wide and 10 m high. It has 6 window, each 3 m × 2.5 m; one door 9.5 m × 6 m and one fire chimney 4 m × 4 .5 m . Find the expenditure of papering its walls at the rate of 70 paise per metre, if the width of the paper is 1.2 m: (a) Rs. 490 (b) Rs. 690 (c) Rs. 1000 (d) None of these 438. When each side of a cube is increased by 2 cm, the volume

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14 cm (d) 12 cm 3 449. A right circular c ylindric al tunnel of diameter 4 m and le ngth 10 m is to be constructed from a sheet of iron. The area of the iron sheet required.

(c)

280  (c) 80p (a)

(b) 40p (d) None of these

cm, the height of the water level rises in it is: 2 2 cm (b) 3 cm 3 3 2 (c) 5 cm (d) None of these 3 458. If h, c, v are respectively the height, the curved surface area and the volume of a cone then the value of 3  vh3 - c2h2 + 9v2 is equal to : (a) 1 (b) 2 (c) 0 (d) None of these 459. If P is the height of a tetra hedron & each side is of 2cm the find the value of 3p2. (a) 6a² (b) 8a² (c) 5a² (d) 7a² 460. If 'h' be the height of a pyramind standing on a base which is an equilateral trinagle of side 'a' units, then the slant edge is:

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(a) 2

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(a) xyz (b) xyz (c) 3xyz (d) None of these 448. A cylindrical cistern whose diameter is 21 cm is partly filled with water . If a rectangular block of iron 14 cm in length, 10.5 cm in breadth and 11 cm in thickness is wholly immersed in water, by how many centimetres will the water level rise? (a) 14 cm (b) 20 cm

450. A conical vessel has a capacity of 15 L of milk. Its height is 50 cm and base radius is 25 cm. How much m ilk can be contained in a ves sel in cylindrical form having the same dimensions as that cone? (a) 15 L (b) 30 L (c) 45 L (d) None of these 451. The height of a metric cylinder is 14cm & the different of its in curved S.A. is 44 cm2. If the cylinder is made up of 99 cm 3 metal the f ind the inne r & outer radius of cylinder. (a) 464 (b) 564 (c) 660 (d) 366 452. If the base rad ius and the height of a right circular cone are increased by 40% then the percentage increase in volume (approx) is: (a) 175% (b) 120% (c) 64% (d) 540% 453. From a circular sheet of paper of radius 25 cm, a sector area 4% is r emov ed. If the rem aining p art is used to make a conical surface, then the ratio of the radius and height of the cone is: (a) 16 : 25 (b) 9 : 25 (c) 7 : 12 (d) 24 : 7 454. A conical tent has 60° angle at the ve rtex. The ratio of its radius and slant height is: (a) 3 : 2 (b) 1 : 2 (c) 1 : 3 (d) can't be determined 455. Water flows at the rate of 5 m per min from a cylindrical pipe 8 mm in radius. How long will it take to fill up a conical vessel whose radius is 12 cm and depth 35 cm? (a) 315 s (b) 365 s (c) 5 min (d) None of these 456. A reservoir is in the shape of a frustum of a r ight circular cone. It is 8 m across at the top and 4 m across at the bottom. It is 6 m deep its capacity is: (a) 224 m3 (b) 176 m3 (c) 225 m3 (d) None of these 457. A conical vessel whose internal radius is 10 cm and height 72 cm is full of water. If this water is poured into a cylindrical vessel with internal radius 30

ERna

444. How many bricks (number near to next hundre d) will be required to build a wall 30 m long, 30 cm thick and 5m high with a provision of 2doors, each 4 m × 2.5 m and each brick being 20 cm × 16 cm × 8 cm when one-ninth of the wall is filled with lime? (a) 13500 bricks (b) 13600 bricks (c) 20050 bricks (d) 18500 bricks 445. A rectangular water reservoir is 15 m by 12 m at the base. Water flows into it through a pipe whose cross-section is 5 cm by 3 cm at the rate of 16 m/ s second. Find the height to which the water will rise in the reservoir in 25 minutes: (a) 0.2 m (b) 2 cm (c) 0.5 m (d) None of these 446. The volume of a wall, 3 times as high as it is broad and 8 times as long as it is high, is 36.864 m 3. The height of the wall is: (a) 1.8 m (b) 2.4 m (c) 4.2 m (d) None of these 447. If the areas of 3 adjacent sides of a cuboid ar e x, y, z respectively, then the voulum of the cuboid is:

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(a) h 2  a 2 / 4 (b) h 2  a 2 / 8 (c)

h 2  a 2 / 3 (d)

h 2  a2

461. Find the v olum e of a tetrahedren whose height is 4 3 cm. (a) 72 (b) 108 (c) 54 (d) 36 462. In a shower 10 cm of rain fall the volume of water that falls on 1.5 hectares of ground is: (a) 1500 m3 (b) 1400 m3 3 (c) 1200 m (d) 1000 m3 463. Find the total surface area of pyramid of heights 12 which is based on a rectangle of length 18 and & breadth 10cm. (a) 117 (b) 564 (c) 120 (d) 456 464. If from a circular sheet of paper of radius 15 cm, a sector of 144° is removed and the remaining is used to make a c onic al surface, then the angle at the vertex will be: 3  (a) sin–1     10 

6  (b) sin–1     5  3  4 (c) 2sin–1     (d) 2sin–1     5   5  465. Find the length of the string bound on a cylindrical tank whose base diameter and

height are 5

1 cm and 48cm 11

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(b) ( 2 + 1) : 7 : 8 (c) ( 2 + 1) : 3 : 4 (d) None of these 467. If l, b, p be the length, breadth and perimeter of a rectangle and b, l, p are in GP (in order), then l/b is: (a) 2 : 1 (b) ( 3 – 1) : 1

35 r 6

r2

(a) p : 3

(b)

: 6

(c) 6 :  (d) 6 :  471. A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their respective volumes is : (a) 1 : 2 : 3 (b) 2 : 1 : 3 (c) 1 : 3 : 2 (d) 3 : 1 : 2 472. A cube of sides 3cm is melted and smaller cubes of sides 1cm each are formed. How many such cubes are possible? (a) 21 (b) 23 (c) 25 (d) 27

(b)

25 r 6

35 r 6 3 476. A sphere of 20cm radius is dropped into a cylindrical vessel of 60cm diameter, which is partly filled with water, then its level rises by x cm. Find x: (c)

(a) 11

21 cm 27

(d)

(b) 12cm

23 cm 27 477. A right circular cone resting on (c) 22.5cm

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(c) ( 3 + 1) : 1 (d) can't be determined 468. The height of a cir cu lar cylinder i s i n cr eased b y 6 t i m es an d base ar ea is decr eased by 1/ 9th tim es. By what factor its later al su r face ar ea is incr eased ? (a) 2 (b) 3 (c) 6 (d) 1.5 469. A pyramid with an equal  based of each side 4 cm while its slant height is twice the height of pyramid. Find its volume8 (a) 8 3 (b) 3 3 (c) 4 3 (d) 3 3 470. A cube and a sphere have equal surfac e ar eas. The ratio of their volume is :

(a)

(d) 11

4 th its height 5 along a parallel to the circular base. The height of original cone is 75cm and base diameter is 42cm.What is the base radius of cut out (top portion) cone ? (a) 4.2cm (b) 8.4cm (c) 2.8cm (d) 3.5cm 478. A solid sphere is melted and recast into a right circular cone with a base radius equal to the radius of the sphere. What is the ratio of the height and radius of cone so formed ? (a) 5 : 2 (b) 4 : 3 (c) 4 : 1 (d) 3 : 2 479. 125 identical cubes are cut from a big cube and all the smaller cubes are arranged in a row to form a long cuboid. What is the per centage inits base is cut at

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cr ease in the total s ur face area of the cuboid over the total surface area of the cube ?

r

1 2 (a) 234 % (b) 234 % 3 3 (c) 117% (d) None of these 480. What is the total surface area of the identical cubes of largest possible size that are cut from a cuboid of size 75cm  15cm  4.5cm ? (a) 20, 250cm2 (b) 20, 520cm2 (c) 22, 250cm2 (d) None of these 481. If the volume of a sphere, a cube, a tetrahedron and a octahedron be same then which of the following has maximum surface area ? (a) sphere (b) cube (c) tetrahedron(d) octahedron 482. A spherical ball of lead 6cm in radius is melted and recast into three spherical balls. The radii of two of these balls are 3cm and 4cm. What is the radius of the third sphere ? (a) 6cm (b) 6.5cm (c) 5.5cm (d) 5cm 483. The base of a right prism is a triangle whose perimeter is 45cm and the radius of incircle is 9cm. If the volume of the prism is 810cm3. Find its height : (a) 5cm (b) 4cm (c) 6cm (d) 4.5cm 484. What is the semi-vertical angle of a cone whose lateral surface area is double the base area ? (a) 30° (b) 45° (c) 60° (d) None of these 485. What is the number of cones of semi-vertex angle  and having r as the radius of the midsection which can be moduled out of a cylinder of base radius r and height 2r cot  : (a) 5 (b) 7 (c) 6 (d) 4 486. A water tank is 30cm long, 20cm wide and 12m deep. It is made of iron sheet which is 3m wide. The tank is open at the top. If the cost of iron sheet is ` 10per meter. Find the total cost of iron required to build the tank ? (a) ` 6000 (b) ` 5000

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(a) ( 3 + 1) : 3 : 4

473. A cuboidal block of 6cm  9cm  12cm is cut into exact number of equal cubes. The least possible number of cubes will be : (a) 6 (b) 9 (c) 24 (d) 30 474. If the cone is cut along its axis from the middle, the new shape we obtain after opening the paper is : (a) isosceles triangle (b) right angle triangle (d) equilateral triangle (d) None of these 475. What is the height of the cone which is formed by joining the two ends of a sector of circle with radius r and angle 60° :

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respectively. The string makes exactly four complete turns round the cylinder, while its two ends touch the tank’s top and bottom : (a) 75cm (b) 70cm (c) 60cm (d) 80cm 466. A cone, a hemisphere and a cylinder stand on equal bases of radius R and have equal heights H. Their whole surfaces are in the ratio:

(c) ` 5500

(d) ` 5800

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40  R1 R1

40  R12 (c) R1

(b)

40  R1 R1 2

(d) None of these

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500. The radius and the height of a cone are in the ratio 4 :3 . The ratio of the curved surface area and total surface area of the cone is : (a) 5 : 9 (b) 3 : 7 (c) 5 : 4 (d) 16 : 9 501. From a right circular cylinder of radius 10 cm and height 21 cm, a right circular cone of same base- radius is removed. If the volume of the remaining portion is 4400 cm 3 then the height of the removed cone is:

22    take    7   (a) 15 cm (b) 18 cm (c) 21 cm (d) 24 cm 502. A right circular cylinder and a cone have equal base radius and equal heig hts . If the ir curved surfaces are in the ratio 8 : 5, then the radius of the base to the height are in the ratio : (a) 2 : 3 (b) 4 : 3 (c) 3 : 4 (d) 3 : 2 503. The curved surface area of a cylindrical pillar is 264 sq.m. and its volume is 924 cu.m. The ratio of its diameter to height is : (a) 3: 7 (b) 7 : 3 (b) 6 : 7 (d) 7 : 6 504. A cube of edge 6 cm is painted on all sides and then cut into unit cubes. The number of unit cubes with no sides painted is: (a) 0 (b) 64 (c) 186 (d) 108 505. There is a pyramid on a base which is a regular hexagon of side 2a cm. If every slant edge of this pyramid is of length

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490. An iron pipe 20cm long has exterior diameter 25cm. If the thickness of the pipe is 1cm, then the whole surface area of the pipe ? (a) 3168cm2 (b) 3186cm2 (c) 3200cm2 (d) 3150cm2 491. The cap acity of two hem ispherical bowls are 64 litre and 216 litre respectively. Then the ratio of their internal curved surface area will be: (a) 2 : 3 (b) 1 : 3 (c) 16 : 81 (d) 4 : 9 492. If the length of a rectangular parallel pipe is three times of its breadth and five times of its he ight. If its volume is 14400cm3, then the total surface area will be : (a) 4230cm2 (b) 4320cm2 (c) 4203cm2 (d) None of these 493. A right angled triangle with its sides 5cm, 12cm and 13cm is revolved about the side 12cm. Find the volume of the solid formed ?

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(a)

(a) 942cm3 (b) 298cm3 (c) 314cm3 (d) 302cm3 494. A hemisphere bowl V 1 and a hollow right circular cylinder V2 (having length equal to its radius) have the same diameter equal to the length of a side of a hollow cubical box V3. Water is filled in all these vessels upto the same level and such that hemispherical bowl is full of water and the volumes of filled water are v1, v2 and v3 respectively in V1, V2 and V3then : (a) V1 7² + 2² The triangle is Obtuse angled triangle The length of the three sides of a right angled triangle are (x – 2)cm, x cm & (x + 2) cm respectively. Then the value of x is: (x + 2)² = (x – 2)² + x²  x² + 4 + 4x = x² + 4 – 4x + x²  x² = 8x  x = 8 If two angles of a right angled triangle other than right angle are (2x + 21)º and (3x + 34)º then x is: 2x + 21° + 3x + 34° = 90° 5x = 35° or x = 7° In a  ABC,  A +  B = 113º,  B +  C = 105º, then find  B.  A +  B +  B +  C = 113° + 105° 180° +  B = 218°  B = 218° – 180° = 38º In a triangle ABC, BC is produced to D so that CD = AC. If  BAD = 111º and  ACB = 80º, then the measure of  ABC is: In ACB, 2x = 180° – 100° = 80°  x = 40º

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B

C 2

1

7.

C

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100º

x

D

C

In  ABD,  ABC = 180º–(111°+40°) = 29º If the sides of a triangle are produced then the sum of the exterior angles i.e.  DAB+  EBC+  FCA is equal to F C

kgei snhe eYari dnag v.iSn ir

1.

E

90º–

2 45º

EXAMPLES

A

B/2

1

60º

3

 ACD = 180 – C 180º–C C  ECD = = 90º – 2 2 In  BEC exterior angle,

2

3

30º

x

A

E

B

D Sol. Sum of exterior angles = 360º \  DAB +  EBC +  FCA = 360°

8.

In the given figure, BC = AC = AD, EAD = 81°. Find the value of x : E A 81° x

C B Sol. In DACD, ACD = ADC = x  CAD = (180°- 2x)  In DABC,

ABC =



BAC =

D

x 2

( ABC + BAC = ACD = x) BAC + CAD + 81° = 180° x + (180°– 2x) + 81° = 180° 2

3 x = 81°  x = 54° 2 9.

In the given triangle ABC, BC = CD and ( ABC – BAC) = 30°. The measure of ABD is: Sol. Given, BC = CD    BDC =  DBC &  ABC –  BAC = 30°

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A

12.

a 3 2 = sin30º sin 45º



4

=  ABC – (  ABD

10.

AC² = 16 + 25 – 40 ×

\ AC =

60º

D

wwM wa. th Les B aryn  ADB = 20°  ADB =  CAD = 20° (  AC = CD) ACB =20°+20°=40°(exterior

angle of  ACD) ABC =  ACB = 40° ( AB =  AC) 11. In  ABC, the bisectors of the internal angle  B & external angle  C intersects at D. If  BDC = 80°, then  A is : Sol.  A = 2  BDC.  A = 2 × 80 = 160°. 80°

B 2

B

AB² = 64 + 36 – 48 \ AB =

15.

D

A

B 2

60º C 6 Find AB = ? Sol. By using Cosine formula, AB²= AC²+BC²–2AC×BC Cos60° B

AB² = 8² + 6² – 2 × 8 × 6 ×

52 = 2 13 A

30º

C/2

C

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R

D

C

S

Q

1 2

A

1 P

2

B

1

As we know that, C

3 2 cm

2 3

45º

B

C

a=? Sol. By using Sine formula,

C/2

4 3 1  2 3 2

\ a = 4 2 cm 17. A, B, C, D are the vertex of a Rhombus and P, Q, R, S are the mid points of AB, BC, CD & DA. Find the largest angle of the rhombus ? & CP  AB Sol. Let the side of rhombus = 2 units

8

D

a 4 3 = sin 45º sin 60º



 a=

31

20°

C

1 2

A

14.



a b = SinA sin B

6

BC² = 25 + 36 – 30 BC² = 31 \ BC =

B

Sol. By using sine formula,

BC² = 5² + 6² – 2×5×6 ×

A

C

a=?

21

R Enak

Sol.

4 3 cm

60º

B

B C Find BC = ? Sol. By using Cosine formula, BC² = AB²+AC²–2AB·AC Cos60º

20°

C

B

A 45º

A

5

16.

1 2

AC² = 41 – 20 AC² = 21 13.

\ a = 3 cm

geisnh eeYa ridna gv.i Sni



+  BAD) 2 (  ABD) =  ABC –  BAD =  ABC –  BAC (   BAD =  BAC) = 30°  ABD = 15°  Con side r ABD su ch that ADB = 20° and C is a point on BD such that AB = AC and CD = CA. Then the measure of ABC is : A

1 2

 a = 3 2× 2 ×

r

 ABD =  ABC –  DBC =  ABC –  BDC (   DBC=  BDC)

60º B C 5 AC = ? Sol. By Cosine formula, AC² = AB² + BC² – 2AB·BC CosB AC² = 4² + 5² – 2×4×5 Cos60º

a b = sin A sin B

90º

P

60º

1

B

 Largest angle =  DAB = 180º – 60º = 120º

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EXERCISE

4.

5.

6.

7.

9.

A man goes 150 m due east and then 200 m due north, How far is he from the starting point ? (a) 200 m (b) 350 m (c) 250 m (d) 175 m In a  ABC,  A :  B :  C = 2 : 3 : 4. Then find the smallest angle is : (a) 45º (b) 40º (c) 60º (d) 65º

10. The sum of two angles of a triangle is equal to its third angle. Determine the measure of third angle : (a) 100º (b) 80º (c) 120º (d) 90º 11. Of the three angles of a triangle, one is twice the smallest and another is three times the smallest. Then find smallest angle : (a) 60º (b) 90º (c) 30º (d) 45º 12. In a ABC, If 2 A = 3 B = 6 C, then A is equal to : (a) 60° (b) 30° (c) 90° (d) 120° 13. The degree measures each of the three angles of a triangle is an integer. Which of the following could not be the ratio of their measures ? (a) 2 : 3 : 4 (b) 3 : 4 : 5 (c) 5 : 6 : 7 (d) 6 : 7 : 8 14. In the given figure below, what is the value of w ? B

(a) (b) (c) (d)

B  

O

x

C

 A

(a)

     (b)     

(c)

     (d)     

17. In the figure below, if s < 50°< t, then A

50°

w

50° O

30°

s

B

t

C

(a) t < 80 (b) s + t < 130 (c) 50 < t < 80 (d) t > 80 18. The sum of two angles of a triangle is 80° and their difference is 20°, then the smallest angle: (a) 50° (b) 100° (c) 30° (d) None of these 19. In the given diagram, equilateral triangle EDC surmounts square ABCD. Find the  BED represented by x. Where,  EBC =  º: E xº

C D

C M

90°

A

D

(a) 100° (c) 120°

(b) 110° (d) 130°

15. AB  BC, BD  AC and CE bisects C, A = 30°. then, what is CED? A

c

16. In the given figure, x = ?

kgei snhe eYari dnag v.iSn ir

3.

8.

ERna

2.

In a triangle ABC, if AB, BC and AC are the three sides of the triangle, then which of the statement is necessarily true? (a) AB + BC < AC (b) AB + BC > AC (c) AB + BC = AC (b) AB – BC > AC The sides of a triangle are 12 cm, 8 cm and 6 cm respectively, the triangle is : (a) Acute Angled (b) Obtuse Angled (c) Right Angled (d) Can't be determined In a  ABC,  BAC > 90º, then  ABC and  ACB must be: (a) Acute (b) Obtuse (c) One Acute and One Obtuse (d) Can't be determined In a  ABC,  A = x,  B = y and  C = (y + 20)°. If 4x – y = 10, then the triangle is: (a) Right-angled (b) Obtuse-angled (c) Equilateral (d) None of these If one angle of a triangle is equal to the sum of the other two, then the triangle is: (a) Right-angled (b) Obtuse-angled (c) Acute-angled (d) None of these If each angle of a triangle is less than the sum of the other two, then the triangle is : (a) Right-angled (b) Acute-angled (c) Obtuse-angled (d) None of these ABC is a triangle. It is given that a + c > 90º, then b is A

wwM wa. th Les aBryn

1.

º A

B

(a) 45º (b) 60º (c) 30º (d) None of these 20. In the given figure, if  ABC = 90°, and  A = 30°, then  ACD = A

b

B a greater than 90º less than 90º equal to 90º can't be said

C

D

30º

E B (a) 30° (b) 60°

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B

C (c) 45° (d) 65°

C

D

(a) 120° (b) 100° (c) 110°

(d) None of these

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21. Value of x and y is :

50° x (a) x = 60°, y = 80° (b) x = 80°, y = 50° (c) x = 50°, y = 80° (d) None of these 22. In the triangle ABC, side BC is produced to D,  ACD = 100° if BC = AC, then find  ABC is : A

B

A

(d)  A + 90° 24. It is given that d = 70°, b = 120°, then : G

F

A

D

E

C

F

B

c

B

a

b C

E

(a) c = 130° (b) a = 110° (c) Both (a) and (b) are correct (d) Both (a) and (b) are wrong 25. It is given that AB  BC. Which one of the following is true :

A

A

(a) 360° (b) 720° (c) 180° (d) 300° 29. In the given figure,  A +  B + C + D + E =

(a) 32° (b) 84°

(c) 64° (d) Can’t be determined 33. In the given figure BC is produced to D and  BAC = 40º and  ABC = 70º. Find the value of  ACD A

D C (b) 40° (d) 110°

(a) 30° (c) 70° 34. In the given figure,  CAG = 120º,  CEM = 25º,  ABC = 40º, Find  EMD = ? G A

D

E

A

E

C

X

40º

B B

C y

x

(a) 900° (c) 180°

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120º

F

D

A

E

B

D

B

d

D

(a) 36° (b) 108° (c) 59° (d) 72° 32. In the given figure below, if AD = CD = BC, and  BCF = 96°, How much is  DBC ?

F C 96°

R Enak

1 (  A + 180°) 2

(c)

1  A then  A is : 2

D

wwM wa. th Les B aryn

B (a)  A + 180° (b) 180°-  A

B =

71°

124 142 (b) 3 3 (c) 39° (d) None of these 28. In the given figure,  A +  B + C + D + E + F =

1 C

C

(b) 80° (d) 70° of ABC is pro ACD = 108° and

C

B

(a)

6

D

C

(c) 80°

5

x B

(a) 90° (c) 100° 31. The side BC duced to D. If

90°

(d) Can’t be determined 23. In the given figure, the side BC of a ABC is produced on both sides, then  1 +  2 is equal to: A

2

E

A

108°

D

(a) 50° (b) 45° (c) 55° (d) 60° 27. In the given figure, if AD = BD = AC, then the value of  C will be :

D

C

B

26. In the given figure, AC  CE and  A:  B :  C = 3 : 2 : 1, find the value of  ECD :

E

r

y

A

geisnh eeYa ridna gv.i Sni

80°

(a) 40° (b) 50°

30. In the given figure, AB divides  DAC in the ratio 1: 3 and AB = DB. The value of x :

(a) x + y = 180° (b) x + y = 270° (c) x + y = 300° (d) can not be said

(b) 720° (d) 540°

25º

B

C

(a) 125° (c) 70°

(b) 140° (d) 110°

M

D

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2  A then  A is : 3 (a) 10° (b) 68° (c) 102° (d) Not 36. In  ABC, value of exterior angle at B and C are 111° & 148° respectively then  A is : (a) 32° (b) 69° (c) 79° (d) 101° 37. In a  ABC,  BAC = 75°,  ABC = 45° BC is produced to D. If  ACD x = x, then % of 60° is: 3 (a) 30° (b) 48° (c) 15° (d) 24°

38. In the given figure which of the following statements is true ?

D F B x A

G

(b) 28°

D

E

Q

(a) PQ = QS (b) PR = RS (c) QR = RS (d) PS < RS 39. In the given figure, AM = AD, B = 63° and CD is an angle bisector of  C, then  MAC = ?

C

(a) 1 : 3 (b) 2 : 5 (c) 3 : 1 (d) 1 : 2 44. In the given figure, ABCD is a rhombus and AR = AB = BP, then the value of  RTP is : T

D

A

wwM wa. th Les aBryn

R

D

B

63° M

(a) 60°

C

(a) 27° (b) 37°

(c) 63° (d) None of these 40. If  A = 44°, BP = BR and CN = RC then  PRN = ?

B

44°

R

N

B

C

(a) 58° (b) 78°

(c) 68° (d) 66° 41. In the given figure, if  B =  C =

78°, BC = EC, CD = BC and DE not parallel to BC, then  EDB = ?

81º

x

B

D C (a) 45º (b) 54º (c) 63º (d) 36º 48. ABC is an equilateral triangle and CD is the internal bisector of  C. If DC is produced to E such that AC = CE then  CAE is equal to :

C P

A

B

(b) 90°

(c) 120°(d) 75°

D

O

x

A B (a) 60º (b) 45º (c) 75º (d) 90º 51. ABCD is a square in which  OBC is an equilateral triangle then find  DOA. A

D O

y°

C

(a) 27° (b) 54° (c) 72° (d) 58° 46. ABC is an isosceles triangle with AB = AC. Side BA is produced to D such that AB = AD. Find BCD : (a) 60° (b) 90° (c) 120° (d) can’t be determined 47. In the given figure, BC = AC = AD and  EAD = 81º. Find the value of x :

Rakesh Yadav Readers Publication Pvt. Ltd.

60º

B

D

C

E

A

27°

(c) 30° (d) 15°

49. In  ABC,  ABC = 120° then relation between sides is : (a) b2 = a2 + c2 + ac (b) b2 = a2 + c2 – ac (c) b2 = a2 + c2 – 2ac (d) b2 = a2 + c2 + 2ac 50. In the figure  ABE is an equilateral triangle in a square ABCD. Find the value of angle x in degrees :

45. In the following figure ADBC, BD = CD = AC, ABC = 27°, ACD = Y. Find the value of y :

A

P

A

(a) 45° (b) 75°

A

ERna

P

E

180 (d) None of these 7 43. In the given figure, if AD = DE = EC = BC then  A :  B =

(c)

B 30° 30°

C

153 7

(a)

S

R

E

(a) 18° (b) 12° (c) 22° (d) None of these 42. In the given figure, if AB = BC = CD = EF = DE = GA = FG, then x =?

kgei snhe eYari dnag v.iSn ir

35. In  ABC, the line BC extend up to D. If  ACD = 170° and  B =

60º

60º

C

(a) 60º (b) 45º (c) 150º (d) 90º 52. In  ABC,  A = 110° and D, E are two points on BC is such that  BDA = 140°,  CEA = 120° and  EAC = 2 ×  BAD. Find the angle  ABD (a) 30° (b) 40° (c) 60° (d) 80° 53. In an obtuse angle  ABC the external angle bisector of  A intersect the extended part of line CB at M and the external angle bisector of  C intersect the ex-

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Z

Y

(c) 30° (d) 20°

P

(b) 5 : 1 (d) 4 : 1

(a)

O. If  QOR =

(c) (b) (d) (c) (c) (d) (b)

15. 16. 17. 18. 19. 20. 21.

(b) (c) (d) (c) (a) (a) (c)

R Enak 22. 23. 24. 25. 26. 27. 28.

3 3 (d) 4 2

b=4

 B

a=5

C

3 4 3 2 (b) (c) (d) 4 5 5 5 64. If the angles of triangle are 60°, 90° and 30°, then what is the ratio of the sides opposite to these angles ?

(a)

(a) 105.5° (c) 106°

(b) 106.5° (d) 105°

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(c)

A

O

R

(b) (a) (c) (b) (d) (b) (a)

3 4

(b)

90º

Q

(b) 37.5° (d) 15°

8. 9. 10. 11. 12. 13. 14.

C

a =4

 A = 90º, Find sin β .

31º , PQ = PR, then 2

P

wwM wa. th Les B aryn (b) (b) (a) (a) (a) (b) (b)

4 3

external angle of  Q is:

57. In  xyz, the bisector of the external  y & internal  z meet at w. If external angle of  y &  z are 100º and 120º respectively then  YWZ is :

1. 2. 3. 4. 5. 6. 7.

b=?

63. In  ABC, AC = 4 cm, BC=5 cm,

(a)

& AC = 5 cm, then find cos β ?

(a) 27.5° (c) 30°

60º

N

M

60. In  ABC, AB = 4 cm, BC = 8 cm

R

P

A

40º

65º

F

45º

62. In  ABC, BC = 4 cm and  A = 60º,  B = 30º, Find AC ?

B

59. In  PQR, the bisector of internal  Q & external  R meet at

Q

(d) 56

30º

140º

S

(c) 120

O

(a) 2 : 1 (c) 3 : 1

(a) 45° (b) 70° (c) 30° (d) 15° 56. In  PQR, the bisector of the external  P & internal  R intersect at S. If external angle of  P = 100º &  PRQ = 45º then  PSR:-

(b) 55

c

D

E

(a) 7

geisnh eeYa ridna gv.i Sni

B C (a) 45° (b) 75° (c) 30° (d) 15° 55. In  DEF, the bisectors of the External  E & internal  F interect at I, If  EDF = 140º, then  I is:

(b)

61. In  ABC, AB = 5 cm, BC = 8 cm and  ABC = 60º, then find AC?

120º

100º

(a) 45° (b) 40°

64 55 (d) 60

(a)

58. In  MNO, the bisector of the external  M & Internal  N are 65º & 40º and meet at point P. Find the ratio of  O :  P is:-

15º

I

55 64 (c) 120

X

W

r

tended part of line AB at N. MA = AC = CN, then find  B = ? (a) 108° (b) 110° (c) 112° (d) 114° 54. In  ABC, the bisectors of the internal  B and external  C at D. If  BDC = 15º, then  A is: A D

3 : 2 :1

(c) 2 : 3 :1

A

(b) 1 : 2 : 2 (d) 3 : 2 :1

65. Three sides of a triangle are 7 cm, c=4

4 3 cm and 13 cm then the smallest angle is : (a) 15° (b) 30° (c) 45° (d) 60°

b =5 

B

C

a =8

ANSWER KEY 29. 30. 31. 32. 33. 34. 35.

(c) (a) (d) (c) (d) (a) (c)

36. 37. 38. 39. 40. 41. 42.

(c) (d) (b) (c) (c) (b) (c)

43. 44. 45. 46. 47. 48. 49.

(a) (b) (c) (b) (b) (d) (b)

50. 51. 52. 53. 54. 55. 56.

(c) (c) (a) (a) (c) (b) (a)

57. 58. 59. 60. 61. 62. 63.

(d) (a) (a) (a) (a) (a) (a)

64. (d) 65. (b)

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SOLUTION 1.

2.

3.

(b) The sum of the smaller sides cannot be equal to or less than the largest side. (b) 12² > 8² + 6² \ the triangle is Obtuse angled triangle. (a) In DABC,

5.

x + 2y = 160° 4x – y = 10°  y = 70°, x = 20°  The angles of the triangle are 20º, 70º, 90º. \ The triangle is right angled. (a) Let, ABC be a triangle.

12



A 

  11

 A = 90º \ The triangle is right angled.

6.

(b) Let  A <  B +  C. Then  A < 180° –  A [   A+  B +  C = 180° ]  2  A < 180º

  

wwM wa. th Les aBryn

  A < 90º

8.

10.

  A+  B+  C = 180º 

7. 





 A +  A = 180º

Similarly,  B 90°

A

   



x=

17.

(d) In DABC,  s + t + 50° = 180° s + t = 130° t = 130° – s  s < 50°  t > 130° – 50° t > 80°





D



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x

D

C M



  



20.

(b)

 EBC +  BEC = 30º 2  BEC = 30º (  BC = EC)  BEC = 15º   DEC = 60º \  DEM = 60º–15º  DEM = xº = 45º (a) A

A 

x

x + 2x = 71°  x =

71 3

  C =  4 = 2x =

142 3

100°

D C B In DABC,  AC = BC \  CAB =  ABC = x  ABC +  BAC =  ACD (exterior angle) x + x = 100°  x = 50°  Hence,  ABC = x = 50° 23. (a)  1 =  A +  5 ..... (i) and  2 =  A +  6 ..... (ii) (exterior angle)  1 +  2 =  A+  5+  A +  6    1 +  2 =  A + (  A +  5 +  6)   1 +  2 =  A + 180° (   A +  5 +  6 = 180°) 24. (c) a = 180°- d =180° - 70° = 110° and c = d +  ACB = 70° + ( 180° - b) = 70° + (180° – 120°) = 70° + 60° = 130° 25. (b)   BCA = 180°- Y  90° + (180°-y) = x (exterior angle) x + y = 270°  26. (d)   A : B :  C = 3 : 2 : 1 \ 3x + 2x + x = 180° 6x = 180°  x = 30°    C = x = 30°   ECD=180° –  ACB –  ACE  ECD = 180° - 90° - 30° = 60°  27. (b)

28.

(a) In AEC,  A +  E +  C = 180° ...(i) In BFD,  B +  F +  D = 180° ...(ii) On Adding (i) and (ii), we get  A +  B + C + D + E +  F = 360° (c)  2 =  E +  B (exterior angle of MEB) and  1 =  A +  D (exterior angle of APD)

r

x

wwM wa. th Les B aryn

A B   EBC+  BEC+  BCE= 180º \  EBC +  BEC = 180º–150º

22.

geisnh eeYa ridna gv.i Sni

(c) Let, ABC be a triangle A + B = 80° ...(i) A - B = 20° ...(ii) On adding equation (i) & (ii), 2 A = 100° A = 50°   B = 80°– A B = 80° – 50° B = 30°  A + B + C = 180°  C = 180°– (50° + 30°)  = 100° Hence, Smallest angle = B = 30° 19. (a) According to the question,  BCD = 90º (  each angle of square is 90°)  ECD = 60º (  each angle of an equilateral triangle is 60°) \  BCE =  BCD +  ECD = 90º + 60º = 150º E

R Enak

18.

29.

D

E

M 2

P

A

C

B

In DPMC,  1 +  2 +  C = 180°   A + B + C + D + E

30.

= 180° (a)  DAC = 180°- 108° = 72° Let,  DAB = y°   BAC = 3y°  y + 3y = 72°  y = 18° Now, DB = AB



 ADB =  DAB = y° = 18°

Now, in ACD,  CAE is exterior angle  108° =  ADC + x° 108° = 18° + x x = 90° Alternate

30º

B

C

D

  B +  A =  ACD (exterior angle) \ 90° + 30° =  ACD  ACD = 120°  21. (c) y = 80° (vertically opposite angle)  x = 180°- 50°- 80° = 50°]

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A

 AD = BD   1 =  2 = x (let)   3 =  1 +  2 = 2x (external angle)   4 =  3 = 2x

y

108° 3y

x D

B

C



 DAC = 180° – 108° = 72°

\

 DAB :  BAC = 1 : 3 (given)  DAB = y and  BAC = 3y



 DAB +  BAC = 72° y + 3y = 72°

( AD = AC)

In  ABC, 71° is an external angle.   1 +  4 = 71°

E

500

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

 

31.

4y = 72° y = 18°  DB = AB (given)  ADB =  DAB = y = 18°   BAC = 3y = 3 × 18° = 54° and  ABC= 2y = 2×18° = 36° In DABC,  ABC+  BAC +  ACB = 180°

34.

37.

(a) According to the question, G A

(d) According to the question, A 75º

120º

F E

B

25º

40º

C M  CAG =  ABC +  ACB = 120º

36° + 54° +  ACB = 180°  ACB = 90° Hence, x = 90° A (d)

C

(   CAG is exterior angle)   ACB = 120º – 40º



  ACB = 80º

x 120 % of 60º = % of 60º 3 3 = 40% of 60º

  ECM = 180º – 80º = 100º 108° B

D

C





32.

108° =

\  EMD = 125º

A + A 2

35.

(c)

A

F

2x

x

A

96°

2x

B

E 

x + y = 180° - 96 = 84 ...(i) Also for  CDB, 4x + y = 180° ...(ii) Subtract eqn. (i) from eqn. (ii), 3x = 96 or x = 32°   DBC = 2x = 64° 33. (d) As we know that,  A +  B +  C = 180   C = 180º – 70º – 40º   C = 70º  A



36.

70º

 

In BDC,  1 is an external angle   1 = 63° + x and in AMC, 2 is an external angle 2=x+y  AM = AD (given)  1 = 2  63°+ x = x + y y = 63° \ MAC = 63° (c) BP = BR BPR = BRP = x (let)  and CN = RC  CRN = RNC = y (let) PBR = 180° – 2x and  NCR = 180°-2y In  ABC,

40. 148º E C

111º



(c)

D

Let,  A = 3x & =  B = 2x  ACD =  A +  B 3x + 2x = 170º 5x = 170º x = 34º  A = 3x = 3 × 34 = 102º (c) According to the question, A

D

B  ABD = 111º

&  ACE = 148º

40º

B

C

D

C   ACD = 180º – 70º = 110º Alternate As we know that  ACD =  A +  B \  ACD = 40º + 70º = 110º



 ACB = 180º – 148º = 32º

&  ABC = 180º – 111º = 69º 

In DABC,



 A +  B +  C = 180º

\

 A = 180º – 69º – 32º

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Hence,  A

= 79º

12

39.

170°

B

wwM wa. th Les aBryn

D

(b)  PSQ = 180°– ( 90° + 60°) = 30°

B 2 = A 3



x y

40  60º = 24º 100

In  PSR,   PSR =  RPS \ PR = RS

(c) According to the question,

 ACD = 170°

3A =108° 2 108  2  72 A = 3 C

38.

  EMD = 100 + 25º

ERna



=

 EMD =  ECM +  CEM

 ACD =  ABC +  BAC

D

 ACD = x =  A +  B (   ACD is an exterior angle) = 75º + 45º  ACD = 120º

D

B

x

45º

kgei snhe eYari dnag v.iSn ir

 



x x

A + B + C = 180° 44°+180°–2x + 180° – 2y = 180° x + y = 112° Hence, PRN = 180° – (x + y) = 180° – 112°= 68°

501

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



 





42.

= 180° 2  CDB + 78° = 180°  CDB = 51° In DBEC,  EC = BC \  EBC =  BEC = 7 8°  EBC +  BEC +  ECB = 180° 78° + 78° +  ECB = 180°  ECB = 24°   ACB = 78° \  DCE +  ECB = 78°  DCE = 54° In DEDC,  DC = EC \  EDC =  DEC  DCE+  EDC+  DEC = 180° 2  EDC = 180° – 54° = 126°  EDC = 63°   EDC =  EDB +  CDB 63° =  EDB +51° Hence,  EDB = 12° (c)

  

Alternate:If the figure is given in the form of zig-zag way then calculate how many sides are equal. In above figure seven sides are equal. 180º DAE = No.of sides equal





43.

(a)

45.

 

180º 7

A x

B

E 1 4 y

2

D

46.

B

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T

x

ARD = x and

2

4 C

 2 + 2 + 4 + 4 = 180°  2( 2 + 4) = 180°  2 + 4 =90°  C = 90° 47.

x

Let

3

1 = 2 ( AB = AC) and 3 = 4 ( AB = AC = AD) In D BDC, B + C + D = 180°  1 + 2 + 4 + 3 = 180°

(exterior angle)

(b)

D

(b)

1

angle. 4 = x + 3 = 3x   y = 4 = 3x ( EC = BC)  A : B = x : y : = x : 3x =1:3

44.

Now in  RTP, RTP = 180° - (x + y) = 180° - 90° = 90° (c)  BD = CD \ DBC =  BCD = 27° In DBDC,   BDC +  DBC + BCD = 180°  BDC + 27° + 27° = 180°  BDC = 126°  ADC = 180° – 126° = 54° In DADC,  CD = AC \  ADC =  CAD  ACD = y° = 180° – (54°+54°) = 72°

C

 DE = EC 3 = 2 = 2x  In  AEC, 4 is an exterior

4 = 2x (exterior angle) [and BC = CD] 9 = 2 = 2x  3 = x ( FG = GA)  1 = x + 3 = 2x  (exterior angle) EF = FG 8 = 1 = 2x  5 = A + 9 = x + 2x

ARD = RDA ( AR = AD) DAB =  ARD + RDA  = x + x = 2x (exterior angle) Similarly, BPC = BCP = y  ABC = y + y = 2y (exterior angle)  2x + 2y = 180° ( ABCD is a rhombus) x + y = 90°

A

3

 AD = DE 1=x  2 = 1 + x = 2x 

x

 AB = BC 4=x   2=x+



180 7

x=

wwM wa. th Les B aryn





R Enak

 BC = EC CEB = EBC = 78°  In  BCD,  CD = BC DBC = CDB  CDB + DBC + DCB 

= 3x (exterior angle)  CD = DE  7+ 8= 5 7 = 3x - 2x = x 10 = A + 8 = 3x (exterior angle)  DE = EF  9 + 6 = 10 6 = 3x - 2x = x Now in  ADE, A + D + E = 180° x + 3x + 3x = 180°

r

(b)

geisnh eeYa ridna gv.i Sni

41.

\  BCD = 90° (b) According to the question, BC = AC = AD  EAD = 81º  ACD =  ADC = x  CAD = 180º – 2x  ABC =  BAC =

BPC = y

x 2

[   ACD =  BAC +  ABC]

502

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E A 81º

51. x

B

C

D

 BAC +  CAD + 81º = 180º

x + 180º – 2x + 81 = 180º 2

 BAO +  ABO +  AOB = 180º 60 + x + 45º = 180 x = 75º (c) According to the question,  BCD = 90º  OBA=  DCO= 90º – 60º = 30º In  OBA,  OB = AB

180º –30º \  BOA =  BAO = = 75º 2

3x = –81 2 \ x = 54º (d) According to question,

30 º 30º

C E

  A =  B =  C = 60°

60º

c

 B = 108º (c) A

D 15º

B

C   BDC = 15º \  BAC = 2 ×  BDC = 2 × 15º = 30º

55.

M

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2y º–

B

x

56.

18

0º

x –2

A

y

In  PQR   P +  Q +  R = 180º 80º +  Q + 45º = 180º  Q = 55º

y

 PSR =

C

In  MAC,  180º–2y + 180º–2y + x + 180–2x = 180º x + 4y = 360º ....(i) In  NAC, 180º – 2x + 180º –2x + 180º –2y + y = 180º 4x + y = 360º ...(ii)

R

 QPR = 180º – 100º = 80º

 –2x 18 0º–2 y

45º P

 180º

Q

(a) S

100º

B

x

F

EDF 140º   70º 2 2

\  EIF =

N

0 18

(c) According to the question,  ABE is an equilateral triangle  BAE =  ABE =  AEB = 60º As we know that,  ABD = 45º [ BD is diagonal of a square] D C E O x

140º

  EDF = 140º

ERna 53.

D

(b) I

E

 ABD = 180º –10º – 140º = 30º (a) According to the question,

b² = a² + c² + ac

A In  ABO,

 54.

C

 ADE = 180º – 140º = 40º  AED = 180º – 120º = 60º  DAE = 180º – 40º – 60º  DAE = 80   EAC = 20º  BAD = 10º In  BDA,

 –1  b² = a² + c² – 2ac ×  2   

50.

 B = 288º – 180º

EAC 2   BAD 1

b

120º B C a b² = a² + c² – 2ac cos 120º



 CEA = 120º   EAC = 2 ×  BAD

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49.

60º

Same in  COD  COD =  ODC = 75º   DOA = 360º–75º–75º–60º   DOA = 150º 52.. (a) According to the question,  BDA = 140º

 ACE = 180° – 30° = 150°  CAE =  CEA (  CE = AC) In  CAE,  CAE +  CEA +  ACE = 180°  2  CAE = 180° – 150°   CAE = 15° (a) By using cosine formula, b2 = a2 + c2 – 2ac CosB A

 B = 2 (x + y)–180º

60º

B

B



O

A

D

 B + 180º–2x +180º– 2y = 180º

D

–

48.



kgei snhe eYari dnag v.iSn ir

A

Solve Eq. (i) and (ii), We get x + y = 144º In  ABC,   A +  B +  C = 180°

57.

Q 55 = = 27.5º 2 2 X

(d) W

120º

100º Y

Z

 YZX = 180º – 120º = 60º

503

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&  XYZ = 180º – 100º = 80º 60.

In  xyz,  x +  y +  z = 180º  x + 80º + 60º = 180º  x = 40º

c=4

a =8 By using Cosine formula,

N

59.

= 75º

Cos β

O 75 2 2 = = P 75 1

O 31° 2

62.

wwM wa. th Les B aryn 31 = 31º 2

In  PQR,  P +  Q +  R = 180º

  PQ = PR  Q = R   

 

31º +  Q +  Q = 180º 2  Q = 149º  Q = 74.5º \ External angle of  Q

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r

64.

5²  8² – b ² 2 5  8

a b c    k (Let) sin A sinB sinC a b c   k sin60º sin90º sin30º

3 2 b = k sin90º = k × 1

a = k sin60º = k ×

A

R

P = 2 × O = 2 ×

a ²  c ² – b² 2ac

4 5 (d) According to the sine formula,  sin β =

1 25  64 – b ² = 2 80  40 = 89 – b²  b² = 49 b = 7 (a) Using sine formula,

c

Q

=

 Cos60º =



\ O : P = 2 : 1 (a) P

C a=5 Using sine formula,

64  16 – 25 28 4

Cos β =

CosB =

O 75º = 2 2

Required Ratio =

B

a b  sin A sinB 5 4   sin90º sinB 5 4  1  sinβ

R Enak

P =

C

55 64 61. (a) By using Cosine formula,



In  MNO  M +  N +  O = 180º 65º + 40º +  O = 180º

O

b=4

geisnh eeYa ridna gv.i Sni

40º

M

90º

a ²  c ² – b² 2ac

CosB =



65º

A

β

O

(a) P

(a)

b =5

B

40 \  YWZ = = 20º 2 58.

63.

= 180º – 74.5º = 105.5º A (a)

60º

c = k sin30º = k × a

b=?

30º

B

a =4

C

a b c   sin A sin B sinC



4 AC  sin60º sin 30º

4  2 AC×2  1 3 4  AC = 3

:

b

:

c

3 k : 2

K

:

k 2

:

2

:

1

3 65.

1 2

(b) By using cosine formula, cos θ =

a ²  b² – c ² 2ab

 cos θ =



 cos θ = θ

49  48 – 13 2 7  4 3 3 2

= 30º

504

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CHAPTER

22

TRANGLES CONGRUENCE & SIMILARITY Same Angles

Triangles are congruent when they have exactly the same three sides and exactly the same three angles. What is ''Congruent'' ..... ? It means that one shape can become another using turns, flips and/or slides:

Does this also work with angles? Not always! Two triangles with the same angles might be congruent;

40°

40°

110°

is congruent to :

Turn!

+

A

is NOT congruent to : 110°

Flip!

30°

40°

110°

ERna

wwM wa. th Les aBryn Slide!

Marking

When two triangles are congruent we often mark corresponding sides and angles like this;

The equal sides and angles may not be in the same position (if there is a turn or a flip), but they are there.

Same sides

When the sides are the same then the triangles are congruent. For example :

5

4

4

5

7

is congruent to :

7

5

7

and

4 Because the two triangles do not have exactly the same sides.

C

is congruent to:

These sides marked one are equal in length. Similarly for the sides marked with two lines. Also for the sides marked with three lines. The angles marked with one arc are equal in size. Similarly for the angles marked with two arcs. Also for the angles marked with three arcs.

F

D

2. SAS Congruence  Side-Angle-Side congruence. When two triangles have corresponding angles and sides equal that are congruent as shown below, the triangles are congruent. B

SSS, SAS, ASA, AAS and HL. These tests describe combinations of congru-

SAS Congruence E  ABC DEF C

A

F

D

3. ASA Congruence  Angle-side-angle congruence. When two triangles have corresponding angles and sides equal that are congruent as shown below, the triangles themselves are congruent. ASA Congruence  ABC DEF

A

B

E

C

D

F

4. AAS Congruence OR SAA Congruence  Angle-Angle-Side congruence. When two triangle s have corresponding angles and sides equal that are congruent as shown below, the triangles are congruent. A

E AAS Congruence  CAB DEF

Congruence Tests for Triangles

Rakesh Yadav Readers Publication Pvt. Ltd.

E

30°

because, even though all angles match, one is larger than the other. So just having the same angles is no guarantee they are congruent.

Translations

SSS Congruence  ABC DEF

30°

Only because they are the same size But they might NOT be congruent because of different sizes: 40°

Reflection

B

30°

110°

Rotation

ent sides and/or angles that are used to determine if two triangles are congruent. 1. SSS Congruence  Side-Side-Side congruence. When two triangles have corresponding sides equal that are congruent as shown below, the triangles are congruent.

kgei snhe eYari dnag v.iSn ir

Congruent Triangles

B

C

F

D

505

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E

F

Similarity of Triangles Two triangles are similar if and only if the corresponding sides are in proportion and the corresponding angles are congruent. There are three accepted methods of proving triangles similar: AA To show two triangles are similar, it is sufficient to show that two angles of one triangle are congruent (equal) to two angles of the other triangle. Theorem If two angles of one triangle are congruent to two angles of another triangle, the triangles are similar. B

E

A

F

C

D B E

Then:  ABC

D

If

Then:  ABC ~

F

B

C

A

{

Then:  ABC ~

C Q

E

Ratio of sides

N

U

R

T

} { =

2.

F

C

Then:  ABC ~  DEF

AB BC AC   DE EF DF Dealing with overlapping triangles : Many problems involving similar triangles have one triangle ON TOP of (overlapping) another ___ triangle. Since DE is marked to be parallel to AC, we know that we have  BDE congurent to DAC by corresponding angles. B is shared by both triangles, so the two triangles are similar by AA

= Ratio of inradii = Ratio of circumradii Ratio o f areas = Ratio of square s of corresponding sides. i.e. If  ABC ~  PQR,

Then,

Area(ABC ) (AB )2 (BC)2 (AC )2    Area(PQR) (PQ)2 (QR )2 (PR)2

*

Perimeter of triangle is:-

P erimeter ABC AB BC AC    Perimeter PQR  PQ QR PR

3.

In a right angled triangle, the triangles on each side of the altitude drawn from the vertex of the right angle to the hypotenuse are similar to the original triangle and to each other too. B

B

D

D

E C

A

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}

Ratio of height (altitudes)

tors

DEF

D

Then:

SSS for Similarity BE CAREFUL!! SSS for similar triangles is NOT the same theorem as we used for congruent triangles. To show triangles are similar, it is sufficient to show that the three sets of corresponding sides are in proportion. Theorem If the three sets of corresponding sides

S

= Ratio of medians = Ratio of angle bisec-

AB AC  DE DF

A

DEF

F

If A   D

F

AB AC BC   DE DF EF

M

D

Once the triangles are similar Theorem : The corresponding sides of similar triangles are in proportion. B E  DEF

C

P

A

1.

E

B

A

If  A

If the two triangles are similar, then for the proportional/corresponding sides we have the following results.

E

D

wwM wa. th Les B aryn

D

B

Properties of Similar triangles

r

B

C

D

There is an additional theorem that can be used when working with overlapping triangles. Additional Theorem : Basic propprtaionality theorem (BPT) or Thales theorem if a time is parallel to one side of a triangle and intersects the other two sides of the triangle, the line divides these two sides proportionality.

geisnh eeYa ridna gv.i Sni

HL Congruence  ABC DEF

A

of two triangle are in proportion, the triangles are similar. SAS for similarity BE CAREFUL!! SAS for similar triangles is NOT the same theorem as we used for congruent triangles. To show triangles are similar, it is sufficient to show that two sets of corresponding sides are in proportion and the angles they include are congruent. Theorem If an angle of one triangle is congruent to the corresponding angle of another triangle and the lengths of the sides including these angles are in proportion, the triangles are similar.

R Enak

5. HL Congruence  Hypotenuse-leg congruence. When two right triangles have corresponding sides equal that are congruent as shown below, the triangles are congruent.

C

i.e.,

A

 BCA ~  BDC ~  CDA. 506

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Some facts on Right angle triangle

F.

In eq. (ii) put the value of AB =

Some facts on Right angle triangle CD²  BD  DA

BC2  AC2 BC × CA = CD Squaring both sides, BC² × CA² = CD² × (BC² + AC²)



(A)

(B)

BC  CA  BA  CD

(C)

BC² AC²

BD × BA

CD² = =

AD × BA

(E)

BD DA



BC² 2 AC

(F)

1 CD2



1 1  BC2 CA 2

4.

If D and E are the mid-points of AB and AC & DE||BC A

B

 BCA~  BDC~  CDA

A

If  BDC~  CDA

E.

BC² = BA × BD  BCA~  CDA

BA CA  CA DA AC² = AB × AD In below equation C divided by equation D

B

E

F

C = E B = D A = F So, option (a) is true.

4.

In the figure AB  QR. Find the length of PB : P

EXAMPLES A Q

12c m

In  APC and  ABC,

R

P

A

9cm

C

B 6 cm

Sol. According to the question,

Q 15cm

3 9

A

P

B

C

D

AB BC CA  = FD DE EF

C

ABC is a triangle and P is any point on AB such that  ACP =  ABC, if AC = 9cm, CP = 12cm and BC = 15cm, then AP is equal to :

BD × BA BC2 2 = AD × BA AC

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A



Sol.

BD BC2 = DA AC2

the following is true? (a)  A =  F and  B =  D (b)  C =  F and  A =  D (c)  B =  F and  C =  D (d)  A =  E and  B =  D Sol. (a)According to the question,

O

Then,  EOD ~  COB

 ABC and  DEF are so related that If

AB BC CA = = , then which of FD DE EF

D

If ED || BC

1.

If the three side of one triangle are equal to the corresponding sides of the other triangle then the triangle are : Sol. Here, triangles are congruent. And congruent triangles are always similar.



B

 BCA~  BDC

BC BA  BD CB

D.

E

BC × AC AB

2.

is parallel to third line i.e. DE || BC Two triangles b/w the two parallel lines will always be similar.

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C.

5.

Area of  ABC 1 1  BC  AC =  AB  CD  2 2 BC × AC = AB × CD CD =

1 BC 2

ERna

θ

BD DC = CD DA CD² = BD × DA B.

DE 

 AD AE    i.e.  then the line  DB EC 

θ

 AP = 7.2cm

3.

Vice versa is also true i.e. If a line divides any two sides i n th e same rati o

D

AP PC AP 12    AC BC 9 15



C

then,

C

E

D

B

A.

BC2  CA 2 BC2  AC2

1 1 1   2 2 CD BC AC2

Explanation of these facts  DCB +  DCA = 90º  DCB +  CBD = 90º   DCA =  DBC 



kgei snhe eYari dnag v.iSn ir

(D)

=

 ACP =  ABC (Given)  A =  A (common)   APC ~  ACB

BC2  AC2

3 9

B 6 cm R

 PQR ~  PAB 507

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8.

AB PB  QR PR

In the given figure, AD = 11cm, AB = 18 cm and AE = 9 cm. Find EC :

3 PB  9 6

and In  DEF,  E = 70°,  F =50° and  D = 60°

 9.

AE AD 9 11    AB AC 18 AC

 AC = 22cm EC = AC - AE = 22 - 9 = 13cm In the given

AO AQ  BO BP 9 AQ  6 8

AQ = 12 cm

If  ABC is similar to  DEF, such that  A = 47° and  E = 63°, then  C is equal to :

47°

63°

B

C E

 ABC ~  DEF   A = 47° =  D  B =  E = 63°   C = 180° - 47° - 63° = 70°

if

F

 AED ~  ABC (by AA)



DE AE 2 12    BC AB 3 AB

r

12.

The lengths of perpendiculars drawn from any point in the interior of an equilateral triangle to the respective sides are 6cm, 8cm, and 10cm. The length of each side of the triangle is :

Sol.

Sol. 14.

2

a=

= 13.

(P1  P2  P3 )

3

=

 AB = 18cm

The corresponding sides of two similar triangles are in the ratio 1 : 3. Their altitude will be in the ratio: Sol. Ratio of altitude = Ratio of corresponding sides = 1 : 3 11. The lengths of perpendiculars drawn from any point in the interior of an equilateral triangle to the respective side of the triangle are P1, P2 and P3 , then the side of triangle is :

2

a

E 105°



Rakesh Yadav Readers Publication Pvt. Ltd.

3



10.

D

A

figure,

R Enak

wwM wa. th Les B aryn

Q

Sol.

1

3 2 a 4

2  2 a(P1  P2  P3 )  4 a

75° 65° B C Sol. In  ABC and  ADE,  BAC =  DAE = 180°- (75° + 65°) = 40° AED = 75° =  ABC 

B

C

1 1 1 a  P3  a  P1  a  P2  2 2 2

A

D

7.



DE 2  and AE = 12. Find AB: BC 3

P

D

Let the side of  ABC be a . O is the point in the interior of  ABC. OD,OE,OF are perpendiculars of BC,AC and AB ar (  OAB) + ar (  OBC) + ar (  OAC) = ar (  ABC)

C

Sol.  ADE ~  ACB (A-A property)   A =  A,  AED =  ABCTh e n  ADE =  ACB

In the given figure, QA and PB are perpendiculars to AB. If AO = 9cm, BO = 6cm and BP = 8cm. Find AQ :

O

B

57° B

E P2

P1

E 123°

D



 In  FED,  F = 50°,  E = 70° and  D = 60°   ABC ~  FED.

Sol.

F P3 O

geisnh eeYa ridna gv.i Sni

In  ABC and  DEF, If  A = 50°,  B = 70°,  C = 60°,  E = 70°,  F = 50°,  D = 60° then : Sol. In  ABC,  A = 50°,  B = 70°,  C = 60° 5.

A

A

A

PB = 2 cm

6.

Sol.

3

2 3

(P1 + P2 + P3 )

(6 + 8 +10)

48 3

= 16 3cm

Two triangles ABC and DEF are similar to each other in which AB = 10cm, DE = 8cm. Then the ratio of the areas of triangles ABC and DEF is : ar (ABC ) AB 2 100 25    ar (DEF ) DE 2 64 16

If G be the centroid of  ABC and the area of  GBD is 6 sq.cm, where D is the midpoint of side BC, then the area of  ABC is :

508

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A

Sol.

to BC. If AD = 6cm and BD = 4cm, then the length of BC is: Sol.

G

19.

In the given figure, DE || BC then the value of x is : A

C

x

D C

AB

AB

16.

=

AB BD   BC AB  AB2 = BC  BD  52 = BC  4  BC = 13cm

36  10  15cm 24

Alternate AD2 = BD × CD 62 = 4 × CD CD = 9 cm  BC = 9 + 4 = 13 cm

Sol.

 ABD ~  CAD (ii)  ABD   CDA (iii)  ADB ~  CAB

Of these statements, the correct ones are : Sol. In  ABD and  CAD

and AD = AD (common)   ADB ~  CAD and

 ABD   CAD In  ADB and  CAB  ADB =  BAC = 90° each and  ABC =  ABD

(common)   ADB ~  CAB Here, (i) and (iii) are correct statements. 17. In a triangle ABC,  BAC = 90° and AD is perpendicular

C

AD AE  (by basic proportionDB EC ality theorem (DE || BC)



10 8  8 NC  NC  6.4  AN = 10 – 6.4 = 3.6  AN : NC = 3.6 : 6.4 = 9:16 Alternate 2 2 AN AB 6 = = 2 2 NC BC 8 64

=

Q

P

B

C

AP AQ 1   BP QC 2

 QC = 2AQ  QC = 2 × 3 = 6  AC = AQ +QC

21.

= 3 + 6 = 9 cm The points D and E are taken on the side s AB and AC of  ABC such that AD = AE =

1 AB, 3

1 AC. If the length of BC 3

is 15cm, then the length of DE is :



36

A

Sol.

AC BC  BC NC

=

x 2

= 1 : 2 and AQ = 3cm, AC is:

10 2  6 2 = 8cm

and BC =

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B

 x2  x  x2  4  x  4 20. In  ABC, PQ || BC, If AP : PB

  ABC ~  BNC

 ADB =  ADC = 90° each  BAD =  ACD = 90° –  B

(x-1)

x

N

C B In  ABC &  BNC,  ABC =  BNC = 90° and  C =  C (common)

wwM wa. th Les aBryn

(i)

ERna

C

(x-2)

 x  2  x 1

In a right angled  ABC,  ABC = 90°; BN  AC, AB = 6cm, AC = 10cm. Then AN : NC is : Sol. A 18.

D

A

AD  BD

2

 ABC ~  DBA

perimeter of ABC

Which of the following is true in the given figure, where AD is the altitude to the hypotenuse of a right angle triangled ABC ? B

=

2

36  16  52cm

36 Sol. PQ  perimeter of PQR = 24

 AB 

B

A

= 6  ar ( BGD) = 6  6 = 36sq.cm 15. The perimeters of two similar triangles  ABC and  PQR are 36cm and 24cm respectively. If PQ = 10 cm, then AB is :

E

kgei snhe eYari dnag v.iSn ir

B D Area of  ABC

(x+2)

D

9 16

 AN : NC = 9 : 16

Sol.

A D

B

E

C

AD AE 1 = = AB AC 3  ΔABC ~ ΔADE DE 1 15  =  DE = = 5cm BC 3 3

509

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ABC is an equilateral triangle. P and Q are two points on

*

AB and AC respectively such that PQ||BC . If PQ = 5cm then the area of  APQ is: Sol.

27.

2

 AB   AC 2 Area of ΔABC    =  Area of ΔPQR  PQ   PR 

A

2

2  BC   Angle Bisector1    =  QR   Angle Bisector2 

Q

P

The Ratio of their area is the square of their sides, Medians, Altitudes, Perimeters, Angle Bisectors.

Sol.  Area of first triangle     Area of second triangle    

2

of  APQ

 Perimeter of Δ ABC 

= 24.

3  (PQ )2 4

23.

3 25 3  25 = sq .cm = 4 4 D is any point on side AC

of  ABC. If P,Q,X & Y are the mid-points of AB, BC, AD and DC respectively, then the ratio of PX and QY is : Sol. (By mid-point theorem) A

X D

P

The corresponding sides of two similar triangle in the ratio 5 : 6. Their area will be in the ratio. Sol. When triangle similar, Area of triangle = (side of triangle)²

C

Q

PX || BD and PX 

1 BD 2

QY || BD and QY 

1 BD 2

 PX : QY = 1 : 1 Area Based question when two trianges are similar  Condition for two triangle to be similar angle of two triangles are equal then the two triangles are similar A P

B

Q

288 12 2  =    A2 15  288 A2 =



28.

25.

Area of two similar triangle are respectively 196 cm² and 289 cm². If shortest side of the shorter triangle 7 cm. Then find the shorter side of the larger triangle. Sol. In two similar triangle the ratio of their area is the square of the ratio of their sides. 26.

Sol.

R

C If  ABC ~  PQR Then

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& Area of

Sol.

11 2

C

So  ADE ~  ABC AD : DB = 2 : 3 Then AB = AD + DB = 5 2 Area of  ADE  AD      Area of  ABC  AB 

 2 2 4 =   = 25 5 

Area of

 DECB = Area of

 ABC –  ADE = 25 – 4 = 21

2

M =

E

DE  BC (Given)

 Altitude of first  2    Altitude of second  

4.5 M

 DECB.

B

= 

9 = 11

AD 2 = then DB 3

A

D

Area of first triangle Area of second triangle

 =   M 

A2 = 450 cm² (A2 = Area of second triangle.) In a  ABC a line DE is drawn

find the ratio of area of  ADE

Area of two similar triangles are respectively 81 cm² & 121 cm². If altitude of first triangle is 4.5 cm. Find the corresponding altitude of the second triangle.

 81      121

16 25

parallel to BC. If

 Area of    5 2 25 1       Area of     36 6  2 

wwM wa. th Les B aryn

Y

B

 Median of first triangle 2    Median of second triangle   

=  Perimeter of Δ PQR   

R Enak

 Area

geisnh eeYa ridna gv.i Sni

 Altitude   Median 2 1 1 =  Altitude  =  Median    2 3

B C PQ || BC   APQ =  ABC = 60° &  AQP =  ACB = 60°

Here M = (Altitude of second triangle) The corresponding medians of two similar triangle are respectively 12 cm & 15 cm. If area of first triangle is 288 cm2. Find the corresponding area of second triangle.

r

22.

 4.5 



29.

Area of  ADE : Area of DECB = 4 : 21 In a  ABC a line DE is Drawn parallal to BC. If

AD 6 = , DB 7

510

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Find the ratio of area of  ADE



and Area of Sol.

31.

A

 DECB is 238

cm² and length of AD & AB respectively 5 cm & 12 cm find the area of  ABC.

E

B

C

DE  BC

(Given)

AB = AD + DB = 6 + 7 = 13 Area of ABC  AB   =   = Area of ADE  AD 

Area of Δ ADF  AD   = Area of Δ ABC  AB 

13 2 169     = 6 36  

2

5 25 =   = 12  144

 DECB = 169 – 36 =

 DECB is 9 : 16 AD DB

 DECB =

144 – 25 =

119 119 units = 238 1 unit = 2 So, Area of  ABC = 144 units = 144 × 2 = 288 cm² 32. In a  ABC a line DE is Drawn  to BC. Which Divide the tri-

wwM wa. th Les aBryn

find the value of

Area of

ERna



Area of  ADE : Area of DEBC = 36 : 133 In a  ABC a line DE is drawn  to BC. If ratio of areas of

angle In equal Area. Then find

A

AD . DB

D

E

B

Sol.

of

 ABC

=

D

A rea

of  ADE+Area of  DECB = 9 + 16 = 25 :  ADE  ABC Area  9 : 25 Side  3 : 5 The ratio of their sides is the square root of the ratio of their area AB 5  AD 3 BD = AB – AD = 5– 3 = 2

So,

E

A

B

C

AB D mid point of AB 2 AC AE = EC = E mid point of AC 2

AD = DB =

Let AB = 2 cm Then AD = 1 cm :  ADE  ABC Side  2 : 1 Area  4 : 1 Area of DEBC = Area of  ABC– Area of  ADE = 4 – 1 = 3 cm²



Now,

Ar. of  ADE 1  Ar. of  DECB 3

34.

In a  ABC a line PQ is Draw parallel to side BC, if AB = 9 cm and AP = 4 cm, find the ratio of shaded and unshaded portion Sol. A

C

DE  BC (Given) So,  ADE ~  ABC Area

A

C

2

Sol.

Ar. of ADE Ar. of  DECB

E

DE  BC (Given) So,  ABC ~  ADE AD = 5 cm, AB = 12 cm

2

 ADE &

In a  ABC, D and E are the mid point of AB and AC, find

D

B

2 –1

kgei snhe eYari dnag v.iSn ir

D

AD 6 = DB 7

30.

33.

1 =1 : 2 –1

Sol.

 ADE ~  ABC

Area of 133

AD  DB

A

Sol.

2 –1

Then,

In a  ABC a line DE draw  to BC. If area of

D

BD = AB – AD =

AD 3  DB 2

Then,

DEBC

B

P

E

C

Area of  ADE : Area of

Side  AD 1  AB 2

Rakesh Yadav Readers Publication Pvt. Ltd.

1 :

B C PQ  BC (Given) So  APQ ~  ABC AP = 4 cm, AB = 9 cm

 DECB =

1: 1 So, Area of  ABC = 1 + 1 = 2 units :  ABC  ADE Area  1 : 2

Q

2  4 2 16 Ar. of APQ  AP           = Ar. of ABC  AB  81 9 



2

= = =

Ar. of PQCB = Ar. of  ABC – Ar. of  APQ 81 – 16 = 65 Area of shaded part : Area of unshaded part 65 : 16

511

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two points on line AB & AC. ME  AC, ND  AB. The line ME and ND intersect at O Find

In a  ABC, D and E are two points on the side BC such that they trisect the line BC. In M and N are the two point on line AB & AC. ME  AC and Sol.

Ar. of DOE + Ar. of  AMON Ar. of ABC

Sol.

M

B

Let BC = 3 cm Then BD = DE = EC = 1 cm In  MEB (  DN  AB)

 ODE ~  EMB

Ar. of ODE = Ar. of EMB

Ar. of

 ED 2  1 2 1          4  BE  2 

 ODBM = Ar. of  EBM – Ar.

2  1 2 1 Ar. of ODE  DE           Ar. of DNC  DC  4 2 

Ar. of OECN = 4 – 1 = 3 cm² Area of  ABC = (BC)² = (3)² = 9 cm² Area of AMON = 9 – (3 + 1 + 3) = 2 cm² Now, Ar. of ODE + Ar. of  AMON = Ar. of ABC 1 2 1 = 9 3 36. In a  ABC, D and E are

AB AP : 3 1

DE 6  EC 5

2 Ar. of APQ  AP      = Ar. of ABC  AB 

2  6 2 Ar. of ODE  DE  36          Ar. of BME  BE  10 100  

=

38.

Ar. of ONCE = Area of  DNC – Area of  DOE = 121 – 36 = 85 units Area of  ABC = (BC)² = (15)² = 225 units Area of AMON = 225 – (64 + 36 + 85) = 40 units Now, Ar. of ODE + Ar. of  AMON Ar. of ABC

= 37.

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36  40 76 = 225 225

In a  ABC Line PQ is draw parallel to side BC where P and Q are respectively lie on side

8 8 100 = 88 % 9 9

In  ABC, AD is perpendicular to the angle bisector of angle B. line DE is drawn parallel to BC. E is a point on line AC. If AC = 12 cm. Find the length of AE.

Sol.

A

Ar. of ODBM = Ar. of  BME– Ar. of  ODE = 100 – 36 = 64 units Same as In  DNC 2  6 2 36 Ar. of ODE  DE            Ar. of DNC  DC  121 11 

 1 2 1     = 9 3 

Area of PQCB = 9 – 1 = 8 cm² Percentage of PQCB in the respect of  ABC

ODE ~  BME

two points on the side BC. BD 2  Such that and DE 3 DE 6  . M and N are the EC 5

Q

B C PQ  BC  APQ ~  ABC AB = 3AP

C

BD : DE : EC 2 : 3 6 : 5 (To make equal ratio) BD : DE : EC 4 : 6 : 5 BE = BD + DE = 4 + 6 = 10 units DC = DE + EC = 6 + 5 = 11 units BC = BD + DC = 4 + 11 = 15 units In  BME , OD  MB (  AB  ND)

wwM wa. th Les B aryn

of  ODE = 4 – 1 = 3 cm² Same as In  DNC

E

R Enak

3 1 3 D –1– E –1– C

–1–

OD  BM

D

BD 2  , DE 3

N O

B

N O

A M

P

r

ND  AB. The line ME and ND interse ct at O. Fi nd

Ar. of DOE + Ar. of  AMON . Ar. of ABC A

AB and AC. If AB = 3AP what is percentage of Area PQCB in the respect of  ABC. A Sol.

geisnh eeYa ridna gv.i Sni

35.

D



E

 B

C

M

A

D

B

 

M

 ABD and  BDM (90º)  ADB =  BDM BD = BD (Common) (BD Angle Bisector)  ABD =  DBM   ABD  BDM

AD = DM =

AM 2

We can say, D is the mid point of AM

512

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A

2  1 2 1 Ar. of  AMN  AM            Ar. of  ADE  AD  4 2

E

M

Ar. of Ar. of Ar. of  AMN :  ADE : ABC

2  AE AM AC

N D

E

E

B

C

C

1 BC. 2

D

E

Side  1 : 2 : 4 Area  1 : 4 : 16 40. In a triangle ABC, D and E are the mid points of line AB & AC. M and N are the mid points of AD & AE. Then find the ratio of Area of  AMN, Area of

M

 AD  Ar. of ADE 1  = =   Ar. of ABC 4  AB 

A

BCED.

A

AB 2 2

Then, BM = DM =

BD 2 = 1 2 2

AM = AD + DM = 2 + 1 = 3 Ar. of Ar. of  ADE : AMN Side  2 Area  4

E

:

: :

Ar. of ABC 3 : 4 9 : 16

In a  ABC, D and E are the mid point of AB & AC. M and N are the mid points of BD & EC then find the ratio of Ar. of  ADE :

C

 D is mid    point AB   

Then, AB = 2 × 2 = 4 M is mid point BD

N

B

C

AD = 2

AD=BD=

42.

D

N

B

B C Let, AM = 1 Then, AD = 2 AM = 2 × 1 = 2 and AB = 2 × AD = 2 × 2 = 4

2

E

M

N

Let,

Sol.

Ar. of ABC

:

2

DENM & Area of

Sol.

Ar. of DENM :

Ar. of MNCB

A

Ar. of Ar. of Ar. of  AMN :  ADE : ABC

N

D E M & N mid points of AD & AE (Given)

Then, MN  DE and

12

A

Sol.

3

D

DE 1 AD AE    Now, BC 2 AB AC

M

: 16 (To make equal ratio)

M

wwM wa. th Les aBryn

We know that when D & E mid point AB and AC then DE  BC and DE =

4

Ar. of Ar. of Ar. of  AMN :  ADE : ABC

A

B

:

4

A

D

4 4

A

Sol.

M

:

Alternate

In the  ABC, D and E are the mid points of line AB & AC. M and N are the mid point of AD & AE find the ratio of Area of  AMN : Area of ADE : Area of  ABC.

:

Ar. of Ar. of  ADE : AMN

: :

ERna

39.

3

Ar. of Ar. of Ar. of  AMN :  ADE : ABC

1

1 AE  2 12 AE = 6 cm

41.

:

Ar. of BCED

So,

In  ADE ~  AMC

AM

1

Ar. of DENM :

In a  ABC, D and E are the mid points of AB & AC. M and N are the mid point BD & EC. Then find the ratio of

1 1

C

AD AE  AM AC

Ar. of  AMN :

kgei snhe eYari dnag v.iSn ir

D

So,

1 MN = DE 2

MN 1 AM AN    DE 2 AD AE

D

1 : 4 : 16 (Discuss in above question) Area of DENM Area of  AMN Area of BCED – Area of  ADE

=Area of  ADE – = 4–1 =3 = Area of  ABC = 16 – 4 = 12

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E

M

N

B

C

Ar. of Ar. of  ADE : AMN

4

:

9

:

Ar. of ABC

:

16

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4

43.

Ar. of DENM :

:

5

Ar. of MNCB

:

7

In a  ABC, D and E are points lie on AB and AC. M and N are points lie on BD & EC. If

AD 2 DM 4   and and Area DM 3 MB 5 of  AMN 800 cm². Find the area of MNCB. A Sol.

M

B

C

AM 1 MD 6   and MD 3 DB 5

AM : MD : DB 1 : 3 6 : 5 (to make equal ratio) AM : MD : DB 2 : 6 : 5 AD = AM + MD = 2 + 6 = 8 AB = AD + DB = 8 + 5 = 13 Ar. of Ar. of AMN : ADE

2

N

B C AD 2 DM 4  ,  DM 3 MB 5

Ar. of Ar. of  ADE : AMN

Side Area

8

:

20

64 :

:

45.

400 :

AM 1  DM 3

an d

DM 6  DB 5

:

13

•

In a  ABC, D and E are points lie on AB & AC. M and N are point lie on BD & EC. If

1225

and

Area of DECB is 210 cm² find the area of DENM. Rakesh Yadav Readers Publication Pvt. Ltd.

M

B

: :

22 484

h D

C B Proof:- Area of  ABC : Area of  ADB 1 1  BC×h :  BD×h 2 2 BC : BD (ii) A

A

Sol.

D

15 225

If two or more than two triangles are based on a same straight line. Then the ratio of their area are same as the ratio of length of their base these triangles have common vertex. (i) A

DENM is 432 cm² Find the area of MNCB.

35

400 units = 800 cm² 1 unit = 2 cm² Area of  ABC = 1225 × 2 = 2450 Area of MNCB = Area of  ABC – Area of  AMN = 2450 – 800 = 1650 cm² 44. In a  ABC, D and E are point lies on AB & AC, and M and N are points lies on AD & AE. If

8

Ar. of ABC

AD 3 DM 6  ,  and Area of DM 2 MB 7

Ar. of ABC

:

:

:

4 : 64 : 169 Area of DECB = Area of  ABC – Area of  ADE = 169 – 64 = 105 105 units = 210 cm² 1 unit = 2 cm² Area of MNED = Area of  ADE – Area of  AMN = 64 – 4 = 60 units = 60 × 2 = 120 cm²

wwM wa. th Les B aryn

AD : DM : MB 2 : 3 4 : 5 (To make equal ratio) AD : DM : MB 8 : 12 : 15 AM = AD + DM = 8 + 12 = 20 AB = AM + MB = 20 + 15 = 35

9 : 81 :

Ar. of ABC

:

Area of DENM = Area of  AMN – Area of  ADE = 225 – 81 = 144 units 144 units = 432 cm² 1 unit = 3 cm² Area of MNCB = Area of  ABC – Area of  AMN = 484 – 225 = 259 units Now, Area of MNCB = 259 × 3 = 777 cm² * (Area of triangles based on a same straight line.)

R Enak

M

Side Area

E

Area

E

Ar. of Ar. of  ADE : AMN

N

D

Side D

AM = AD + DM = 9 + 6 = 15 AB = AM + MB = 15 + 7 = 22

A

r

Ar. of  ADE :

Sol.

geisnh eeYa ridna gv.i Sni

(Discuss in above solution) DENM = Area of  Area of  AMN – Area of  ADE = 9 – 4 = 5 units MNCB = Area  Area of of  ABC – Area of  AMN = 16 – 9 = 7 units Now,

E N C

h

AD 3 DM 6  ,  DM 2 MB 7

AD : DM : MB 3 : 2 6 : 7 (To make equal ratio) AD : DM : MB 9 : 6 : 7

B

D

E

F

G

C

Ar. of Ar. of Ar. of Ar. of  ABD : ADE : AEF : AGC

46.

BD : DE : EF : GC In a triangle ABC, D is a point on BC. If area of  ABD is 80

514

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cm² and length of BD & DC are respectively 4 cm & 5 cm. Find the area of  ABC.

Sol.

Sol.

2

Sol.

D

E G

A B

A2 = 100 cm² (here A2  Area of  ABD) 47. In  ABC, D is a point on BC Area of  ABD 60 cm² and Area of  ADC 75 cm². Find the ratio of BD and DC. Sol.

B

C

D

6

E

6

B D Area of  ABE = Area  BED = 6 units (E is mid point of AD) E 1 G

A

Area of  ADE : Area of  ABC Side  2 : 5 Area  4 : 25 25 units = 150 cm² 1 unit = 6 cm² Area of  ADE = 4 units = 4 × 6 = 24 cm² In  AEB, * When two triangle are based on a same straight line (AB). Then

the ratio of their area are same as the ratio of length of their base have triangles have common vertex.

B D Let AD = 6 units

AD 6   3 (E 2 2 point of AD)

C

ED =

60 BD  75 DC

AG =

BD : DC = 4 : 5 In figure find the area of  ABC ? If BD = 6 c m, DE = 5 cm, EC = 7 cm A

*

40 cm²

B

6

D

5

E

7

2

m id

2 2 AD =  6 = 4 units 3 3

Ar. of BEG EG 1   Ar. of BGD GD 2 (When two triangles are based on a same straight line then the ratio of their area are same as the ratio of there base)

Then, Area of  BEG = 6 

24

D

Area of  BED = 6 units

C

Sol. In  ADE, 5 units = 40 cm² 1 unit = 8 cm² Area of  ABC, BC = BD + DE + EC = 5 + 6 + 7 = 18 units 18 units = 18 × 8 = 144 cm² 49. In  ABC, AD is median, E is the mid-point of line AD and G is centroid find the ratio of area of  BEG and  ABC.

is

B

2 units = 24 cm² 1 unit = 12 cm² Area of  BDE = 3 units = 3 × 12 = 36 cm² 51.

In a  ABC a line DE is parallel to BC. If Area of  ABC 192 cm² and AD : DB = 3 : 5 find the area of  DEB.

Sol.

A

1 = 3

3 D

2 units

Ar. of BEG 2 1   Ar. of ABC 24 12

50.

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Ar. of  BEG : Ar. of  ABC = 1 : 12 In a  ABC a line DE is draw parallel to BC. If area of  ABC 150 cm² and AD : DB = 2 : 3 find the area of  DBE.

E

3

1 1  2 units GD = AD = 6 × 3 3 EG = AG – AE = 4 – 3 = 1 unit

wwM wa. th Les aBryn

48.

A

ERna

Area of ABD Base of ABD BD   Area of ADC Base of ADC DC

(Given)

 ADE ~  ABC AD : DB = 2 : 3 AB = AD + DB = 5 units

2

D

C

DE  BC

kgei snhe eYari dnag v.iSn ir

C D 80 4 Area of ABD BD   = A2 5 Area of ADC DC

E

3

Let area of  ABC = 24 units then area of  ABD = 12 units A

B

B

A

A

E

5 B

C

DE  BC (Given)  ADE ~  ABC AD 3  DB 5 AB = 3 + 5 = 8 units

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2 2 Area of ADE  AD 3  9           Area of ABC  AB  8  64

64 units = 192 cm² 1 unit = 3 cm² Area of  ADE = 9 × 3 = 27 cm²

6 2  9 3 Ar. of  ABC : Area of  BCD = 2 : 3

53.

 ABC and  BCD have common base. If the length of their median 7 cm and 11 cm. Find the ratio of their areas.

A

Sol.

54. In a triangle ABC, AD is median. If area of  ABC 120 cm² find the Area of  ABD. Sol. According to the Question A

A

3 E

B

G

B

Area of ADE AD 3   Area of DEB DB 5 3 unit = 27 1 unit = 9

Draw AM & DN are altitude on BC. AG and GD are medians

 AMG and  DGN  AMG =  DNG (90º) (Vertically  DGN =  AGM opposite) AG = GD

So,

C

Area of  ABC : Area of  BCD

1 1  BC × h1 :  BC × h 2 2 2 h1 : h2  ABC and  BCD have common base BC. If the lenght of their altitude 6 cm and 9 cm. Find the ratio of their areas. A

6

 BEA and  ABC.

Sol.

A

E

AM 7  DN 11

wwM wa. th Les B aryn

D

55. In a  ABC, D is the midpoint of line BC and E is the mid point of AD. then find the ratio of area of

AM DN

We know that when two triangle have common base. Then the ratio of their area are same as the ratio of length of altitude. So,

h2

1 × 120 = 60 cm² 2

 AMG ~  DGN

R Enak

h1

Sol.

1 ar of  ABD = Ar. of  ABC 2

D

A

52.

C

=

Area of  DBE = 5 unit = 5 × 9 = 45 cm² * Area of triangle which have common base If base of two or more triangles have common base then the ratio of their area are same as the ratio of length of altitude drawn on same base.

B

N

M

B

r

5

Proof

C

D AD is median then ,

geisnh eeYa ridna gv.i Sni

D

Area of  ABC : Area of  BDC = AM : DN = 7 : 11

B C D Let the area of  ABC = 100 units AD is median Then,

1 Area of  ABD = × area of  ABC 2

=

1 × 100 = 50 units 2 A

E

Median

A line from corner to the midpoint of the opposite side. It is called median. It is also called Area Bisector of triangle A

B

D

E is mid point of AD So, BE is median.

cm

B

C

median/Area bisector

9 cm

D

Ar. of ABC Altitude of ABC  = Ar. of BCD Altitude of BCD

B C D When AD is median, BD = DC ar  ABD = ar of  ADC = 1 Ar. of  ABC 2

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1 Ar of  BED = Ar of  BEA = × 2 ABD

Ar of  BEA =

1 × 50 = 25 units 2

Ar of  BEA : Ar of  ABC = 25 : 100 = 1 : 4

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In a  ABC, BD and CE are two median which intersect each other at 'O'.  Some Important results (i) Ar of  DOE : Ar of  ABC = 1 : 12 (ii) Ar of  DOE : Ar of  DOC = 1 : 2 (iii) Ar of  DOE : Ar of  BOC = 1 : 4 A

F

B

=

o

1 Ar .of ABC  6

56. In a  ABC, BD and CE are two medians which intersects each other at 'O'. AO intersect the line ED at M. find the ratio of AM : MO

C D

E

C

D

O

A

Sol.

E

C

E and D mid point Then,

D

o

1 BC 2

B

C

K

1 1 (AO  OC)  AC  AE 3 3

58. A, B, C and D are the vertex of a Rhombous M and N are the midpoints of AB and BC. P is a point on diagonal BD In such a way BP

Let AO = 4 unit (Large median)

So, 2

then AK =

wwM wa. th Les aBryn

2   Ar OED  ED   = BC /2  = 1    BC  Ar BOC  4    BC 

Ar of  OED : Ar of  BOC = 1 : 4 (iv). In a  ABC, O is centroid then

1 1 AO and OF = OC 3 3

 OE =

=

ERna

ED||BC and ED =

(a) AE (b) BE (c) CE (d) DE Sol. AE : EO = 2 : 1 and CF : FO = 2 :1

 EF = OE + OF

M

B

57. In the adjoining figure ABCD is a ||gm and E,F are the centroids of D ABD and DBCD respectively, then EF equals :

Ar. of  BOD = Ar. of  DOC = Ar. of  OEC = Ar. of  OAE = Ar. of  AOF = Ar. of  OBF.

D

B Proof

E o

kgei snhe eYari dnag v.iSn ir

E

OM = AO – AM =4–3=1 So, AM : MO 3 : 1

A

=

3 AO 2

3 = × 4 = 6 units 2 A

3 BD 5

Find Sol.

U

E

B

C

Ar. of  AOB = Ar. of BOC = Ar. of  AOC =

(v). In a  ABC, AD, BE and CF are medians and then all trianges have same Areas.

1 BC 2

AE median of AED AM = = AB median of ABC AK

1 Ar .of ABC  3

1 AM = 2 6 AM = 3 units

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C

P N

o R

A

B C E and D are mid points (Because CE and BD are median) Then ED||BC ED =

S

D

c

T

D

A

o

Ar of BMN Ar of MNP

B M We know BR : RO : OU : UD (By property) 1 : 1 : 1 :1 (when M, N, S and T are midpoints)

BP =

3 BD (given) 5

1 BR= BD 4 Let BD =20 cm then BR = OR = OU = UD = 5 cm

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3 = 12 cm 5 RP = BP –BR = 12 – 5 = 7 cm

60.

P N

R

= 4 + 2 + 4 = 10 units

 BMN : Ar. of  MNP = 3 :

Ar. of 7

and BP = 20 ×

In the given figure, ABCD is a rectangle. P and Q are the midpoints of sides CD and BC respectively. Then the ratio of area of shaded portion : area of unshaded portion is :

 area of unshaded portion = area of  APQ =

 61.

B Ar of BMN BR 5 = = Ar of  MNP RP 7 When two triangles have common base then the ratio of their Area is same as the ratio length of their Altitude. So, Ar of  BMN: Ar of  MNP = 5 : 7 59. A,B,C and D are the vertex of a square M and N are the mid point of line AB and BC. P is a point on

the Diagonal BD, if BP =

Sol.

Sol. (c) Let total area of rectangle ABCD = 16 units

D 4 M

C

o

N

A

M

R

Q

X

C 10

S

10

20

R

S

Let total area of || ABCD = 40 units

4

4

D

20

4

R

B

C

P

wwM wa. th Les B aryn

S

T P U

Sol. (a)

C

P

A

D

 are a of

16 = 8 units 2

AXPD =

5 BD. 6

Ar. ofBMN Ar. ofMNP

8

8

R Enak

Find

geisnh eeYa ridna gv.i Sni

r

M

16 – Area of shaded portion = 16 – 10 = 6 units Required ratio = 10 : 6 = 5 : 3 In the ||gm ABCD, P,Q,R and S are mid-points of sides AB, CD, DA and BC respectively. AS, BQ, CR and DP are joined. Find the ratio of the area of the shaded region to the area of the ||gm ABCD.

2

R

B

R

1 BD 4 5 BP = BD 6 L.C.M. of 4 and 6 = 12 Let BD = 12 cm.

BR =

Ar of ΔBMN BR 3 = = Ar of ΔMNP RP 7

In  DMC, Q is the mid point of DC and QN || DM [ DMNQ is also a ||gm)  N is the mid-point of MC

Q

 area of  PCQ = 2 units

BR : OR : OU : UD (When M,N, S, and T are mid 1 : 1 : 1 : 1 points) So,

1 BR = ×12 = 3 4 5 BP = × 12 = 10 6 RP = BP – BR = 10 – 3 = 7 cm

2

4

 area

of

 ar (quadrilateral DMNQ) = 4 - 1 = 3 units

 ADP = 4 units

8 8

D

Similarly,

4

In

4

 area of  ABQ = 4units  total area of shaded portion = area of  ADP + area of  ABQ + area of  PCQ

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ar (  QCN) : ar(  DMC) = 1:4 Let ar (  QCN) = 1unit

 4

 DAX

ar (  DRM) : ar (  DAX) = 1 : 4 ar (  DRM) = 1 unit

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= 1 +1 = 2units and ar(  DRC) = 4 + 1 = 5units but from (i) ar (  DRC) = 10 unit (2 times)  ar (  DRM) + ar (  QNC) = 2  (2 times) = 4 units Similarly, ar (  APX) + ar (  BOS) = 2  2 = 4units  total shaded area = 4 + 4 = 8 units & area of ||gm ABCD = 40units Required ratio =



8 1  40 5

ABCD is 20cm2, then the area of quadrilateral PQRS is :

and area of rectangle ABCD = 3xy

S

 Required ratio = P

(area of 

=

=

1 1 (AP  AD) + (PB  AD) 2 2 1 AD (AP + PB) = 2

1 (AD  AB) 2 =

1 (area of 2

=

1  20 = 10cm2 2

ERna Q

M

N

3

1

1 1 (AP  AD) + (PB  BC) 2 2

wwM wa. th Les aBryn PQRS) = 8 units

C

D

BPRC)

( BC = AD)

2

1

1 PQR) = 2

APRD)

1 (area of 2

=

ABCD)

A

E

O B

 E and H are mid points of OA and OB respectively 1 AB 2

similarly, GH =

2

FG =

Sol. Let BC = x and FB = EF = AE = y  AB = CD = 3y

1 x  2 y  xy 2

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1 BC, 2

1 1 CD, EF = AD 2 2

Peri meter o f parallelogram = (AB+BC+CD+DA) Perimeter of quadrilateral EFGH = EH + HG + FG + EF

1 xy 2

of  CBE

H

A

 EH =

area

B

G

E

1

and

F

ar(  CAE) = ar(  CEF) = ar(  CFB) = 1 unit  required ratio = 1 : 6 65. ABCD is a parallelogram in which diagonals AC and BD intersect at O. If E,F, G and H are the mid points of AO, DO, CO and BO respectively, then the ratio of the perimeter of the quadrilateral EFGH to the perimeter of parallelogram ABCD is: Sol. D C F

64. In the given figure, ABCD is a rectangle with AE = EF = FB. What is the ratio of the area of the  CEF to that of the rectangle ?

Now, Area of  CBF =

1

1

Let ar( ABCD ) = 6 units Base and height are same

2

S R Area of PQNM = area of  PNQ + area of  PNM = 2 + 1 = 3 units  area ( PQ NM) : area ( PQRS) = 3 : 8 63. ABCD is a ||gm. P, Q, R and S are points on sides AB, BC, CD and DA respectively such that AP = DR. If the area of the ||gm

1 xy : 3 xy = 1 : 6 2

Alternative:

Sol. Area of (  PRS + 

62. PQRS is a square, M is the midpoint of side PS and N is the intersecting point of its diagonals. Then the ratio Area ( PQNM) : Area ( PQRS) is :

Sol. Let ar. ( P

 area of  CEF 1 1 = xy  xy = xy 2 2

kgei snhe eYari dnag v.iSn ir

 ar (  DAX) = 4 units  ar (  RMXA) = 4 – 1 = 3 units  ar(  DRM) + ar (  QNC)

=

= 

1 (AB+BC+CD+DA) 2 Perimeter of EFGH =1:2 Perimeter of ABCD

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C

S o

Q

T A

B

P R

D

4

4

4

S



O

S

P

S 2 2

P R

C

o T 2 2

A

2 2

Z A

P

S

wwM wa. th Les B aryn

2

1

Q B

1

P In  ASP Area of  ASP = 2 units T is the mid point of PS The Area of  ATS = 1 units

Q

2

O

B

P

2

2

2

A

2

P

B R

2 O

1 3

1 RP 2

Now, Area of  SOZ : area of ABCD = 1 : 16 69. A, B, C and D are the vertex of a  gram. M and N are the mid points of CD and MD respectively. Find the ratio of area of  ADN and ABCD



1 1

Sol. Let

S N Ar of  SRN = 2 units O is midpoint on RS

2

Q

 OS 2  1 2 1 Area of ΔSOZ       =   Area of SRP 4  SR  2 

2 Q S Ar of  SRQ = 4 units N is the mid point on RQ Then R

S

4

2

2

OZ  RP and OZ =

N

2

4

P Let SR = 2 cm SO = 1 cm O and Z are mid points SR and SP. Then,

R

A

4

Z

2

A

T

A

B

2

1

2

4

Let the ar. of Parallelogram ABCD = 16 units. (P, Q, R and S are the mid points of AB, BC, CD and DA) D R C

S

B C

C

S

S

R

2

B

P S

T

S

M

A D

R

D

2

8

Q

C

O

8

8

S

R

Q

R Enak

D

D

Sol.

A B P Let Area of ABCD = 16 units C D

O

A

N

Z

B

A, B, C and D are the vertex of || gram P, Q, R and S are the mid point AB, BC, CD and DA. O and Z are the mid points RS and SP. Find ratio of area of  of SOZ and ABCD



1  4  2 units 2

Now, Ar of  ATS : Ar of  PTQ = 1 : 2 67. A,B,C and D are the vertex of || gram P,Q,R and S are the midpoint AB, BC, CD and DA. M, N, O and Z are the mid points PQ, QR, RS, and SP. find ratio of area of  SON and ABCD. Sol. R D C

Q

A

=

1  Ar.ofPSQ 2

C

4

68.

r

R

D

T is midpoint of SP Then, Ar of  PTQ

geisnh eeYa ridna gv.i Sni

66. A,B,C and D are the vertex of a || gram. P,Q, R and S are the mid points of side AB,BC,CD and DA. T is the midpoint of line PS. then find the ratio of Areas of  ATS and  PTQ Sol. Let the area of parallelogram = 16 units

16

area D

N

  AB CD

of M

=

C

So, P

Area of  PSQ = 4 units

Ar of  SON = 1 unit area of  SON : area of ABCD = 1 : 16

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 A

B

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D

M is the mid point of CD Ar. of  ADM = 4 units D M C 4

N 2

N is mid point of DM Then area of  ADN = 2 unit So, area of  ADN : area of ABCD = 2 : 16 =1:8

M

2



8 4

A

B

kgei snhe eYari dnag v.iSn ir

A

EXERCISE

1.

In the given figure, AD : DC = 3 : 2, then  ABC :

6.

For a triangle ABC, D, E, F are the mid - points of its sides. if area of  ABC = 24 sq. units, then area of  DEF is (a) 4 sq. units (b) 6 sq. units (c) 8 sq. units (d) 12 sq. units

A 9cm

B

(a) 30° (b) 40° (c) 45° (d) 50° Triangle ABC is such that AB = 9 cm, BC = 6 cm, AC = 7.5 cm. Triangle DEF is similar to  ABC. If EF = 12 cm then DE is: (a) 6 cm (b) 16 cm (c) 18 cm (d) 15 cm ABCD is a rhombus. A straight line through C cuts AD produced at P and AB produced at Q. If DP 1 AB, then the ratio of the 2 length of BQ and AB is (a) 2 : 1 (b) 1 : 2 (c) 1 : 1 (d) 3 : 1 D and E are two points on the sides AC and BC respectively of ABC such that DE = 18 cm,

=

4.

CE = 5 cm and DEC = 90°. If tan  A B C = 3.6, then AC : CD is (a) BC : 2 CE (b) 2CE : BsC (c) 2BC : CE (d) CE : 2BC 5.

In the adjoining figure (not drawn to scale) AB, EF and CD are parallel lines. Given that EG = 5 cm, GC = 10 cm and DC = 18 cm. Calculate AC, if AB = 15 cm: A

In  ABC and  DEF, AB = DE and BC = EF, then one can infer that ABC  DEF , when

AB AO  CD OC

(b)

AB BO  CD OD

(c)  AOB ~  COD (d) All of these

10.

In the figure  ACB ~  APQ. If BC = 8 cm, PQ = 4 cm, AP = 2.8 cm, find CA : B

D

B

8.

C

G

F

P

Q

P

y

z

1

O

A

B

(a) x + y = z 1

11.

(a) 8 cm (b) 6.5 cm (c) 5.6 cm (d) None of these In the figure QA and PB are perpendicular to AB. If AO = 10 cm, BO = 6 cm and PB = 9 cm. Find AQ :

R

A

1

Q

1 1 1 (d) x  y  z  0

In the given diagram AB  CD, then which one of the following is true ? O

12.

(a) 8 cm (b) 9 cm (c) 15 cm (d) 12 cm In the given figure AB = 12 cm, AC = 15 cm and AD = 6 cm. BC  DE, Find the length of AE: B

(a) BAC  EDF

12

(b) ACB  EDF (c)  ACB   DFE (d)  ABC   DEF

Rakesh Yadav Readers Publication Pvt. Ltd.

C

E

cm

A

D C

A

B

C

(b) xy = 2z

(c) x  y  z

9.

Q C

(a) 21 cm (b) 25 cm (c) 18 cm (d) 28 cm In the adjoining figure PA, QB and RC are each perpendicular to AC. Which one of the following is true :

x

P

A

E

ERna

3.

7.

wwM wa. th Les aBryn

2.

6cm

D 130° 30° C

(a)

B

m 15c

6

cm

D

521

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13.

(b) 7.5 cm (d) 10 cm

18.

DE 2  In the given figure, if BC 3

In the given figure, AB || QR. Find the length of PB : P P A

and AE = 10 cm. Find AB:

4cm

26.

B 5cm

A

75º

B

14.

105º 65º

C

19.

(a) 10cm (b) 15 cm (c) 8 cm (d) 9 cm In the figure AD = 12 cm, AB = 20 cm and AE = 10 cm. Find EC:

20.

A

57º

B

(a) 14 cm (c) 8 cm 15.

E 123º

C

21.

(b) 10 cm (d) 15 cm

In a right angle triangle

 BAC

in which AB = BE = 5 cm and AC = 12 cm. DE is perpendicular on BC. Then find DE?

22. D12

B

16.

17.

the area of  ABC is 60sq.cm then the area of the quadrilateral BDGF is : (a) 10 sq.cm (b) 15 sq.cm (c) 20 sq.cm (d) 30 sq.cm The areas of the similar triangles are in the ratio of 25 : 36. What is the ratio of their respective heights ? (a) 5 : 6 (b) 6 : 5 (c) 1 : 11 (d) 2 : 3 Given that : ABC  PQR , if

wwM wa. th Les B aryn

A

E

C

(a) 3.3 cm (b) 4.3 cm (c) 5.3 cm (d) 6.3 cm The side AB of a parallelogram ABCD is p roduced to E in such way that BE = AB, DE intersects BC at Q. The point Q divides BC in the ratio (a) 1 : 2 (b) 1 : 1 (c) 2 : 3 (d) 2 : 1 ABCD is a trapezium whose side AD is parallel to BC, Diagonals AC and BD intersect at O. If AO = 3, CO = x –3, BO = 3x –19 and DO = x – 5, then the value(s) of x will be : (a) 7, 6 (b) 12, 6 (c) 7, 10 (d) 8, 9

(a)

27.

23.

3

1 2

(c)

(a) 12 2 cm (b) 15.5 cm (c) 16 cm (d) 15.75 cm Two isosceles triangles have equal vertical angles and their areas are in the ratio 9 : 16. then the ratio of their corresponding heights is (a) 4.5 : 8 (b) 3 : 4 (c) 4 : 3 (d) 8 : 4.5 Given that the ratio of altitudes of two triangles is 4:5, ratio of their areas is 3 : 2, The ratio of their corresponding bases is

(d)

2 1

In the given figure,  BAC =  BCD, AB = 32cm and BD = 18cm, then the ratio of

B

28.

C

(a) 4 : 3 (b) 8 : 5 (c) 5 : 8 (d) 3 : 4 In the given figure In  ABC,  BAC =  BCD and AD = 14 cm, AB = 32 cm, A

D

C

B

Perimeter of ABC Find Perimeter of BDC ?

(a)

29.

30.

4 3

(b)

5 3

(b)

6 3

(d)

7 3

ABC i s a righ t-angl ed triangle. AD is perpendicular to th e h ypo te nuse BC. If AC = 2A B, the n the val ue of BD i s :

cm, then AC is equal to?

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1

D

256 area ΔABC = 441 and PR = 12

25.

(b)

A

area ΔPQR

24.

2 3

is equal to :

perimeter of  BCD and  ABC is :

R Enak

D

Q 8cm R (a) 2.5 cm (b) 2 cm (c) 3 cm (d) 3.5 cm The areas of two similar triangles are in the ratio 9 : 16. Their corresponding side s will be in the ratio : (a) 3 : 5 (b) 3 : 4 (b) 4 : 5 (d) 4 : 3 If G is centroid and AD, BE,CF are three medians of  ABC with area 72cm2 , then the area of  BDG is : (a) 12 cm2 (b) 16 cm2 2 (c) 24 cm (d) 8 cm2 The three medians AD, BE and CF of  ABC intersect at G. If

Area of Δ QBC Area of Δ PAC

geisnh eeYa ridna gv.i Sni

D

E

(a) 5 : 8 (b) 15 : 8 (c) 8 : 5 (d) 8 : 15 ABCD is a square. Draw a triangle QBC on side BC considering BC as base and draw a triangle PAC on AC as it base such that  QBC   PAC then,

r

(a) 6 cm (c) 9 cm

(a)

BC 2

(b)

BC 3

(c)

BC 4

(d)

BC 5

The difference between altitude and base of a right angled triangle is 17 cm and its hypotenuse is 25 cm. What is the sum of the base and altitude of the triangle is :

(a) 30 cm (b) 31 cm (c) 32 cm (d) con't be determined

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33.

(a) 6 cm

(b) 6 2 cm

(c) 4 cm

(d) 3 2 cm

34.

38.

In right angled  ABC,  B = 90º, If P and Q are points on the sides AB and BC respectively, then : (a) AQ² + CP² = PQ² + AC² (b) AB² + CP² = PQ² + AC² (c) 2(AQ² + CP²) = PQ² + AC² (d) AQ² + CP² = 2(PQ² + AC²) If ABC is right angled triangle at B and M, N are the mid-points of AB and BC, then 4 (AN² + CM²) is equal to: (a) 4AC² (b) 6AC² (c) 5AC²

37.

 ABC = 90°, AB = 3, BC = 4, CA = 5; BN is perpendicular to AC, AN : NC is (a) 3 : 4 (b) 9 : 16 (c) 3 : 16 (d) 1 : 4

39 .

40.

5 AC² 4

(d)

In a right angled  ABC,  C = 90º and CD is the perpendicular on hypotenuse AB if BC = 15 cm and AC = 20 cm then CD is equal to : C

(a) 12 cm (b) 18 cm (c) 4 cm (d) 6 cm In a triangle ABC, BAC  90 and AD is perpendicular to BC. If AD = 6 cm and BD = 4 cm then the length of BC is : (a) 8 cm (b) 10 cm (c) 9 cm (d) 13 cm In a right angle d  ABC ,

41.

In ABC A  90 and AD  BC where D lies on BC. If BC = 8 cm , AC = 6 cm , then area of ABC : area of ACD = ? (a) 4 : 3 (b) 25 : 16 (c) 16 : 9 (d) 25 : 9 A point D is taken from the side BC of a right-angled triangle ABC, where AB is hypotenuse. Then (a) AB2+ CD2 =BC2+AD2 (b) CD2+ BD2 =2AD2 (c) AB2+ AC2 =2AD2 (d) AB2= AD2+BC2 BL and CM are medians of ABC right- angled at A and BC

35.

wwM wa. th Les aBryn

= 5 cm. If BL =

A B D (a) 18 cm (b) 12 cm (c) 17.5 cm (d) can't be determined ABC is a right angle triangle at A and AD is perpendicular to the hypotenuse. Then

(a) 2 5 cm

42.

BD is equal CD

2

 AB     AC 

AB AB (d) AC AD Suppose  ABC be a right-angled triangle where A  90 and

(c) 36.

B

45.

3 5

cm, then

2

46.

47.

48.

(b) 5 2 cm

Rakesh Yadav Readers Publication Pvt. Ltd.

(a) 6 cm (c) 5.5 cm

AX : AB A X

(a) (c)

A

49. D

(b) 7 cm (d) 7.5cm

C

Y

B

E

B

C

(c) 16 2 cm2 (d) 27cm2 In the given figure, the line segment XY || AC and it divides the triangle into two parts of equal areas. Find ratio

(d) 4 1 34 In the given figure, AB = 18cm, AC = 11cm and AD = 9 cm and BC || DE, find the length of AE

AD  BC . If ar(  ABC ) = 40cm ,

(b) 16 3 cm2

(b) 5 2

2

ar(  ACD ) = 10 cm² and AC = 9 cm,then the length of BC is

C

(a) 9cm (b) 8cm (c) 6cm (d) 7cm A straight line parallel to base BC of the triangle ABC intersects AB and AC at the points D and E re spe ctive ly. I f the are a of the  ACD is 36sq cm, then the area of  ABE is : (a) 36 sq.cm (b) 18 sq.cm (c) 12 sq.cm (d) None of these In  ABC, P and Q are the mid points of the sides AB and AC respectively. R is a point on the segment PQ such that PR : RQ = 1 : 2, If PR = 2cm, then BC = (a) 4 cm (b) 2 cm (c) 12 cm (d) 6 cm Find the maximum area that can be enclosed in a triangle of perimeter 24cm : (a) 32 cm2

(c)

43.

Q

P

(c) 10 2 cm (d) 4 5 cm Q is a point in the interior of a rectangle ABCD, if QA = 3 cm, QB = 4 cm and QC = 5 cm then the length of QD ( in cm) is (a) 3 2

2

 AB     AD 

(b) 

In the given triangle ABC, BP = 3 AP, QC = 3AQ and BC = 36cm. Find the value of PQ ? A

the length of CM is

to :

(a) 

44.

kgei snhe eYari dnag v.iSn ir

32.

One side other than the hypotenuse of right angle isosceles triangle is 6 cm. The length of the perpendicular on the hypotenuse from the opposite vertex is :

ERna

31.

1 2

2?1 2

(b)

1 2

(d)

2?1 2

D and E are the mid-points of AB and AC of  ABC, BC is produced to any point P; DE, DP and EP are joined. then, area of :

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1 ABC 4 

D

50.

55.

52.

D

57.

(a) 19 cm (b) 20cm (c) 25 cm (d) 17.5cm The triangle is formed by joining the mid-points of the sides AB, BC and CA of  ABC and

the area of  PQR is 6 cm², then the area of  ABC is (a) 36 cm² (b) 12 cm² (c) 18 cm² (d) 24 cm² ABC is a triangle and DE is drawn parallel to BC cutting the other side at D and E. If AB = 3.6 cm, AC = 2.4 cm and AD = 2.1 cm, then AE is equal to: (a) 1.4 cm (b) 1.8 cm (c) 1.2 cm (d) 1.05 cm

In  ABC, AC = 5 cm. Calculate the length of AE where DE  BC. Given that AD = 3 cm and BD = 7 cm :

D

B

wwM wa. th Les B aryn

53.

58.

(c) 9cm 54.

(d)

62.

C

(b) 1 cm (d) 2.5 cm

63.

3 2 cm and PR = 10 cm, AB  QR. Find the length of BR : P

18 cm 5

14 cm 5

61.

In  PQR, AP = 2 2 cm, AQ =

A

Q

(b)

60.

E

(a) 6 2 cm

59.

In the given figure, EC || AM || DN and AB = 5cm, BC = 10 cm. Find DC:

Rakesh Yadav Readers Publication Pvt. Ltd.

4

C

D

(a) 2.5 (b) 3 (c) 2.25 (d) 2 ABC is a triangle in which  A = 90º, AN is perpendicular to BC, AC = 12 cm and AB = 5 cm. Find the ratio of the areas of  ANB and  ANC : (a) 125 : 44 (b) 25 : 144 (c) 144 : 25 (d) 12 : 5 Area of  ABC = 30 cm². D and E are the mid-points of BC and AB respectively. Find Ar (  BDE) : (a) 10 cm (b) 7.5 cm (c) 15 cm (d) None of these ABCD is a trapezium in which AB  CD and E and F are the point on DA & BC, DC = 40 cm, AB = 105 cm. The ratio of

64.

In  ABC, D and E are midpoint of side AB and BC. P is point on line AC in such a way that R and S are the mid-point of AP & PC find DR : ES (a) 1 : 1 (b) 2 : 1 (c) 4 : 1 (d) 3 : 1 In the given figure AE BD 2   , AC = 760 cm ED DC 3

B

Then find AE = ?

3 2

36 cm 5

3

DE 2 = . Then find EF. EA 3 (a) 66 cm (b) 70 cm (c) 72 cm (d) 73 cm

2 2

(a)

B

A

(a) 2 cm (c) 1.5 cm

A N B E (a) 5 : 9 (b) 9 : 4 (c) 4 : 5 (d) 4 : 9 In the given figure, AB || CD || PQ, AB = 12cm, CD = 18cm and AC = 6cm. Then PQ is:

5

C

M

R Enak

51.

18cm

B

N

56.

(a) 12.5cm (b) 15cm (c) 17.5cm (d) 20cm In a  ABC, D is the mid-point of BC and E is the mid-point of AD. The line BE is extended and it intersects AC at T. If AB = 18cm, BC = 17cm and AC = 15cm. Find TC ? (a) 8cm (b) 9cm (c) 10cm (d) 7cm In the given figure, DE || BC and EC || ND, AE : EB = 4 : 5, then EN : EB is : C

A

15cm

r

(b)  PED =  BEC (c)  ADE =  BEC (d)  BDE =  BEC In the given figure, ABCD is a trapezium such that AD || BC and P, Q are the points on AB and CD respectively such that PQ || AD and AP : PB = 5 : 3. Then PQ is :

A

E

geisnh eeYa ridna gv.i Sni

(a)  PED =

A

R

(b) 6 cm

(c) 5 2 cm (d) None of these In the adjoining figure the  BAC and  ADB are right angles. BA = 5 cm, AD = 3 cm and BD = 4 cm, what is the length of DC ?

F E B

D

(a) 80 cm (c) 100 cm

C

(b) 90 cm (d) 120 cm

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67.

68.

69.

In  ABC , PQ is parallel to BC. If AP : PB = 1 : 2 and AQ = 3 cm; AC is equal to (a) 6 cm (b) 9 cm (c) 12 cm (d) 8 cm In  ABC , D and E are points on AB and AC respectively such that DE  BC and DE divides

(a) 1 : 1

70.

1. 2. 3. 4. 5. 6. 7. 8.

(a) 71.

3 5

72.

73.

that AD =

74.

(c)

75.

(d) (c) (c) (b) (b) (a) (a) (b)

17. 18. 19. 20. 21. 22. 23. 24.

(d) (b) (b) (a) (c) (a) (d) (b)

5

25. 26. 27. 28. 29. 30. 31. 32.

(c)

3 4

(d)

4

76.

5

1

4

AB, AE =

1

4

77.

(c) always similar (d) always congruent In triangle ABC a straight line parallel to BC intersects AB and AC at D and E respectively. If AB = 2AD, then DE : BC is (a) 2 : 3 (b) 2 : 1 (c) 1 : 2 (d) 1 : 3 In a ABC , D and E are two points on AB and AC respectively such that DE  BC . DE bisects the ABC in two equal areas. Then the ratio BD : AB is

31 cm 3

(d)

(b) (c) (d) (a) (d) (b) (d) (a)

Rakesh Yadav Readers Publication Pvt. Ltd.

(a) 1 : 2 (c)

78.

BD

79.

80.

41 cm 3

81.

(b) 1 : 2

 2 – 1 :

2 (d)

2 :1

If in a triangle ABC, D and E are on the sides AB and AC, such that, DE is parallel to BC and AD

AC. If

In ABC , two points D and E are taken on the lines AB and BC respectively in such a way that AC is parallel to DE. Then ABC and DBE are (a) similar only If D lies outside the line segment AB (b) congruent only If D lies out side the line segment AB

(b) 1 : 2 – 1

9. 10. 11. 12. 13. 14. 15. 16.

2

BC = 12 cm, then DE is (a) 5 cm (b) 4 cm (c) 3 cm (d) 6 cm In PQR , S and T are point on sides PR and PQ respectively such that PQR = PST , If PT = 5 cm, PS = 3 cm and TQ = 3 cm, then length of SR is (a) 5 cm (b) 6 cm

(c) 1 : 2 (d) 1 : 2  1 ABCD is a rectangle where the ratio of the length of AB and BC is 3 : 2 . If P is the mid- point of AB,

(b) (c) (a) (a) (d) (b) (b) (c)

(b)

Inside a triangle ABC, a straight line parallel to BC intersects AB and AC at the point P and Q respectively. If AB = 3 PB, then PQ : BC is (a) 1 : 3 (b) 3 : 4 (c) 1 : 2 (d) 2 : 3 In ABC , DE  AC , D and E are two points on AB and CB respectively. If AB = 10 cm and AD = 4 cm, then BE : CE is (a) 2 : 3 (b) 2 : 5 (c) 5 : 2 (d) 3 : 2 For a triangle ABC, D and E are two points on AB and AC such

wwM wa. th Les aBryn

the  ABC into two parts of equal areas. Then ratio of AD and BD is

then the value of sin  CPB is

kgei snhe eYari dnag v.iSn ir

66.

The length of the diagonal BD of the parallelogram ABCD is 18 cm. If P and Q are the centroid of the and  ABC  ADC respectively then the length of the line segment PQ is (a) 4 cm (b) 6 cm (c) 9 cm (d) 12 cm A straight line parallel to BC of  ABC intersects AB and AC at points P and Q respectively. AP = QC, PB =4 units and AQ = 9 units, then the length of AP is : (a) 25 units (b) 3 units (c) 6 units (d) 6.5 units D is any point on side AC of  ABC . If P, Q, X , Y are the mid-points of AB, BC, AD and DC respectively, then the ratio of PX and QY is (a) 1 : 2 (b) 1 : 1 (c) 2 : 1 (d) 2 : 3

ERna

65.

=

3

5

. If AC = 4 cm, then AE

is (a) 1.5 cm (b) 2.0 cm (c) 1.8 cm (d) 2.4 cm ABC is a triangle in which DE  BC and AD : DB = 5 : 4. Then DE : BC is (a) 4 : 5 (b) 9 : 5 (c) 4 : 9 (d) 5 : 9 D and E are mid-points of sides AB and AC respectively of the  ABC. A line drawn from A meets BC at H and DE at K. AK : KH = ? (a) 2 : 1 (b) 1 : 1 (c) 1 : 3 (d) 1 : 2 If ABC be an equilateral triangle and AD perpendicular to BC, Then AB² + BC² + CA² = ? (a) 3AD² (b) 5AD² (c) 2AD² (d) 4AD²

ANSWER KEY 33. 34. 35. 36. 37. 38. 39. 40.

(c) (b) (a) (b) (d) (b) (c) (a)

41. 42. 43. 44. 45. 46. 47. 48.

(a) (a) (c) (a) (a) (c) (b) (d)

49. 50. 51. 52. 53. 54. 55. 56.

(a) (b) (c) (d) (a) (c) (d) (a)

57. 58. 59. 60. 61. 62. 63. 64.

(c) (b) (c) (b) (b) (a) (a) (a)

65. 66. 67. 68. 69. 70. 71. 72.

(b) (c) (b) (b) (b) (d) (d) (d)

73. 74. 75. 76. 77. 78. 79. 80. 81.

(c) (c) (c) (c) (c) (a) (d) (b) (d)

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SOLUTION AD 3 AB 3  and  , DC 2 BC 2

AP AQ  BC BQ

AP

1.

(b)



AD AB  DC BC



BD is the bisector of  B

(  AQ = AB + BQ)

Now,  CBD = 180° – (130° + 30°) = 20°

AB 3 1  BQ 2

E

C

6



4.

r C E

Note: Two triangles are congruent if two sides and the included angle of one are equal to the corresponding sides and the included angles of the other triangle (SAS criterion). 6. (b) According to question

D

B

R Enak tanC =

wwM wa. th Les B aryn C

tanC =

5 2

2

3

EC = 5 cm

DE EC

18 = 3.6 5

 tan ABC  tan ACB

4

B

E5 C

tan ABC  3.6

P

D

F

Given: DE = 18 cm

1 1

Q

AB = BC= CD =DA 1 DP = AB 2

1

=

area of  ABC

4

Area of  DEF

1

AF AC FC   DE DC EC

C

3

5

D E 5

15

A

4

3

Q

B

4 Q

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BC AC FC  (  FC = )  2 CD EC

18 cm G 10

B

F

C

E= C  EFG =  D 

2

 24

4

A

 APQ  BCQ P

1

(b) According to the question,

In a rhombus 2  3

( Q is common angle 2  3 )

=

= 6 sq. units

7.

DP 1 = AB 2

2

E

B As we know that Given: area of  ABC = 24 square units D,E and F are the midpoints of AB, AC and BC Area of  ADE = area of  DBF  = area of  DEF = area of  EFC Area of  DEF 

N o te: - In a is os c ele s tr iangle perpendicular bisects the opposite sides  AFC  DEC

ABCD is a rhombus



F

ABC  DEF

18

DE = 18 cm (a) According to question, Given:

2

B

D

9 6 = DE 12

2

D

A

AB BC AC   DE EF DF

A

(d) According to question A

BQ 2 = AB 1 (a) According to question

F

12

 ABC~  DEF

3.

5.

AB 1  BQ 2

7.5

B

 AC : CD = BC : 2EC

geisnh eeYa ridna gv.i Sni

D

A 9

2

AB 3  –1 BQ 2

(c) According to the question,

AC BC = CD 2EC

3

AB  BQ 3  BQ 2

  B = 2(  CBD) = 40° 2.

BC

=

 GEF ~  GCD EF EG  CD CG

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EF 5  18 10

EF = 9 cm

AB  BC y  AB z

D

E 5 G

18

G

BC y  AB z

1+

10 C

E 9 C

B



15

AB

F

15 15  x = 9 15

 z² = xy – xz – zy + z²

(divide xyz both sides)

y C

10.

B

2. 8

4

C

(sides are propor-

Q

14.

BC CA  PQ AP

tional)

AB x 1  BC z

11.

20 D

CA = 5.6 cm (c) According to the question, P 9

...(i)

A

Similarily,

 CRA ~ BQA Rakesh Yadav Readers Publication Pvt. Ltd.

Q

3 AB  2 10 AB = 15 cm (a) According to the question, A 12

8 CA  4 2.8

AB  BC x  BC z

O 10

C

6

B

 ADE

 ABC ~ BC AB  DE AE



P A

65º

 ADE =  ACB

(c) According to the question,

8

75º

 A = 40º  AED =  ABC

 ACB~  APQ

 APC~  BQC (AA similarity) So,

105º

 AED = 180º – 105º = 75º  A = 180º – 75º – 65º

CD OC OD   AB AO OB

 C =  C (common angle)

AB x x–z  – 1= BC z z

B

 COD ~  AOB

wwM wa. th Les aBryn

B

 PAC =  QBC (both 90º)

or

(Common)

 OCD =  OAB



E

D

 ODC =  OBA

Q

 APC &  BQC

AC x  BC z

A

 DOC =  AOB

z



13.

AE = 7.5 cm (b) According to the question,

(d) According to the question,

R

A

15 12  AE 6

1 1 1   z y x

9.

cm

 ACB ~  AED

xy = zx + zy

x = 10 cm AC = 15 + x AC = 15 + 10 = 25 cm

6

AC AB  AE AD

x –z z  z y–z

x

In

......(ii)

AB AC  EF EC

(c) P

m 15 c

D

z

ERna

8.

C

From (i) & (ii)

 CAB ~  CEF

E

cm

A

or BC  y – z

C

135 + 9x = 225 9x = 90 

12

BC y – z  AB z

A 15

B

BC y  –1 AB z

A = E B = F 15 +x

12.

AQ OA  BP OB AQ 10  9 6 AQ = 15 cm (b) According to the question,

kgei snhe eYari dnag v.iSn ir

F

 OAQ ~  OBP

AC y  AB z

So,

10

E 123º

x

57º

B C  AED = 180º – 123º = 57º D = C E = B   ABC ~  AED AB AC  AE AD

527

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AOD  COB

20 10  x  10 12

24 = 10 + x x = 14 cm (a) According to the question, AB = BE = 5 cm AC = 12 cm  BAC =  DEC [each 90º]  ECD =  ACB [common]

B

OB OD

18.

E

8 cm

3

23.

19.

(d)

r

Q

20.

(a)



9 3  16 4



A

F

G

R Enak



C

1  Ar ea 6

BDG 

wwM wa. th Les B aryn 1 unit





A

E

AD|| BC and AB||DC Point B is the midpoint of AE 1  2

3  4

1   =  EA ED AD 2 QC : QB = 1 : 1 (d) According to question,

17.

EQ

B

22.

C

2

2 56 4 41

441

16

=

21 4

2

1 2  2 A C  2



 12       AC 

12 AC

=

21

 AC =

E

²

P R  ² AC 

3 AC 63 4

 AC = 15.75 cm. (b)

24.

 BDGF

D

A

1   60 = 20 sq.cm. 3 (a) According to the question,

QB

B

As we know that A

P

25

36

C o

A

G

B C D Required area of

( Corresponding Angle) ( Corresponding Angle)  EQB  EDA

F



2

1   72  12cm 2 6

(c)

=

 ABC

16        21 

2

B

=

E

of  ABC

21.

Q

EB

 PQR

 ABC  PQ R

D

Area of

C

1 unit

B

Area of triangle1  Corresponding side1   Area of triangle2  Corresponding side 2 2



1

A

R

 We know that in similar triangle

(b) Required ratio =

B

4

x

  PQR ~  ABC

256

D

A

=3:4

DE = 3.3 cm (b) According to question

3

P 12

4 PB   PB  2.5cm 8 5

C

AB AC   DE EC 5 12  DE 8 10 DE = 3

16.

AB 5  PQ 6

x 3

(a) AB  PB QR PR

5

C

OA

9x – 57 = x2 –8x +15 x2 – 17x + 72 = 0 x(x – 8) – 9(x – 8) = 0 (x – 8) (x – 9) = 0 x = 8 or 9

D

12

25 36



By using pythagoras theorem.

A

OC

x –5

 ABC ~  EDC In  ABC &  EDC

BC² = AB² + AC² BC² = (5)² + (12)² BC² = 25 + 144 BC = 13 cm  BC = BE + EC EC = 13 – 5 EC = 8 cm

AB  PQ

D



3 x – 19



B

CA



AB2 Area of ABC  PQ2 Area of PQR

O

geisnh eeYa ridna gv.i Sni

15.

O

B D

Rakesh Yadav Readers Publication Pvt. Ltd.

C

Q

R

M

C

E

N

F

If two isosceles triangles have equal vertical angles then both triangle is similar So,  ABC ~  DEF We know, In similarity case

528

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27.

Area of  ABC Area of  DEF 2

=

 AB 2  DE 

and  B =  B (common)   ABC ~  CBD

corresponding sides square

AB BC = BC BD

2

=

A M  2 D N 

9



=

height

 A M 2  D N

=

9

= R a tio of the ir he igh t

16



Ratio of height = 3:4

25.

area of triangle1 3 (b) area of triangle  2 2

28.

A

18

Q

1

B

wwM wa. th Les aBryn D

P



 QBC ~  PAC

=

 Let each side of square =1

Area of PAC

2

=

 QC  2  PC 

2

=

QB 2 PA 

=

29.

2

=

C

BC 2  AC 2 1 2

 2

=

1 2

Rakesh Yadav Readers Publication Pvt. Ltd.

31.

(d)

In  ABC

 Sum of altitude and base = x + x + 17 = 7 +7 +17 = 31 cm (d) According to the question,

D

6 B

6

C

AB = BC = 6 cm By using pythagoras theorem. AC² = BC² + AB² AC² = 6² + 6² AC² = 72 AC =

C

A

C

A

24 4 = 18 3

 =

   QBC ~  PAC



θ

AB AC = BC CD

 then diagonal of square = 2

Area of QBC

B

Perimeter of ABC BC = Perimeter of BDC BD

Given



C

x + 17

AC² = AB² + BC² (25)² = x² + (x+17)² 625 = x² + x² + 289 + 34x 2x² + 34x – 336 = 0 x² + 17x – 168 = 0 x² + 24x – 7x – 168 = 0 x (x + 24) –7 (x + 24) = 0 (x + 24) (x – 7) =0 x=7

18

BC BD  AB BC BC² = AB × BD BC² = 32 × 18 BC² = 576 BC = 24 cm

C

1

B

C

D

θ

25

x

 ABC ~  CBD



B

2

30.

Let  BAC =  BCD = θ  B is common in both triangle

A

BC 5 (b) Let the altitude of a triangle =x According to the question,

BD =

A

θ

B

32 1

= 24 : 32 = 3:4 (a) According to the question,

ERna

1

 BC 2     = BD × BC  5

14 θ

B1 : B2  15 : 8 Ans.

BC 5

AB² = BD × BC

D

B1 3 5 15    B2 2 4 8

(c)

AB =

A

1  B1  4 3 2  1 B 5 2 2 2

26.

32 BC = BC 18



 BC2 =18 × 32  BC = 24cm  perimeter of  BCD : perimeter of  ABC = BC : AB

2

16

By using pythagoras theorem AC² + AB² = BC² (2AB)² + AB² = BC² 5AB² = BC² BC = 5 AB

(d) In  ABC and  BDC  BAC =  BCD (given)

kgei snhe eYari dnag v.iSn ir



6 2

AB  BC AC 6 6 BD = 6 2

 BD =

D B

BD = 3 2 cm

529

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32.

(a) According to the question, By using Pythagoras theorem,

34.

B

(b) According to the question, C

B

A

 C = 90° BC = 15 cm AC = 20 cm By using pythagoras theorem, AB² = AC² + BC² AB² = (20)² + (15)² AB² = 400 + 225 AB = 25 cm As we know that,

B

Q

In  AQB, AQ² = BQ² + AB²

...(i)

 PCB PC² = BC² + BP²

...(ii)

In

 ACB

AC² = BC² + AB² In

...(iii)

 PQB

CD =

PQ² = BQ² + BP² ...(iv) Adding eq. (i), (ii), (iii) & (iv) AQ² + PC² + AC² + PQ² = BQ² + AB² + BC² + BP² + BC² + AB² + BQ² + BP² AQ² + PC² + AC² + PQ² = 2(BQ² + AB² + BC² + BP²) AQ² + PC² + AC² + PQ² = 2(PQ² + AC²)

B

C

C

...(i)

 ADB ~  CDA

40

AD AB BD   CD AC AD

10

B

40

In  MBC, MC² = BM² + BC²

A

10

AN² + CM² =

37.

A

D

BC2 AB2 + 4 4

4AC2  AC2 4

4 AN2  CM2  5AC2



Rakesh Yadav Readers Publication Pvt. Ltd.

AD

2 2



2

AC 2 = 2 DC 2 AC

BC

2 2

9

 81 = BC 2

BC = 18 cm (d) According to Question

C

...(i)

 BAC ~  ADC

Given: BAC is a right angle triangled

AB BC AC   AD AC CD B A

A C D AC² = BC × CD

BC 

AD² = BD × CD

...(ii)

In  ABC, AC² = AB² + BC² ...(iii) Adding eq (i) and (ii) AN²+CM² = AB² + BN² + BM² +BC² AN² + CM² = AC² + BN² + BM² AN² + CM² = AC² +

D

AB

(In similar  ratio of their area is s quare of r atio of corresponding sides)

 ABN,

AN² = AB² + BN²



area of ADC

  DAB =  DCA

wwM wa. th Les B aryn

M



ABC ~ DAC

D



N

(b) According to Question Given: AC = 9cm

area of ABC = 40 cm2 area of ADC = 10 cm2

 DBA =  DAC

A

In

36.

20 15 CD = 25 CD = 12 cm (a) According to the question,

A

(c) According to the question, By using Pythagoras theorem,

B

2 BD  AB      CD  AC 

area of ABC

AQ² + CP² = PQ² + AC²

33.

35.

AB² BC  BD  AC² BC  CD

BC  AC AB

R Enak

In

B

D

geisnh eeYa ridna gv.i Sni

C

A

..(iii)

r

P

A A C D AB² = BC × BD eq. (iii) divide by eq. (ii)

AD  BC

AD = 6 cm BD =4 cm BC = ? In BAD

C

...(ii) 2

 AD

2

 BAC ~  BDA

AB 

BD

BA BC AC   BD BA AD

AB 

2 2 4 6 =

52 cm

BAC  BDA

530

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NC = 3.2 AB

=

AN

ΒC

3

NB

AN

4

=

2.4

AN = 1.8 AN





AB

BD

BC





AB

52 4

BC =

52

A

4

BC = 13 cm Alternate:AB² = BD.BC 2

= BC,

4

38.

41.

39.

3 5 2

According to figure, when two medians intersect each other in a right angled triangle then we use, this equation.  4 (BL2 + CM2) = 5BC2

C

2

3 5    2 2 4 ×   2  + 4CM = 5(5)  

x2 AN 32 9 = y2 = 2  NC 4 16  AN : NC = 9 : 16 (c) According to question

 45 + 4CM2 = 125  CM2 =

125 – 45 = 20 4

 CM = 2 5 cm

8c m

42. (a) According to question

=?

ERna

NC

9 = 3.2 16

y

B

 BC = 13 cm (b) According to question Given: ABC = 90° AN

1.8

N

x

 BD²+AD² = BD.BC 2  4²  6² = 4.BC 52

=

Alternate:In such cases use the following method to save your valuable time.

 52

NC

kgei snhe eYari dnag v.iSn ir

BC

AD² = AC² + CD² AC² = AD² – CD² .......(ii) Put the value of AC² in equation (i) AB² = AD² – CD² + BC² AB² + CD² = AD² + BC² (a) According to question

CAB  CDA area of CAB

wwM wa. th Les aBryn ABC  BNC ABC  ANB

area of CAB

 ABC  BNC  ANB AB = 3, BC = 4, AC = 5

area of CDA

area of CDA

area of CAB area of CDA

40.

BC



NC

AB  BC

34



AC

=

Given: QA = 3 cm QB = 4 cm QC = 5 cm QD = ? As we know that QD² + QB² = QA² + QC² (By using Pythagoras theorem) QD² + (4)² = (3)² + (5)² QD² + 16 = 9 + 25 QD² = 34 – 16 QD² = 18

64 36

=

16 9

(a) According to question

QD = 18 , QD = 3 2

43.

 2.4

5

AB NB

4

3 

NC

2

2 8 2 6

=

area of CAB

BN =

BC

 area of CDA = 2 AC

2.4

Rakesh Yadav Readers Publication Pvt. Ltd.

In ABC AB² = AC² + BC² In ACD

........(i)

(c) BC || DE   ABC =  ADE  ACB =  AED and  BAC =  EAD   ABC ~  AED 18 11 AB AC    9 AE AD AE AE = 5.5cm 

531

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47.

AB 4  &  AP 1 AC = AQ + QC = AQ + 3AQ = 4AQ

AC 4  AQ  1

 Area =

In  ABC and  APQ  BAC =  PAQ (common) and

48.(d)

ar(ΔABC)

 AB 

ar(  ABC) = 2 ar(  XYB) (given)

Also,

A

ar(ΔABC)  ar(ΔXBY) = 2

 DBC &  EBC lie on the

........(ii)

Now In  APR and  ABC  APR =  ABC ( PQ || AD || BC)

and  ARP =  ACB ( PQ || BC)   APR ~  ABC



 AB  2   =  XB  1 2

wwM wa. th Les B aryn

same base BC and between same parallel lines .  ar (  DBC) = ar (  EBC)

AP PR AP   PR   BC AB BC AB

AB 2 = XB 1

 ar (  ABE) = ar (  ACD)  ar (  ACD) = ar (  ABE)

= 36sq.cm (c) A

B

(b)

.......(i)

R Enak

E C

P

1  ar ABC  4

2

therefore from (i) & (ii)

46.

and  BDE =  PED [  both triangles lie on the same base DE and between two parallel lines DE and BP]  ar (  PED)

50.

BC 4 36    4  PQ  9cm PQ 1 PQ

B

1  ar ABC  4

  ABC ~  XBY

 ar(ΔXBY) =  XB 

D

=

=

AB AC BC    4 AP AQ PQ

(a)

1 BC 2

 ar (  BDE)

3 2 3 a =  (8)2 4 4

XY || AC (given)   BXY =  A and  BYX =  C

  ABC ~  APQ

45.

DE =

= 16 3cm 2

AB AC  AP AQ



=2×6 = 12cm. (b) For the given perimeter of a triangle the maximum area is enclosed by an equilateral triangle.  3a = 24 cm  a = 8cm

r

(a) AB = AP + BP  AP + 3AP = 4AP

geisnh eeYa ridna gv.i Sni

44.

or



1 XB 1- 1 XB = or 1= 2 2 AB AB

5 45  PR  18  cm 8 4 and

2 -1 AB - XB  AB = 2

Q

R

PR 1 2 1 =  =  RQ = 4 RQ 2 RQ 2  PQ = PR + RQ = 2 + 4 = 6cm The line joining the mid-points of two sides of a triangle is parallel to and half of the third side. BC = 2PQ

49.

Rakesh Yadav Readers Publication Pvt. Ltd.



(a) (By mid-point theorem) A

D

B DE || BC

RQ RC = AD AC

 RQ =

E

C

AP AR 5   PB RC 3

similarly,  RCQ ~  CAD

2 -1 AX  AB = 2

C

AP  BC AP  PB

P

=

RC ×A D AR + RC

3 15 ×10 = cm 8 4

 PQ = PR +RQ =

45 15 + 4 4

= 15cm

532

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(c)

D

A

B

P

T

E

A

C





AC = 15cm, AT = TF = FC =

  52.

= 5cm TC = TF + FC =5+5 = 10cm (d) C

 

CD



AQ × CD .........(ii) AC from (i) & (ii) -

PQ =

ERna 

5

Let AE = 40  EB = 50 and

5 9

EN = 40   EN : EB 

53.

200 9

200 : 50  4 : 9 9

B

(a)

P A

Q

DN

55.

Rakesh Yadav Readers Publication Pvt. Ltd.

15

DC

A

P

Area of

C

DC

DC =

In  ABC and triangle PQC

R

C

Q

 PQR = 6 cm² There are 4 congruent triangles.  APR~  BPQ~  PQR~  RQC Hence area of  ABC = 4× area of  PQR = 4 × 6 =24 cm²

 PQ = 2x.......(i)

18 x= 5

AM

15 ×15 = 25cm 9 (d) According to the question,



B

In  ACD and Triangle AQP PQ AQ  CD AC PQ 6 – x  18 6 .........(ii)  PQ = 18 – 3x From (i) and (ii), 2x = 18 – 3x

5

 AM = AC  9 = 15

Alternatively Let QC = x  AQ = 6 – x

PQ x  12 6

AM

AB

PQ QC  AB AC

 AN = AD = 4 = 4 : 5 .....(ii)



from (i)

(c) In  ABM and  BEC  BAM =  BCE  BMA =  BEC ( AM || EC)   ABM ~  BEC

12 18 36 PQ    cm 6 5 5

In  AEC EC || ND DC

18 5

36 5

 BC = EC  10 = 18  AM =9 cm  AM || DN   AMC ~  DNC

AQ AB × CD = × QC AC AC  AQ  CD = AB  QC  (AC - QC)  18 = 12  QC  (6 - QC)  18 = 12  QC  (6 - QC)  3 = 2QC

 5QC = 18  QC =

AD AE 4    4 : 5 ......(i) DC EB 5

NE

AC

 PQ = AQ

wwM wa. th Les aBryn



54.

ACD ~ AQP



1 1 5 + = 12 18 36

PQ =

PQ||CD

C

In  ABC, DE || BC

=

AB PQ = × QC .........(i) AC



AC 3

B

E

 ABC ~  PQC AB AC = PQ QC



D

A

C

PQ || AB

Draw DF || ET In  ADF, E is the mid-point of AD and DF || ET T will be the mid-point of AF. i.e. AT = TF ............(i) Now, In  BTC, D is the mid-point of BC & DF || ET || BT F will be the mid-point of TC i.e TF = FC ............(ii) from (i) and (ii) AT = TF = FC



Q

18 = 5

36 cm 5 Alternate When AB|| PQ || CD 1 1 1 = + PQ AB CD

F D

From (i), PQ = 2x = 2 ×



kgei snhe eYari dnag v.iSn ir

51.

56.

(a) According to the question, AB = 3.6 cm AC = 2.4 cm AD = 2.1 cm AE = ? A

D B

E C

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A

A

A



BE BD ED 1    BA BC AC 2



Area of BED 1  1      Area of BAC  2  4

2

B

AB BC AC   AD DE AE

(c) According to the question, A

C

AC = 5 cm AD = 3 cm BD = 7 cm By using midpoint theorem,

5

2

15 4

C D

AD AE DE   AB AC BC

3

A

AE = 1.5 cm (b) According to the question,

60.

In  ADB, E & M are the mid points of AD & BD. By using B.P.T. theorem. D 2

C

A

3 2 Q

A

 PAB =  PQR

 PBA =  PRQ



61.

 PAB ~  PQR

x 2 2  10 5 2

59.

C

12

2

Area of ANB AB2 5 2  2 = Area of ANC AC 12 25 = 144

R

x = 4 cm  BR = 10 – x = 10 – 4 = 6 cm (c) According to the question, As we know that,

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(b) According to the question, E & D are the midpoint of AB & BC.

E A

A

3 B

105

DE DM 2   EA MB 3 DE DM EM   DA DB AB 2 EM  5 105

EM = 42 cm N

10 cm

B 10–x

M

3

AC AB  DC AD 15 5  4DC 3 9 DC = 4 DC = 2.25 (b) According to the question,

5

2

E

P

x

B

105

DE 2  EA 3

B

2 2

F

M

AB = 105 cm DC = 40 cm

 CAB ~  CDA

wwM wa. th Les B aryn

3 AE  10 5

C

40

E

3

A 

 Area of  BED = 7.5 cm² (a) According to the question,

D

R Enak

AD AE  BD EC

58.

62.

In  BAC and  ADC B = A A = D B A

E

B

D

geisnh eeYa ridna gv.i Sni

2.1  2.4 AE = 3.6 AE = 1.4 cm

D

4

4 units = 30 1 unit = 7.5

C

1 1 1   AD² AB² AC² 1 1 1   2 2 2 3 5 AC 1 1 1  – 2 AC 9 25 1 25 – 9  AC2 225 15 AC = 4

3.6 2.4  2.1 AE

57.

3

E

r

5

B C D DABC ~ ADE

B

In  BDC, M & F are the mid points of BC & BD By using B.P.T theorem B 3 F

3 M

2

2 D

C BF FM  BC CD 3 FM  5 40

D C

FM = 24 cm

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 EF = EM + FM

63.

EF = 24 + 42 = 66 cm (a) Let ES = x According to the question, In  BPC, E is the midpoint of BC and S is the midpoint of PC. B

AF 2×2= 4

:

4

:

FM 3×2= 6 2×3= 6 6

:

MC

:

3×3= 9 9

PX||BD [mid point theorem] 1  PX = BD 2 Similarly,QY||BD

Length of AC = AF + FM + MC = 4 + 6 + 9 = 19 units 19 units = 760



 QY 

760 ×2 = 80 cm 19

2 units = C R P S By using midpoint theorem. A



ES =

65.

BD

2

PX : QY = 1 : 1 (b) According to question Given:

D

C

. Q 2

1

.

1

DR =

1  BP 2 1  2x 2

P

A

18 6

66.

Let

F

2

E

M

3

D

3

1



AC = 9 cm (b) A

1 E 1

.Q

C

B C

B

AQ

ar (  ADE) = ar DECB

arADE AD2 = arABC AB2

QC

9



x

x = 36 x=6  AP = 6 cm (b) According to question

In  CBF By using B.P.T.

1 AD  2 AB 

P

D

Q

AD  DB

1 2 1

 DB  AB  AD 

Y B

2 1  AD     2   AB 

A X

CD CM 3   DB FM 2

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69.

D

4

2

 to make FM same

3

AC

ABC = 2unit2

4

67.

PQ

BC



3

9 P.

=

AC

To apply similarity property

x

AE AF 2   ED FM 3

AQ

So, ar  ADE= 1 unit2 and ar

BP

In  ADM. By using B.P.T.

=

APQ  ABC AP

C

AP

AB

A

A

2

AQ = 3 To app ly s im ilar tr iangle property.

=3

2 units = 3 × 2 = 6  PQ = 6 cm (c) According to Question Given: AP=QC AQ=9cm BP= 4cm AP =xcm

1



PB

Given: BD = 18 cm Note: Centroid is the point where m ed ians inter s ec ts and it divides median in 2 : 1 BD = 6 units, PQ = 2 units 6 units = 18 cm

wwM wa. th Les aBryn

ES x 1   DR x 1 (a) According to the question, Let draw a imaginary line DM which is parallel to EF.

AP

B

1 unit =

DR = x  ES = DR

2

ERna

DR =

2

2

1

(b) According to question,

1 BP 2

In  BPA, D is the midpoint of AB and R is the midpoint of AP By using midpoint theorem,

B

BD :

 AE = 2 units = 80 cm

BP = 2x

64.

68.

1

kgei snhe eYari dnag v.iSn ir

E

BD

2

 PX : QY =

760 1 unit = 19 D

1

C

So, AD : BD = 1 :

2 1



2 -1

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70. (d) According to question

BE



CE

3

AC||DE

2

D  A

73. (c) According to question

E 3 C AE  AC DE



CP = 2.5

AB

1

BC

AB = 2AD AB

,

12

2.5

Sin CPB =

AD

.

71. (d) According to question

T

Q

Given:

AB PB

1

wwM wa. th Les B aryn AP



AB

AQ AC

PQ



BC

PQ

Q

BC

PR



R

PT

3

PR =

72. (d) According to question

3

5

S

T

PR

 E  C

8 



PS

SR =

C

SR =

Given: AB = 10 cm

DE||AC

5

D  B

3

 ABC  ADE

40 3

area of ABC

40

–3

3

=

40 – 9 3

, SR=

31 3

.

D

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AB ² AD ²

 B

E

BC ² 

AE ²

2

DE ²

AB ² 

1

AB



AC ² 

AD ²



AD

BD

6 4



area of ADE

AB

2 1

 BD =

.

 CE

cm

A

 ABC   DBE

BE



75. (c) According to question Given: 'D' and 'E' are the points on AB and BC

BD BE   AD CE

C

DE ||BC Area of  ABC = 2 (Area of  ADE)

 SR = PR – PS

AD = 4 cm

E

1

B

P

PQ



2

B

1

D

PQR  PST

8

To apply B.P.T

2

77. (c) According to question Given: A

P

3 

AC

1

PS = 3 cm SR = ?

R Enak

C

AE

=

R

PT = 5 cm, TQ = 3 cm,

B

BC

=

BC

.S

DE

=

DE

5

1

By applying B.P.T

AB

4

2

=

AD

DE = 3 cm 74. (c) According to question Given:

2

C

B

,

DE



E

1

BC

BC

4

CP

D

DE



r

AD AB AD

Sin CPB 

1

By using B.P.T

6.25

 Sin CPB 

A

B

geisnh eeYa ridna gv.i Sni

CP =

  BDE   BAC 76. (c) According to question Given:

1

1 D 3

In CBP CP² = BP² + BC² CP² = (1.5)² + (2)² CP² = 2.25 + 4 CP² = 6.25

E  C

=

2 –1 2–1 2

C

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(a) According to question

Here,

A



3 D

E

AD AB

4

DE

=

Let, the side of equilateral triangle is = a

5

=

BC

=

5

Then AB = BC = CA = a

9

 DE : BC = 5 : 9

5

BD = DC =

A

80. (b)

a 2

C

B

Given: AD = 3, BD = 5 AB = 8, AC = 4 AE = ? By applying B.P.T AD

AE



AB 3



AC

mid point

DE

B

BC

 4

3 AE = = 1.5 cm 2 (d) In  ABC , DE||BC A

9

D

5

...(i)

AC 2 = AD2 + CD2

...(ii)

Add eq. (i) & (ii)

AB 2 + AC2 = 2AD2 + BD2 + CD2

C

Add BC² both sides

sides AB and AC respectively Then DE will be parallel to BC [by thales theorem]  And DE, always cuts the two equal part  Therefore AK : KH 1 : 1

AB 2+AC 2+BC 2

= 2AD²+BD²+CD² + BC² Put the value of side in R.H.S 2

A

E

9

=

AD AB

3 a² = 4AD² 2

a

a

C

=

2

a  a  2AD² +   +   + a² 2  2 

2AD² +

B

AC

AB 2 = AD2 + BD2

 Point D and E are mid points of

4

AE

E K

H

81. (d)

5

mid point D

AE

8

79.

AD DB

kgei snhe eYari dnag v.iSn ir

78.

K

DE BC B

a 2

D

a

a 2

C

wwM wa. th Les aBryn

ERna

(Basic Prop. theorem)

2 2  a² – a  AD2  3a   4 4   

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CHAPTER

23

CENTRES OF TRIANGLES

Orthocentre Circumcentre

r

V1

I

I

V3

Q

Properties

2.

Incentre of any triangle lies inside that triangle. Inc entr e is the only centre which is equiperpendicular distance from all the sides of a triangle. Ge nerally angle bisector doesn't intersect the opposite side perpendicularly.

3.

wwM wa.th Les B aryn

Circumcircle

P

R Enak geisnh

Where is the centre of triangle ? There are actually thousands of centres ! Here are the 4 most popular ones:

The Incentre is also the centre of the circ le inscr ibed in the

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R 2

 PIR = 90º +

Q 2

Q/2

R/2

R

In  QIR, I + Q/2 + R/2 = 180º Q R = 180º – I = 180º – – 2 2

Q  R     2 

180º– P  P  = 180º – 90º + 2  2 

I = 180º –  I = 90°+

6.

P 2

The angle between the external bisectors of two angles of a triangle is difference between right angle and half of the third angle.  QSR = 90º –

V1

P 2

P

ri

I

ri ri I

V2

 PIQ = 90º +

Q

Note:- The angle x as given in the figure above is a right angle only in the case of a isosceles and equilateral triangle. 4. The radius of the circle inscribed in a triangle is known as inradius (ri) of the triangle.

The incentre is the point of inters ection of the three ang le bisectors. The angle bisectors of a triangle are each one of the lines that divide an angle into two equal parts.

P 2

I

centroid

INCENTRE

 QIR = 90º +

P

Incentre

For each of those, the "centre" is where special lines cross, so it all depends on those lines!

R

Proof

Incircle

Centroid, Circumcentre, Incentre and Orthocentre

The angle betwen line s egments drawn from the two vertices to the incentre is equal to the sum of a right angle (90º) and half of third vertex.

V2

1.

Triangle Centres

5.

triangle.

eeYa ridn agv .iSn i

In geometry, a triangle centre is a point in the plane that is in some sense a centre of a triangle and can be obtained by simple constructions. Centre of triangle always occupy the same position (relative to the vertices) under the operations of rotation, reflection and dilation. Ancient Greeks discovered the classic centres of triangle (centroid, cir cumc entr e, incentre and orthocentre)

Q

Q/2

R/2 180°– R 2

180°– Q 2

R

V3 S

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9.

Inradius of a right angled triangle triangle PQR.

Proof

sum of two adjacent sides and opposite side.

P

P

Q 180º–Q IQS = 2 + = 90º 2

Z

R 180º–R IRS = 2 + = 90º 2 In Quadrilateral QIRS,

ri X

IRS +IQS= 180º

l1 AreaQIR

l2 AreaPIR

:

:

l3 =1:1:1 AreaPIQ l1 AreaQIR l3 AreaPIQ



=

ri

Proof

R

PQ

QX

In other words  = PR XR

Ar QIR : Ar  PIR : Ar PIQ 

1 1 1  l1  ri : l2  ri : l3  ri 2 2 2 = l1 : l2 : l3

[Because altitude of all triangles is same and equal to inradius (r)] i

8.

PQ + QR – PR 2

wwM wa.th Les aBryn

l1

ri 

Area of a triangle can be obtained by multiplying inradius (r 1 ) of triangle to the semiperimeter of that triangle.

Area  = ri × s

Proof

Proof

 ri × S

l2+l3 l1

l1×l2 l2 + l3

XR =

12. EXTERNAL ANGLE BISECTOR THEOREM. The external angle bisector theorem states that the external bisector of an angle of triangle divides the opposite side externally in the ratio of the sides containing the angle. In other words. PQ QX = PR RX

Given that PX is the bisector ofQPR.

Y P

 

Z



XPR = PRY =   alternative interior angles

 PRY is isosceles and PY = PR

(asPRY =PYR =  ) Now as QPX ~ QYR So, PQ QX = PY XR

 Q 

Since PY = PR, we have PQ QX = PR XR

 Ar QIR + Ar  PIR + Ar  PIQ 1 1 1  × l1 × ri + × l2 × ri + × l3 × ri 2 2 2 1 (l +l +l ) 2 1 2 3

l2×QR l1×l2 = l2 + l3 l2+l3

In  PXR l2× (l2+l3) PI PR l = = 2 = IX XR XR l1 × l2

 Draw a line parallel to PX from R. Which intersects extended QP at Y. Now, QPX = XPR =   bisected angles QPX = PYR =   corresponding angles

Area of triangle PQR

 ri ×

QX + XR l3+l2 QR l +l  = 3 2 = XR XR l2 l2

XR =

10. A NGLE B ISEC TOR TH EOREM:The angle bisector theorem states that an angle bisector divides the opposite side of a triangle into two segments that are proportional to the triangle's other two sides.

I

Q

QX l3 QX l =  +1= 3 +1 XR l2 XR l2



=

R

By Interior angle bisector theorem if PX is angle bisector ofP, then

Proof

PX = PZ, QX = QY = ri, RY = RZ (Tangents from external points)

( PX = PZ, ZR = YR)

l2 AreaPIR

l2

ri

ri

PXQ is tangent at X and QYR is tangent at Y So that IX  PQ and IY  QR



X l1

PI : IX = l2 + l3 : l1 QI : IY = l1 + l3 : l2 RI : IZ = l1 + l2 : l3

Proof

= 2 × ri

P

l3

Q

PQ+QR–PR ri = 2

The ratio of sides of a triangle is equal to the area of triangle formed by corresponding sides and incentre. l1 : l2 : l3 = Area  QIR : Area  PIR : Area  PIQ



l2

eYrai dnag



Y

R

Y

kgei snhe

 P = 90º– 2

I

ERna



I

v.iSn ir

Q

= 180º– 90º  P  2 

l3 Z

ri

ThenQSR + QIR = 180º QSR = 180º –QIR

7.

ri

11. Each angle bisector divided by Incentre is divided in the ratio equal to the ratio of length of

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

R

X

Given that PX is the external bisector of  RPY Draw a line parallel to PX from R, which intersect PQ at Z.

Now  RPX =  XPY = θ  bisected angle

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tive interior angles  RZP =  XPY = θ = corresponding angles  PRZ is isosceles and PZ = PR as  PZR =  PRZ Now as  QPX ~ QZR So,

3.

O is the incentre of  ABC and  A = 30°, then BOC is Sol. According to the Question, Given: A

o

1

B

PQ QX = PR RX



I is the incentre of  ABC ,  ABC = 60° and  ACB = 50°.

4.

Then  BIC is Sol. According to the question

C

BOC = 105° Let O be the in-centre of a triangle ABC and D be a point on the side BC of ABC , such that

OD  BC . If BOD = 15°, then

A

2

1

1 BOC  90  A 2

1 = 90° +  30 = 90° + 15° 2

Examples

1 A 2

BIC = 90 +

1  60 = 90 + 30 2

1

 B + 2  C +  BIC = 180°

eeYa ridn agv .iSn i

Since PZ = PR, we have

BIC = 90 +

BIC = 120° Alternate In  BIC,

A=30°

PQ QX = PZ RX

1.

We know that

BIC = 110°

r

 ZRP =  RPX = θ  alterna-

2

(65° + 55°) +  BIC = 180°

 BIC = 180° – 60° = 120°

6.

The internal bisectors of ABC and ACB of ABC me et e ach othe r at O. If BOC = 110°, then BAC is equal to Sol. According to the question A

I 60°

50°

B

C

B I and CI ar e the angle bisector  CBI  30

CBI  BCI  BIC  180 30° +25° + BIC = 180°

2.

BIC = 125° I is the incentre of ABC , If ABC = 60°, BCA  80 , then

the BIC is Sol. According to the question A

I

B

C

Given: ABC =60°

BCA = 80° BIC = ?  BAC = 40°



BIC = 90°+

o 110° B C Given: BOC = 110° BOC = 90°+

Given: BOD  15 In DBOD  BDO  DOB  DBO  180 DBO = 180 – (90 + 15) = 75°

wwM wa.th Les B aryn

BCI  25 In  BIC

R Enak geisnh

 ABC = Sol. According to the Question,

ABC = 2  DBO ABC = 2 × 75°

5.

ABC = 150° I is the incentre of a triangle ABC. If ACB = 55°, ABC =

65° then the value of  BIC is Sol. According to the question Given: A ACB=55° ABC=65° BIC= ? I



1 × 40° 2

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B C ACB  ABC  BAC = 180° BAC = 180° – 55° – 65°

BAC = 60°

110° = 90° + 20° = 7.

A 2

A 2

A , 2

 A = 40°

In

ABC , B  60 and C  40 . If AD and AE be respectively the internal bisector of A and perpendicular on BC, then the measure of DAE is Sol. According to the question,

Given:

B

= 60°

C = 40° As we know that

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A = 80°

In AEB A  B  E = 180° A = 180°– 60° – 90° A = 30°

The n, DAE  DAB – EAB = 40 – 30 DAE = 10° Alternate

 DAE =

1 A = 30° 2

A = 60°
10. The point in the plane of a triangle which is at equal perpendicular distance from the sides of the triangle is : (a) circumcentre (b) centroid (c) incentre (d) orthocentre Sol. A

C

B

BOC = 102°

102°= 90°+

1 A 2

r

D

C

A = 12°, A = 24° 2 9. The angle between the external bisectors of two angles of a triangle is 60°. Then the third angle of the triangle is Sol. According to the question

O = 90 –

1 A 2

AO 33  57 90 3 = = = OD 30 30 1

The cir cumc entr e is the point where the perpendicular bisectors of all 3 sides of a triangle intersect. It is also the centre of the triangle’s circumcircle.

P

C Q

R

PROPERTIES

P Z

Y

rc

rc

rc

Y

Q

C

X

R

Z

O

1. C

B

= 90° +



C

D 30

A

x

A  ZOY =  BOC = 90° + 2

Given: BOC = 60° As we know that

33

o

B

kgei snhe

Incentre is a point in the plane of a triangle which is at equal perpendicular distance from the sides of the triangle Here, OD  BC, OE  AC, OF  AB and OD = OE = OF (Inradius) 11. In  ABC, the internal bisector of the  A,  B and  C intersect the circumcircle at x, y and z respectively. If  A = 50º,  CZY = 30º then  BYZ will be Sol. According to the question

wwM wa.th Les aBryn

B

O

E

ERna

o 102°

r

r

8.

A

57

CIRCUMCENTRE

F

B – C 60 – 40 = = 10° 2 2

Internal bisectors of B and C of ABC intersect at O. If BOC = 102°, then the value of BAC is Sol. According to the Question, Given:

A

eYrai dnag

80 = 40°   BAD = 2

meets BC at D, and the bisector passes through incentre O. The ratio AO : OD isSol. By 7 property of incentre

1 A = 90° – 60° 2

v.iSn ir

A  B  C = 180° A = 180° – 60° – 40°

50 = 115º 2

In  ZOY,  BYZ = 180 – (30 + 115) = 35º 12. Three sides of a  ABC are, a =30 cm, b = 33cm, c =57 cm. The internal bisector of  A

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2.

3.

The length from all 3 vertices to the circumcentre is equal. and it is called circumradius. PC = QC = RC = Rc If we draw a circle joining all 3 vertices of the triangle, it will be called the circumcircle of that tr iang le. And circumcentre of that triangle will also be the centre of its circumcircle. The angle between line segments joining the circumcentre and two vertices is double the angle of the third vertex.

541

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Proof:-

7.

P 

= half of hypotenuse (Note:- QC is also the median of  PQR) The distance (d) between the circumcentre (r c) and Incentre (r i ) of a triangle can be expressed as rc2 – 2rcri

d=

or

2

4.

1 1 1 rc  d + rc – d = ri

R

d=

Because of the property of chord of a circle, an angle subtended by a chord at the centre is twice the angle subtended by the chord at any point on major arc of the circle. Circumradius of a triangle (rc) is

P

Examples

ri

Z S

C I  rc

 Q

rc

R

l1



l1  l 2 l 3 rc = 4  (AreaPQR )

If  PQR is an obtuse angled tr iang le its circumcentre lies outside the triangle PQR. P



Y



C

N

6.

Let C be the circumcentre of triangle PQR and I be the incentre The extension of PI meets circumcircle at X, then X is the midpoint of arc QR Join XC and extend it so that it intersects the circumcircle at Y. From I construct a perpendicular to PQ and let Z be its foot. so IZ = ri Now  PZI ~ YQX or  PZI =  YQX = 90º (   YQX = 90º, an angle in a semicircle)  ZPI =  QYX =  (  angles made by a chord)

wwM wa.th Les B aryn

5.

R Enak geisnh

l2

rc

Q

R Q X If  PQR is a right angled triangle, then its circumcentre lies on the mid-point of its hypotenuse and circumradius (rc) is half of its hypotenuse. P



(M id po int of PR C )

IZ PI So XQ = YX IZ × YX = PI × XQ

PI × XQ = 2rcri

 Q

R

X

rc

R

PC = QC = RC = r c (radius of circumcentre)

rc2 –2rc ri

Y

T

P l3

eeYa ridn agv .iSn i

Q

[   RQX =  P/2 angles from same chord XR] We have  QIX =  IQX So IX = QX Now extend IC, so that it intersects the circumcircle at S and T. Now, TI × SI = PI × IX (r c + d) × (rc – d) = 2rcri rc2 – d2 = 2rcri d² = rc² – 2rcri

r

 QCR = 2  P  PCQ = 2  R  PCR = 2  Q



IZ  ri      XY  2 rc 

Now join QI.  QIX =  QPI +  PQI =  P/2 +  PQR/2  IQX =  IQR +  RQX =  PQR/2 +  P/2

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13. The equidistant point from the vertices of a triangle is called its : (a) centroid (b) incentre (c) circumcentre (d) orthocentre Sol. The Perpendicular bisector of sides meet at a point called ‘circumcentre’. A

F

B

O

D

E

C

OA = OB = OC (Circumradius) 14. The circumcentre of a triangle ABC is O. If BAC = 85° and

BCA = 75°, then the value of  OAC is Sol. According to the Question Given: A

BAC =85° BCA =75° OAC = ?

o B

C

In DABC ABC  BCA  CAB =180° ABC = 20°  COA = 2 ABC COA = 2 × 20 = 40° In  AOC

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× × B

C

×

16. The equidistant point vertices of a triangle its: Sol. The equidistant point vertices of a triangle circumcentre

from the is called from the is called

A

F

C

Here OA = OB = OC (Circum Radius) 17. The radius of the circumcircle of a right angled triangle is 15 cm and the radius of its incircle is 6 cm. Find the sides of the triangle.

b = 24

c = 30 a+b = 18 + 24 = 42 cm 18. If the length of the three sides of a triangle are 6 cm, 8 cm and 10 cm, then the length of the median to its greatest side is Sol. According to question Length of the three sides of a triangle are 6 cm, 8 cm and 10 cm, this is right angle triangle.

5

(a) 30, 40, 41 (b) 18, 24, 30 (c) 30, 24, 25 (d) 24, 36, 20 Sol. According to question A

c

P O N b

C

D

5

B

P B

5 6

Median

C

v.iSn ir

a b –c and Ir = 2 a + b – c = 12 a + b = 12 + c a + b = 12 + 30 a + b = 42 cm Now check the option, any one option is satisfied option:(b) Here a = 18

8

B

O

A

D

a

AC , 2

AC = 2 × IR = 2 × 15 = 30 cm

wwM wa.th Les aBryn

B

E

O

A

  

 BOC = 50º ×2 = 100º OB= OC  BOC =  OCB 180º –100º = 40º 2

eYrai dnag

×

O

IR=

kgei snhe

× ×

A

e nc e

Given: PC = 15 cm = IR( circumradius) ON = 6 cm = Ir (Inradius) As we know that

ERna

We know OC = OA (Circum Radius)  OAC  OCA  OAC  OCA  COA = 180° 2OAC =180°– 40° 2OAC = 140° OAC = 70° 15. If the circumcentre of a triangle lies outside it, then the triangle is Sol. C ir cumc entr e of a triangle lies outside it, then triangle is obtuse angled triangle.

=

In  BPC,  BPC +  PBC +  PCB = 180º

or,

 BPC +

1 1 ×40º + ×40º = 180º 2 2

  BPC = 180º –20º –20º = 140º Alternate   BPC = 90º +

BOC = 90º + A = 2

90º + 50° = 140º

CENTROID Draw a line (called a "median") from the vertices to the midpoint of the opposite sides. When all three medians intersect is called the centroids which is also the centre of mass.

C

In right angle triangle median divides the hypotenuse in two equal parts



BD =

H 2

10 2 BD = 5 cm 19. O is the circumcentre of a triangle ABC whose  A = 50º. If bisector of  OBC and  OCB intersect at P then what is the measure of  BPC? Sol. Since angle subtended at the centre of the circle is double the angle subtended at circumfer-

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BD =

Q

Median

not a right angle S

P

R

In the above figure the  PSR is 90º only in the case of an equilateral triangle or if P is the top vertex of isosceles triangle.

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P

of 2 sides is Z

angle Area  XYC = Area  YZC = Area  ZXC

Centroid C

Q

1 th of area of tri12

Y

X

=

R

1 Area  PQR 12 P

CS : SP = CT : TQ = CU : UR = 1 : 3 PS : SC = QT : TC = RU : UC = 3 : 1 Proof Let PX = 6n then PC = 4n and CX = 2n  PZS ~  PQX   PZS =  PQX [  ZS||QX corresponding angle]   ZPS =  QPX, [  common angle]

Properties

P

Y

Z

Z

Y

C

[S is the mid point of PX]

X

Q

R

 CZY ~  CQR  CZY =  CRQ  CYZ =  CQR

2

alternate interior angle. CY ZY = CQ QR

1   QR    2  =  (QR)²

then SC = PC – PS = 4n – 3n = n



6.

PS 3n 3 = = SC n 1

In a triangle the ratio of sum of medians to the perimeter is always s malle r than PX+QY+RZ

1 3

1

P



×  Area PQR 

wwM wa.th Les B aryn

When we draw all 3 medians in a triangle. then it divides the triangle into six small triangles of same area.

P

Z



Z

C

1

Area  CZY = Area  PQR 12 C is also the centroid of  XYZ The line segment drawn via joining the mid points of two sides divides the line joining from centroid to the common vertex of two sides in the ratio 1:3

R X Area  PCZ = Area  ZCQ = Area  QCX = Area  XCR = Area  RCY = Area  CPY = Q

1 Area  PQR 6

Area of triangle formed by joining the centroid to mid points

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Y

QR > PX 2

...... (i)

Similarly in  QYR QR +

PR > QY 2

...... (ii)

PQ > RZ 2

...... (iii)

3 (PQ + QR + RP) > (PX + QY + RZ) 2

U X

R

adding (i), (ii) and (iii), we get

S C

Q

PQ +

PR +

T

X

Proof:App ly triangle prop erty in  PQX PQ + QX > PX

Similarly in  RZP

P

Z

Y C

Q

So,

4. 5.

Y

3

PQ+QR+RP 2 > PX+QY+RZ 3

  1  Area CQR = Area  PQR   3  

CY 1 = CQ 2

3 , or 2

vice-versa PQ+QR+RP < or 2

1

So

3.

1 = 4

PX 6n = = 3n 2 2

Area  CZY = × Area  CQR = 4 4

 ZY 1    (by mid point theorem)    QR 2

2.

So, PS =

(ZY)² AreaCZY = (QR)² AreaCQR



R Enak geisnh

X

R

Proof: CZY ~  CRQ [ZY||QR = Mid-point theorem] So,  YZC =  CRQ and  ZYC =  CQR

C Q

ZS PS 1 = QX = PX 2

r

Each median is divided by the centroid in the ratio 2 : 1 i.e. PC : CX = QC : CY = RC : CZ =2:1 Proof

eeYa ridn agv .iSn i

1.

R

(PX+QY+RZ) 3 (PQ+QR+RP) < 2

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2

1 PX = MP = 2

2 2

2 1

2l12  2l 32 – l 22

1 RZ = MR = 2

l3

Y

2 3

2M Q2  2M R2 – M P2

l2 =

2 3

2M P2  2M R2 – M Q2 2 P

2 Q

2M  2M – M

l2

P

Z

R

X

C

l1

Proof:-

n

'



n

X

R

l1

Let PX = d and QX = XR = n So l1 2

Now In  PQX l32 = n² + d² – 2dn cos θ ....... (i) [  using cosine formula] In  PXR,

l22 = n² + d² – 2dn cos θ ' = n² + d² – 2dn cos (180º – θ ) [ θ + θ ' = 180º] = n² + d² + 2dn cos θ ....... (ii) [cos(180º – θ ) = – cos θ ] adding (i) and (ii) l22 + l32 = 2 (n² + d²) ....... (iii) Multiply (iii) by 2. 2l22 + 2l32 = 4n² + 4d² 2l22 + 2l32 = l12 + 4d² [4n² = (2n)² = l12] 2 2 4d² = 2l2 + 2l3 – l12 2d = d=

1 2

R

4 (PX² + QY² + RZ²) = 3(PQ² + QR² + RP²) Proof: Apply appolonium theorem for all three medians  4MP2 + 4MQ2 + 4MR2 = (2l22 + 2l32 – l12 ) + (2l12 + 2l32 – l22) + 2l12 + 2l22 – l32 4(MP2 + MQ2 + MR2) = 3 (l12 + l22 + l32) 9. De te rm ining the ar ea of a triangle by Heron's formula for medians, providing medians of a triangles are given let MP, MQ and MR are the three medians of a triangle, and

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n=

Y

kgei snhe

Q

d

2 R

Q

SM =



2l 22  2l 32 – l12 2l 22  2l 32 – l12

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R

2 P

5(PR) = 4 (M + MR2) Points To Remember:MP2 + MR2 = (XY)2 + (PR)2 4(MP2 + MR2) = 5(PR)2 MP2 + MR2 = 5(XY)2

XY =

PR PR 2 So, XY2 = 2 4

PR 2 + PR2 4 4(MP2 + MR2) = 5PR2 Now as PR = 2XY, PR2 = 4XY2 4(MP2 + MR2) = 5 × (4XY2) MP2 + MR2 = 5XY2 Acc or ding to pr op er ty of circumcentre of right angled triangle

MP2 + MR2 =

MP + MQ + MR 2 Area DPQR =

PR 2 2 MP + MR2 = 5MQ2 The line segment joining the m id -p oint of a me dian to vertex divides opposite side in the ratio 1 : 2. Y is the mid-point of PX, then MQ =

11.

P

...(i) ...(ii)

...(iii) MP2 + MR2 = 5MQ2 ...(iv) Proof:MP2 + MR2 = (PQ2 + QX2) + (YQ2 + QR2) MP2 + MR2 = XY² + PR2 [PQ2 + QR2 = PR2 & YQ2 + QX2 = XY2] Ac cord ing to m id -p oint theorem.

4 SM (SM – MP ) (SM – MQ ) (SM – MR ) 3

Q

X

2

ERna

l2

l3

X

Q

P

P

In a triangle four times of sum of squares of median is equal to thre e times the sum of squares of sides.

C

Q

× MQ In a right angle triangle five times square of hypotenuse is equal to four times of sum of square of two medians (not right angle vertex median)

Y

8.

Z

10.

l1 =

2 l3 = 3

2l12  2l 22 – l 32 P

2l 22  2l 32 – l12

Note:- We can also find the length of sides if medians are given

2l  2l – l

1 2

QY = MQ =

2 3

1 2

v.iSn ir

7.

PX =

eYrai dnag

(PQ+QR+RP)

or (PX+QY+RZ) > 3 The length of medians can be obtained from Apollonius theore m

P Z R

Y

Note:- If MP2 + MQ2 = MR2 then area of triangle Q

1  4 2 = ×  × MP  × MP × MQ   = 2 3 3  

X

R

PZ : ZR = 1 : 2 Proof:-

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P

13. S

X

R

In DPQR draw XS||QZ as PX is a median then X is the mid-point of QR, then S is also the mid-point of ZR. (According to mid point theorem) So, ZS : SR = 1 : 1 In DPXS, YZ||XS and Y is the mid point of PX, then Z is also the midpoint of PS. PZ : ZS = 1 : 1 PZ : ZR = 1 : 2 12. I f 2 me dians ar e perpendicular to each other then five times the square of common side is equal to sum of square of other two sides

P

Z

l2

X

R

14.

...(iii) Adding (i), (ii) and (iii) 2(PQ + QR + RP) > 2(PX + QY + RZ) PQ + QR + RP > PX + QY + RZ Sum of three medians of a tr iang le is always g re ater

R

l1

2 2 2 2 2     M M + Q R  = l1   3  3 

9 2 l 4 1 by appolonius theorem

MQ2 + MR2 =

2l12  2l32 – l22 4

1

B

3 (l + l2 + l3) 4 1

2l12  2l22 – l 32 + 4

9 = l12 4 l22 + l32 = 5l12 Note:- If median form pythagoras triplet with each other, i.e., MQ2 + MR2 = MP2 then also result will be the same i.e., l22 + l32 = 5l12 If two m ed ians are  pe rp endicular then all me dians will f or m

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

l2 Y

C

Q

2 PC = MP 3 QC =

2 MQ 3



[  PX = MP] [  QY = MQ]

2 MR [  RZ = MR] 3 Apply inequality of triangle

RC =

In  PCQ 2 2 MP  MQ 3 3

> l3

...(i)

In  PCR 2 2 MP  MR 3 3

6  2 = 4 cm 3

BG = 4 cm

 B GD is a r ight angle triangle BD2 = BG2 + GD2 BD2 = (4)2 + (3)2 BD2 = 16 + 9 BD =

25 BD = 5 cm 21. AD is the median of a triangle ABC and O is the centroid such that AO = 10 cm. The length of OD (in cm) is Sol. According to question A

> l2

...(ii)

In  QCR 2 2 MQ  MR 3 3

unit C

D G is t he c e ntr oid whic h divides the median in 2 : 1  AD = 3 units = 9 cm 3 units = 9 cm 9 1 unit = = 3 cm 3 GD = 3 cm BE = 3 units = 6 cm 3 units = 6 cm 6 1 unit = 3

2 units =

R

X l1

units Median

G

P

l 3Z

E

2

Proof:-

wwM wa.th Les B aryn

Proof If MQ  MR then In DQCR, QC2 + CR2 = QR2

20. Two medians AD and BE of ABC intersect at G at right angles. If AD = 9 cm and BE = 6 cm, then the length of BD (in cm) is Sol. According to question A

3 than of perimeter 4 i.e., MP + M Q + MR >

Q

...(i) ...(ii)

R Enak geisnh

X C

Examples

PQ + PR > 2PX PQ + QR > 2QY PR + QR > 2RZ

P l3 Y

Y

C

Q



r

Q



eeYa ridn agv .iSn i

Z Y

Adding (i), (ii) & (iii), We get, 4 MP  MQ  MR > l1+ l2 + l3 3 3 MP + MQ + MR > l1  l 2  l3 4

pythagorean triplets. The sum of any two sides of a triangle is greater than twice the median drawn to the third side.

O

> l1

...(iii) B

D

2 units centroid 1 unit C

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2AB ²  2AC ² – BC ² 4

AD =

32  50 – 4 = 4

=

x² + y² =

78 4

39 2 25.. BL and CM are medians of ABC right- angled at A and

AD =

3 5 cm, then 2 the length of CM is Sol. According to question

BC = 5 cm. If BL =

eYrai dnag

3 5 2

2

=

A

4cm

B

C

=

11 E

11

1x

O

2y

D 1y 2x

B

19 2

C

In right angle  COD, OC² + OD² = CD² 4x² + y² =

361 4

361 5x + 5y = + 121 4

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2

2

...(i)

...(ii)

845 = 5

484 + 361 5

169 = 13 cm

27. If AD, BE and CF are medians of ABC , then which one of the following statements is correct?

(a) (b) (c) (d) Sol.

AB + BC + CA < AD + BE + CF AB + BC + CA > AD + BE + CF AB + BC + CA = AD + BE + CF None of these According to question A

E

F

B 19 2

In right angle  BOE, OB² + OE² = BE² x² + 4y² = 121 Add Both Eq. (i) & (ii)

5cm

D 2cm

A

AB2  AC2 5

(22)2 + (19)2 = 5

26. In a  ABC, BD&CE are the two medians which intersect each other at right angle. AB = 22, AC = 19, find BC = ? Sol. According to the question,

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BC AD = 2 AD = BD = DC (AD is median) In DABD, BD = DA  DBA =  DAB = q1 In DADC DC = DA  DAC =  DCA = q2 Now, In DABC  A +  B +  C = 180° q1 + q2 + q1 + q2 = 180° 2(q1 + q2) = 90° q1 + q2 = 90°   A = q1 +q2 = 90° So, Triangle will b e Right Angle triangle 24. In a  ABC, three sides are 5cm, 4cm. and 2 cm. Find the length of median from smallest angle vertex. Sol. Smallest angle is opposite to smallest side. By Apollonius theorem.

According to figure, when two medians intersect each other in a right angled triangle then we use, this equation.  4 (BL2 + CM2) = 5BC2 2 3 5     4 ×  2  + 4CM2 = 5 × (5)2    45 + 4CM2 = 125 125 – 45 = 20  CM2 = 4  CM = 2 5 cm

kgei snhe

In DABC, AD is median on base BC According to the question,

ERna

1

....(iii)

In right angle  BOC, OB² + OC² = BC² 4y² + 4x² = BC² 4(y² + x²) = BC² From Eq. (iii) BC² = 169 BC = 13 Alternate: In this type of question we use direct formula. 5BC2 = AB2 + AC2 BC =

1 2

169 4

v.iSn ir

AD is the median and 'O' is the centroid  AO = 10 cm 2 units = 10 1 units = 5  OD = 5 cm 22. If the median drawn on the base of a triangle is half its base the triangle will be Sol. According to question If the median drawn on the base of a triangle is half its base of the triangle will be right angled triangle.

D

C

Points D, E & F are midpoint of BC,CA and AB. ........(i)  AB + AC > 2AD AB + BC > 2 BE .......(ii) BC + CA> 2CF ......(iii) Adding to equation (i),(ii) and (iii) we get 2(AB + BC + CA) > 2(AD + BE + CF)  (AB + BC + CA) > (AD + BE + CF) 28. ABC is an isosceles triangle and AB  AC = 2a unit, BC = a unit. Draw AD  BC , and find the length of AD .

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Sol. According to the question Given A

2a

Sin60° =

AB = AC = 2a BC = a

BD =

BC 2

BD =

a 2

In ADB us ing p y thag oras theorem . AB 2 = BD2 + AD2 2 a     + AD2  (2a)2 =  2  a2 + AD2  4a2 = 4

3 From APD AP Sin75º = AD



AD2 = 4a2 –

a

2

Q

A

F

E

D

AD =

C

By using this formula, We can calculate the area of

15a units 2

 ABC.

29. An isosceles triangle ABC is right-angled at B.D is a point inside the triangle ABC. P and Q are the feet of the perpendiculars drawn from D on the side AB and AC respectively of ABC . If AP = a cm , AQ = b cm and BAD  15 , Find the value of sin 75° ? Sol. According to the question,

4 Area of S S a S bS c ABC = 3 Here, a, b & c are the length of the median.



S=

 Area =

A

=

30° b a 15° Q P 75° 60° D

B

PX, QY & RZ are the altitudes of  P QR & O is the orthocentre.

Properties

The sum of angle between line segments joining the orthocentre and two vertices and the third angle is always a supplementary angle.  QOR +  QPR = 180º =  ZOY +  QPR [  QOR =  ZOY, Opposite angle]   POR +  PQR = 180º =  ZOX +  PQR [  POR =  ZOX, Opposite angles]   POQ +  PRQ = 180º =   YOX +  PRQ [  POQ =  YOX, Opposite angles] Proof:P Y Z

a  b  c 8  6  10 = = 12 2 2

O

Q

4 12(12 – 6)(12 – 8)(12 – 10) 3

4 12  6  4  2 3

4  24 = 32 cm² 3 Alternate: When m12 + m22 = m32 The n =

C

DABC is a isosceles triangle So,  B = 90°,  C =  A = 45° In DAQD

Area of D =

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R

X

Here,

a 3a  3 Sin75º = 2b 2b 30. In  ABC, BE, AD, CF is a me dian on AC, BC and AB respectively. AD = 10, BE = 6 and CF = 8 cm. Then find the area of  ABC ? Sol. According to the question,

B

4

O

wwM wa.th Les B aryn



Y Z

R Enak geisnh



P

2b

AD =

AD  BC In isosceles triangle perpendicular sides bisects the opposite side of the length

Orthocentre A point where all 3 altitudes of a tr iang le m ee t is c alle d the orthocentre.

b 3 = AD 2

C

D a

2 × 6 × 8 = 32 cm2 3

r

B

AQ AD

=

eeYa ridn agv .iSn i

2a

=  BAC –  BAD =  DAC 45 – 15 = 30° and  QDA = 180° – (90 + 30) = 60° From AQD

2 × m1 m2 3

2.

X

R

In Quadrilateral PYOZ  Z =  Y = 90º So,  P +  O = 180º  QPR +  ZOY = 180º  QPR +  QOR = 180º Orthocentre of triangles Orthocentre of  PQR = O Orthocentre of  QOR = P Orthocentre of  POQ = R Orthocentre of  POR = Q

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3.

P

Pair of similar triangles when all three altitudes are drawn

angled triangle PQR. The points P and O are joined and produced to meet the side QR at S. If

P

 PQS= 60° and QCR  130 °,

Y Z O

In this figure Orthocentre (O) lies outside the triangle PQR. Sum of three altitudes of a triangle is less than sum of three sides of the triangle.

8.

 OXQ ~  OYP It must be noted that, In triangles PQR  OXQ,  OXR,

P

q  T Z

X

In any right angled triangle Hypotenuse > side (altitude) In PXQ, PQ > PX ...(i) In  QYR, QR > QY

Y 



q



R

On Base, QS  QPS =  QRS On Base, RS  RPS =  RQS

6.

On Base, QR  QTR =  QUR etc. The orthocentre of a right angled triangle is that point where triangle forms the right angle.

QPR 

1 130  65 2

Now, PQS  PSQ  QPS  180 60  90  QPS  180 QPS  30

31. In  ABC, AD, BE and CF are the altitudes in the ratio 1:2 : 3 respectively, then the ratio of AB : BC : CA is: (a) 3 : 2 : 1 (b) 1 : 2 : 3 (c) 1 : 4 : 9 (d) 2 : 6 : 3 Sol. A

= 65° – 30°

E

F

B

D

C

1 1 AC×BE = AB×CF 2 2 1 1 1 : :  AB : BC : AC = CF AD BE 1 1 1 = : : = 2 : 6 :3 3 1 2 32. O and C are respectively the or thoc entr e and the cir cumc entr e of an acute-

=

Orthocentre of  PQR

7.

1 QCR 2

RPS = 35° 33. The perpendiculars drawn from the vertices to the opposite sides of a triangle, meet at the point whose name is Sol. According to question Dr aw a line (called the "altitude") at right angles to a side and going through the opp osite corner . Where all thr ee lines inte rsect is the "orthocentre" O is Orthocentre. A

1 Area of  ABC = BC×AD 2

P

Q

QCR  130

RPS  QPR – QPS

wwM wa.th Les aBryn

S

 QPR 

...(ii)

ERna

Q

60°

Examples

O X

130°

R

In  PRZ, RP > RZ ...(iii) adding (i), (ii) & (iii) We get, PQ + QR + RP > PX + QY + RZ

P

U

O

kgei snhe

O is the orthocentre of  PQR. Dr aw a circumcircle to triangle PQR. Since angles in the same segments of a circle are equal.

Z

Q

o c

Q S R Given PQS  60

Y

 OY P,  OY R,  OZ P &  OZQ are right angle triangles.

v.iSn ir

O

 OXR ~  OZP

5.

R

R

X

 OYR ~  OZQ

4.

then RPS  ? Sol. According to question P

eYrai dnag

Q

Q

R

The orthocentre of an obtuse angled triangle lies outside the triangle.

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F

E o

B D 34. The orthoce ntre of angled triangle lies Sol. The or thoce ntr e of angled triangle lies at angular vertex

C

a right a r ight the right

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A

35. In ABC , draw BE  AC and CF  AB and the perpendicular BE and CF intersect at the point O. If  BAC = 70°, then the value of BOC is Sol. According to question

A 70° E B

C

R Enak geisnh

O



 ABC = 110º =  NBP (Vertically opposite) In quadrilateral BNOP  N +  B +  P +  O = 360º  B = 360º – 90º – 90º – 110º

C  Area of DABC = × Base × 2 height

B

is R is R ² – 2rR where circumradius and r is inradius. In an equilate ral triangle, length or radius of the circumcircle is equal to twice the radius of its incircle ie., if  ABC is equilateral then R = 2r Ceva Theorem:- If O is any point inside the triangle ABC and AO, BO,CO meet sides BC,CA,AB respectively at point D,E,F then

BD CE AF   =1 DC EA FB

A

1.

In an equilateral triangle all the four centres are coincident i.e ., c entr oid, inc entr e, circumcentre and orthocentre of an equilateral triangle lie at the same point. Centroid (G), orthocentre (P) and circumcentre (O) of a triangle are always collinear (i.e, lie in a straight line) and PG: GO = 2 : 1. The orthoce ntre of a right angled triangle lies at the right angled ver tex while its circumcentre is mid point of hypotenuse. Circumcentre and orthocentre of an obtuse angled triangle always lie outside the triangle. The sum of diameters of circumcircle and incircle of right angled triangle is equal to the sum of its perpendicular sides.

2.

3.

4.

5.

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E

F O

Mixed properties of centres of a triangle

wwM wa.th Les B aryn Y

c

 B = 70º

A

X

C

P

In the given figure

ABC is a right angled triangle with  A = 90º. If radius of circumcircle and incircle of the triangle be respectively R and r then 2(R +r) = b + c 6. The distance between incentre and circumcentre of a triangle

8.

Given: A = 70° AEOF is a quadrilateral  In a quadrilateral sum of all angles are 360°  A +  F +  O +  E = 360° 70° + 90° +  O + 90° = 360°  O = 360° – 250°  O = 110°  BOC = 110° (Vertically Opposite angle) 36. Let ABC be an equilateral triangle and AX, BY, CZ be the altitudes. Then the right statement out of the four given responses is (a) AX = BY = CZ (b) AX  BY  CZ (c) AX = BY  CZ (d) None of these Sol. ABC is an equilateral triangle and AX, BY and C Z b e the altitude so AX = BY = CZ

B

110º

N

B

Z

A

7.

A

O

b

r

C

F

C

eeYa ridn agv .iSn i

Orthocentre

O

  × AB × CZ = × BC × AX 2 2  = × AC × BY 2 CZ = AX = BY (Side of  AB = BC = C A equilateral) so, we can say if three ltitudes are equal then the triangle is equilateral triangle 37. In obtuse angle  , obtuse angle is 110º. Find the angle made on its ortho centre: Sol. According to the question, =

B

9.

C

D

Since Ceva Theorem is true for any point inside the trian g le , it is t he r e f or e al s o true for centroid, incentre, orthocentr e and cir cum centre of the triang le . C onve rs e of Ce va Theore m is also true. Menelaus Theorem:- If a transverse cuts the sides BC,CA and AB (or its produced part) of a triangle at D,E, F then

BD CE AF   = –1 DC EA FB A F

E B

C

D

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

4 6 =  4AC = 18 3 AC

Interior Angle Bisector Theorem



AC =

Interior Angle Bisector Theorem : The angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. Given: In ABC in which AD is the internal bisector of  A and meets BC at D. Prove that : BD /DC = AB/AC E 4

18 = 4.5 cm. 4

Exterior Angle Bisector Theorem Exterior angle bisector theorem: The external bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle.

Given

Statements 1) CE||DA

In  AB C, in which AD is the

C Reasons 1) By Construction

2) 2 = 3 2) Alternate interior angles 3) 1 = 4 3) Corresponding angles 4) AD is the bisector 4) Given 5) 1 = 2

5) Definition of angle bisector

6) 3 = 4 7) AE = AC

6) From (2),(3) and (5) 7) In ACE, side opposite to equal angles are equal

8) BD BA = DC AE

8) In BCE DA||CE and by BPT theorem 9) From (7)

K

A

E

B

Examples

38. In the given figure, AD is the Internal bisector of  A, If BD =4cm, DC = 3 cm and AB = 6 cm, find AC. A

6cm

B

4cm D 3cm

C

Sol. In  ABC, AD is the bisector of A 

intersects BC produced at D. Prove that : BD / CD = AB / AC Construction : Draw CE || DA meeting AB at E.

wwM wa.th Les aBryn

9) BD AB = DC AC

bisector of the exterior  A and

kgei snhe

D

ERna

B

39. In the given figure, AE is the bisector of the exterior  CAD meeting BC produced in E. If AB = 10 cm, AC = 6 cm and BC = 12 cm, Find CE. Sol. D A 10cm

6 cm

B

C x cm E Given : AB = 10 cm,AC = 6 cm and BC = 12 cm, By exte rior ang le b isec tor theorem CE = x (Let) 12cm

eYrai dnag

A 1 2 3

Examples

v.iSn ir

Converse of the Theorem is also true.

Statements

2

1

4

BE AB = CE AC

12  x 10 = x 6 6(1 2 + x) = 10x [by cross multiplication] 72 + 6x = 10x 72 = 10x – 6x

3

72 = 4x D

C

x = 18 CE = 18 cm

Reasons

1) CE||DA

1) By Construction

2) 1 = 3 3) 2 = 4

2) Alternate interior angles 3) Corresponding angles (CE ||DA

and BK is a transversal)

4) AD is a bisector of A

4) Given

5) 1 = 2

5) Definition of angle bisector

6) 3 = 4 7) AE = AC

6) From (2),(3) and (5) 7) If angles are equal then side opposite to them are also equ al

8) BD BA = CD EA

8) By Basic proportionality theorem (EC||AD)

9) BD AB = DC AE 10) BD AB = CD AC

9) BA =AB and EA =AE

10) AE = EC and from (7)

BD AB = (Angle bisector theorem) DC AC

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EXERCISE

C of ABC (where AB and AC extended to E and F respectively ) me et at p oint P. If BAC  100 , then the mea-

3.

4.

sure of BPC is (a) 50° (b) 80° (c) 40° (d) 52° Internal bisectors of angles B and C of a triangle ABC meet at O. If BAC = 70°, then the value of BOC is (a) 125° (b) 140° (c) 110° (d) 130° In ABC , the internal bisectors of ABC and ACB meet at I and BAC = 50°. The mea-

6.

7.

(a) 2 3 cm 8.

R of PQR intersect at O. If ROQ = 96° then the value of RPQ is : (a) 12° (b) 24°

(c) 36° (d) 6° 10. O is the incentre of PQR and QPR = 50°, then the measure

of QOR is:

(a) 125° (b) 100° (c) 130° (d) 115° 11. AD is perpendicular to the inte rnal bis ector of  ABC of  ABC . DE is drawn through D and parallel to BC to meet AC at E. If the length of AC is 12 cm, then the length of AE (in cm.) is (a) 8 (b) 3 (c) 4 (d) 6

12. If any two sides of a triangle are produced beyond its base and the exterior angles thus obtained are bisected, then these bisectors will include : (a) half the sum of the base angles (b) sum of the base angles (c) half the difference of the base angles (d) difference of the base angles 13. If I is the in-centre of  ABC and A = 60°, then the value of BIC is: (a) 100° (b) 120° (c) 150° (d) 110° 14. In the given figure, A = 80°, B = 60°, C = 2x and BDC = y°, BD and CD bisect angles B and C respectively. The value of x and y respectively are :

wwM wa.th Les B aryn

5.

sure of BIC is (a) 105° (b) 115° (c) 125° (d) 130° If I be the incentre of  ABC and  B=700 and  C=500 , then the magnitude of  BIC is (a) 1300 (b) 600 0 (c) 120 (d) 1050 ABC is an equilateral triangle and CD is the internal bisector of C . If DC is produced to E such that AC = CE, then  CAE is equal to (a) 45° (b) 75° (c) 30° (d) 15° The radius of the incircle of the equilateral tr iangle having each side 6 cm is

Internal bisectors of Q and

(b)

3 cm

(c) 6 3 cm (d) 2 cm The internal bisectors of the angles B and C of a triangle ABC meet at I. If BIC =

A + 2

X, then X is equal to (a) 60° (b) 30° (c) 90° (d) 45°

B

A 80° D y

x°

x°

C

(a) 15° and 70° (b) 10° and 160° (c) 20° and 130° (d) 20° and 125°

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15. The sides of a right angle triangle containing the right angle measure 3 cm and 4 cm. The radius of the incircle of the triangle is : (a) 3.5 cm (b) 1.75 cm (c) 1 cm (d) 0.875 cm 16. In the given figure, BO and CO are the bisector of CBD and BCE respectively and A = 40°, then BOC is equal to :

r

meet at O. If BAC = 80°, then the value of BOC is (a) 120° (b) 140° (c) 110° (d) 130° The external bisector of B and

9.

eeYa ridn agv .iSn i

2.

Internal bisectors of angles B and C of a triangle ABC

R Enak geisnh

1.

A

B

D

C E

0

(a) 60° (b) 65°

(c) 75° (d) 70°

17. In the given figure, AM  BC and AN is the bisector of A. What is the measure of MAN ?

65°

30°

(a) 17.5° (b) 15.5° (c) 16° (d) 20° 18. If the bisector of an angle of  bisects the opposite side, then  is : (a) Scalaene (b) Isosceles (c) Right triangle (d) None of these. 19. Incentre of a triangle lies in the interior of : (a) an isosceles triangle only (b) an equilateral triangle only (c) every triangle (d) a Right-triangle only. 20. In  ABC, the bisectors of B and C intersect each-other at a point O, then BOC is equal to :

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A

(c) 120° +

1 2

A

(d) 120° -

1 2

A

21. In  ABC, the sides AB and AC are produced to P and Q respectively. The bisectors of PBC and QCB intersect at a point O, then BOC is equal to : (a) 90° -

1 2

of a  ABC intersect at O, if BOC = 102°, then the value of BAC is : (a) 12° (b) 24° (c) 48° (d) 60° 28. The in-radius of an equilateral triangle is of length 3cm, Then the length of each of its medians is : (a) 12 cm

1 A 2 1 (c) 120° + A 2 1 (d) 120°– A 2

and ACB of  ABC meet each-other at O. If BOC = 120°, then BAC is equal to : (a) 80° (b) 50° (c) 60° (d) 90° 24. O is the incentre of  ABC and A = 30°,then BOC is : (a) 100° (b) 105° (c) 110° (d) 90° 25. In  ABC, AD is the internal

bisector of A, meeting the side BC at D. If BD = 5cm, BC = 7.5cm, then AB : AC is : (a) 2 : 1 (b) 1 : 2 (c) 4 : 5 (d) 3 : 5 26. The external bisector of B and

C of  ABC meet at point P. If BAC = 80°, then BPC is: (a) 50° (b) 40° (c) 80° (d) 100°

9 cm 2

30.

31.

AB at P. Then  BPO is: (a) right angled isosceles triangle (b) isosceles but not a right angled (c) equilateral triangle (d) None of these If O be the circumcentre of a triangle PQR and QOR = 110°, OPR = 25° , then the measure of PRQ is (a) 65° (b) 50° (c) 55° (d) 60° For a triangle circumcentre lies on one of its sides. The triangle is (a) right angled (b) obtused angled (c) isosceles (d) equilateral In ABC , ABC  70 , BCA = 40°, O is the point of intersection of the perpendicular bisectors of the sides, then the angle BOC is (a) 100° (b) 120° (c) 130° (d) 140° ABC is an equilateral triangle and O is its circumcentre, then the BOC is (a) 100° (b) 110° (c) 120° (d) 130° O is the circumcentre of ABC . If BAC = 85°, BCA = 75°, the OAC is equal to: (a) 70° (b) 60° (c) 50° (d) 40° If O is the circumcentre of a triangle ABC lying inside the triangle, the  OBC +  BAC is equal to

wwM wa.th Les aBryn

22. O is the incentre of  ABC and BOC = 130°. Find BAC: (a) 80° (b) 40° (c) 150°(d) 50° 23. The internal bisector of ABC

(b)

(c) 4cm (d) 9 cm 29. O is any point on the bisector of the acute angle  ABC. From O a line parallel to CB meets

A

(b) 90 ° +

C

(a) 120° (c) 90°

(b) 110° (d) 60°

36. The circum-centre of a triangle is always the point of intersection of the: (a) Medians (b) bisectors (c) Perpendiculars (d) Perpendiculars bisector of the sides 37. The radius of circum-circle of an equilateral triangle of side 12cm is : (a) (4/3) 3

(b)

(c) 4 3

(d) 4

eYrai dnag

1 2

(b) 90° +

B and

kgei snhe

A

v.iSn ir

27. Internal bisectorsof

1 2

ERna

(a) 90° -

32.

33.

34.

35.

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4 2

38. In  ABC, B is a right angle, AC = 6cm, and D is the mid-point of AC. The length of BD is: (a) 4 cm

(b)

6 cm

(c) 3 cm (d) 3.5 cm 39. If P and Q are the mid- points of the sides CA and CB respectively of a triangle ABC, rightangled at C, then the value of 4(AQ2 + BP2) is equal to : (a) 4BC2 (b) 2AC2 2 (c) 2BC (d) 5AB2 40. The length of the two sides forming the right angle of a right angled triangle are 6cm and 8cm. The length of its circum-radius : (a) 5 cm (b) 7 cm (c) 6 cm (d) 10 cm 41. In  ABC,  B = 90°,  C = 45° and D is the mid-point of AC. If AC =

4 2 units, then BD is:

(a)

2 2 units (b) 4 2

(c)

5 units 2

(d) 2 units

42. In a triangle ABC, median is AD and centroid is O, AO = 10 cm. The length of OD ( in cm ) is (a) 6 (b) 4 (c) 5 (d) 3.3 43. G is the centroid of the equilateral ABC . If AB = 10 cm then length of AG is (a)

5 3 3

cm

(c) 5 3 cm

10 3 cm 3 (d) 10 3 cm

(b)

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49. In  ABC,the medi an BE intersects AC at E, if BG = 6cm, where G is the centroid, then BE is equal to : (a) 8 cm (b) 10 cm (c) 7 cm (d) 9 cm 50. If in the given figure, PQR = 90°, O is the centroid of  PQR, PQ = 5cm and QR = 12cm, then OQ is equal to: P

(a) 90° (b) 45°

M

Q

1 (a) 3 2 1 (c) 4 2

R

1 3 1 (d) 5 3 (b) 4

A

M

N

r

H

L

B

C

(a) N (b) A (c) L (d) M 60. In the given figure  QPR = 90°, QR = 26 cm, PM = 6 cm, MR = 8cm and  PMR = 90°, find the area of  PQR.

(c) 30° (d) 60°

55. In  ABC, G is the centroid, AB = 15cm, BC = 18cm, and AC = 25cm. Find GD, where D is th mid-point of BC : (a)

1 86 2

7 86 cm (c) 3

(b)

1 86 3

2 86 cm (d) 3

56. If G is the centroid of  ABC and AG = BC, then  BGC is : (a) 75° (b) 45° (c) 90° (d) 60° 57. I and O are respectively the incentre and circumcentre of a triangle AB C. The line AI pr oduc ed inte rsec ts the circumcircle of ABC at the point D. If ABC = x°, BID = y° and BOD = z°, then

O

then the orthocentre of  HBC is (figure given) :

1 BC. If  BAD = 30° is, 2 then  ACB is : AD =

wwM wa.th Les B aryn

ABC is equal to (a) 1 : 12 (b) 12 : 1 (c) 4 : 3 (d) 3 : 4

59. If H is the orthocentre of  ABC,

eeYa ridn agv .iSn i

The area of  ABC is 60 cm². The area of  GBC is (a) 30 cm² (b) 40 cm² (c) 10 cm² (d) 20 cm² 48. The medians CD and BE of a triangle ABC intersect each other at O. The ratio of ODE :

51. The medians AD, BE CF of a triangle ABC intersect in G. Which of the following istrue for any  ABC ? (a) GB + GC = 3GA (b) GB + GC < GA (c) GB + GC > GA (d) GB + GC = GA 52. If G is the centroid and AD be a median with length 12cm of  ABC, then the value of AG is: (a) 4 cm (b) 6 cm (c) 10 cm (d) 8 cm 53. Two medians AD and BE of  ABC intersect at G at right angles. If AD = 9cm and BE = 6cm, then the length of BD, in cm is : (a) 10 (b) 6 (c) 5 (d) 3 54. In  ABC, AD is the median and

R Enak geisnh

44. If the three medians of a triangle are same, then the triangle is (a) equilateral (b) isosceles (c) right- angled (d) obtuse-angle 45. In ABC , D is the mid-point of BC. Length AD is 27 cm. N is a point in AD such that the length of DN is 12 cm. The distance of N from the centroid of ABC is equal to (a) 3 cm (b) 6 cm (c) 9 cm (d) 15 cm 46. G is the centroid of  ABC. Th e m e d i ans A D an d B E intersect at right angles. If the lengths of AD and BE are 9 cm and 12 cm respectively; then the length of AB (in cm) is? (a) 11 (b) 10 (c) 10.5 (d) 85 47. The centroid of a  ABC is G.

z x y =?

(a) 3 (b) 1 (c) 2 (d) 4 58. In an obtuse-angled triangle ABC, A is the obtuse angle and O is the orthocentre. If BOC = 54°, then BAC is : (a) 108° (b) 126° (c) 136° (d) 116°

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(a) 180 cm2 (b) 240 cm2 (c) 120 cm2 (d) 150 cm2 61. In the given figure, if AD is bisector of  BAC then AB is: A 7cm

B 3cm D 4cm C

(a) 6cm (b) 5 cm (c) 5.25cm (d) 5.75 cm 62. In  ABC, D is a point on BC such that

AB BD  . If  B = AC DC

70°,  C = 50°, then the value of  BAD : (a) 30° (b) 60° (c) 40° (d) 50° 63. In the figure AD is the external bisector of  EAC, intersects BC produced to D. If AB =12 cm, AC = 8 cm and BC = 4 cm, find CD:

554

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B D C (a) 10cm (b) 6 cm (c) 8 cm (d) 9 cm 64. In triangle ABC , DE || BC where D is a point on AB and E is point on AC. DE divides the area of  ABC into two equal parts. Then DB : AB is equal to (a)

2 : ( 2  1)

(b) ( 2 – 1) : 2

(c)

2 : ( 2 – 1)

(d) ( 2  1) : 2

(a) 10º (c) 100º

(b) 35º (d) 115º

67. In an isosceles  ABC,AB = AC and  A is two times of  B. If AB =3 cm. then ratio of inradius to the circumradius is(a) 1 : 2

(b)

(c) 2 2 – 1:1

(d) 1 : 2 2 – 1

2 – 1:1

v.iSn ir

A

65. Let ABC be a triangle with AB = 3cm. and AC = 5 cm. If AD is a median drawn from the vertex A to the side BC, then which one of the following is correct? (a) AD is always greater than 4cm. but less then 5 cm (b) AD is always greater than 5 cm. (c) AD is always less than 4 cm (d) None of the Above 66. In  ABC, the internal bisec-

68. If O be the orthocentre of ABC, OF  r AB and OE  r AC. If OE = 2cm and BE = 5 cm then find the value of OF × OC. (a) 10 (b) 3 (c) 6 (d) 3

tor of the  A,  B and  C intersect the circumcircle at x,y and z respectively. If  A= 50º,  czy = 30º then  Byz will be

eYrai dnag

E

ANSWER KEY 8. 9. 10. 11. 12. 13. 14.

(c) (a) (d) (d) (a) (b) (c)

15. 16. 17. 18. 19. 20. 21.

(c) (d) (a) (b) (c) (b) (a)

22. 23. 24. 25. 26. 27. 28.

(a) (c) (b) (a) (a) (b) (d)

29. 30. 31. 32. 33. 34. 35.

(b) (d) (a) (d) (c) (a) (c)

36. 37. 38. 39. 40. 41. 42.

(d) (c) (c) (d) (a) (a) (c)

kgei snhe

(d) (c) (a) (b) (c) (d) (b)

43. 44. 45. 46. 47. 48. 49.

(b) (a) (a) (b) (d) (a) (d)

50. 51. 52. 53. 54. 55. 56.

(b) (c) (d) (c) (d) (d) (c)

57. 58. 59. 60. 61. 62. 63.

(c) (b) (b) (c) (c) (a) (c)

64. 65. 66. 67. 68.

(b) (c) (b) (b) (c)

ERna

1. 2. 3. 4. 5. 6. 7.

wwM wa.th Les aBryn

SOLUTION

1. (d) According to question



BPC

= 90° –

1 A 2

= 90° –

1 × 100° 2

A

o

B

Given:

C

= 80°

BAC

1 A 2 1 × 80 BOC = 90 + 2 = 130° 2. (c) According to question Given:

BOC

= 90° +

A

B

= 40° BPC 3. (a) According to question.

C

70°

B

C BAC = 70° BOC = ? 1  BOC = 90 ° + A 2

Given:

1 × 70°= 90° + 35° 2 BOC = 125°

4. (b) According to question

B

C

Given: A = 50° I is the incentre 1 A 2

BIC = 90 

1  50 2

BIC

= 90 

BIC

= 90° + 25°

BIC = 115° 5. (c) According to question

BOC = 90 °+ F

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B

C

As we know that

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A = 60°

1 unit =

1  BIC = 90° +  60 2

BIC = 120° 6. (d) According to question Given: ABC is an equilateral triangle

CD is the angle bisector of C



6

rin =

= 30°

=

12.(a)

BOC = 90° -

A ........(ii) 2

Value of  ROQ = 90 +

B

P 2

wwM wa.th Les B aryn 

C B D Given AB = BC = CA = 6 cm AG = IR = Circumradius = 2 units GD = Ir = Inradius = 1 unit AD = height = 3 units As we know that height of the

equilateral triangle is h =

3 a, 2

2



P 2

 P = 12° Therefore  RPQ = 12° 10. (d) According to the question, 

P

 QOR = 90º+

A 2

 QOR = 90º +

50º 2

 QOR = 90º + 25º = 115º

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A = 90° -

1 2

1 ( B + C) 2 A ...(ii)

From Eq. (i) = (ii)

O R

1 2

C)

1 1 ( B + C) = 90° 2 2





50º

Q

where'a' is the sides of a triangle 3 × 6 = 3 3 cm 2

6=

C

multiply both sides by

P

6

....(i)

A + B + C = 180°  A = 180º – ( B +

R

96 = 90 +

A

O

O 96° Q

1 2

A

x

1unit

AD =

C

Compare equation (i) and (ii) X = 90° 9. (a) According to the question,



C

M

BD  AM BDA = BDM = 90º It happens only in equilateral and isosceles triangle AD = DM i.e. AD = AM/2 Given DE || BC From thales theorem E will be mid point of AC.  AC = 12 cm. So, AE = 6 cm.

3 cm

2 3 8. (c) According to question Given:

2units

cm .

q

P

CAE = 15° 7. (b) According to question A

E

ABD = MBD =  (angle bisector)

a

R Enak geisnh

CAE  CEA  ECA = 180° (CE = CA)  2CAE = 180° – 150°

G

q B

2 3

BIC = 90° +

= 180° – 30° = 150°

In CAE

6

D

A + X ......(i) 2 As we know that

CAE  CEA ACD = 30°

2CAE

12

3

3

BIC 

AC = CE

 ECA

=

3

 GD = Ir = Alternate rin =

A

3 3

B



11. (d) According to the question,

3 units = 3 3

r



eeYa ridn agv .iSn i

= 180° A  B  C  A = 180° – 70° – 50°

13.(b)

BOC =

1 ( B + C) 2

BIC = 90° +

A = 90° + 30° 2

= 120° A + B + C = 180°  C = 180° - ( A+ B)  C = 180° - (80° + 60°) = 40°  2x = 40  x = 20°

14.(c)

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( C = 2x)

1 y = 90° + 2

*

1 A = 90° + (80) 2

= 130° 15.(c) According to the question A

This theorem apply for equilateral triangle because equilateral has also all sides equal (AB = BC = CA) 19.(c) Incentre:- meeting point of angle bisector. Every triangle:- lies incentre of a triangle in the interior

From Eq. (i) =180°–(90°–

C

4

32  4 2  5cm

O

1 A = area of  ABC =  4  3 2

B

= 6cm

C Equilareral triangle A

C

A

34–5 =1 2 16.(d) According to the question

Q

0

1 17.(a)  MAN = (  B –  C) 2

ERna

Right Angle

=

1 1 [  B +  C) = (180° -  A] 2 2

1 1 (65° - 30°) = (35°) 2 2

1 2

= 90° – C B Obtuse angle triangle

20.(b)

C

Since  1 =  2

( BD

= CD given)

 AB  AC ( the given  is isosceles)



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1 ( BAC) 2    BAC = (130 - 90)  2 = 80°

A 23.(c)

A

0

1 (180° - A) 2

1 A 2 

22.(a)  BOC = 90° +

0

1 2 1 2 B C A + B + C = 180° A + 2  1 + 2  2 = 180°

  1+  2 =

(ii)

  1 = 180° – 180 – B  C 2  

A

A

(i)

1  1 +  2 = 180° – 2 (B + C)  BOC = 180° – (  1 +  2)

C

= 17.5°

1 B 2 

Add both (i) and (ii)



B

wwM wa.th Les aBryn

1 1  BOC = 90 2 A  90 2 (40)  70

AB BD AB   1 AC CD AC

P

C

2

1  2 = 90° - 2  C

Scalene triangle

r=

D

2

similarly,

kgei snhe

B

AB + BC – CA 2

B

B 1 1

  1 = 90° -

Alternate:For right angle triangle

18.(b)

A

2  1 +  B = 180°

345 S= = 6cm 2  r = A/S = 6/6 = 1cm

r=

1 A 2 

21.(a)

eYrai dnag

AC =

= 90° +

1 A) 2 

v.iSn ir

B





A

3

=

1 A ...(i) 2  BOC = 180° - (  1 +  2 ) = 90° -

120° B



C

1  BOC = 90° + 2 (  BAC)  BAC = (120° - 90°)  2 = 60° 557

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A

B

D

From (i) and (ii) PB = PO  ABC is an acute angle or  ABC < 90°

 



C



AB BD  AC DC

1  ABC < 45° 2  POB =  PBO < 45°

×

×

 BPO > 90° Hence,  PBO is isosceles



P

A 2

25°

.

o

= 90° – 40° = 50°

A

Q

1 27.(b)  BOC = 90°+ (  BAC) 2   BAC = (102° - 90°)  2 = 24° A O

R

Given: QOR = 110°

32. (d) According to question Given:

OPR = 25° 'O' is the circumcentre then OP = OR = OQ  OPR = ORP = 25°

In OQR

wwM wa.th Les B aryn

 2ORQ = 180° – 110° (OQ = OR)

D B C In equilateral triangle centroid, incentre, orthocentre coincide at the same point. in-radius = OD = 3 cm

AD = 3cm  AD  9cm 3

= median.

A

B



C

Scalene Triangle

 OQR + ORQ + QOR = 180°

P

O

B

110°

R Enak geisnh

26.(a)  BPC = 90° -

29.(b)

×

30. (d) According to question

= 2:1



C

Obtuse angle triangle

 but not a right-angled triangle.

AB 5 = = AC 2.5



×

O

B

DC = BC – BD = 7.5 – 5 = 2.5

28.(d)

× ×

A

r

25.(a)

1 × 30° = 105° 2

eeYa ridn agv .iSn i

= 90° +

BD = Median at hypotenuse = Same as circumradius = Half of hypotenuse For an obtuse angled triangle, circumcentre lies outside the triangle

 PBO =  POB  OB is the bisector of  B. .....(ii)  PBO =  OBC



1 24.(b)  BOC = 90° + (  A) 2

 2ORQ = 70°  ORQ =

 OP || BC  POB =  OBC

C

....(i) (Alternate angle)

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C

A + B + C = 180° A = 180° – 40° – 70°

70 2

A = 70°As we know that

 ORQ = 35°  PRQ = PRO + ORQ PRQ = 60° 31. (a) Only right angled triangle where circumcentre lies on hypotenuse (side) For a right angled triangle, circumcentre is midpoint of hypotenuse and ircumradius is half of hypotenuse.

O



B C =40° B = 70° In ABC

 BOC = 2A (O is a circumcentre) BOC = 2 × 70° = 140° 33. (c) According to question

A (Mid point of AC) D

B C Right angle Triangle

B

C

 A =  B = C = 60°

BOC = 2A BOC = 2 × 60° BOC = 120°

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34. (a)

A

40.(a)

A

A Ci um rc re nt ce

O R

A

a

A

90°

B

C

Hence, BD =

wwM wa.th Les aBryn x

D 45° B

C

 BD= 42. (c)

1 (4 2)  2 2 units. 2 According to question

1 AC = 3cm 2

A 2 units

.

A

O 1 unit

C

OB = OC (circum radius) OBC = OCB = x (Let) then,  BOC = 180º – 2xº BOC 180 – 2x = 2 2

 BAC = 90 – x   OBC + BAC = 90º–x + x = 90º

36.(d) Perpendicular bisectors Draw a line (called a "perpendicular bisector") at rig ht angles to the midpoint of each side.Where all three lines inter sect is the centre of a triangle's "circumcircle", called the "circumcentre":

B

P

In  BOC

 BAC =

A

BD = AD =CD (mid-point of hypotenuse is circumcentre.)

D

39.(d)

O

B

41.(a)

12

10. = 5cm. 2

circum radius =

i.e. mid-point of hypotenuse.

C

= =4 3 3 3 38.(c) In the right-angled triangle the length of median to the hypotenuse is half the length of the hypotenuse.

180º-2x

x



a = 12cm

R=

6 2  82  10cm.

AC =

12 cm

B

90º- x

90°

B

C

Q

AQ2 = AC2 + QC2 .......(i) BP2 = BC2 + CP2 .......(ii) add (i) and (ii)  AQ2 + BP2 = (AC2 + BC2) + (QC2 + CP2)  



2

 BC   AC     AB2 +   2   2 

 AB2 +

2

D

= AB2 +

A

1 ( BC 2  AC 2 ) 4

1 AB 4

2



5 AB 4

C

Let AO = 2 units OD = 1 unit Given:AO = 10 cm  2 units = 10 cm 1 unit = 5 cm  OD = 5 cm 43. (b) According to question

2 units 10

G

 (AQ² + BP²)

1unit

2

 4(AQ2 + BP2) = 5 AB2

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C

6cm

B

cm 12

A

37. (c)

b=

 O is circumcentre So, OA = OB = OC = R (Radius)  BAC = 85º, BCA = 75º Then, [ABC = 180º – (BAC + BCA)] ABC = 180º – (85º + 75º)  ABC = 180º – 160º  ABC = 20º [Angle made by same chord at the centre is doubled than that of any other part of the circumference at same sector] Then,AOC = 20º×2 = 40º  OA = OC = R So, AOC = Isosceles triangle Then, OCA + OAC + AOC = 180º x + x +AOC = 180º 2x + 40º = 180º x = 70º Therefore OAC = 70º 35. (c)

C

B C

c=

75º

B

ERna kgei snhe

R

8cm

O

v.iSn ir

R

eYrai dnag

85º xº

B 10

D

C

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the equilateral triangle is =

3 2

3 units = 5 3 5 3

1 unit =

3

2 units =

3

10 3

AG =



5 3

3

×2=

10 3

9

cm

12 90º

10 3

B

AB a  3

=



10 3

cm

3

44. (a) According to question A

E

C

B

D Given: AD = BE = CF = median The n AB = BC = CA  The triangle is an equilateral triangle. 45. (a) According to question Given:A

D AD = 27 cm, DN = 12 cm As we know that AG = 2 units,

2 × BE 3

AG =

Centroid

C

C

B 2

 OE  2  OB 

=

1 4

=

B

3

Area of  ABC 1 of  ABC 3

Area of  ODE=

50.(b)

3 3 BG = 6 2 2

G

PR = PQ 2  QR 2  52  122  13cm



O is the centroid  QM is median and M is the midpoint of PR .



QM = PM =



centroid divides median in ratio 2:1



OQ =

C 2

By Pythagoras theorem,

=

1 2× × 60 = 20 cm2 6

13 2

2 QM 3

2 13 13 1    4 cm 3 2 3 3

51.(c)

A 2

[A Median divides a triangle in two equal parts]

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1

49.(d) We know that the centroid of a triangle divides each median in the ratio of 2 : 1  BG : BE = 2 : 3

A

1  Area of GBC = 2 ×  of ΔABC  6  



Area of ODE Area of BCO

= 9 cm.

2 AG = ×9  3 AG = 6 cm  In right angled triangle AB will be a hypotenuse Using pythagoras theorem (AB)² = (AG)² + (BG)²  (AB)² = 6² + 8²  AB = 10 cm  Therefore, length of AB = 10 cm. 47. (d)

Area of ODE Area of BCO

Area of  BCO =

 BE =

2 × AD 3

Area of  ABC = 60 cm

1unit

B

C

2 × 12 3 BG = 8

2units

N G

E O

Area of  ABC = 12 ×area of  ODE  ODE :  ABC 1 : 12

G

D

Then BG =

wwM wa.th Les B aryn

F

D

BG =



Median

E

R Enak geisnh

AG =

A

4

3

Alternate AG (rc) =

48. (a) In  ODE &  BCO

eeYa ridn agv .iSn i

i.e  AGB = 90º and  AGB will be a right angled triangle We know, In a triangle centroid divides the medians in 2 : 1 Ratio A

× 10 = 5 3



GD = 1 unit  AD = 3 units = 27 cm 3 units = 27 cm 1 unit = 9 cm  GD = 9 cm  GN = DN – GD = 12 – 9 = 3 cm 46. (b) Medians AD and BE intersect at G on 90º

r

Given: AB = BC = CA = 10 cm G = Centroid AG = 2 units GD = 1 unit  AD = 3 units = Height As we know that the height of

F 2 B

1

G 1 1 D

E 2 C

560

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52.(d)

AB2 + AC2 = 2(AD2 + BD2)  225 + 625 = 2(AD2 + 81)  AD2 = 344

A

G



Ang le s ubte nded on the ci rcum circ le i s half the ang le s ubte nded on the centre of circle

AD = 2 86 and GD =

2 2 AD =  12 = 8cm. 3 3

AG =

 GD =

A

53.(c)

BAD 

1 AD 3

A



90° B

G

C

D

AD = 9cm

1  9 = 3cm 3



BE = 6cm

2  6 = 4cm 3

 BG = 

2

BD =



2

3  4 = 5cm.

 

A

54.(d)

B C D AG = BC = 2x (let) GD = x ( centroid divides median in 2 : 1) Now in  BDG, BD = GD = x

 DBG =  BGD =  1 (let) Similarly in  DGC, CD = GD = x  DCG =  DGC =  2 (let)  BGC = 1   2 Now in  BGC =  1 +  2 +

ERna

30°

B

58.(b) BAC = 180° - BOC = 180° - 54° = 126° 59.(b) In  ABC,

BD = CD =AD  BAD = 30° Now in  ABD

BN  HN

Thus, the two altitudes HL and BN of  HBC, intersect at A. 60.(c) PR= PM 2 MR 2



 BGC = 90°



 AD = DC   DAC =  DCA =  (let)

I

x/2

D

Given:



ABC = x° BID = y°, BOD = z° 'I' is the incentre

55.(d)



ABI =

25cm G

B

D 18cm

C

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= C

 30°+  + 30° +  = 180°  2  = 120°   =  ACB = 60°

15cm

1 ABC 2 1 x ABI  x  2 2 BAD 

ar (  PQR) =

1 (PR) (PQ) 2

o z°

x/

2

y°

 A+  B +  C =180°

A

QR 2 – PR 2

A

 ABD = 30° ( BD = AD)

B

PQ =

 26 2 10 2 24cm

57. (c) According to question

Now, In  ABC

HL  BC and

 36 64 10cm

1   2 = 90°

wwM wa.th Les aBryn



 

C

D

x  z 2

z º  x º 2y º  =2 yº yº

(  1 +  2) = 180°



y° =

2yº = xº +zº Now,

kgei snhe

 GD =

z 2

x z  (Exterior angle) 2 2

 y° =

E

G

BAD =

2 86cm 3

56.(c)

1 BOD 2

v.iSn ir

C

D

eYrai dnag

B

1 BOD 2

61.(c)

1  10  24 = 120cm2 2 AB BD 3   AB   7 AC DC 4

21  5.25cm 4 A 62.(a) 

70° 50° B C D  A + 70° + 50° = 180°

  A = 60° 561

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AD is the bisector of  BAC.

 BAD =

1 1 BAC   60  30 2 2

63. (c) According to the question, AD is the external bisector of  CAE By using external bisector theorem.

 2  1 :

DB : AB =

33 – 3 2 2

66. (b)

3 2– 2





2

Circumradius(R) = r 2– 2 = = R 2

A

y

z

3 2 2

2 –1 : 1

68. (c) In  OBF and  OCE,

A

C

F

x

C

D

x

12 4  x  8 x 12x = 32 + 8x 4x = 32 x = 8

  OBF   OCE

A

wwM wa.th Les B aryn

C

B

ar of  ADE = 1, ar of  ABC= 2 ar of Δ ABC

ar of Δ ADE

=

1

S id e o f  A B C (A B )



x

x

C

2

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OB OF  or, OC OE

OB OC  OF OE

or, OB × OE = OF × OC

x = 45º

OF × OC = OB × OE

 A = 2 × 45 = 90º

= (BE – OE) (OE)

2

S id e o f Δ A D E (A D )

Hence,

3

x +x + 2x = 180º

2

BC =

C

 OFB =  OEC = 90º

 BYZ = 180– (30 + 115) = 35º

3

E

D

and  BOF =  EOC

2x

D 1 1

B

In  ZOY,

67. (b)

A

64. (b)

50 = 115º 2

R Enak geisnh (Property 2)

E

O

 ZOY =  BOC = 90 +

AB BD  AC CD

=

Inradius(r) =

2

65. (c) By property 12. AB + AC> 2AD 3 + 5 >2AD  AD PQ (d) PT = PQ 87. The length of the tangent drawn to a circle of radius 4 cm from a point 5 cm away from the centre of the circle is

wwM wa.th Les aBryn

73. Two circles of same radius 5 cm, intersect each other at A and B. If AB = 8 cm, then the distance between the centre is ; (a) 6 cm (b) 8 cm (c) 10 cm (d) 4 cm 74. AD is the chord of a circle with centre O and DOC is a line segment orginating from a point D on the circle and intersecting AB produced at C such that BC = OD. If BCD =20°, then AOD = ? (a) 20° (b) 30° (c) 40° (d) 60° 75. In a circle of radius 17 cm, two parallel chords of length 30 cm and 16 cm are drawn. If both chords are on the same side of the centre. then the distance between the chords is (a) 9 cm (b) 7 cm (c) 23 cm (d) 11 cm 76. Two circles touch each other internally. Their radii are 2 cm and 3 cm. The biggest chord of the greater circle which is outside the inner circle is of length

7

cm

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(b) 4 2 cm

(c) 5 2 cm (d) 3 2 cm 88. From a point P, two tangents PA and PB are drawn to a circle with centre O. If OP is equal to diameter of the circle, then  APB is (a) 45° (b) 90° (c) 30° (d) 60° 89. The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle and BD is a tangent to the smaller circle touching it at D and the bigger circle at E. Point A is joined to D. The length of AD is (a) 20 cm (b) 19 cm (c) 18 cm (d) 17 cm 90. PQ is a chord of length 8 cm of a circle with centre O and radius 5 cm. The tangents at P and Q intersect at a point T. The length of TP is

v.iSn ir

(c) 65

(c) 3

5

(a) 3 cm

eYrai dnag

65

(a) 6

kgei snhe

(a)

1 65 (b) 2 65 (d) 2

chord PB is 12 cm , the distance of the point N from the point B is

ERna

72. Chords AB and CD of a circle intersect at E and are perpendicular to each other. Segments AE, EB and ED are of lengths 2 cm, 6 cm and 3 cm respectively. Then the length of the diameter of the circle (in cm) is

(a)

20 cm 3

(b)

21 cm 4

(c)

10 cm 3

(d)

15 cm 4

91. The maximum number of common tangents drawn to two circles when both the circles touch each other externally is (a) 1 (b) 2 (c) 3 (d) 0 92. The radius of two concentric circles are 17 cm and 10 cm. A straight line ABCD intersects the larger circle at the point A and D and inte rsec ts the smaller circle at the points B and C. If BC = 12 cm, then the length of AD (in cm) is (a) 20 (b) 24 (c) 30 (d) 34 93. Two chords AB , CD of a circle with centre O intersect each other at P. ADP = 23° and APC = 70°, then the BCD is (a) 45° (b) 47° (c) 57° (d) 67° 94. In the following figure, AB is the diameter of a circle whose centre is O. If  AOE = 150°,  DAO = 51° then the measure of  CBE is : D E A

150° 51° O

(a) 115° (c) 105°

B

C

(b) 110° (d) 120°

585

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OM  2 11 cm, then radius of

O

15° M

Y

50° N

(a) 30° (b) 40° (c) 20° (d) 70° 101. In the adjoining figure 'O' is the centre of circle AC and BD are the two chords of circle which meeets at T outside the circle. OT bisects CD, OA = OB = 8cm and OT = 17cm. What isthe ratio of distance of AC and BD from the centre of the circle ? B

A

K

R

O

S

A

90

P

900 S

O

R

D

(a)

Q

B

3 2 r 2

(b)

E

r

C

F

 3 3   –  2  

(a)

r 2

(b)

r2  3 3   –  2 4 

(c)

r2  2 3   –  3 4 

(d) data insufficient 106. ABCD is a cyclic quadrilateral. The angle bisector of  A,

 B,  C and  D intersect at P, Q, R and S as shown in the figure. These four points form a quadrilateral PQRS. Quadrilateral PQRS is a : C R A

R O

B

o

60°

3 3r 2

2

D

60°

D

(c) (d) none of these 3r 104. ABC is an isosceles triangle and AC, BC are the tangents at M and N respectively. DE is the diameter of the circle.  ADP =  BEQ = 100°. What is the value of  PRD? P Q

A

A

2

C

0

=  OCB = 60°. What is the area of the shaded region? E D

B

(a) 8 cm2 (b) 10 cm2 2 (c) 12 cm (d) none of these 103. In the adjoining figure O isthe centre of the circle.  AOD = 120°. If the radius of the circle be 'r', then find the sum of the areas of quadrilaterals AODP and OBQC:

wwM wa.th Les B aryn

the circle is (a) 15 cm (b) 12 cm (c) 10 cm (d) 11 cm 100. In the given figure, ONY = 50°and OMY =15°. Then the value of the MON is

A

105. In the adjoining figure O is the centre of the circle with radius 'r' AB, CD and EF are the diameters of the circle.  OAF

eeYa ridn agv .iSn i

(c) 2 : 1 (d) 2 : 1 97. Each of the circles of equal radii with centres and B pass through the centre of one another circle they cut at C and D then DBC is equal to (a) 60° (b) 100° (c) 120°(d) 140° 98. The three equal circles touch each other externally. If the centres of these circles are A, B, C, then ABC is (a) a right angle triangle (b) an equilateral triangle (c) an isosceles triangle (d) a scalene triangle 99. ‘O’ is the centre of the circle, AB is a chord of the circle, OM  AB . If AB = 20 cm,

(a) 15 : 17 (b) 8 : 15 (c) 8 : 9 (d) none of these 102. In the adjoining figure O is the centre of the circle. The radius OP bisects a rectangle ABCD, at right angle. DM = NC = 2cm and AR = SB = 1cm and KS = 4cm and OP = 5cm. What is the area of the rectangle? P D M C L N

R Enak geisnh

95. The angle in a semi-circle is (a) a reflex angle (b) an obtuse angle (c) an acute angle (d) a right angle 96. The angle subtended by a chord at its centre is 60°, then ratio between chord and radius is (a) 1 : 2 (b) 1 : 1

Q

S

P B

(a) Square (b) rhombus (c) rectangle (d) cyclic quadrilateral 107. In the adjoining figure the diameter of the larger circle is 10 cm and the smaller circle touches internally the larger circle at P and passes through O, the centre of the larger circle at R and OR is equal to 4 cm. What is the length of the chord SP?

B

O

M

O

N

P R

S

C

D T

C

(a) 60° (b) 50° (c) 20° (d) Can't be determined.

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s

(a) 9cm

(b) 12 cm

(c) 6cm

(d)

8 2 cm

586

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111. ABCD is a square, in which a circle is inscribed touching all the sides of square. In the four corners of square 4 smaller circles of equal radii are drawn, containing maximum possible area. What is the ratio of the area of larger circle to that of sum of the areas of four smaller circles?

E 30° D

B

2 3 (c) 5 (d) can't be determined 110. In the adjoining figure ABCD is a rectangle in which length is twice of breadth. H and G divide the line CD into three equal parts. Similarly points E and F trisect the line AB. A circle PQRS is circumscribed by a square PQR S which pass es through the points E,F,G and H. What is the ratio of area of circle to that of area of rectangle? S

C

D

H

G

P

R

E

F B

A Q

(a) 3 : 7 (c) 25  : 72

B

3 ) : (2 +

(c) 12 : 17 2 (d) None of these 112. In the adjoining figure  ACE is a right angle there are three circles which just touch each other and AC and EC are the tangents to all the three circles. What is the ratio of the smallest circle?

(b) 3 : 4 (d) 32  : 115

A

E

D

B

3

3):2

(d) 2(1 + 3 ) : 2 3 : 3 114. Through any given set of four points A, B, C, D it is possible to draw:(a) atmost one circle (b) exactly one circle (c) exactly two circles (d) exactly three circles 115. The number of common tangents that can be drawn to two given circles intersect each other is :(a) one (b) two (c) three (d) four 116. The radius of a circle is 13 cm and the length of one of its chords is 10 cm. Find the distance of chord from the centre. (a) 8 cm (b) 10 cm (c) 9 cm (d) 12 cm 117. In the given figure O is the centre of the circle. If  BAC = 38°, then  OCD is D

A C

38° O

(a) 17 : 12 2 (b) 1 : (17 – 12 2 )

(c) 12 : 17 2 (d) none of these 113. In the adjoining figure three congruent circles ar touching each other. Triangle ABC circumscribes all the three circles. Triangles PQR is formed by joining the centres of the circle. There is a third triangle DEF. Points A, D, P and B, E, Q and C, F, R lie in the same straight respectively.

Rakesh Yadav Readers Publication Pvt. Ltd.

3):

eYrai dnag

(b) 1 : (17 – 12 2 )

wwM wa.th Les aBryn

4 5

B

kgei snhe

C

A (a) 1 : (68 – 48 2 )

ERna

BD = 6cm and CE = 5 3 cm, What is the ratio of AC : AD?

(b)

E

(c) 2(1+ 3 ) : (2 +

A (a) 4 cm (b) 6 cm (c) 5.6 cm (d) data insufficient 109. In the given figure ADEC is a cyclic quadrilateral, CE and AD are extende d to mee t at B.  CAD = 60° and  CBA = 30°.

3 4

Q

P

(b) 2 (4 +

D

(a)

R

(a) 3 2 : 2 2 : 1

O

60°

F

A

C

A

C

D

C

D

What is the ratio of perimeters of  ABC :  DEF :  PQR

v.iSn ir

108.A smaller circle touches internally to a larger circle at A and passes through the centre of the larger circle. O is the centre of the larger circle and BA, OA are of the diameters of the larger and smaller circles respectively. Chord AC intersects the smaller circle at a point D. If AC = 12 cm, then AD is : B

x° C

B

(a) 76° (c) 38° 118. In the given centre of the

(b) 52° (d) 19° figure, O is the circle. If  OBC =

20°, the  BAC:A

O B

C

587

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(a) 80° (b) 70° (c) 100° (d) 140° 119. In the given figure PQ = 12 cm, BQ = 8cm, then the length of chord AB:-

124. The value of x :-

Q

B

T

C

P

x

B

A

5 cm

B

(c) 4 cm (d) 18 cm 120. The value of x will be:A

9 cm

x°

85°

P

20° C

D

= 90°, then  APB will be:-

A

B

(a) 10 cm (b) 9 cm (c) 7.5 cm (d) 11 cm 126. In the given figure ABCD is a cyclic quadrilateral and O is the centre of the circle. If  BOC =

A

B

 BOC = 110°, then  ABC is B

A

O

x°

D

133. 45°

(a) 50° (c) 120°

O

x°

C

D

(a) 40° (b) 90° (c) 45° (d) 30° 128. If O is centre of the circle, then x is equal to

C

B

(b) 60° (d) 70°

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132.

A

B

120°

A

B

C

(a) 112° (b) 111° (c) 109° (d) None of these 127. In the given figure, O is the centre of the circle, then the value of x will be:-

C

(a) 75° (b) 60° (c) 80° (d) 70° 123. ABCD is a cyclic quadrilateral, then the value of x will be :-

P

138° D

wwM wa.th Les B aryn

(a) 90° (b) 60° (c) 45° (d) 30° 122. In the given figure, O is the centre of the circle  AOB = 90°,

A

(a) 40° (c) 39°

D

C

x

40° O

134.

135. B

(b) 45° (d) 35°

B

6 cm

O

B

7

A

A

O

A

x

(a) 16 cm (b) 9 cm (c) 12 cm (d) 7 cm 131. In the given figure, PA and PB are tangents from a point P to a circle such that PA = 6 cm and  APB = 60°. What is the length of the chord AB?

138°, then  BDC will be

P

T

12

R Enak geisnh

(b) 90° (d) 75° figure, O is the circle and  AOB

x

T

(a) 144° (b) 132° (c) 48° (d) 96° 130. Find the value of x in the given figure:

P

7 cm

D

(a) 70° (c) 60° 121. In the given centre of the

5 cm

C

B

A

r

(b) 4 5 cm

(a) 10 cm

(a) 2.2 cm (b) 1.6 cm (c) 3 cm (d) 2.6 cm 125. The value of x :-

D

P

eeYa ridn agv .iSn i

A

48°

6 cm

P

O

129. In the given figure,  ADB:-

(a) 12 cm (b) 8 cm (c) 9 cm (d) 6 cm ABC is a right angled triangle AB = 3 cm,BC = 5 cm and AC = 4 cm, then the inradius of the circle is (a) 1 cm (b) 1.25 cm (c) 1.5 cm (d) None of these T he nu m b e r o f c o m m o n tangents that can be drawn to two given circles is at the most (a) 1 (b) 2 (c) 3 (d) 4 Two circle of radii 12 cm and 7 cm touch each other internally. Find the distance between their centres. (a) 6 cm (b) 13 cm (c) 9 cm (d) 5 cm Three circ les touc h each other externally.The distance between their centres is 5 cm, 6 cm and 7 cm. Find radii of the circles:-

588

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(a) 2 cm, 3 cm, 4 cm (b) 3 cm, 4 cm, 1 cm (d) 1 cm, 2 cm, 4 cm (d) None of these 136. In the given figure, O is the centre of the circle and  ACB = 30°. Find  AOB.

gent is (a) 4 cm (b) 6 cm (c) 8 cm (d) 10 cm 142. In the given figure, AB and CD are two c ommon tangents to the two touching circle. If CD = 7 cm, then AB is equal to

146. In the given figure, ABCD is a cyclic quadrilateral and diagonals bisect each other at P. If  DBC = 60°, and  BAC = 30° then

 BCD is D

C P

C D

B

A

(a) 30° (b) 90° (c) 60° (d) 50° 137. In the given figure, AB = AC and  ABC = 50°, Find  BDC: A D

A

C

B

(a) 14 cm (b) 10.5 cm (c) 12 cm (d) None of these 143. O and O’ are the centres of two circles which touch each other externally at P. If AB is a common tangent. Find  APO.

(a) 60° (c) 100° 138. In the given centre of the

C

O

E

(b) 80° (d) 90° figure, O is the circle.  AOB =

70°, find  OCD. D

B

(a) 90° (b) 120° (c) 60° (d) data insufficient 144. In the given figure,  ABC is an equilate r a l tr iang le . Find  BEC.

wwM wa.th Les aBryn

A

70° O

B

C

(a) 70° (b) 55° (c) 65° (d) 110° 139. If the diagonals of a cyclic quadrilateral are equal, then the quadrilateral is (a) rhombus (b) square (c) rectangle (d) None of these 140. The quadrilateral formed by angle bisec tors of cyclic quadrilateral is a (a) rectangle (b) square (c) parallelogram (d) cyclic quadrilateral 141. The distance between the centres of equal c ircles each of radius 3 cm is 10 cm. The length of a transverse tan-

O'

P

ERna

B

kgei snhe

A

50°

B

(a) 90° (b) 60° (c) 80° (d) None of these 147. In the given figure, AD||BC, if  ABC = 72°, then  BCD = ? D

C

A 72°

B (a) 108° (b) 36° (c) 90° (d) 72° 148. In the given figure, O is the centre of the cir cle. PA and PB are tangents if  AOB :  APB = 5 : 1, then

 APB

A

A O

D

P B

C

B E

(a) 60° (c) 80°

(b) 120° (d) 90°

145. In the given figure,  AOC =

Rakesh Yadav Readers Publication Pvt. Ltd.

30°

eYrai dnag

A

v.iSn ir

60°

O

130°. Find  CBE, where O is the centre. D

C 130° O

A

(a) 130° (c) 115°

B

E

(b) 100° (d) 105°

(a) 150° (b) 30° (c) 60° (d) 90° 149. R and r are the radius of two circles (R>r). If the distance between the centre of the two circles be d, then length of com mon tang ent of two circles is : (a)

r 2 d 2

(b)

d 2 (R r )2

(c)

(R r )2 d 2

(d)

R 2 d 2

589

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O

A

B

C

(a) 5 (b) 6 (c) 3 (d) 4 One chord of a circle is known to be 10.1 cm. The radius of this circle must be: (a) 5 cm (b) greater than 5 cm (c) greater than or equal to 5 cm (d) less than 5 cm The length of two chords AB and AC of a circle are 8 cm and 6 cm and  BAC = 90°, then the radius of circle is : (a) 25 cm (b) 20 cm (c) 30 cm (d) 5 cm If a chord of length 16 cm is at a distance of 15 cm from the centre of the circle, then the length of the chord of the same circle which is at distance of 8 cm from the centre is equal to: (a) 10 cm (b) 20 cm (c) 30 cm (d) 40 cm PR is tangent to circle, with centre O and radius 4 cm, at

R Enak geisnh

160.

161.

162.

163.

Rakesh Yadav Readers Publication Pvt. Ltd.

point Q. If  POR = 90°, OR = 5

20 cm, then, in 3 cm, the length of PR is : 16 (a) 3 (b) 3 23 25 (c) (d) 3 3 cm and OP 

164. Circumcentre of  ABC is O. If  BAC = 85°,  BCA = 80°, then  AOC is : (a) 80° (b) 30° (c) 60° (d) 75° 165. If O is the circum centre of  ABC and  OBC = 35°, then

r

the  BAC is equal to : (a) 55° (b) 110° (c) 70° (d) 35°

eeYa ridn agv .iSn i

157. The radius of a circle is 13 cm and XY is a chord which is at a distance of 12 cm from the centre. The length of the chord is : (a) 15 cm (b) 12 cm (c) 10 cm (d) 20 cm 158. SR is a direct common tangent to the circles of radii 8 cm and 3 cm respectively, their centres being 13 cm apart. If the points S and R are the respective points of intersect, then the length of SR is : (a) 12 cm (b) 11 cm (c) 17 cm (d) 10 cm 159. In the following figure, if OA = 10 and AC = 16, then OB must be :

wwM wa.th Les B aryn

150. Two circles of radii 8cm and 2 cm respectively touch each other externally at the point A. PQ is the dir ect comm on tangent of those two circles of centres O1 and O2 respectively. Then length of QP is equal to : (a) 2 cm (b) 3 cm (c) 4 cm (d) 8 cm 151. PQ is a direct common tangent of two circles of radii r 1 and r2 touching each other externally at A. Then the value of PQ2 is : (a) r1r2 (b) 2r1r2 (c) 3r1r2 (d) 4r1r2 152. Two circles with radii 5 cm and 8 cm touch each other externally at a point A. If a straight line through the point A cuts the circles at points P and Q respectively, then AP : AQ is : (a) 8 : 5 (b) 5 : 8 (c) 3 : 4 (d) 4 : 5 153. The radius of a circle is 6 cm. An external point is at a distance of 10 cm from the centre. Then the length of the tangent drawn to the circle from the external point upto the point of contact is : (a) 8 cm (b) 10 cm (c) 6 cm (d) 12 cm 154. A triangle is inscribed in a circle and the diameter of the circle is its one side. Then the triangle will be : (a) right-angled (b) obtuse-angled (c) equilateral (d) a square 155. Two circles of radii 4 cm and 9 cm respectively touch each other externally at a point and a common tangent touches them at the points P and Q respectively. Then the area of square with one side PQ, is : (a) 97 sq.cm (b) 194 sq.cm (c) 72 sq.cm (d) 144 sq.cm 156. The length of the chord of a circle is 8 cm and perpendicular distance between centre and the chord is 3 cm. Then the radius of the circle is equal to : (a) 4 cm (b) 5 cm (c) 6 cm (d) 8 cm

166. If I is the incentre of  ABC and  BI C

=

135°,

then

the  ABC is : (a) Acute angled (b) equilateral (c) right angled (d) obtuse angled 167. If S is the circum centre of  ABC and  A = 50°, then the

value of  BCS is : (a) 20° (b) 40° (c) 60° (d) 80° 168. The distance between the centres of two equal circles, each of radius 3 cm, is 10 cm. The length of a transverse common tangent is : (a) 8 cm (b) 10 cm (c) 4 cm (d) 6 cm 169. A unique circle can always be drawn t hrough x number of given non-collinear points, then x must be : (a) 2 (b) 3 (c) 4 (d) 1 170. The length of radius of a circumcircle of a triangle having sides 3cm, 4cm and 5cm is: (a) 2 cm (b) 2.5 cm (c) 3 cm (d) 1.5 cm 171. AB and CD are two parallel chords of a circle such that AB = 6cm and CD=8cm. If the chords lie on the same side of centre O and radius 5cm the distance between AB and CD is : (a) 2 cm (b) 1 cm (c) 2.5 cm (d) 3 cm 172. In the given figure PKQ is a tangent and LN is the diameter of the circles. If  KLN=30º then

 PKL will be :

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P

0  POR = 120 then  QPO will be:

K Q N

O

(a) 30° (b) 50° (c) 60° (d) 70° 173. In the giv en figure  ADC 120º and AOB is the diameter of the c ircle, then  BAC :

(b) 30° (d) 50°

178. In the given figure  ABC = 55º, the  CDT is:

O

35

T

30

1

0

D

0

A

wwM wa.th Les aBryn

(a) 60° (b) 80° (c) 100° (d) 120° 175. AB and CD are two parallel chords of a circle such that AB =10cm and CD = 24cm, If the chords are on the opposite sides of the centre and the distance b etwe en them is 17cm, then the radius of the circle is : (a) 8 cm (b) 15 cm (c) 11 cm (d) 13 cm 176. In the given figure, O is the centre of the circle then ACB will be : C

0

25

B

30

C

0

70

O

(a) 40° (b) 50° 180.  QSR is :-

S

0

B

(c) 35°(d) 90°

Q

O

60

0

Rakesh Yadav Readers Publication Pvt. Ltd.

0

20

B

O

x

30

D 30

0

C

0

O 0

10 0

0

x

C

(a) 400 (b) 450 (c) 500 (d) 600 184. If a circle is provided with a measure of 190 on centre, is it possible to divide the circle into 360 equal parts ? (a) Never (b) Pos sible when one measure of 200 is given (c) Always (d) Pos sible when one measure of 210 is given 185. I n the ad joining fig ur e  A=6 0ºand  ABC = 8 0º,  BQC = ? P D

A

B

P

(a) 550 (b) 900 0 (c) 70 (d) 600 181. Find the value of  x in the given figure :-

(a) 105° (b) 230° (c) 115° (d) 100° 177. In the given figure, ROQ is the diamete r of the cir cle. If

B

R

O

A

T

D

(c) 15° (b) 20° (c) 25° (d) 30° 179. In the given figure if AB is the diameter of the circle, then  ACD will be :

C

A

C

kgei snhe

B

A

0

(a) 300 (b) 600 (c) 500 (d) 700 183. In the following figure, find the value of x

eYrai dnag

B

O

30

A

B

(a) 30° (b) 40° (c) 50° (d) 60° 174. OAB = 25°, OCB = 35º then  AOC will be :

A

(a) 40° (c) 60°

ERna

A

0

P R

C

D

T

P

v.iSn ir

L

25

Q

(a) 1200 (b) 1300 0 (c) 110 (d) 1000 182. In the given figure AB is the diam eter of the circle and  PAB = 30º, Find  TPA

A

C B

Q

(a) 400 (b) 800 (c) 200 (d) 300 186. Two circles of radius 37cm and 20cm intersect each other at A and B.O and O' are the centres of the circles. If the length of AB is 24cm , then OO':(a) 50cm (b) 51cm (c) 40cm (d) 57cm

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187. Two circles of radius 4cm and 6c m touch each other internally. Find the longest chord of the bigger cir cle which is outside of the smaller circle?

ADC=1300 find CAB .

C

Y

r

A

0

B

M

197. In the given figure,  CAB = 30º and  AKB = 115º, find  KCD :D

(a) 5 r

(b)

(c) 3 r

(d)

5

2

x  y

A

r

194. In the given figure, O is the centre of the c ircle. Then is equal to-

P

O

100

0

A

C

B

(a) 650 (b) 350 (c) 400 (d) 720 198. In the given figure,the chords AC and BC are equal. The radius OC intersect AB at M then AM : BM :-

O

(b)

(a) 1:1

Z 2

(c) Z (d) None of these 195. In the figure , AB C is a triangle in which AB = AC. A circle through B touches AC at D and intersects AB at P. If D is the mid-point of AC, Find the value of AB :A

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(a)

(a) 2AP (b) 3AP (c) 4AP (d) None of these

2 :3

3 :2

(b)

3 :1

5 :1 (d) None of these 200. ABCD is a cyclic trapezium whose sides AD and BC are parallel to each other; if ABC = 75° then the measure of BCD is: (a) 75° (b) 95° (c) 45° (d)105° (c)

D B

(b)

(c) 3: 2 (d) None of these 199. If two circles are such that the centre of one lies on the circumference of the other, then the ratio of the common chord of two circles to the radius of any of the circles is :-

P

B

D

B

C

(a) 2 Z

(a) 1400 (b) 500 0 (c) 40 (d) 1300 192. In the given figure, TP and TQ are tangents to the circle. If  PAQ = 50º,what is  PTQ ?

M

A

R

0

P

0

T

Q

(a) 40 (b) 50 (c) 300 (d) 1300 191. In the given figure, O is the centre of the circle find  CBD

30

115 0

x

B

0

C

K

5 r

y

O

D

(a) 120° (b) 90° (c) 360° (d) None of these

D

wwM wa.th Les B aryn A

B C

M

0

13

A

X

S

C D

Q

(a) 800 (b) 700 0 (c) 100 (d) 900 193. Two circle s X and Y with centres A and B intersect at C and D. If Area of circle X is 4 time area of circle Y, then AB = ?

r

(c) 6 2 cm (d) 3 2 cm 188. In a circle of radius 17cm two parallel chords are present on the opp osite s ide of the diamete r. I f the distance between them is 23cm and the length of one chord is 16cm then the length of other chord is:(a) 15 cm (b) 20cm (c) 18 cm (d) 30cm 189. AB is a chord of the circle (centre O). P is a point on the circle such that OP  AB a n d OP intersect AB at point M. If AB = 8cm, MP = 2cm then radius (r):(a) 7 cm (b) 5cm (c) 6 cm (d) 4cm 190. In the given figure, ABCD is a cyclic quadrilateral whose side AB is a diameter of the circle thr ough A,B and C. If

196. In the given figure, CD is a direct common tangent to two cir cles interse cting each other at A and B, then  CAD + CBD = ?

eeYa ridn agv .iSn i

(b) 4 2 cm

T

50°

A

R Enak geisnh

(a) 8 2 cm

P

C

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APC  80 0 and  ADP = 30º ,

C

207. A

B

(a) BD = AC (b) AB + BC = CD + AD (c) AB + BC = AC (d) AB + CD = BC + AD 202. In the given figure, Tangents TQ and TP are drawn to the larg er cir cle ce ntre O and tangents TP and TR are drawn to the smaller circle (centre O'). Find TQ : TR :-

208.

T Q

P R

O

O

209.

ang le b isector of

AOC ,

0

COD  60 . Find ABC :-

O

A

60

0

D

C

(a) 1200 (b) 600 0 (c) 30 (d) 900 204. If O is the centre of the circle and PA and PB are two tangents drawn from a point P on the circumference of the circle. If APB  68 0 , then POA = ? (a) 680 (b) 340 0 (c) 56 (d) 900 205. In a circle, AB is the diameter of the circle, and CD is a chord such that CD||AB .P is any point on the circle such that

and  CAT = 44°. The angle subtended by BC at the centre of the circle is: (a) 84° (b) 92° (c) 96° (d) 104° 215. BC is the chord of a circle with centre O. A is a point on major arc BC as shown in the figure . What is the value of  BAC +  OBC? A

O

110

C

B

0

50

B

O

E

(a) 1200 (b) 1000 (c) 1150 (d) None of these 211. AB is the diameter of a circle whose center is O and CD is a chord in the circ le and

1 AB. AC and BD on 2 producing meet at P. Find APB ? (a) 300 (b) 400 (c) 500 (d) 600 212. In the given figure, AB is the diameter of the circle and O is the centre, Find APB ? CD=

A C O

BPC  48 0 , then BCD  ? (a) 480 (b) 420

BC produced at T,  BTA = 40°

A

wwM wa.th Les aBryn

B

210.

ERna

(a) 1 : 1 (b) 5 : 4 (c) 8 : 7 (d) 7 : 8 203. 'O' is the centre of the circle, line segment  BOD is the

then BCD  ? (a) 300 (b) 800 0 (c) 100 (d) 500 AB CD is a cyclic quadr ilateral. Side AB and DC when produced meet at P and side AD and BC when Produced meet at Q. If  APD = 40º,  ADC = 85º, then  AQB is equal to :(a) 300 (b) 400 0 (c) 50 (d) 550 AB is the d iame ter of the circle with centre O. DC is a chord such that DC||AB. If  BAC = 20º, then  ADC is equal to :(a) 1000 (b) 900 0 (c) 110 (d) 1200 In question above, find  COD? (a) 500 (b) 1000 0 (c) 25 (d) 900 Find  O B E ?

eYrai dnag

D

(a) 1490 (b) 74.50 (c) 620 (d) None of these 213. O is the circum centre of the triangle ABC with circumradius 13 cm. Let BC = 24 cm and OD is perpendicular to BC. Then the length of OD is: (a) 7 cm (b) 3 cm (c) 4 cm (d) 5 cm 214. A, B, C are three points on a circle. The tangent at A meets

v.iSn ir

206. AB and CD are two chords of a circle intersect at a point P. If

kgei snhe

201. A c ircle touche s a quadrilateral ABCD. Find the true statement :-

(c) 240 (d) 960

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31

0

P D

B

C

(a) 120° (b) 60° (c) 90° (d) 180° 216. AB and CD are two parallel chords drawn on two opposite sides of their parallel diameter such that AB = 6 cm, CD = 8 cm. If the radius of the circle is 5 cm, the distance between the chords, in cm, is : (a) 2 (b) 7 (c) 5 (d) 3 217. A chord AB of length 3 2 unit subtends a right angle at the centre O of a circle. Area of the sector AOB (in sq. units) is: (a)

9 sq. units 4

(b) 5sq. units (c) 9sq. units (d)

9 sq. units 2

218. AB and BC are two chords of a circle with centre O. If P and

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D

C

(a) 12 

(b)

36 

(c) 12 (d) 18 220. Two tangents are drawn from a point P to a circle at A and B. O is the centre of the circle. If  AOP = 60°, then  APB is : (a) 120° (b) 90° (c) 60° (d) 30° 221. If the length of a chord of a circle, which makes an angle 45° with the tangent drawn at one end point of the chord, is 6cm, then the radius of the circle is : (b) 5 cm

(c) 3 2 cm

(d) 6 cm

(a) 2 2 cm

222. Two equal c ircles p ass through each other’s centre. If the radius of each circle is 5 cm, what is the length of the common chord? (a) 5

(b) 5 3

5 3 2 223. PA and PB are two tangents drawn from external point P to a circle with centre O where the points A and B are the points of contact. The quadrilateral OAPB must be: (a) a rectangle(b) a rhombus (c) a square (d) concyclic 224. The radius of two concentric circles are 9 cm and 15 cm. If the chord of the greater circle (c) 10 3

(d)

(b) 4 3 cm

(c) 2 3 cm (d) 8 cm 229. ABCD is a cyclic parallelogram. The  B is equal to : (a) 30° (b) 60° (c) 45° (d) 90° 230. From four corners of a square sheet of side 4 cm, four pieces, each in the shape of arc of a circle with radius 2 cm, are cut out. The area of the remaining portion is : (a) (8 -  )sq.cm. (b) (16 - 4  )sq.cm. (c) (16- 8  )sq.cm. (d) (4 - 2  ) sq.cm. 231. If a chord of a circle of radius 5 cm is a tangent to a circle of r ad ius 3 c m , b oth the c ir c le s b e ing c onc e ntr ic , then the length of the chord is : (a) 10 cm (b) 12.5 cm (c) 8 cm (d) 7 cm 232. Two circles touch each other

wwM wa.th Les B aryn

(a) 6 2 cm

r

B

externally at point A and PQ is a dire ct c omm on tange nt which touches at P and Q respectively. Then  PAQ = ? (a) 45° (b) 90° (c) 80° (d) 100° 233. AB is a chord to a circle and PAT is the tangent to the circle at A. If  BAT = 75° and  BAC = 45°, C being a point on the circle, then  ABC is equal to : (a) 40° (b) 45° (c) 60° (d) 70° 234. O is the centre of a circle and arc ABC subtends an angle of 130° at O. AB is extended to P. Then  PBC is : (a) 75° (b) 70° (c) 65° (d) 80° 235. The circumcentre of a triangle ABC is O. If  BAC = 8 5° and  BCA = 75°, then the value of  OAC is: (a) 40° (b) 60° (c) 70° (d) 90° 236. The length of each side of an

eeYa ridn agv .iSn i

A

be a tangent to the smaller circle, then the length of that chord is : (a) 24 cm (b) 12 cm (c) 30 cm (d) 18 cm 225. The length of a chord of a circle is equal to the radius of the circle. The angle which this chord subtends in the major segment of the circle is equal to : (a) 30° (b) 45° (c) 60° (d) 90° 226. The ratio of the areas of the circumcircle and the incircle of an equilateral triangle is: (a) 2 : 1 (b) 4 : 1 (c) 8 : 1 (d) 3 : 2 227. AB = 8 cm and CD = 6 cm are two parallel chords on the same side of the centre of a circle. The distance between them is 1 cm. The radius of the circle is : (a) 5 cm (b) 4 cm (c) 3 cm (d) 2 cm 228. Two equal circles of radius 4 cm intersect each other such that each passes through the centre of the other. The length of the common chord is :

R Enak geisnh

Q are the mid-points of AB and BC resp ectively , then the quadrilateral OQBP must be: (a) a rhombus (b) concyclic (c) a rectangle (d) a square 219. If the area of the circle in the figure is 36 sq. cm, and ABCD is a square, then the area of  ACD, in sq. cm, is :

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equilateral triangle is 14 3 cm. The area of the incircle, in cm2, is : (a) 450 (b) 308 (c) 154 (d) 77 237. If the radii of two circles be 6 cm and 3 cm and the length of the transverse common tangent be 8 cm, then the distance between the two centres is : (a)

154 cm

(b)

(c)

145 cm

(d)

140 cm

135 cm 238. Two parallel chords are drawn in a circle of diameter 30 cm. The length of one chord is 24 cm and the distance between the two chords in 21 cm. The length of the other chord is : (a) 10 cm (b) 18 cm (c) 12 cm (d) 16 cm 239. If two equal circles whose centres are O and O’ , intersect each other at the points A and B, OO’ = 12 cm and AB = 16cm, then the radius of the circles is : (a) 10 cm (b) 8 cm (c) 12 cm (d) 14 cm

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(a)

2 1

(b)

E C

F

D

(a) (b) (c) (d)

AD 2AD 3AD 4AD

= AB = AB = AB = AB

+ + + +

BC BC BC BC

+ + + +

CA CA CA CA

241. PP1 and QQ1 are two direct com mon tang ents to two circles intersecting at points A and B. The common chord on produced intersects PP1 in R and QQ1 in S. Which of the following is true ? P

R

P

1

A

1

(a)

a2 b2 c2 2

(b)

a 2  b2  c2

a2  b2  c2 2 (d) None of these 247. AB and CD are two chords of a circle which intersect at right angle at E. If AE = 2cm, EB = 6cm, ED = 3cm, then radius (r) is equal to:(c)

wwM wa.th Les aBryn

(a) RA² + BS² =AB² (b) RS² =PP'2+ AB² (c) RS² + PP'2 =QQ'2 (d) RS2 = BS2+PP'2 242. If two equal circles of radius 5cm have two common tangent AB and CD which touch the circle on A,C, and B,D respectively and if CD=24cm, find the length of AB.

ERna

S

Q

A

B

C

(a)

D

(a) 27cm (b) 25cm (c) 26 cm (d) 30cm 243. If ABC is a Quarter Circle and a circle is inscribed in it and if AB=1 cm, find rad ius of smaller circle. A

B

C

C B

(d) 1  2 2

244. Find the length of the common chord of two circles of radius 15c m and 20 cm if their centres are 25cm apart? (a) 12cm (b) 20cm (c) 18cm (d) 24cm 245. AB and AC are two chords of a circle such that AB=AC=6cm. If radius of the circle is 5cm, then BC is:(a) 4.8cm (b) 9.6cm (c) 2.4cm (d) 8.4cm 246. '2a' and '2b' are the length of two chords which intersect at right angle. If the distance between the centre of the circle and the intersecting point of the chords is 'C' then the radius of the circle is:-

B Q

2

O P

A

(a) 7cm (b) 5cm (c) 4cm (d) 6cm 250. Two tangents PA and PB are drawn to the circle(centre O) from a point P. CD is another tangent on the circle which intersects PA and PB at C and D respectively. If APB = 34° then COD:(a) 146° (b) 68° (c) 73° (d) None of these 251. Two tangents PA and PB are drawn from a point P to the circle. If the radius of the circle is 5 cm and AB=6cm and O is the centre of the circle. OP cuts AB at C and OC = 4cm, then OP:-

eYrai dnag

B

2 1

2

kgei snhe

A

(c)

2 1

v.iSn ir

240. In the given figure, AD, AE and BC are tangents, then:-

65 2

(b)

65

(c) 2 65 (d) None of these 248. AB is a chord of a circle (centre O) and DOC is a line segment originating from a point D on the circle and intersecting AB on producing at C such that

25 cm (b) 25cm 4 (c) 13 cm (d) None of these 252. If in the given figure, AB=a, AC=4cm, while O is the centre of the circle and D is a point between O and B such that AD  BC . Find the length of OD. (a)

A 4

a B

O

D

(a)

4 a 4

(b)

(c)

4a  16 16a  a 2

(d)

16  a 2 2 a 2  16 2 a 2  16 16  a 2

253. In the given figure, PQ = 24cm. M is the mid-point of QR. Also, MN  PR , QS = 7cm and TR = 21cm, then SN = ?

BC=OD. If  BCD = 20°, then

 AOD :(a) 30° (b) 40° (c) 100° (d) 60° 249. In the given figure, O is the centre of the circle. If BA = 7cm, OP = 13cm & AP = 9cm then radius (r):-

Rakesh Yadav Readers Publication Pvt. Ltd.

C

24 P

(a) 50 cm (c) 31 cm

Q 7 S

M

R 21

N

T

(b) 12.5cm (d) 25 cm

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C

D d

c

a

b B

A

(a) 179 (b) 89 (c) 357 (d) 358 255. Three equal circle of unit radius touch each other. Then , the area of the circ le circum scribing the thr ee circle is :-

259. ABC and MNC are two secants of a circle whose centre is O. AN is the d iame ter of the cir cle if  BAN = 38 ° and  ACM = 20° then  MBN :(a) 38° (b) 42° (c) 28° (d) 32° 260. PT is a tangent of a circle at T and AB is a chord. If AB =18cm and PT = 2AP then find PT ? (a) 12cm (c) 6 cm C A

(a) 6  (2  3 )2

 (2  3 )2 6

(c)

 (2  3 )2 3

D

(a) 220° (b) 110° (c) 55° (d) 70° 262. In the given figure, AB = 27, AD = 38cm, ED = 24cm and  E = 90°, then radius of the circle is equal to :-

x

C

O

B

(a) 11 cm (b) 15 cm (c) 13 cm (d) 17 cm 263. Two cir cles hav ing radius 'a'cm and 'b'cm touch each other e xter nally. another circle whose radius is 'c'cm, touches both the circles and also their common tangent. Then which statement will be true :-

D

z

P

B

33

(a) 230° (b) 237° (c) 337° (d) None of these

C

A

F

(a)

a b  c

(b)

a  b c

(c)

a b  bc  ac

(d)

1 1 1   a b c

O

A

D

E

wwM wa.th Les B aryn

256. In  ABC, AB=4cm, BC=3.4cm and AC=2.2cm. Three circles are drawn with centre A, B and C in such a way that each circle touches the other two. Then the diameter of the bigger circle is: (a) 5.2 cm (b) 2.6 cm (c) 2.8 cm (d) None of these 257. The angle bisectors of angle A, B and C of a  ABC intersect the circum fere nce of the circumcircle at X, Y and Z respectively. If A = 50°, CZY = 42°, then BYZ is equal to :(a) 46° (b) 42° (c) 23° (d) 21° 258. In the given figure, chord BE = BD, CBD = 33°, & OB  AC then x+y+z is equal to

y

I O

110

(d) 3  (2  3 )2

E

O

B

R Enak geisnh

(b)

(b) 18cm (d) 9cm

261. Find  BO'D ?

264. In a  ABC, I and O are the ince ntre and cir cum-centre respectively. The line AI is produced to a point D on the circumcircle. If  BOD = Z,

Rakesh Yadav Readers Publication Pvt. Ltd.

 BID = Y and  ABC = X, then

xz 3y

is equal to :-

2 3 4 (c) 3

(a)

(b)

1 3

(d) 1

265. P and Q are the middle points of two chords (not diameters) AB and AC respectively of a circle with centre at a point O. The lines OP and OQ are produced to meet the circle respectively at the points R and S. T is any point on the major arc between the points R and S of the cir cle. If  BAC = 32°,  RTS = ? (a) 32° (b) 74° (c) 106° (d) 64° 266. O and C are respectively the orthocentre and circumcentre of an acute-angled triangle PQR. The points P and O are joined and produced to meet the side QR at S. If  PQS = 60° and  QCR = 130°, then  RPS = ? (a) 30° (b) 35° (c) 100°(d) 60° 267. Two chords AB and CD of circle whose centre is O, meet at the point P and  AOC = 50°,  BOD = 40°. Then the value of  BPD is : (a) 60° (b) 40° (c) 45° (d) 75° 268. The tangents are drawn at the extremities of a diameter AB of a circle with centre P. If a tangent to the circle at the point C intersects the other two tangents at Q and R, then the measure of

r

In the given figure, AB || CD if a,b,c and d are integers, what is the number of possible value of (a+b–c–d)?

eeYa ridn agv .iSn i

254.

the  QPR is : (a) 45° (b) 60° (c) 90° (d) 180° 269. Two circles of radii 9 cm and 2 cm respectively have centres X and Y and XY = 17cm. Circle of radius r cm with centre Z touches two given circles externally. If  XZY = 90°, find r : (a) 18 cm (b) 3 cm (c) 12 cm (d) 6 cm 270. A circle (with centre at O) is touching two intersecting lines AX and BY. The two points of contact A and B subtend an angle of 65° at any point C on the circumference of the

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AD  BC ,

if

ABC  70 , then the value of

 BCD is : (a) 60° (b) 70° (c) 40° (d) 80° 272. ABCD is a cyclic trapezium whose sides AD and BC are parallel to each other. If  ABC = 72°, then the measure of the  BCD is (a) 162° (b) 18° (c) 108° (d) 72° 273. If an exterior angle of a cyclic quadrilateral be 50°, then the interior opposite angle is : (a) 130° (b) 40° (c) 50° (d) 90° 274. ABCD is cyclic parallelogram. The angle B is equal to : (a) 30° (b) 60° (c) 45° (d) 90° 275. ABCD is a cyclic quadrilateral and O is the centre of the circle. If COD = 140° and  BAC =

30°, then  BCD is : (a) 75° (b) 90° (c) 120°(d) 60° 277. ABCD is a cyclic trapezium

with AB  DC and AB is a diam eter of the circ le. If CAB  30 , then  ADC is (a) 60° (b) 120° (c) 150°(d) 30° 278. ABCD is a cyclic quadrilateral. AB and DC are produced to mee t at P. I f  ADC = 70 ° and DAB  60 , the n  PBC +  PCB is (a) 130° (b) 150° (c) 155° (d) 180°

(b)

 2

(c) 2 (d) 3

280. A quadrilateral ABCD circumscribes a circle and AB = 6 cm, CD = 5 cm and AD = 7 cm. The length of side BC is (a) 4 cm (b) 5 cm (c) 3 cm (d) 6 cm 281. In a cyclic quadrilateral ABCD, A + B + C + D = ? (a) 90° (b) 360° (c) 180° (d) 120° 282. A square ABCD is inscribed in a circle of 1 unit radius. Semicircles are inscribed on each side of the square. The area of the region bounded by the four semi-circles and the circle is (a) 1 sq. unit (b) 2 sq. unit (c) 1.5 sq. unit (d) 2.5 sq. unit 283. All sides of a quadrilateral ABCD touch a circle, If AB = 6 cm, BC = 7.5 cm, CD = 3 cm, then DA is (a) 3.5 cm (b) 4.5 cm (c) 2.5 cm (d) 1.5 cm 284. ABCD is a cyclic quadrilateral. The side AB is extended to E in such a way that BE = BC, If ADC = 70° , BAD = 95°, then

wwM wa.th Les aBryn

40°, then the value of  BCD is equal to (a) 70° (b) 90° (c) 60° (d) 80° 276. ABCD is a quadrilateral inscribed in a circle with centre O. If COD = 120° and  BAC =

(a) 

in whic h  A  4 x , ,  C  5 y  ,  D  y , then B  7x x : y is (a) 3 : 4 (b) 4 : 3 (c) 5 : 4 (d) 4 : 5 287. ABCD is a cyclic quadrilateral and AD is a diameter. If DAC = 55°, then value of ABC is (a) 55° (b) 35° (c) 145° (d)125° 288. A square is inscribed in a quarter-circle in such a manner that two of its adjacent vertices lie on the two radii at an equal distance from the centre, while the other two vertices lie on the circular arc. If the square has sides of length x. then the radius of the circle is:

v.iSn ir

that

eral

eYrai dnag

suc h

angle  ABC = ?

kgei snhe

the measure of  APO is: (a) 25° (b) 65° (c) 90° (d) 40° 271. ABCD is a cyclic trapezium

279. A cyclic quadrilateral ABCD is such that AB = BC , AD = DC, AC  BD , CAD   , then the

ERna

circle. If P is the point of intersection of the two lines, then

DCE is equal to

(a) 140° (b) 120° (c) 165° (d) 110° 285. In a cyclic quadrilateral A  C  B  D = ?

the

D

C O ´

A

B

(a) 270° (b) 360° (c) 90°(d) 180° 286. If ABCD be a cyclic quadrilat-

Rakesh Yadav Readers Publication Pvt. Ltd.

(a)

(c)

2x

16 x

(b)

 4

5x

(d)

2

x

2x

289. ABC is a cyclic triangle and the bisectors of BAC , ABC and BCA meet the circle at P, Q and R respectively. Then the angle RQP is : (a) 90° –

(c) 90° –

B 2

A 2

(b) 90° +

(d) 90 +

C 2

B 2

290. ABCD is a cyclic quadrilateral. AB and DC when produced meet at P, If PA = 8 cm, PB = 6, PC = 4 cm, then the length (in cm) of PD is (a) 10 cm (b) 6 cm (c) 12 cm (d) 8 cm 291. AB is a diameter of a circle with centre O. The tangents at C meets AB produced at Q. If CAB = 34°, then meas ure of CBA is (a) 56° (b) 68° (c) 34°(d) 124°

597

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ANSWER KEY 61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. 80. 81. 82. 83. 84. 85. 86. 87. 88. 89. 90.

(d) (a) (b) (a) (a) (c) (b) (c) (c) (c) (d) (a) (a) (d) (b) (d) (b) (b) (d) (d) (d) (a) (b) (c) (d) (d) (a) (d) (b) (a)

91. (c) 92. (c) 93. (b) 94. (c) 95. (d) 96. (b) 97. (c) 98. (b) 99. (b) 100. (d) 101.(d) 102.(b) 103.(c) 104.(c) 105.(a) 106.(d) 107.(c) 108.(b) 109.(a) 110.(c) 111.(a) 112.(* ) 113.(c) 114.(a) 115.(b) 116.(d) 117.(c) 118.(b) 119.(a) 120.(d)

121.(c) 122.(c) 123.(b) 124.(a) 125.(d) 126.(b) 127.(c) 128.(a) 129.(b) 130.(b) 131.(d) 132.(a) 133.(d) 134.(d) 135.(a) 136.(c) 137.(b) 138.(b) 139.(c) 140.(d) 141.(c) 142.(a) 143.(a) 144.(b) 145.(c) 146.(a) 147.(d) 148.(b) 149.(b) 150.(d)

151.(d) 152.(b) 153.(a) 154.(a) 155.(d) 156.(b) 157.(c) 158.(a) 159.(b) 160.(b) 161.(d) 162.(c) 163.(d) 164.(b) 165.(a) 166.(c) 167.(b) 168.(a) 169.(b) 170.(b) 171.(b) 172.(c) 173.(a) 174.(d) 175.(d) 176.(c) 177.(b) 178.(c) 179.(a) 180.(d)

181.(d) 182.(b) 183.(c) 184.(c) 185.(d) 186.(b) 187.(a) 188.(d) 189.(b) 190.(a) 191.(b) 192.(a) 193.(b) 194.(c) 195.(c) 196.(d) 197.(b) 198.(a) 199.(b) 200.(a) 201.(d) 202.(a) 203.(b) 204.(c) 205.(b) 206.(d) 207.(a) 208.(c) 209.(b) 210.(a)

211.(d) 212.(b) 213.(d) 214.(d) 215.(c) 216.(b) 217.(a) 218.(b) 219.(b) 220.(c) 221.(c) 222.(b) 223.(d) 224.(a) 225.(a) 226.(a) 227.(a) 228.(b) 229.(d) 230.(b) 231.(c) 232.(b) 233.(c) 234.(c) 235.(c) 236.(c) 237.(c) 238.(b) 239.(a) 240.(b)

241.(b) 242.(c) 243.(a) 244.(d) 245.(b) 246.(c) 247.(a) 248.(d) 249.(b) 250.(c) 251.(a) 252.(b) 253.(d) 254.(a) 255.(c) 256.(a) 257.(c) 258.(b) 259.(d) 260.(a) 261.(b) 262.(c) 263.(d) 264.(a) 265.(b) 266.(b) 267.(c) 268.(c) 269.(d) 270.(a)

r

(d) (b) (d) (b) (c) (c) (a) (b) (d) (a) (a) (a) (a) (d) (a) (a) (c) (c) (c) (b) (d) (c) (b) (b) (b) (b) (b) (b) (b) (a)

eeYa ridn agv .iSn i

31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60.

R Enak geisnh

(d) (b) (a) (b) (a) (c) (a) (c) (b) (a) (b) (a) (b) (a) (c) (c) (a) (a) (d) (a) (b) (d) (d) (a) (b) (b) (a) (c) (a) (b)

271.(b) 272.(d) 273.(c) 274.(d) 275.(a) 276.(b) 277.(b) 278.(a) 279.(c) 280.(a) 281.(b) 282.(b) 283.(d) 284.(a) 285.(d) 286.(b) 287.(c) 288.(c) 289.(a) 290.(c) 291.(a)

wwM wa.th Les B aryn

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

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598

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SOLUTION   POB = 180 – 120°  POB = 60°

12

(d)

5 O

 OBP =  OPB In  OPB,

P Q = 12cm, O Q = 5cm By using pythagoras theorem OP² = OQ² + PQ² OP² = 5² + 12² OP² = 25 + 144 OP = 1 6 9 , (b)

 PBO = 60° 5.



(a) ATQ

9. B

O

A

r

60º x P

C

  AOP = 120º

kgei snhe

(c)

B

O

wwM wa.th Les aBryn

4.

3 N

3

B

OM = 3 cm In  ONB OB2 = ON 2 + NB2 (5)2 = ON 2 + (3)2 ON = 4cm  MN = ON – OM MN = 4 – 3 MN = 1 cm. (b) According to the question,

7.

A

120°

A

O

 POA = 120°

B

2 132 – 8 – 3

=

169 – 25

O C

5 M

radius =

B

A

10 = 5cm = OA 2

OB = 4 By using pythagoras theorem OA² = OB² + AB² 5² = 4² + AB² AB² = 25 –16 AB

O 4

5 B

C

AB AM = (  bisect chord 2 form centre) AO² = OM² + AM² OM² = 9 , OM = 3 cm

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=

4

(a)

P

2

00 ' – R – r 

A

D

 According to figure BC will be a chord of circle having centre ‘O’  OD will be perpendicular on BC And BD = DC  Therefore  BDO = 90º

D

2

=

11. (b)

5

M4

3 O'

13

= 144 = 12 cm 10. (a) Required ratio = 5 : 3

 PBO  60º 6.

O

 Length of the direct common tangent PQ

 x  x  60º  180º  x = 60º

O

A

B

  POB = 180º – 120º = 60º  OP = OB = R  So Let PBO  x   BPO

ERna

AB = BC = AC = 2r So, ABC is an equilateral  3. (a) ATQ CM = MD = 4 cm AN = NB = 3 cm In OMC OC² = MC² + OM² (5)² = (4)² + OM²

5

x

r

Q

8

=

9 AB = 3cm AC = 2 × AB = 2 × 3 = 6cm 12. (a) A 6 cm

r

r

4

AB||PQ out of given option only cyclic trapezium follow the property. (b) P

r

120º

C

A =B AB  PQ

O

8 cm

6 cm

A r

OP = 13cm

 POB +  OPB +  OBP = 180° 2  OBP = 180 – 60°

eYrai dnag

2.

(c) ATQ

OB = OP

P

13

8.

v.iSn ir

Q 1.

M

28

O'

B

599

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O'M =

28 O O' = OM + O'M = 8  28  13.3 cm 13. (b) A

angle is 180º] 15. (c)

9

3

According to the question PT = 5 cm. PA = 4 cm. PB = (4+x) cm. As we know that PT2 = PA × PB 25 = 4 (4 + x) 25 = 16 + 4x x

A

8

B

D

Angle subtended by an arc at the centre is twice the angle subtended by the arc on the circle.  C = 30º D = 150º [Be c aus e in a cy c lic quadrilateral sum of opposite

4P

A B 2 r r 1 2

AB = 2 8  4

AB = 2  32

PQ =

169 – 25

PQ =

144 ,

19. (d)

PQ = 12cm

In right ΔCOQ ,

17 cmC

cm 25

O 8 cm

S R

16cm

QC² = OQ² + OC² (By pt) 17² = 8² + OC² OC = 15 cm In right ΔCOP CP² = OP² + CO² 25² = OP² + 15² OP = 20cm PS= 2× OP = 2 × 20 = 40 cm

20. (a)

In ΔACB C

AB = 8 2 cm

17. (a)

D

A

wwM wa.th Les B aryn 60º

60º

B

We know that,

30º O

(13)2 – (11 – 6)2

Q

Q

C

PQ =

P

A

O

B

60º

9 cm . 4

16. (c)

According to the question Let OA = x = OP AB = 2x OM = x – 3 In OMP, x2 = (9)2 + (x – 3)2 x2 = 81 + x2 + 9 – 6x 90 = 6x x = 15  AB = 2×15 = 30 cm. Alternate:PM × MQ = MB × AM 9 ×9 = 3 ×x x = 27  AB = AM + BM= 27 + 3 = 30cm 14. (a)

B

r

4 A

R Enak geisnh

P

(radius – radius )2 1 2

P

O M 9

T

5

eeYa ridn agv .iSn i

O O' = 8 + 28 = 8 + 5.29 = 13.3 cm N ote:  AMO = R ig ht ang le d triangle In, AMO  AM = 6, AO = 10 then, OM = 8 In, AMO'  AM = 6, AO' = 8 then,

B

5

. . A P

AB = 13 cm, r2 = 6cm

.Q .B r1 = 11 cm

PQ= (Distance between centres)2

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34º O

56º B

Q

O

According to the question Let AD = DB = x OA = 5 OD = 3 AB = 2x In ODA, OA2 = OD2 + AD2 (5)2 = (3)2 + (x)2 x = 4  AB = 2×4 = 8 cm.

18. (a)

A

3

 ACB = 90º (Angle formed by semicircle is 90º)

 ACB +  CAB +  CBA = 180º 90º + 34º +  CBA

= 180º

 CBA = 56º

21. (b) In APQ , P = Q = x P x 48º

O

A

x Q

xº + xº + 48 = 180º 2x = 132º x = 66º

 APQ = 66º

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(d) According to question Given:



 AOB = 110°  APB = ?

AOPB is a quadrilateral  O +  A +  P +  B = 360° 110° + 90° + 90° +  P = 360°  P = 360° – 290°

OB =10 cm, OA = radius=6 cm AB = tangent

A

6 cm.  P



B

3 cm.

 

 

AP

CQ

=

wwM wa.th Les aBryn

50°

According to the question AP = 6 cm (Radius1) QC = 3 cm (Radius2) As we know, any line draw from centre to the tangent be perpendicular So, PAB = QCB = 90º APB = CQB =  [same alternative angle] So, APB CQB

Q

R

 RPT = 50°

= 9 0° Semicircle

(Angle

6



in

 PQR = 50°

(Alternate segment theorem ) In  PQR

 P + Q +  R = 180°  R = 180° – 90° – 50°  R = 40°

 SPQ = 40° (Alternate segment theorem) 25. (b) According to question

3



8

Q

T

So  OTP = 90° length of PT = 8 cm 3,

4,

6

× 2 8

2×

CB

5 ×2 10

29. (a)

8

CB

CB = 4 cm. In right angled triangle PAB (PB)2 = (PA)2 + (AB)2 (PB)2 = 6 2 + 8 2 PB = 10 cm. Again, right angled triangle CQB  BQ2 = (BC)2 + (CQ)2  (BQ)2 = 3 2 + 4 2  BQ = 5 cm.  Therefore PQ = PB + BQ  10 + 5 = 15 cm. 27. (a) 24 cm.

a

     

M 12 cm.

C

A

D

5

O 12

5 cm.

N

B

10 cm.

According to the question

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r 6

10

Let PTQ is the Tangent of circle having centre O and Radius = r and point T, touches the circle  We know that any line draw on tangent's touching point from centre, always makes a perpendicular

AB

17 cm.

 ABC = ACB AB = AC = 17 cm 24. (a) According to question Given: S T P

O

P

kgei snhe

AB = 17 cm AC = ?

 DAB = ACB (By alternate segment theorem  DAB = ABC (Alternate angle)

Q

C

 

C

 QPR

8 cm.

E

B

DE||BC,

A

ERna

D

Therefore APB = 70º 26. (b)

eYrai dnag

In right angle  OAB By using pythagoras theorem OB² = OA² + AB² (10)² = (6)² + (AB)² (AB)² = 100 – 36 (AB)² = 64 AB = 8 23. (d) According to question Given:

 APB = 70°

OMD and ONB are r ig ht angled triangle And, OD = OB = R In OMD, from triplets 5, 12, 13 OM = 5 cm, OD = 13 cm, MD = 12 cm. Again, In ONB, from triplets 5, 12, 13 OB = 13 cm, ON = 12 cm  OB = OD = 13 cm Radius = 13 cm. 28. (c)

v.iSn ir

22

a

O 60º

A

90º a

D b C

a a B

 AOB = 60°

  

 COD = 90° length of chord AB = a length of chord CD = b AO = OB = AB = OD = OC = a

 

In ΔODC OD² + OC² = CD² a² + a² = b²



b=



2a 30. (b) Let smallest circle radius = R Then biggest circle radius = (14 – R)

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AOD = 100° BOC = 70°  ACD  ACP =

According to the question π (14 – R)2 + π R2 = 130 π (14 – R)2 + R2 = 130 196 + R2 – 28R + R2 = 130 R = 3 cm Radius of smallest circle R = 3 cm X 31. (d) 9c m

     

z 15 cm

y

100 = 50° 2  The angle subtended at the centre is twice to that of angle at the circumferance by the same arc

.

BOC = 70°

In  APC,  PAC +  ACP +  APC = 180°  APC = 180° – 50°– 35°  APC = 95° 34. (b)According to question Given:

T

xy = 9 cm, tx = 15 cm  We know Length of tangents drawn from a point to the circle are equal Therefore xy = xz = 9cm, tz = rt tx = 15 cm xz + zt = 15 zt = 15 – 9 = 6 RT = ZT = 6 cm 32. (a) (b) M A B

o

B

C

 OAB =  OBA

D

 

   

=

C

O

According to figure PA = AM (equal tangent drawn from a external point) PD = OD MB = BN OC = CN

CN+NBAP+DP 

=1

33. (d) According to question Given : A

AOB

In right angle BEC  BEC + CBE + ECB = 180°

CBE = 180° – 65°– 90° = 25° 35. (c) According to question Q

D

o

B

A

C C2

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P

D B

BDC = 120°  AB P = ?  CDP = 180° –  BDC

CDP = 180° – 120° CDP = 60° CD||AB  CDP =  AB P = 60° 38. (b) According to question

y x

y

P

Let the radius of the circle be = r O

 NQP =  NPQ  NRP =  NPR In PQR

C1

R

N x

QR is the common tangent and N O is als o the com mon tangent.  QN = NP = NR

70° P C

2

In QPN

100°

 OO '  O ' E  O'D = O'E Clearly CD = EF = 4.5 cm 37. (a) According to question Given:

= 25°

130 = = 65° 2

AB+CD CB+AD

AM+BMOD+OC

CD= 4.5 cm EF = ? Let O and O' be the centre of a circle  OC = OF = radius of circle B

AOB = 130°

wwM wa.th Les B aryn

N

E

BF

 AOB = 180° – 25° – 25°

1

o

o

OAB = 25° OA = OB = r

 ACB =

P

D

E

R Enak geisnh

R

AC D

70  = 35°  BDC = BAC = 2

A

S

x+y+x+y = 180° 2x + 2y = 180° x+y = 90° As shown in the figure x + y =  P = 90° 36. (c) According to question Given :

r

R

eeYa ridn agv .iSn i

–R) .(14 14

 P + Q +  R = 180°

 DO = OC =

r 2

In right angle  AOD By using pythagoras theorem AD² = OD² + AO² r² =

r² 4

+ AO²

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BC = 10 – 5 BC = 5 units 41. (a) According to question Given:

r² 4

3r ² AO² = 4

7 13 cm

O

3 r × 2, 2

AB = 3r units 39. (d) According to question

AP = 5 cm,

2

OB = 3 cm

PQ is  bisector PO = OQ  AO = 1,

OO' = 13 cm, OA = 7 cm O'B = 2 cm Leng th of dir e ct com mon tangent



AB =

OO ' ² – R – r  ²

AB =

13 ² – 7 – 2 ²

AB =

169 – 25

AB = 1 4 4 AB = 12 cm 42. (a) According to question

r

r

o

C

Let AB is the chord and 'O' is the centre of circle Given : The length of AB is equal to radius  OA = OB = AB = r

wwM wa.th Les aBryn

PQ = 2 × 2 6

B

ERna

= 2 6 cm  PQ = 2 × OP

r

A

In right angle  POA AP² = OA² + OP² (5)² = (1)² + (OP)² (OP)² = 25 – 1 (OP)² = 24 (OP)

O

eYrai dnag

AB =

AB = 2 × AO

B

kgei snhe

AO =

A

3r 2

PQ = 4 6 cm 40. (a) According to question Given:

 AOB is an equi lateral triangle  AOB = 60°

which chord   ACB subtends in the major segment 60

is =

OA = radius = 5 units

2

= 30°

43. (a) According to question, Given:

AB = 5 3 units BC = ?

In right angle OAB OB² = AB² + OA² OB² = 5 3

2

 

r

+ (5)²

OB² = 75 + 25 OB² = 100 OB = 10 units  BC = OB – OC

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In, OMA OA2 = OM2 +AM2 r2 = (x –1)2 + (4)2 r2 = (x – 1)2 + 16 .........(i) In OND OD2 = ON2 + ND2 r2 = x2 + (3)2 r2 = x2 + 9 ........(ii) Comparing equations (i) and (ii) (x – 1)2 +16 = x2 +9 x2 +1 – 2x + 16 = x2 +9 17 – 2x = 9 2x = 8 x=4 Put the value of 'x' in equations(ii) r 2 = (4)2 + 9 r 2 = 16 + 9 r 2 = 25 r = 5 cm Alternate:AM = 4 cm CN = 3 cm {3,4,5 } = formed a triplet  Radius = 5 cm 44. (d) According to question, AB and AC is a chord AB = 8 AC = 6 A



v.iSn ir

AO² = r² –

In  BAC  A = 90°  BC 2 = AB2 +AC 2 BC 2 = 82 + 62 BC 2 = 64 + 36 BC 2 = 100 BC = 10 cm Here BC is the diameter of a circle because only subtended on the arc of semi circle is 90° BC



O

r 4 x A 1cm M C 3 N

r

r

3

4

B D

AB and CD are chord AB = 8 cm Let ON = x cm

C

B

2

= radius=

10 2

= 5 cm

45. (a) According to question Given: A

M

B r

o r C

N

D

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In PCR and RBP PC = PB (radius) RC = RB PR is common PCR  RBP

CPR  RPB = x (CPCT) Similarly, (CPCT)  APQ =  CPQ = y 2y + 2x = 180°  x + y = 90° QPR =90°  49. (c) According to question, Given:

AO= 15 cm OD= 9cm

P

B

D

In ODA OA2=OD2 + AD2 (15)2 = (9)2 + AD2 AD2 = 225 – 81 AD2 = 144 AD = 12 cm  AB = 2 × AD AB = 2 × 12 = 24 cm 47. (c) According to question Let'AB' is the chord of biggest circle and 'O' be the centre of a circle

D

C

A

y

yx p

9

25 cm 3 52. (c) According to Question

PR =

45°

D

P

B

AOC  BOD  2ABC  2BCD

R Enak geisnh O

C

(Angle formed on major arc is half the angle formed on centre) = 2ABC  2BCD = 2BPD [ Exterior angles of triangles AOC  BOD  2BPD 2BPD = 50° + 40°

 ABC = 60° 50. (b) According to Question, AO is perpendicular to PQ P

o

A

Due to Alter nate Seg m ent theorem)  BCA = 75°

Q

OA = OP = OQ

A

1

 BPD = 2 × 90° = 45° 53. (b) According to question AB is a common chord O and 0' is the centre of the circle.

OAQ = 45°

B

 PAQ = 90°

51. (d) According to Question, Given: Q

P 20 3

90°

B



20 3

cm

In  POR PR2 = PO2 + OR2

Rakesh Yadav Readers Publication Pvt. Ltd.

R

4 5



 POR = 90° OR =5 cm OQ = 4 cm OP =

In ODA AO2 = AD2 + OD2 (4)2 = (AD)2 + (2)2 AD2 = 12

o

R

x

625

OPA  OAP  OQA  OAQ

In  ODA OA2 =OD2 + AD2 (5)2 = (3)2 + AD2 AD2 = 25 – 9 AD2 = 16 AD = 4 cm  AB = 2 × AD =2×4 AB = 8 cm 48. (c) According to question Q

400  225 9

B

 Then BAC  BCA  ABC =180° 45° + 75° +  ABC = 180°

o

A



(

wwM wa.th Les B aryn



PR2 = PR2 =

75°

C

BAT  BCA



OA= 5 cm OD= 3 cm

400  25 9

T

o A

PR2 =

r



BAT=75° BAC=45° A



2

 20  2 PR2 =    5 3

eeYa ridn agv .iSn i

AB = CD = 8 cm r = 5 cm  In OMB OB 2 = OM2 + MB2 r2 = OM2 +(4)2 (5)2 = OM2 + 16 25 – 16 = OM2 OM2 = 25 – 16 OM2 = 9 OM = 3  MN = 2 × OM MN = 2 × 3 = 6 cm 46. (a) According to questions, Let 'O' be the centre of a circle and 'AB' is the chord of the biggest circle

AD = 2 3 AB = 2 × AD

AB = 2 × 2 3 = 4 3 54. (b) According to question OBA is a right angle triangle

B5.05 A r o

604

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AB = 24 cm AE= EB = 12 cm

D4 3

=

B r

o



In BDO , using pythagoras 2

= 152 – 122

2

2

BO = OD + BD r2 = (4)2 + (3)2 r2 = 16 + 9 r2 = 25 r=5 56. (b) Let OD = x and DO' = y According to question

225 –144 = 81 = 9 cm OF = 21 – 9 = 12 cm 152 – 122 = 9 cm

also FD =

 CD = 2 × 9 = 18 cm 60. (a) According to Question A 12 C

O

O

B

AB = 16 AC = BC = 8 cm OC = OC' = 6 cm

A O x

12 D y 12

O

62  82 =

OA =

B

OA =

= x2 = = =

OD2 + AD2 +(12)2 225 – 144 81 9 DO'2

A

6

B

C

3

D

x

y2 144

to ques tion   2r length of arc = 360 72 22  2   21 = 360 7 = 26.4 cm 58. (b) one and only one circle can pass through 3 non- collinear points. 59. (b) According to question A

12 E 12

B

15

Given: AB = 6, CD = 3. PD = 5 Let PB = x Note: If chords AB and CD intersect externally at point, p then PB × PA = PD × PC x × (6 + x) =5×8 x2 + 6x – 40 =0 x2 + 10x – 4x – 40 =0 x(x +10) – 4(x+10) =0 (x + 10) (x – 4) =0 x = 4 , – 10 ( –10 neglected)  PB = 4 cm 62. (a) According to question.

d A 8 3

6 B

AB =

d2 – R  R 1 2



2



AB2 = d2 –(R1 +R2)2 (8)2 = d2 – (6 + 3)2 64 = d2 – 81 d2 = 145 d=

AB =

d 2 – R1  R2

AB =

10



2



EOB 15

F

Given: PAB = 35° As we know that APB = 90° Therefore,  PAB  APB  ABP = 180° ABP = 180° – 90° – 35° ABP = 55° 64. (a) According to question Let le ngth of transv e rs e common tangent = AB = 8 cm Distance between them = d

145 65. (a) According to Question Let le ngth of transv e rs e common tangent be AB Distance between them = 10 cm

AE = EB = 5 cm CN = DN = 12 cm

o

C

P

5

wwM wa.th Les aBryn

In ADO ' (AO')2 = AD2 + (13)2 = (12)2 +y2 169 = 144 + y2 = 169 – y2 = 25 y =5 x + y =9+5  = 14 57. (b) Ac c or d ing

36  64

100 = 10 cm 61. (d) According to Question

ERna

In ADO AO2 (15)2 = x2 x2 x

OC 2  CA2

OA =

13

kgei snhe

15

eYrai dnag

4

A

OB2 – EB2

OE =

(13)2 = (OE)2 + (5)2 169 = (OE)2 + 25 OE2 = 169 – 25 OE2 = 144 OE = 12 cm In  OND OD2 = ON 2 + ND2 2 (13) = ON 2 +(12)2 169 = ON 2 + 144 ON2 = 169 – 144 ON2 = 25 ON =5 = OE + ON  EN EN = 12 + 5 = 17 cm 63. (b) According to question

v.iSn ir

OA is a hypotenuse Hypotenuse is always greater than other two sides  Radius is always greater than 5 cm 55. (b) According to question.

 

D

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r =

OE 2  EB 2

2

2 – 3  3

605

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AB = 100 – 36

BOC  AOD  2BAC  ACD



AB =

= 2BPC

64 AB = 8 cm 66. (c) According to question Length of the transver se common tangent

AB =

24



2



2 – 5  3

AB =

576 – 64

AB =

512

A

x o

9 Q

= 90° PR= 2 + x PQ= 17 RQ = 9 + x By using pythagoras theorem PQ2 = PR2 + RQ2 2 (17) = (2 + x)2 + (9 + x)2 289 = 4 + x 2 + 4x + 81 + x 2 + 18x x2 + 11x – 102 = 0 x2 +17x – 6x – 102 = 0 x(x + 17) –6 (x + 17) = 0 (x + 17) (x – 6) = 0

In BOE r² = x² + 5² ....(i) In DOF r² = (17– x)² + (12)² ....(ii) Compare equation (i) and (ii) x² + 25 = 289 + x² – 34x + 144 25 = 433 – 34x 34x = 408 x = 12 ....(iii) Put the value of x in equation (i) r² = (12)² + (5)² r² = 144 + 25 r² = 169 r = 13 cm Alternate:Apply triplet 5,12,13 r = 13 cm 70. (c) According to question

wwM wa.th Les B aryn

PRQ

x = 6 as x  –17  x = 6 cm Alternate  PRQ = Right angle  QR(B)

A

D

B

o

PR(P)

(9 +x)cm (2 + x) cm

AD = DB = x

Triplet = (17,15,8)  x = 6 cm 68. (c) According to question A D

o

2

B

BOC  2BAC

 AOD  2ACD

Rakesh Yadav Readers Publication Pvt. Ltd.

Let AO = OB = x MO = y ON = 50 – y AM = 30 cm AN = 40 cm

In AOM AM² = OA² +OM² (30)² = OA² + y² x² = 900 – y² ....(i) In AON AN² = ON² + OA² (40)² = (50 – y)² + x² (x)² = 1600 – (50 – y)²......(ii) Compare equation (i) and (ii) 900 – y² = 1600 – (50 – y)² 900 – y² = 1600 – (2500 +y² – 100y) 900– y² = 1600 – 2500 –y² + 100y y = 18 ............... (iii) put the value of y in equation (i) x² = 900 – 324 x² = 576 x = 24 cm OA = 24 cm AB = 2 × 24 AB = 48 cm Alternate:30,40,50 (triplet)  AMN = Right triangle

 AMN =

1 2

 3  1   3 – 1

4  2 3  4 – 2 3  x²

x² = 4 3

N

o

 MAN = 90°

In BOD C12 = C 22 + BD2

P

C

 3 – 1 cm =  3  1 cm

C2 = C1

A

B

R Enak geisnh

2

17cm

D



P

AB = 44 3 cm 71. (d) According to question

M

AE = EB = 5 cm CF = FD = 12 cm BO = OD = r cm

R x 90° x

 AB = 2 × BD

B

r

r 17–x C 12 F 12

AB = 16 2 67. (b) According to question

PQ(H)

2

5 E 5

 24 3

AB = 2 × 24 3

50

BPC = 25° 69. (c) Accrding to question

2

2

r

d2 – R  R 1 2

BPC =

 

eeYa ridn agv .iSn i

AB =

30° + 20° = 2BPC

1

x= 4 3

2  x²

1 2

×b ×h

× 30 × 40 =

1 2

×50 × AO

AO = 24 AB = 2AO  AB = 48 cm

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72. (a) According to questions C

3.5 2

P 0.5 A

2

O

E R

OM = 3 cm  OO' = 2 × OM OO' = 2 × 3 OO' = 6 cm 74. (d) According to question

B

4

Alternate:17,15,8 (triplet) d is tanc e on sam e s id e between chords = (15 –8) = 7 cm 76. (d) According to question D

D

3

o

D

2

65 4

65 2 Diameter = 2r 

 OBA = 40°  OBA =  OAB = 40°

ERna 17

wwM wa.th Les aBryn

65

Alternate:Diameter = =

AE ²  EB ²  EC ²  ED ²

2²  6²  3²  4² =

65

73. (a) According to question A

r1

O

r2

M

B

r1 = r 2 = 5 cm AM = MB = 4 cm



In  BDO O'D² = DB² + BO'² BD² = (3)² – (1)² BD² = 9 – 1 BD² = 8,

BD = 2 2  DE = 2 × BD

In  AMO OA2 = OM² + AM² 25 = OM² + 16 OM² = 25 – 16 OM² = 9

DE = 2 × 2 2

DE = 4 2 cm 77. (b) According to question

O

A

15

C

o 17

M

8 N 8

15

B

D

OA = OD = 17 cm AM = MB = 15 cm CN = ND = 8 cm

In OMA OA² (17)² 289 OM² OM² OM

= = = = = =

AM² + OM² (15)² + OM² 225 + OM² 289 – 225 64 8

In  OND OD² = ON² + ND² (17)² = (8)² +ON² 289 = 64 + ON² ON² = 289 – 64 ON² = 225 ON = 15 MN = ON – OM  MN = 15 – 8 MN = 7 cm

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r1

r2

Given: r1 + r 2 = 7 cm r1 = 4 cm r2 = 7 – 4  r2 = 3 cm 78. (b) According to question A

B 90° o

0° 11

65 2

radius is equal to

O'A = 3 cm OA = 2 cm O'D = 3 cm O'B = 1 cm

 DOA = 60° 75. (b) According to question

65 2

A

O

v.iSn ir

 OCB =  COB = 20° In  AOB  OBA = 20°+ 20°

B

E

DOC is a line  COB +  AOB +  DOA = 180° 20° + 100° +  DOA = 180°

r=

D=

BC = DO = OA = OB = r In  OBC

 A +  O +  B = 180° 40° +  O + 40° = 180°  O = 100°

49 4

D = 2

C

C

kgei snhe

r² =

B

In  AOB

7  r² = (2)² +   2 

r² = 4 +

20°

A

eYrai dnag

Given:AE = 2 cm EB = 6 cm ED = 3 cm As we know that AE × EB = EC × ED 2 × 6 = EC × 3 EC = 4 cm  In OPC OC² = CP² + PO²

C

OA = OB = OC



In  OAB OAB  OBA  AOB = 180° 2OAB = 180° – 90 2OAB = 90° OAB = 45°

In  OAC OAC  OCA  AOC = 180°



2OAC = 180° – 110° OAC = 35° = 45° + 35° BAC = 80°

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P A

B

AB = 2r = 14 cm PB = 12 cm = 9 0° (ang le in the APB semicircle) Let AN = x and NB = (14 – x)  In APB AB² = PB² + AP² (14)² = (12)² + (AP)² 196 = 144 + (AP)² (AP)² = 196 – 144 (AP)² = 52 AP =

52

In  APN AP² = PN² + AN² 2

 52

= x² + PN²

A

D 5

x=

104 28

Let AD = DB = x OA = 5 cm OD = 3 cm  In ODA OA² = OD² + AD² (5)² = (3)² + (AD)² 25 = 9 + (AD)² (AD)² = 25 – 9 (AD)² = 16 AD =4  AB = 2 × AD AB = 2 × 4 = 8 cm Alternate:In  AOD, 3,4,5 (triplet)

72 NB = 7

NB

2 = 10 cm 7

Alternate:PB² = PN × AB 12² = 14 × BN BN = 12 × 12 ÷ 14 BN = 72 ÷ 7



A

D

AC H = 2 2 85. (d) According to question, N

o P

20°

Rakesh Yadav Readers Publication Pvt. Ltd.

S

Given:

 PSQ = 20°  PRQ = ?

OPSQ is a quadrilateral

 OPS =  OQS = 90°

o



OPS  OQS  POQ  QSP

= 360°

OPS  OQS  POQ  QSP = 360°

AD = DB = 6 cm

In  ODB OB² = DB² + OD² OD² = OB² – DB² OD² = (10)² – (6)² OD² = 100 – 36 OD² = 64 OD = 8 cm 83. (b) According to question

80. (d) According to question

Q R

B

Let OD = x, OB = 10 cm

2 BN = 10 7

BN = circumradius N = circumcentre BN = AN = NC BN =

AD = 4 cm AB = 2AD = 2 × 4 = 8 cm 82. (a) According to question

26 x= 7 NB = 14 – x 26 NB = 14 – 7

B

3 o

wwM wa.th Les B aryn

PN² = 52 – x² ................(i) In PNB PB² = PN² + NB² (12)² = PN² + (14 – x)² PN² = 144 – (14 – x)²..........(ii) 52 – x² = 144 – 196 – x² + 28x 28x = 104

C

Given: BEC = 130° = 180° – 130° = 50°  DEC = 180° – 50° – 20 = 110°  EDC 110°  BAC  EDC = (Angle on the same arc are equal) 81. (d) According to question

eeYa ridn agv .iSn i

N

R Enak geisnh

A

D

E B

Let AB is the chord and 'O' is the centre of a circle Given:OA = OB = AB  All sid es ar e e qual the n triangle is equilateral triangle.  Then the angle subtended by the chord is 60° 84. (c) In a right angled triangle the cir cum ce ntre of the triangle lies on mid point of the hypotenuse

r

79. (d) According to question

A

POQ = 360° – 90° – 90°– 20° POQ = 160°



PNQ 

 

B



.o

PNQ = 1 2

1 2

POQ

160 = 80°

NPRQ is a cyclic quadrilateral sum of opposite angles of cyclic quadrilateral is 180° PNQ  PRQ = 180°

 PRQ = 180° – 80°  PRQ = 100°

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T

Q

 

APB  OPA  OPB

 APB  APB

= 30° + 30° = 60° 89. (b) According to question E

A

r

BD =

 DE = BD =

105 cm

 In AED AD² = DE² + AE² 2

P

AD² =

wwM wa.th Les aBryn

o

2r

B

Given: OA = OB =r.(radius) OP = 2r (diameter) In  OAP OP² = OA² + AP² (2r)² = r² + AP² AP² = 4r² – r² AP² = 3r²

 105 

+ (16)²

AD² = 105 + 256 AD² = 361 AD = 19 cm 90. (a) According to question OT is the pe r p end ic ular bisector of chord PQ. let TR = y

PT² =

256 + 16 9

PT² =

400 , 9

PT =

20 cm 3

Alternate:In  POR, OP2 = OR2 + PR2 52 = OR2 + 42 OR2 = 25 – 16 = 9  OR = 3cm In  POR and  POT  PRO =  TOP 

 POR~  POT



PR OR = PT OP



4 3 = PT 5

20 cm 3 91. (c) Maximum no. of tangent are  PT =

A

E

B

T

3r

In  OAP

4

P

3r tan

105 cm

ERna

A

2

kgei snhe

P

 ORP is a right angle triangle  By using pythagoras theorem. OP² = OR² + RP² (5)² = (4)² + (RP)² 25 = 16 + (RP)² (RP)² = 25 – 16 (RP)² = 9 RP = 3 cm 88. (d) According to question

32 16 = cm 6 3

y=

16  PT² =   + (4)² 3

B

o

Given: OA = OB = 13 cm OD = 8 cm  AE = 2× OD AE = 2 × 8 = 16 cm In ODB OB² = OD² + BD² BD² = OB² – OD² BD² = (13)² – (8)² BD² = 169 – 64 BD² = 105

4 5

TO² = PT² + OP² .......(i) PT² = TR² + PR² ........(ii) Put the value of PT² in equation (i) TO² = TR² + PR² + OP² (y + 3)² = y² + (4)² + (5)² y² + 9 + 6y = y² + 16 + 25 9 + 6y = 41, 6y = 32

In right angle TRP PT² = TR² + PR²

D

R

o

1 3

eYrai dnag

B

As shown in the f ig ur e Tangent are equal  PT = PQ Alternate:PQ2 = PA × PB ..... (i) PT2 = PA × PB .....(ii) From both equation, PT2 = PQ2 PT = PQ 87. (a) According to question

tan30º 

 OPA = 30° Similarly in  OPB  OPB = 30°

 A

AP =

1 3

tan

P

v.iSn ir

86. (d) According to question

OA AP

tan

r 3r

5

R 4 o

Q

In right angle PRO PO² = PR² +RO² (5)² = (4)² + RO² (RO)² = 25 – 16 (RO)² = 9 , RO = 3 cm Right angle TPO and TRP

Rakesh Yadav Readers Publication Pvt. Ltd.

C

F

D

5

92. (c) According to question

o

A

B

N

C

D

609

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



 OAB =  OBA = 60°

 

 OAB is a equilateral triangle OA = OB = AB

AB 1  OB 1 97. (c) According to question

B

r D

ONY = OYN = 50°

 ADB is a equilateral triangle  DBA = 60°



Similar In  ABC  ABC = 60°



=  ABC = 23°

 APC

= 70° =  DPB

 APD

= 180° – 70° = 110º

R Enak geisnh

  DBC = 120° 98. (b) According to the question

 ADP

 APD =  BP C (Ver tically opposite angle)

r

r

 

B

r

r

wwM wa.th Les B aryn

Also BCD = 180° – 23° – 110° = 47° 94. (c) According to figure  DAC = 51°  EOB = 180° – 150° = 30° OB = OE = radius   OEB =  OBE then  OEB +  OBE +  EOB = 180° 2 OBE = 180° – 30°  OBE = 75°   CBE = 180° – 75°  CBE = 105° 95. (d) The angle in a semi-circle is a right angle 96. (b) According to question Given : A

r

A

r

C

NOY = 180° – 50° – 50° NOY = 80°

 MON = 150° – 80°

 MON = 70° 101. (d) Since CS = SD The two chords must be equidistant from the centre 'O'. Thus, the required ratio is 1 : 1 102. (b) (OS)2 = (OK)2 + (KS)2 25 = OK2 + 16  OK = 3 and (OS)2 = (OL)2 + (LN)2  OL = 4 cm KL = OL – OK = 1cm Area of rectangle= 1×10 =10cm2 103.(c) OQ = OB = OC = r (say) C

Let radius of the circle be = r AB = 2r,BC = 2r, CA= 2r All these sides are equal T r ia n g l e ABC is an equilateral 

99. (b) According to the question Given: A

M

B

OM  AB OB = radius

60° o

In right angle  OMB By using pythagoras theorem

Rakesh Yadav Readers Publication Pvt. Ltd.

S

Q

B

 AOD =  BOC = 120°  BOQ =  COQ = 60°

OM = 2 11 cm

B

O

o

AB = 20 cm AM = MB = 10 cm



 MOY = 180° – 15° – 15°

In  ONY

AB = AD = DB = r



In  OMY

 MOY = 150°

 DBC = 60° + 60°







r

+ (10)²

 OMY =  OYM = 15°

r

A



OB² = 44 + 100 OB² = 144 OB = 12 cm 100. (d) According to the figure. OM = OY = ON

C

OA² = ON² + AN² (17)² = (8)² + AN² AN² = 289 – 64 AN² = 225 AN = 15 cm AD = 2 × AN AD = 15 × 2 = 30 cm  93. (b) According to question

2

OB² = 2 11

r

In right angle ONB OB² = ON² + BN² (10)² = ON² + (6)² ON² = 100 – 36 ON² = 64 ON = 8 cm In right angle ONA

OB² = OM² + MB²

 BOA = 60° OA = OB = r

eeYa ridn agv .iSn i

 

Given:BC=12cm, OA = 17 cm OB = 10 cm BN = NC = 6 cm



SB 3 = sin60° = OB 2

 SB =

r 3 2

BC = 2SB = r 3 Area of quadrilateral BQCO =

1 × BC × OQ 2

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=4 3 +5 3 =9 3 Now, since AB and CB are the secants of the circle. BD × BA = BE × BC

104. (c)  PDB =  QEA = 80°

 PED = QDE = 10° (   DPE =  DQE = 90°)

6 × BA = 4 3 × 9 3 BA = 18cm

 DRE = 180–(10+10) = 160°  PRD = 180 –  DRE = 20° 105. (a) Notice that all the given triangles are equilateral Area of shaded region

6a DH = HG = GC = = 2a 3 2a HM = MG = = a = SM 2 NQ = a SQ = SM + MN + NQ = a + 3a + a = 5a Since diagonal of square SQ = 5a But, diameter of circle SQ = diagonal of square SQ

107.(c) Note:  ORP = 90° ( OP is a diameter ofsmaller circle) OS = 5cm and OR = 4cm

ERna

 SR = 52 – 42 = 3cm  SP = 2(SR) = 6cm (Since, OR passes through centre O and perpendicular to SP therefore OR bisects SP.)

wwM wa.th Les aBryn

108.(b)  ADO is a right angle (angle of semicircle) O



90°

D



Radius of the circle =

Here

A

r = R( 2 – 1)2

r = R(3 – 2 2 )

111. (a)



1 =4 3–2 2



=



2



=

2

R

2

1



4 17 – 12 2



1 68 – 48 2

112. Apply the same logic as in the previous problem. 113.(c) Let the radius of each circle be r unit then PQ = QR = PR = 2r

 PDM =  QEN = 30°

2r Q

P D

Area of circle Area of rectangle

 =

2r

M

N

E

DM  cos 30 DP

25 72

DM = DP ×

O

C

DM =

3 [DP = QE = (r)] 2

r 3 2

DE = DM + MN + NE

 BED = 60°

P

 EDB = 90°

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R2

R 2 = = 4 3–2 2 4 r 2

2

25 2 a  4 = 3a×6a

BD = cos30° BE

Area of larger circle  Area of 4 smaller circle

5a 2

109. (a)  CED = 120 °(ACED is cyclic)A 9.

R 2 –R=r+r 2

Area of the circle  5a  =  ×   2 

Again when OD is perpendicular on the chord AC and OD passes through the centre of circle ABC, then it must bisect the chord AC at D. AD = CD = 6cm

and BP = OB – OP = R 2 – R

eYrai dnag

106. (d)

kgei snhe

=

AC 9 3 = = AD 12 4 110.(c) Let AD = 3a and DC = 6a

 3 3  –    2  

then BQ = r 2

R ( 2 –1) = r ( 2 + 1)

Again  ACB is a right angle triangle (   C = 90°) AC = AB sin30° (sin30° = 1/2) AC = 9cm (alternatively apply Pythagorus theorem). and AD = AB – BD = 12cm

 60 3 2  3 r 2 – r   360 4   r2 2

circle) then OB = R 2 Similarly PQ = MQ = QR = r (radius of the smaller circle)

BE = 4 3 cm BC = BE + CE

r 2 3    = 2  2  = r2 3 cm2  

=

OA = AB = BC = OC = OP (radius) Let OA = R (radius of the larger

6 3 = BE 2

v.iSn ir

1 r2 3 = ×r 3 ×r= cm2 2 2 Area of both the quadrilateral

=

Q R

A

M

B

r 3 r 3 + 2r + = (2 + 2 2

DE = DF = EF = (2 +

3 )r

3 )r

 PAM =  QBN = 30°

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Again,

  OCB =  OBC = 20°   BOC = 180° – (20 + 20)

1 PM = tan30° = AM 3

1

r 1  AM 3

  BAC = 2  BOC = 70° 119. (a) PQ2 = BQ × AQ  (12)2 = AQ × 8

AB = AM + MN + NB = r 3 + 2r + r 3

 AQ = 18 cm  AB = AQ – BQ

AB = BC = AC = 2r ( 1  3 ) Ratio of perimeter of equilateral triangle = ratio of their sides Ratio of perimeter of  ABC :

 DEF :  PQR 3 ) : (2 +

120. (d)  ACB =  ADB = 20° (made by same arc AB)

 In  ACB,

 x° = 180° – 85° – 20° = 75° 1 121.(c)  APB=   AOB 2

3):2

1 1 AOB   90 2 2 = 45°

114. (a) 115. (b)

116. (d)

O

R Enak geisnh

122.(c)  AOC = 360° – (90° + 110°) = 160°

If two cir cle inter sect each other t h en m axi m u m t w o t an gen t s will be dr awn.





B

L



The perpendicular from the centre of a circle to a chord bisects the chord.

1  AL = LB = AB = 5 cm 2 In  OAL  OA2 = OL2 + AL2

  

2

2

13 = OL + 5

2

OL2 = 132 – 52 OL = 12 cm

117.(c)  ODC =  BAC = 38°(Angle made by same arc BC) and OC = OD = Radius



 OCD =  ODC = 38°

118.(b) OB = OC

2

 AOC = 80°

36 = 5(5 + x)

wwM wa.th Les B aryn A

13

1

123.(b)  ADC = Opposite exterior angle = 120°  x° = 180° – 120° = 60° 124..(a) PT2 = PA × PB



cm

 ABC =

5+x=

130.(b) (PT)2 = PA × PB  144 = x × (7 + x)

 x2 + 7x – 144 = 0  (x + 16)(x – 9) = 0  x = 9 or –16 –16 cannot be the length, hence this value is discarded thus, x = 9 cm. 131.(d) PA = PB

eeYa ridn agv .iSn i

= 18 – 8 = 10 cm

= 2r ( 1  3 )

  C +  D = 180°   D = 180° – 48° = 132°

r

 AM = r 3 = BN

= 2 (1 +

= 140°

129.(b) ABCD is a cyclic quadrilateral.

36 = 7.2 5

  PAB =  PBA In  APB

Also,  PAB +  PBA +  APB = 180°

  PAB +  PBA = 120°   PAB =  PBA = 60° i.e.  PAB is an equilateral triangle.  AB = 6 cm

1 132.(a) A = Area of  ABC = ×3× 2 4 = 6 cm2 S = Semiperimeter of  ABC = 354 = 6 cm 2

A

 inradius =

=

S

6 6

= 1 cm

133.(d)

x = 2.2 cm

125. (d) PA × PC = PB × PD

 

14 × 9 = (7 + x) × 7

18 = 7 + x

126. (b)  BAC =



x = 11 cm

1 × 138° = 69º 2

134.(d) O O' 7

  BDC = 180° – 69° = 111° ( ABCD is a cyclic quadrilateral)

127.(c) In  OBC OB = OC

  B =  C = 45°  D = C ( made by same arc AB)   D = x = 45°

128.(a) x = 40° ( made by same arc AB)

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12

 OO’ = 12 – 7 = 5 cm 135.(a)

A x y

B y

x z zC

AB = 5 cm = x + y

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136.(c)  AOB = 2  ACB = 2 × 30° = 60° 137.(b) AB = AC

  ACB =  ABC = 50°   BAC = 180° – (50 + 50) = 80°   BDC =  BAC = 80° (angle by same arc BC) 138.(b)  DOC =  AOB = 70°  OD = OC = radius

  OCD =  ODC =

1 2

(180° –

70°) = 55° 139.(c) It is necessarily a rectangle. 140.(d) 141.(c) Length of transverse tangent =

here d = 10 cm, R1 = R2 = 3 cm 2

2

10 – 6

= 8 cm 142.(a) CD = 7cm  AC = 7 cm and BC = 7 cm ( AC = CD and BC = CD. Two tangents from the same point are always equal)  AB = 7 + 7 = 14 cm 143.(a) Tangent is always perpendicular to the radius.

  BCD = 180° – (30° + 60°)

P

147.(d)  D = 180° – 72° = 108°

  BCD = 180° – 108° = 72° ( AD||BC) 148.(b)  OAP =  OBP = 90° In

AOBP,  O +  P = 180°

  AOB +  APB = 180° 1

  APB =

6

2

= d 2 – R – r  150.(d)

P

2

Q

155. (d)

P

r1

r2

PQ  2 r r  2 4  9 12

12cm

Required area (side PQ) = (12)² = 144 156. (b) O

O2

O1

A

A

151.(d) PQ2 = (r1 + r2)2 - (r1 - r2)2 = 4r1r2 152.(b)

OA  OC 2 CA 2

P

 32  4 2

A

= 9  16  25 = 5 cm 157.(c) Q

O

 AP : AQ = 5 : 8

153.(a)

B

X

O

 OBA = 90°;

From  OAB,

AB  OA 2  OB2

Rakesh Yadav Readers Publication Pvt. Ltd.

M

Y

XM = MY OM = 12 cm OX = 13 cm

A

 XM  OX 2 – OM 2  132  122 =5  XY = 2XM = 10 cm S

158.(a)

R

8

 10 2  6 2

 AOC = 65°

B

C

AC = CB = 4 cm OC = 3 cm

 PQ = 2 r1r2  2 8  2  8cm

144.(b)  BAC = 60°   BEC = 180° – 60° = 120° 145.(c)  CBA =

i.e  ACB = 90° ABC is a right angled triangle.

Q

OA = 10 cm, OB = 6 cm

1



× 180° = 30°

[  AOB :  APB = 5 : 1] 149.(b) Length of common tangent

O’

B

B

AB = diameter of circle. Angle of a semi-circle is a right angle.

= 90°

A

O

A

180°

wwM wa.th Les aBryn

 length =

90°

  BCD +  BDC +  DBC =

2

d 2 – R1  R 2 

C

v.iSn ir

5 + z = 9  z = 4 cm x = 7 – z = 3 cm and y = 6 – z = 2 cm x = 3 cm, y = 2 cm, z = 4 cm

146.(a)  BDC =  BAC = 30° In  BCD

eYrai dnag



x+y+z=9

154.(a)

ERna

  

  CBE = 180° – 65° = 115°

kgei snhe

BC = 6 cm = y + z AC = 7 cm = z + x  2(x + y + z) = 5 + 6 + 7 = 18

O

3 13

O'

 100  36  64  8 cm

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SR =

(OO')² – (R1 – R 2 )²

132  52  188 12 cm

=

OR = 5cm

 B =  C = Incentre is angle bisector So,  ABC is a right angled  167.(b) A

20 3

OP =

 POR = 90º

159.(b)

In triangle POR PR² = PO² + OR²

50° S

O 2

 20      5²   3 

C

B

AB = BC = 8 OA = 10

400  25 9

In  OAB

25 625 = 3 9

 OB  OA 2  AB2

164.(b)

 BAC = 50°   BSC = 100° BS = SC = radius

A

168. (a)

160.(b) The largest chord of circle is its diameter.

O

C

B

 ABC = 180° - 80° – 85º = 15°   AOC = 2  ABC = 2  15° = 30°

165. (a)

A

B

8

 BAC = 90°  BC is the diameter of the circle.

A

O

C

wwM wa.th Les B aryn

 BC  64 36  100 10 cm

 Radius of the circle = 5cm 162.(c)

16 cm E A 15 O 8 D F

B

C

B

Q

= Distance

2

betw een centres  ²  r1  r2 

 10 2  6 2  16 4  8cm 169.(b) One and only one circle can pass through three non-collinear points. 170.(b) 32 + 42 = 52

 ABC is a right angled triangle.  B = 90° = angle at the circumference  Diameter of circle = 5 cm

  OBC =  OCB = 35°  BOC = 180° – 35° – 35° = 110° 

 BAC =

1  BOC = 55° 2 A

OB = 17 cm

In  OFC OC² = OF² + FC² (17)² = 8² + FC² FC = 15 cm DC = 2 × FC = 30 cm. 163.(d)

R

Rakesh Yadav Readers Publication Pvt. Ltd.

C

B

 Circum-radius = 2.5 cm 171.(b) In OPC

OC2  OP2  CP2 2

8   5 2  OP2     OP2  5 2  4 2 2 

I

C

B

 OP2  9  OP  3cm

 BIC = 135°  BIC = 90°+

1 A 2

1  A = 135 – 90 2

O

Q

Y

A

OB² = BE² + OE² OB² = 15² + 8²

P

S

Transverse common tangent

OB = OC = radius

166.(c)

In  BEO

R

R Enak geisnh

6

P

X

10.1 R= = 5.05 2

C

1 (180 100) = 40° 2

  BCS =

 10 2  8 2  36  6

(Greater than 5) 161.(d)

C

B

r

A

eeYa ridn agv .iSn i

PR =

 A = 45 × 2 = 90°

O C

D

P

A

Q

B

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r²  x 2  (5)2 ..................(ii)



 (17  x )2  (12)2  x 2  5 2

OA 2  OQ 2  AQ 2 2

6   5 2  OQ 2     OQ 2  5 2  3 2 2 

 OQ  4cm  Distance between chords AB and CD = PQ = OQ - OP = 4-3 = 1cm 172.(c) LKN  90 0 (angle semicircle)

in

 289  34 x  x 2  144  x 2  25  34 x  408  x  12cm  from(ii)

r 2  (12)2  (5)2  (13)2  r  13 cm 176. (C) OA = OB = radius

 OAB  OBA  25 0

 x = 1800-(100+300) =500

(angle in alternate segment) 173.(a)  ADC = 120º

1 1   A C B   A O B  2 3 0 1 1 5 0 2 2 177.(b) QOP  180 0  120 0  60 0

 ABC = 180º -120 = 60º and ACB  90 0 (angle semicircle)

and PQO  90 0

in

0

174.(d) In AOB, AO = OB (radius)

0

 QPO  180  (90  60 )  30

kgei snhe

 BAC  180 0  (90 0  60 0 )  30 0

0

0

178.(c) ADC  180  55  125

0

0

 CDT  180 0  (125  30 O )  25 0

B

179.(a) In cyclic  AB CD,

ADC  ABC  180 0 O

ERna

 ADC  180 0  70 0  110 0

A

C

now In ADC,

= 60º  ABC = 80;

  CBQ = 180º – 80º = 100º In  BCQ,

 Q = 180º – (100 + 60º) = 20º 186.(b) AB=24cm  AM=MB=12cm In  AMO, (OM)2 =(AO)2-(AM)2 (OM)2=(37)2-(12)2

 OM  35cm

wwM wa.th Les aBryn

180.(d) PQO  PRO  90 0

similarly In OBC,

37

 In PQOR

0

 AOC  2  ABC = 2 × 60º = 120º 175.(d) Let OP = x cm cm  OQ=(17-x ) radius = r cm  In  OQC, (OC)² = (OQ)² + (QC)²

A

( PQ and PR are tangents)

0

 ABC  25  35  60

  BCQ = 180º – 120º

 ACD = 180° – (30°+110°) =40°

 OBA  OAB  25 0

 OBC  OCB  35

184.(c) It will always be possible to divide a circle into 360 equal par ts, because the sum of angle that can be subtended at the centre = 3600 185.(d) AB CD is a cyclic quadrilateral. Therefore  DCB = 180º –  A = 180º – 60º = 120º

eYrai dnag

AOB  360 0  130 0  230 0

 PKL  LNK  60 0

  PBA =180º–(90º + 30º) = 60º   TPA =  PBA = 60º (by alternate segment theorem) 183.(c)  BAC=  BDC=300 ( made by same arc BC) In  ABC,

  AOB = 180° – 50° = 130°  Major

 LNK  180 0  (90  30)  60 0

  BOC = 2 ×  BAC   X = 100º 182.(b)  APB = 900 (angle in a semicircle = 900)

v.iSn ir

In OQA

O

ROQ  360 0  (90 0  90 0  60 0 )  120 0

1 1 QOR  120   60 º 2 2

QSR 

and

181.(d) In AOB OA=OB=radius

In  AMO';

 OAB  OBA  20 0

(O'M)² = (AO')² – (AM)² (O'M)2 = (20)2 - (12)2

A

C

D

B

0

20

O

x

 O'M = 16cm 30

 OO' = OM + O'M = 35 + 16 = 51cm

0

C

O

A

P

B

and In  OAP, (OA)² = (OP)² + (AP)²

Rakesh Yadav Readers Publication Pvt. Ltd.

1

O

M

B

 r²  (17  x )2  (12)2 .......(i) Q

20

similarly In AOC,

 OAC =  OCA = 30º   BAC = 20º + 30º = 50º

187. (a) OM = 4cm = radius of smaller circle and O'M = 6cm = radius of bigger circle



O'N = 8–6 = 2cm In  O'NB,

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B

also,  ACB = 90º

ABK  180 0  (115  30)  35 0

 In ABC,

 KCD  ABK  35 0

 CAB = 180º – (90º + 50º) = 40º M

O' O

191.(b) AOC  2 APC

198.(a) AOC  BOC

 APC  50 0 Als o, ABCP quadrilateral

C 2

2

2

(O'B) =(O'N) +(BN)  (BN)2 = 36- 4 = 32

 OC is the per pend icular bisector of AB  AM = BM IInd -Method

eeYa ridn agv .iSn i

16 AM = MB = = 8cm 2

In AOM and BOM OM = OM (common)

 TP  TQ  TQP  TPQ  50 0

193.(b)  ACB = 90 0 [angle at the point of intersection to the centres of the circles. ] BC = r AC = 2r (as area of x=4 area of y)

17 O D

 AB 

(CN)2  (17)2  (8)2  CN  15cm  CD  2CN  30cm 189.(b) AB = 8cm  AM = MB = 4cm N

 OA  OB 

A

Z 2

z z  180 0   x  y  360 0 2 2 xyz 195.(c) AB = AC and AD = CD  AB = 2AD Now, since AD is a tangent

A

M

2

B

2

3 r  r2     r 2 2

CD 2CO  3r 

AM  MB  PM  MN

 4  4  2  (2r  2)

 CAD  CAB  DAB

 4  r 1  r  5 c m 19 0.(a) AB CD is a cyclic quadrilateral   ADC +  ABC = 180º    ABC = 180º – 130º = 50º

CD 3.r 3   AC r 1

200. (a)

A

196. (d) CAB  BCD and DAB  BDC (alternate segment theorem)

P

 BCD  BDC

BDC  CBD  180 197.(b)

Rakesh Yadav Readers Publication Pvt. Ltd.

0

D

75º

xº

B

C

According to figure 

 CAD  CBD  BCD 

B

 OC  (AC)2  (OA)2 

180 0 

 AB     AP  AB  AB  4 AP  2 

O D

 AD 2  AP  AB

O

r 2

C

Also,  QMR = SMT  y  In quadrilateral PSMT

wwM wa.th Les B aryn

In ONC,

AM BM  AM : BM  1 : 1 199.(b) AB = r (say) then AC = BC = r, also

Z 2

 PSR  PTQ  180 0 

 ON  23  15  8cm

 AOM BOM

r 2  4 r 2  5r

194.(c) QSR  QTR 

 In  AMO, (OM)2 = (17)2 - (8)2  OM = 15cm

AOM  BOM ( AC  BC) OA = OB = radius

(  s in the alternate segment)

 PTQ  180 0  50 0  50 0  80 0

B

N

192.(a) TPQ  PAQ  50 0

R Enak geisnh

M

C

( AC  BC)

cyc lic

 CBD  180 0  130 0  50 0

 BC  4 2  4 2  8 2cm 188.(d) MN=23cm

17

a

 ABC  180 O  50 0  130 0

 NC  BN  4 2

A

is

 ABC  APC  180 0

 BN = 4 2

(  s made by same arc AD )

r

N

AD  BC

  ABC = 75 The n   ABC +  ADC = 180º

616

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75º +  ADC = 180º

205.(b) BAC  BPC  48 0

  ADC = 180º – 75º

( AB is diameter)

 ADC  90 0  20 0  110 0

P 0

209. (b)

48

(AD || BC, corresponding angle) A

105 +  DCB = 180º

  DCB = 75º 201.(d) AP = AS, BP = BQ, CQ = CR and DR = DS  AB = AP + BP = AS + BQ R

C

 ABC  180 0  (90 0  48 0 )  42 0

B

kgei snhe

CD = CR + DR = CQ + DS  AB + CD = (AS + DS ) + (BQ + CQ) = BC + AD 202.(a) TQ = TP and TP = TR  TQ = TP = TR

C

1  2 A  220 0

 2  360 0  220 0  140 0

B

 PAD  180 0  (100 0  30 0 )  50 0

A

BCD  BAD  50 [  s by same arc BD]

ERna

203.(b) AOC  2  60 0  120 0

207.(a) ABC  180  85  95

120 ABC   60 0 2

0

0

110

C

wwM wa.th Les aBryn

In AOP and BOP PA = PB

D

85

0

A

A

1 2

3 4

0

B

0

50

2

0

O

Q

204.(c) APB  3  4  68 0

1

E

 ABO  360 0  (110 0  50 0  140 0 )

 60 0

C 0

 OBE  180 0  60 0  120 0

P

40

1 211. (d) CD  AB  AO OB  r adiu s 2

B

In ADP ,

P

DAP  180 0  (85  40)  55 0

B

 In AQB,

OA=OB= radius

O

A

AQB  180 0  (95 0  55 0 )  30 o

OP = OP (common) and

210. (a)

0

 TQ : TR  1 : 1

O

 COD  2  DAC  2  50 0  100 0

P

0

C

D

 DAC = 180°– (110° + 20°) = 50º

0

30

80

B

20º

eYrai dnag

D A

P

O

A

206. (d) APD  180 0  80 0  100 0

Q

S

ADC  110 0 (fr om Q above)

D

 BCD  ABC  42 0

C

ACD  BAC  20 0

( AB  DC) an d

B

( AB  CD)

A

BDC  BAC  20 0 [  s by same arc BC]

 ADC +  DCB = 180º

D

ADB  90 0 [sem i cir cle]

same arc BC) and ACB  90 0

  ADC = 105º  As we know in a cyclic quadrilateral



(by

v.iSn ir



C

B

D

208. (c)

  AOP BOP

 1  2 and 3  4

0

0

O

A

68  3   34 0 2 0

 1  POA  180  (90  34 )  56

0

Rakesh Yadav Readers Publication Pvt. Ltd.

20

D

B

0

C

P

 CD  OC  OD  OCD is an equ ilat er al tr i angle

COD  OCD  ODC  600

617

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ACB  90 0 (angle in sem icir cle)  BCP  90 0 1 COD  30 0 2 (m ade by sam e ar c CD) CBD 

 In BCP,

APB  180 0  90 0  30 0  60 0



2

 18

 r2 = 9  r = 3 units 

Area of the sector AOB

  OBC =  OCB

A

A

r

ACB  90



 2r 2  3 2

1 1 9  r 2    9  sq.units 4 4 4 218.(b)

215.(c)  BOC = 2  BAC OB = OC

0

OA2 + OB2 = AB2

 BCP  90 0

O

A

O

C

eeYa ridn agv .iSn i

212. (b)

  BAC = 180°- 84° - 44° = 52°  Angle subtended by BC at centre (  BOC) = 2  52° = 104°

C

P

Q

B

O

C

B

P



D

In  BOC

B

2  OBC = 180º –  BOC

1 CBP  COD  15.5 0 2

APB  180 0  90 0  15.5 0  74.5 0 A

213. (d)

F 8

C

BC BD  12 cm 2 OB = 13 cm In  OBD,

E

B

6

r

C

wwM wa.th Les B aryn OF =



52 – 4² = 3 cm

OE =

C

44°

A

40°

T

5² – 3² = 4 cm  EF = OE + OF =4+3 = 7 cm 217.(a)

 ACT = 180°- 44°- 40° = 96°

 CAT =  CBA = 44°  BCA = 180°- 96° = 84

A

In  OAB,  AOB = 90°

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12 

cm

12 cm 

1  Diagonal 2

1 12 6 2   cm 2   Area of  ACD = =



1 6 2 6 2 36    sq.cm 2    A 220.(c)

 CAT = 44°  BTA = 40°

cm

Side of square =

S

In  OBE,

B



Diagonal of square =

In  ODF

132 12 2  169 144

6

 AC = Diameter =

and, OF  CD  FD = CF = 4 cm

OB2  BD2

O

D

Area of circle = r 2  36 36 r2  

OE  AB  BE = AE = 3 cm

 25  5 cm 214. (d)

D

O

A

=

B

 OBC = 90° -  BAC

C

D

OD =

A

  BAC +  OBC = 90° 216. (b)

O B

BOC 2

hence, OQBP must be concylic quadrilateral 219.(b)

R Enak geisnh

OBC  90 

 In BPC,

 OPB =  OQB = 90°  OPB +  OQB = 180° and,  PBQ +  POQ = 180°

O P

B

60°

O

B

In right  OAP and  OBP, AP = PB, OA = OB = radius OP = OP

618

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and  APO =  OPB In  AOP,

 APO = 180° - 90° - 60° = 30°   APB = 2  30 = 60° 221.(c)

In quadrilateral OAPB,



 AOB+  OAP +  APB +  OBP = 360°

In  OCF, (x + 1)2 = r2 - 9 By equation (ii) - (i) (x + 1)2 - x2 = r2 -9 - r2+16  2x + 1 = 7

  APB +  AOB = 180° The quadrilateral will be concyclic. 224.(a) 

O

D

x = 3 cm



 From equation (i), 9 = r2 - 16  r2 = 25  r =5

C

A

x2 = r2 - 16

B

v.iSn ir

  OAP   OBP   AOP =  POB

A

228.(b) O

B

BD = 3 cm  In  OBD 3

1

BO = OC = 15 cm. OD = 9 cm

O

BD= 15 2  9 2  24 6 12 cm  BC = 2  12 = 24 cm. 225.(a) C

3

OB = 3 2 cm

O

C A

A

  AOB = 60°   ACB = 30°

ERna

D

AO = OB =

B

AO = OB = AB

B

O

kgei snhe

222.(b)

226.(b) Side of the equilateral triangle a,

5 2

AC = 5 In  OAC

wwM wa.th Les aBryn

In radius 

5 25 2  OC  5    25  2 4

O’

D

B

OC = 2 cm OA = 4 cm

AC  42 - 22  16- 4  12  2 3

cos 45° = OB  2  OB



C

eYrai dnag

 ABC = 45°   ABO = 45° (  OBC = 90°)

C

D

 AB  4 3 cm 229.(d) ABCD is cyclic parallelogram.   B +  D = 180°  2  B = 180°   B = 90° A

D

B

C

a

2 3

2

Circum-radius 

a 3

230.(b)

Required ratio

10025 75 5 3    4 4 2

 CD= 2× OC  2 

5 3  5 3 cm 2

2

223.(d) OA  AP and OB  BP

 OAP = 90° and  OBP = 90°   OAP +  OBP = 90°+ 90° =180°

B

O

A P

2

 a   a  1 1    4 :1  :    :  3  2 3  3 12 227.(a)

A

C

r 2  4 sq.cm. O E F

B D

Let OE = x cm  OF = (x + 1)cm OA = OC = r cm AE = 4 cm, CF = 3 cm In  OAE, OA2 = AE2 + OE2  r2 = 16 + x2

Rakesh Yadav Readers Publication Pvt. Ltd.

Area of sectors =

Area of square = 4  4 = 16 cm. Area of the remaining portion = (16 - 4  )sq.cm. 231.(c)

5cm

D

O 3cm

A

AC =

C

B

AO 2 -OC 2 = 5 2 -3 2

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 25  9  16  4 cm  AB = 2  4 = 8 cm 232.(b) P

Q

O

 OAC =

A

1800 – 400 = 70º 2

236.(c)

225144  81 9 cm  OF = 21 - 9 = 12 cm In  OCF

r

B

β   (   ) 180

 CD  2  9  18 cm 239. (a)

C

D

BD = DC = 7 3 cm

   β = 90° AD 

  PAQ = 90° 233. (c)

2

AB -BD

2

O

2



 (14 3 ) – 7 3



Q

B

75°

 OD =

 ACB =  BAT = 75° (angles in the alternate segment) In  ABC,

1  21 = 7 cm 3

 62  82  36 64

22 77 154 sqcm . . 7

237.(c)

S

P

wwM wa.th Les B aryn

 ABC = 180°– 45° –75° = 60° 234.(c) D

=

x

y Q

R

C

Length of common transverse tangent

A

B

P

 XY 2  r1 r2 2

 AOC = 130°  ADC =

 PBC =  ADC = 65° (exterior angle is equal to the opposite interior angle) 235.(c)

 100 = 10 cm 240.(b) Tangents drawn from any external point are of s ame length  AD = AE, BD = BF and CE = CF AD = AB + BD = AB + BF and AD = AE = AC + CE = AC + CF  2AD = AB + AC + BF + CF 2AD = AB + BC + CA 241.(b) 242.(c) A

B C

 8  XY 2  9 2

1 130 65 2

1

2

 XY  64  81 145

O

C

A

A

O

OC = O' D = 5cm (radius) CE = ED (CD = 24cm)

 COE   EO'D E

O

F

O C B

B

Rakesh Yadav Readers Publication Pvt. Ltd.

'

E 2 D

 XY  145

238.(b)

B

OA  OC2  CA 2

2  Area of circle = r

T

O'

C

AB = 16, AC = BC = 8cm OC = CO’ = 6 cm

 21 3  7 3  21 cm

R Enak geisnh

45°

A

2

 (14 3  7 3)(14 3  7 3)

C

152 12 2  9 cm

eeYa ridn agv .iSn i

O

Now In  PAQ,

P

152 12 2

=

 CF =

 OQA =  OAQ = β

O

OE =

A

OA = OP and OA = OQ  OA = OP = OQ Let  OPA =  OAP = 

A

AO = OC  5 cm (radius) AB = 24 cm AE = EB = 12 cm In  AEO

 ABC = 180° – 85° – 75° = 20°   AOC = 2×20° = 40°

D

(  C =  D =90°and  1=  2)  OE = O'E and CE= ED=12cm In  COE ;

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OE ²= CE² + OC² OE² = 12² + 5² = 169

245.(b) Let BM = x cm

 OMQ, r2 = x2 + b2 In  ONA, r2 = y2 + a2 In

B



OE=13  OO1 = OE + EO' = 13 + 13 = 26cm = AB 243.(a) Let radius of smaller circle=r

O

M 6

r 

A C

r

N B



=

C

E

ON  AB

AB and BC are tangents to smaller circle



In

& OD=OE=r= radius BM=1cm OB=(1-r)cm ODBE is a square



=

× = ( 2 -1)cm 2 +1 ( 2 -1)

(5) 2  (3) 2  4cm

× AB × ON =

1

2

= 12  x =

wwM wa.th Les aBryn 20 cm

cm 15

M



' O

× 6 × 4 = 12cm

24

=

P

 OAO'



OA OO'

=

AM O'A

 AM =

N D

y O x M

O

Q

B

25

 AB = 2AM = 24cm

= 12cm



OM  DN  x then ON  DM  y let

Let radius = OQ =r cm

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A

200

B

C

0

 NOM  900



65 cm 2

D

C

it mean DMON is a rectangle

15 × 20

2 7  49 65 r² = (2)2 +   = 4 + = 2 4 4  

248.(d) BC = OD (given)  BC = OD = OB = BA = radius

A

 D   M   N  90

 O (common)}

8  4cm 2

(OC)2 = (ON)2 + (NC)2

r 

In quadrilateral DMON,

O

C

N

PM = MQ=

OO'=25cm  (25)2=(15)2 +(20)2   OAO' is a rig ht ang le

{  OAO' =  AMO = 90° and

O

 EM = 4-2 = 2cm  ON = EM = 2cm ( ONEM is a rectangle) Now In  ONC,



2b b 2 2a a and AN = NB = 2

M 3cm E

 AM  BM 

= 4.8cm

B

triangle AB is common chord

B

A

OM  PQ and ON  AB

A

 OMA 

D

2 5  BC = 2x = 9.6 cm 246.(c) Let AB = 2a and PQ = 2b & 'O' is the centre

244.(d)

O

OM  AB

ERna

r=

2

5x

 ( 2 -1)

1

2×6 = 4cm 3

Let O is the centre

Again Area of  AOB



2r = 1 - r  r( 2 +1) = 1

1

 ANO,

ON=

OB= 2r

 CE =

6 AN=BN=  3 cm 2



 OD  AB and OE  BC   DOE = 90°

 

1 5x ×5× x = cm 2 2

247.(a) AE × EB = DE ×CE

eYrai dnag

O

1 ×OA×BM 2

kgei snhe

r

D

Area of  AOB =



a2  b2  c2 2

v.iSn ir

M

(ii)

(i)+(ii) 2r2= a2 + b2 + (x2 + y2) 2r² = a2+b2+c2

N A

(i)

In  BOC, BC = OB   BOC=  OCB = 200

  ABO = 200 + 200 = 400 In  OAB, AO = OB

621

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249.(b) Draw a tangent (PT) from P-

13 c m

A

In  OAP and

P

9

PO

OC = OC (common) OA = OE = radius and CA = CE

OC

 OP =

 OP =

25 4

i,e,  180 0 2

=

(5)

OC

C can take any value 1 to 179  (a + b - c -d) cm take one value for each value of C, i,e,179 values. 255.(c) AB = BC = AC = 2cm ( radius of each circle=1cm)



4

a 2  42



2

=

OA

2

cm

252.(b) BC = 2(OB) =

a  16

( A  90 )

OA =

  ABD   CBA

BD AB   BD.BC  a 2 AB BC



a2 a2 BD    BC a 2  16

C 1 2 3 4

E

D

a2

B

a 2  16

  OAC   OEC   AOC =  COE   1=  2 Similarly  OBD   OED  3 =4

253.(d)

251.(a)

As

C

B

P

PR

PS PT

=

7 21

=

1 3

 PR  3  PQ  72cm and

A

RT

=

PT = 3×PS= 75cm

 ST = PT– PS = 50cm  SN = 25cm

254.(a)If a pair of sides of a cyclic quadrilateral are parallel, it become an isosceles

Rakesh Yadav Readers Publication Pvt. Ltd.

+1

3

cm

O

C

P

OM is the radius of the larger circle.

2 a 2  16

PQ

2

3

16  a 2

= 625

=

2+ 3

B

 PQS   PRT QS

cm

3

A

PS = 25cm



3

2

× 3=

–

 PQS   PMN   PRT



2

M

 N is the mid-point of ST Also In  PQS, PS2 = (24)2 + (7)2

 In AOB

 1+  2+  3 +  4 =1460   2+  2+  3+  3 =1460   2+  3 = 730   COD = 730



3 ×2= 3cm 2

OM = OA + AM =

=

a 2 16 2

 OD = OB – BD =

wwM wa.th Les B aryn

P

AP=

Let O is the centroid, then

0



A

O



R Enak geisnh

 PT2 = PA × PB  PT2 = 9 × 16  PT = 12cm In  OTP,  T=90 0  (OT)2 = (13)2 -(12)2 = 25  OT = r = 5cm 250.(c) In  AOBP (  B =  A = 900)   AOB = 1800- 34 =1460 In  OAC and  OEC

OA

=

AO

O

 OCA

(cyclic quadrilateral) a = b and c = d (Isosceles trapezium) a + b - c - d = (a + b + c + d) - 2( c + d) = 3600 - 4c ( c=d) Since, no angle of the quadrilateral ABCD is reflex

eeYa ridn agv .iSn i

O

7

Here, a + c = b + d = 1800

6  AC = BC = =3cm 2

 OAP=  OCA=900  AOP=  AOC   OAP   OCA

T

B

trapezium.

AB = 6cm

r

  OAB =  ABO = 400   AOB = 1800-(400 + 400) = 1000   AOD = 1800-(1000 + 200) = 600



Are a of the cir cums crib ing circle=  R 2 2

 2+ 3  =    3     (2+ 3)2 3

256.(a)  r1 + r2 = 4

r2  r3  3.4 , r1  r3  2.2 r1

A

r2

B r2

r1 r3

r3 C

622

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 2(r1  r2  r3 )  4  3.4  2.2  9.6

 r1  r2  r3  4.8

263.(d)

semicircle}

  MNB = 180º – 70º = 110º

 MBN = 180º – 38º – 110º = 32º

d 2 =5.2cm  CBY=  CZY=420

(  S by same arc YC)

  ABC = 42 + 42 = 84   C = 1800 - (84 + 500) = 460

(PT)2 = PA × PB  (2AP)2 = PA × PB  4AP2 = AP × BP  4AP = BP  4AP = (18 + AP)  3AP = 18  AP = 6cm  PT = 2AP = 12 cm 261.(b) B C

C  230   BCZ= 2   BYZ =  BCZ = 230 (  S by

A

x  CBD  330

(by alternate segment theorem)

Z  1800  x  1470

(BPDE is a cyclic quadrilateral) and

 BDE =  BED= x  330

{ BE=BD}

E

{by alternate segment theorem}   y=900-330=570

N

x  y  z  33  57  147  2370  M

N

0

O 0 38

20

C

A

 ABN = 90º

{

ang le

in

A

C

D

z

  BAI = 2 [ angle made by arc are at the circumferences is half of the angle made by the arc at the centre.

 ABI =  IBC =

X 2

[ I is the incentre] Now, In  ABI,

C

 BID = y (exterior angle) O

 y =  BAI +  ABI A B

OM  DE and ON  EF  ONEM is a square (  EM = EN)  OM = ON = radius = 13cm

Rakesh Yadav Readers Publication Pvt. Ltd.

O

  BOD = Z

D

M

F

B

1 1 1 + = a b c

B

  2 =  BO'D = 360º–250º = 110º 262.(c) AB = 27 cm  AC = 27 cm  CD = 38 – 27 = 11 cm  DM = CD = 11 cm  EM = 24 – 11= 13 cm  EN = EM = 13 cm

  ABE=  BDE= 330

259.(d)



z

ERna

wwM wa.th Les aBryn X

abc

1 1 1 = + c b a

I y

  1 = 2  BCD = 250º

258.(b)

on dividing both sides by

O

 1250

C

ab = ac + bc

264.(a)

  BCD = 180º – 55º

B



1  ACD =  AOD = 55º 2

Y

0

1

and

2 ab = 2 ac + 2 bc [ AB = AM + BM]

D

A 42

2

O 110 0

same arc BZ) Z

kgei snhe

&

AB = 2 ab

eYrai dnag

B

Y

BM  2 bc



A

circle



Sim ilar ly AM  2 ac

T P

d3  2r3  1.6cm

0’

v.iSn ir

260.(a)

d 2  2r2  5.2cm

257.(c)

O

AB is a common tangent of X and Y

 In  BMN

 diameters d1  2r1  2.8cm

B

b

and  BMN =   BAN = 38º

bigger

a

X

  BNC = 180º – 90º – 20º = 70º

r3  0.8cm,

is

M Z c

 In  CBN

 r1  1.4cm, r2  2.6cm, and

 d2

A

  CBN = 90º

=



Z X X+Z + = 2 2 2

x + z = 2y

x z

2y

2

 3 y =3y = 3 623

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S Q

O P

A

B

R

 OQA =  OPA = 90°  QOP +  QAP = 180°   QOP = 180°- 32° = 148°

90°

9

R

 PQS = 60°  QCR = 130°



O B

In  PBC,  BPD =  BCP +  CBP = 20° + 25° = 45° 268.(c) C

Y

X

271. (b) According to question Given:

B

C 70°

ABC = 70°

B

D

A o

A

D

 BAD = 180 – 70° = 110°

Rakesh Yadav Readers Publication Pvt. Ltd.

E

In cyclic quadrilateral, sum of opposite angles is 180° CDA = 180° – 130° CDA = 50° 274. (d) According to question ABCD is a cyclic parallelogram In a cyclic quadrilateral, sum of opposite angle is 180° But In cyclic parallelogram opposite angle is same

AD||BC P

B

ABC = 130°

P

 APO = 180°- 90°- 65° = 25°

R

A

ABC = 180° – 50°

 OAP = 90°,  AOP = 65°

B

C

C

ABC  CBE = 180°

 AOB = 2  65° = 130°

P

D CBE=50°

 ACB = 65°

D

50° O 40°

= 72° 273. (c) According to question, Given:

C

1 AOC  25 2

A

CDA = 180 – 72 = 108° AD||BC ADC  BCD = 180° (  Sum of corresponding angle of parallel line is 180°) = 180° – 108° BCD BCD

 r = 6 cm

A

C

ABC  CDA = 180°

 r2 + 17r - 6r - 102 = 0

1 BCP  BOD  20  2

72°

B

 r2 + 11r - 102 = 0

270.(a)

D

A

 2r2 + 22r - 204 = 0

wwM wa.th Les B aryn

30° = 35° 267.(c) Join BC

 BCD =  ABC = 70°

AD||BC ABC = 72°

 (r-6) ( r + 17) = 0

 QPS = 180° – (60° + 90°) = 30°  RPS =  QPR –  QPS = 65° –

B = C

272. (d) According to question Given:

 r(r + 17) –6 (r + 17) = 0

 PSQ = 90°

A



R Enak geisnh

In  QSP,  PQS = 60°

Q

and

r

 289 = 81 + 18r + r2 + r2+ 4r + 4

1 QPR   130  65 2

and CBP 

A = D

XZ = (9 + r)cm, YZ = (r + 2) cm XY = 17 cm  XY2 = XZ2 + ZY2  172 = (9 + r)2 + (r + 2)2

C

S

In cyclic trapezium

Y

X

 XZY = 90°

Q

Alternate

2

P

O

N ote: I n tr ap e zium s um of opposite angles are 180°:

Z r

148  74 2

266.(b)

 BCD = 180 – 110° = 70°

  PCR   RPB Similarly,  CPQ   QPA   QPR = 90° Because  APB = 180° 269.(d)

 QOP =  SOR = 2  STR   RTS =

(  C or r es p ond ing ang le s between two parallel line sums to 180°)

In  PCR and  RBP, PC = PB (radii) RC = RB PR is common.

T

r

C

eeYa ridn agv .iSn i

265.(b)

B

C

But B + D =180°

624

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2B  180 B 

277. (b) According to question D

180

C

2 30°

A

B = 90°

B

275. (a) According to question ABCD is a cyclic quadrilateral Given: AB is a diameter CAB = 30° As we know that ACB = 90°

B o

140° D

 ACB  CAB  CBA =180° CBA = 180° – 90° – 30°

C

CAD 

1 2

COD

(The angle subtended by an arc of a c ircle at the centre is double the angle subtended by it at any p oint on the remaining part of the circle) CAD 

1 2

CBA = 60° Note: In a cyclic trapezium sum of opposite angle is 180°  D  B = 180°

eYrai dnag

40°

D = 180° – 60°

 D= 120°

D

C

A

B

278. (a) According to question

kgei snhe

A

We know tangents drawn to c ir c le fr om s am e e xte r nal point are equal  AM = AQ = x  MB = BN = 6 – x QD = DP = 7 – x Let NC = PC = z Now 7 – x + z = 5 (consider side DC) –x + z = – 2 ....(i) BC = 6 –x + z .........(ii) Put the value of equation (i) in equation (ii) BC = 6 – 2 BC = 4 cm Alternate:AB + CD = BC + AD 6 + 5 = BC + 7 11 – 7 = BC 4 cm = BC 281. (b) According to question

v.iSn ir

BCD = 90°

B  D  180

140

D

P

CAD = 70°

60°

 C = 180º – 110º = 70º

ADC = 70°

ABC = 180° – 70° = 110° PBC = 70°



 PCB = 60°  PBC  PCB = 70° + 60° = 130°

30°

279. (c) According to question A

B

D

C

D

Given:

 COD = 120º  BAC = 30º  BCD = ?

1  CAD =  COD 2

2

120 = 60°



 B +  D = 180° 180° – 2  +  B = 180° B = 2

280. (a) According to question

BCD = 180° – 90°

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x

A

M 6–x

B 6–x

x

N

Q

Z

7–x

D

7–x

P z

C

BD = 2 units

C

In  ADC  A +  D +  C = 180°  D = 180° – 2 

 BAD = 30° + 60° = 90° Note: In cyclic quadrilateral sum of opposite angle is 180° BAD  BCD = 180°

B



120°

1

D

A

B

o

CAD 

ABCD is a cyclic quadrilateral  A  C =180°  B   D = 180° A  B  C  D = 360° 282. (b) According to question

BCD = 180° – 60° = 120°

wwM wa.th Les aBryn

276. (b) Acc or d ing to que s tion, ABCD is cyclic quadrilateral with centre'O'.

B

A

ERna

 DAB = 70 + 40 = 110° In cyclic quadrilateral sum of opposite angles are 180° A  C =180°

A

C

70°

C

 AB =

2 units

Area of square = 2 units Area of four semicircles r ²

= 4

2

= 2r2

 2 2 2     = units = 2   = 2  4  2 Area of circle = r² = 1² =  Required area of shaded portion = Area of square – Area of circle + area of semicircle = 2 +–  = 2 sq. units

625

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A G

E B

F

C

AE = AH .......(i) BE = BF .......(ii) DG = DH .....(iii) GC = FC ......(iv) Add equation (i) ,(ii),(iii) and (iV) AE + BE + DG + GC = AH +BF + DH + FC AB + CD = AD + BC  6 + 3 = AD + 7.5 AD = 9 – 7.5 = 1.5 cm Alternate:AB + CD = DA + BC 6 + 3 = 7.5 + DA DA = 1.5 cm 284. (a) According to question Given: B

A

E

95°

D

C

BE = BC ADC = 70°,  BAD = 95° DCE = ? In cyclic quadrilateral sum of opposite angle is 180°  BCD = 180° – 95° = 85°

ABC = 180° – 70° = 110°  EBC = 180° – 110°= 70° BE = BC = BEC  BCE In BCE BCE + BEC + EBC = 180° 2BCE = 180° – 70° BCE = 55° DCE = BCE +  BCD = 55° + 85°= 140° 285. (d) In a cyclic quadrilateral sum of opposite angle is 180°  A  C = B  D = 180°

i.e OD = MB =

D

C

x y

=

4 3

MB  OD,MB = OD = x/ 2

BD  MO,MO = BD =

289. (a)

55°

A

D

[Semi-circle]

In  CAD DAC + DCA + CDA = 180° CDA = 180°– 90° – 55° = 35° As we know that in a cyclic quadrilateral sum of opposite angle is 180°  D + B = 180° B = 180° – 35° B = 145°

288. (c)Let ABCD is a square of x unit side B

x A 2

O

2 x

x

5x

=

2

Ans.

A

Q

O

B

C

P

As we know that Angle subtended by the chord at the centre is double to the angle subtended by the same chord at the circle  BOP = 2BAP

A  A 2 Similarly BOR = 2BCR = 2×C/2 =C The re fore , R QP = 1/2 ( R OB +BOP) = 1/2 (C + A) And also we know that A + B C = 180º A + C 180º– B RQP = 1/2 180– B)

C

290. (c)

D

x

C P B

A

PA.PB = PC.PD 8 × 6 = 4 × (PC + CD) 48 = 4 (4 + x) 12 = 4 + x x = 8, PD = 12 291. (a) In ΔACB

   

C

D

2

B 2

 RQP = 90 –

x

x

2 x

2

DAC = 55° DCA = 90°

x

2

 2 x

R

C

90°

x/ 2

2

 x  +   2

R=  

287. (c) According to question Given : B

M

x

2  then MBDO will be a Rectangle become

 B = 7x°  A = 4x° = 5y°  D = y° C As we know that in a cyclic quadrilateral sum of opposite angle is 180°  A + C = 180° 4x° + 5y° = 180° ......(i) + = 180° B D 7x° + y° = 180° ....... (ii) From equation (i) and (ii) 4x + 5y = 7x +y 4y = 3x

wwM wa.th Les B aryn

70°

A

r

H

Line MB inscribed  OD

B

eeYa ridn agv .iSn i

D

286. (b) According to question

R Enak geisnh

Alternate : In such type of questions the area of the shaded portion is equal to the area of the regular fig ure on which the sem icircles are made. In this case the area of the square. 283. (d) According to questions

A

34º O

56º B

Q

Then AOD = 90° then OD =

x

2 diagonal of square ABCD =

Rakesh Yadav Readers Publication Pvt. Ltd.

2 x

 ACB = 90º (Angle formed by semicircle is 90º) ACB + CAB + CBA = 180º 90º + 34º +  CBA = 180º CBA = 56º

626

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CHAPTER

25

CO-ORDINATE GEOMETRY Sol.

20 15

12 3 4

X

–1 IV –2 Quadrant –3 (+,–1) –4

Y' Region Quad- Nature of rant X&Y

0

X'

I

x >0,y>0

(+,+)

YOX'

II

x 0

X'OY'

III

x 0 and the point (a, b) lies in the third quadrant in which the point (–b, –a) lies is :

84. The shaded region in the given figure is the solution set of the inequalities :

x+

a b c 1 1 1 (c)(( a  b  c 2 2 2

76. The area of triangle formed by the lines 3x – 6y = 12, 3x – y = 3 and x -axis is : (a) 5.4 sq. units (b) 2.7 sq. units (c) 4.5 sq. units (d) 3.6 sq. units 77. The shaded region represents :

R Enak

a1 b1 c1 (b)((( a  b c 2 2 2

Rakesh Yadav Readers Publication Pvt. Ltd.

85.

86.

87.

88.

89.

90.

91.

(a) x + y  2, x + 3y  3, x  0, y0 (b) x + y  2, x + 3y  3, x  0, y0 (c) x + y  2, x + 3y  3, x  0, y0 (d) x + y  2, x + 3y  3, x  0, y0 Area of the rectangular region 2  x  5, –1  y  3 is : (a) 9 sq. units (b) 12 sq. units (c) 15 sq. unit (d) 20 sq. units Graph of the inequation 2x – 5y  5 in cartesian plane is : (a) above the line 2x – 5y = 5 (b) below the line 2x – 5y = 5 (c) on & below the line 2x – 5y = 5 (d) on & above the line 2x – 5y = 5 Find the area bounded by |x| + |y| = 6 (a) 72 sq. units (b) 48 sq. units (c) 54 sq. units (d) 84 sq. units The area of the region bounded by y = |x| –1 and y = 1 – |x| (a) 3 sq. units (b) 4 sq. units (c) 2 sq. units (d) 1 sq. unit The point (–5, 7) lies in the quadrant: (a) First (b) Second (c) Third (d) Fourth The point (7, –5) lies in the quadrant: (a) First (b) Second (c) Third (d) Fourth Find the distance between the points (–6,2) and (2 , 4) : (a) 2 17

(b) 4 17

(c) 2 5 (d) 10 92. The distance between the points A (b, 0) and B (0, a) is : (a)

a 2  b2

(b) a 2  b2

(c)

ab

(d) a + b

636

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 11 (a)   9 , 

 0  

 11  (b) 0,  9   

11   11     (d)  0,  4  4  101. The slope of the line 3x + 7y + 8 = 0 is : (a) 3 (b) 7 (c)  0,

3 3 (d) 7 7 102. The slope of the line joining P(–4, 7) and Q(2, 3) is : (c) 

(a) 

2 3

(b)

2 3

3 3 (d) 2 2 103. The equation of a line parallel to x -axis at a distance of 6 units and above x -axis is : (a) x = 6 (b) y = 6x (c) x = 6y (d) y = 6 (c) 

114. If the point (x, y) is equidistant from the points (a + b, b – a) and (a–b, a + b) then bx = ? (a) a2y (b) ay2 (c) ay (d) a2 y2 115. If the sum of the square of the distance of the point (x, y) from the point (a, 0) and (–a, 0) is 2b2, then : (a) x2 + a2 = b2 + y2 (b) x2 + a2 = 2b2 – y2 (c) x2 – a2 = b2 + y2 (d) x2 + a2 = b2 – y2 116. P (– 4, a) and Q(2, a + 4) are two points and the co-ordinates of the middle point of PQ are (–1, 4). The value of a is : (a) 0 (b) 2 (c) –2 (d) 3 117. If the points P(2, 3), Q(5, a) and R (6, 7) are collinear, the value of a is : (a) 5/2 (b) – 4/3 (c) 6 (d) 5 118. The equation of a line parallel to x-axis and passing through (– 6 ,– 5 ) is : (a) y = – 5 (b) x = – 6 (c) y = – 5x (d) y = – 6x – 5 119. The equation of a line parallel to y-axis and passing through (2 ,– 5 ) is : (a) x = 2 (b) y = –5 (c) y = 2x (d) x = – 5y 120. Two vertices of a triangle PQR are P(–1, 0) and Q(5, –2) and its centroid is (4, 0). The co-ordinates of R are : (a) (8, –2) (b) (8, 2) (c) (–8, 2) (d) (–8, –2) 121. The co-ordinates of the point of intersection of the medians of a triangle with vertices P(0, 6), Q(5, 3) and R(7, 3) are : (a) (4, 5) (b) (3, 4) (c) (4, 4) (d) (5, 4) 122. The ratio in which the line segment joining A(3, –5) and B(5, 4) is divided by x-axis is : (a) 4 : 5 (b) 5 : 4 (c) 5 : 7 (d) 6 : 5 123. The ratio in which the line segment joining P(–3, 7) and Q (7, 5) is divided by y-axis is : (a) 3 : 7 (b) 4 : 7 (c) 3 : 5 (d) 4 : 5

ERna

kgei snhe eYari dnag v.iSn ir

104. The equation of a line parallel to y -axis at a distance of 5 units to the left of y-axis, is : (a) y = –5 (b) x = –5 (c) x + 5y = 0 (d) y + 5x = 0 105. The equation of a line parallel to x -axis and at a distance of 7 units below x -axis is : (a) y = –7 (b) x = 7 (c) x = –7 (d) y = –7x 106. The area of the triangle whose vertices are P (4, 5), Q(–3, 8) and R (3, –4), (in square units) is : 1 (a) 66 (b) 16 2 (c) 33 (d) 35 107. The points A(0, 0), B(0, 3) and C(4, 0) are the vertices of a triangle which is : (a) Isosceles (b) Right angled (c) Equilateral (d) None of these 108. The co-ordinates of the centroid of  PQR with vertices P(–2, 0), Q(9, –3) and R (8, 3) is : 19  (a) (1, 0) (b)  3 , 0    (c) (0, 5) (d) (5, 0) 109. The equation of a line passing through the points A (0, –3) and B (–5, 2) is : (a) x + y + 3 = 0 (b) x + y – 3 = 0 (c) x – y + 3 = 0 (d) x – y – 3 = 0 110. The length of perpendicular from the origin to the line 12x + 5y + 7 = 0 is : (a) 2 units (b) 1 unit

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93. The distance between the points A (7, 4) and B(3, 1) is : (a) 6 units (b) 3 units (c) 4 units (d) 5 units 94. The co-ordinates of point situated on x-axis at a distance of 5 units from y-axis is : (a) (0, 5) (b) (5, 0) (c) (5, 5) (d) (–5, 5) 95. The co-ordinates of a point situated on y-axis at a distance of 7 units from x -axis is : (a) (0, 7) (b) (7, 0) (c) (7, 7) (d) (–7, 7) 96. The co-ordinates of a point below x-axis at a distance of 6 units from x -axis but lying on y-axis is : (a) (0, 6) (b) (–6, 0) (c) (0, –6) (d) (6, – 6) 97. The distance of the point (6, –8) from the origin is : (a) 2 units (b) 14 units (c) 7 units (d) 10 units 98. The point of intersection of the lines 2x + 7y = 1 and 4x + 5y = 11 is : (a) (4, –1) (b) (2, 3) (c) (–1, 4) (d) (4, –2) 99. The line 4x + 7y = 12 meets x axis at the point : (a) (3, 1) (b) (0, 3) (c) (3, 0) (d) (4, 0) 100. The line 4x – 9y = 11 meets yaxis at the point :

7 7 units (d) units 13 11 111. The angle which the line joining (c)

the points



3,1 and



 15, 5 

makes with x–axis is : (a) 30° (b) 45° (c) 60° (d) 90° 112. The lines whose equations are 2x – 5y + 7 = 0 and 8x – 20y + 28 = 0 are: (a) parallel (b) perpendicualr (c) coincident (d) intersecting 113. If the distance of the point P(x, y) from A(a, 0) is a + x, then y2 = ? (a) 2 ax (b) 4ax (c) 6ax (d) 8ax

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2y –

12 x – 9 = 0 and 7 = 0, is: (a) 30° (b) 45°

3y–x+

1º 2 130. If P(3, 5), Q (4, 5) and R(4, 6) be any three points, the angle between PQ and PR is : (a) 30° (b) 45° (c) 60° (d) 90°

(d) 22

11 21  and B  2 , 2  in the ratio 2 : 3  

(c) 3x + y + 7 = 0 or x – 3y – 31 =0

131. Given a  PQR with vertices P (2, 3), Q (– 3, 7) and R (– 1, –3). The equation of median PM is : (a) x – y + 10 = 0 (b) x – 4y –10 = 0 (c) x – 4y + 10 = 0 (d) None of these

(b) (4, 5)

3 7   5 (c)  4,  (d)  2 , 2  2     133. The length of the portion of the straight line 8x + 15y = 120 intercepted between the axes is : (a) 14 units (b) 15 units (c) 16 units (d) 17 units 134. The equation of the line passing through the point (1, 1) and perpendicular to the line 3x + 4y – 5 = 0, is : (a) 3x + 4y – 7 = 0 (b) 3x + 4y + k = 0 (c) 3x – 4y – 1 = 0 (d) 4x – 3y – 1 = 0 135. The equation of a line passing through the point (5, 3) and parallel to the line 2x – 5y + 3 = 0, is : (a) 2x – 5y – 7 = 0 (b) 2x – 5y + 5 = 0 (c) 2x – 2y + 5 = 0 (d) 2x – 5y = 0 136. The sides PQ, QR, RS and SP of a quadrilateral have the equations x + 2y = 3, x = 1, x – 3y = 4, 5x + y + 12 = 0 respectively, then the angle between the diagonals PR and QS is: (a) 30° (b) 45° (c) 60° (d) 90° 137. The equations of two equal sides of an isosceles triangle are 7x – y + 3 = 0 and x + y – 3 = 0 and its third side passes through the point (1, –10). The equation of the third side is : (a) x – 3y – 31 = 0 but not x – 3y – 31 = 0

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(d) 3x + y + 7 = 0 but not x – 3y – 31 = 0 138. If P1 and P2 be perpendicular from the origin upon the straight lines x sec θ + y cosec θ = a and x cos θ – y sin θ = a cos 2 θ respectively,

r

are : (a) (4, 3)

wwM wa. th Les B aryn

(c) 60°

(b) neither 3x + y + 7 = 0 nor x – 3y – 31 = 0

then the value of 4 P12  P22 is :

geisnh eeYa ridna gv.i Sni

 10   P 1,  divides the join of the 3  point A(–3, 2) and B(3, 4) is : (a) 2 : 3 (b) 1 : 2 (c) 2 : 1 (d) 3 : 1 125. The equation of a line with slope 5 and passing through the point (–4, 1) is : (a) y = 5x + 21 (b) y = 5x – 21 (c) 5y = x + 21 (d) 5y = x – 21 126. The value of 'a' so that the lines x + 3y – 8 = 0 and ax + 12y + 5 = 0 are parallel is : (a) 0 (b) 1 (c) 4 (d) – 4 127. The value of P for which the lines 3x + 8y + 9 = 0 and 24x + py + 19 = 0 are perpendicualar is : (a) –12 (b) – 9 (c) – 11 (d) 9 128. The value of a so that line joining P(–2, 5) and Q (0, –7) and the line joining A (–4, –2) and B(8, a) are perpendicular to each other is : (a) –1 (b) 5 (c) 1 (d) 0 129. The angle between the lines represented by the equations

132. The co-ordinates of the point P which divides the join of A(3, –2)

R Enak

124. The ratio in which the point

(a) a2

(b) 2a2 2

(d) 3a 2 a2 139. Find the equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9 ? (a) x + 2y – 6 = 0 but not 2x + y – 6=0 (b) neither x + 2y – 6 = 0 nor 2x +y–6=0 (c) 2x + y – 6 = 0 but not x + 2y – 6=0 (d) x + 2y – 6 = 0 or 2x + y – 6 = 0 (c)

140. Romila went to a statomeru shop and purchased 2 pencils and 3 erasers for ` 9. Her friend Sonali saw the new variety of pencils and erasers with and she also bought 4 pencils and 6 erasers of the same kind for ` 18. rep rese ntationis situation a l g e b r a i c a l l y a n d graphically. 141. Aftab tells his daughter, "Seven years ago, I was seven times as old as you were. Also, three years from now, I shall be three times as old as you will be." interesting?) Represent this situation algebebraically and graphically. 142. 5Pens and 7 Pencil together cost ` 50. Whereas 7 pens and 5 pencils together C ost ` 46. find the cost of one pencil and that of one pen.

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ANSWER KEY (c) (c) (c) (d) (c) (b) (c) (a) (b) (b) (a) (b) (a) (a)

15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28.

(a) (c) (b) (c) (a) (b) (a) (a) (d) (c) (d) (c) (b) (a)

29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42.

(b) (b) (c) (a) (b) (b) (c) (b) (b) (c) (b) (b) (a) (d)

43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56.

(d) (b) (a) (b) (c) (a) (c) (d) (b) (c) (a) (b) (b) (a)

57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70.

(c) (a) (d) (b) (a) (c) (b) (d) (b) (b) (d) (c) (a) (b)

71. 72. 73. 74. 75. 76. 77. 78. 79. 80. 81. 82. 83. 84.

(a) (c) (a) (c) (a) (b) (c) (a) (d) (c) (b) (d) (a) (d)

85. 86. 87. 88. 89. 90. 91. 92. 93. 94. 95. 96. 97. 98.

(b) (c) (a) (c) (b) (d) (a) (b) (d) (b) (a) (c) (d) (a)

99. (c) 100.(b) 101.(c) 102.(c) 103.(d) 104.(b) 105.(a) 106.(c) 107.(b) 108.(d) 109.(a) 110.(c) 111.(a) 112.(c)

113.(b) 114.(c) 115.(d) 116.(b) 117.(c) 118.(a) 119.(a) 120.(b) 121.(c) 122.(b) 123.(a) 124.(c) 125.(a) 126.(c)

kgei snhe eYari dnag v.iSn ir

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.

127.(b) 128.(d) 129.(a) 130.(b) 131.(c) 132.(a) 133.(d) 134.(d) 135.(b) 136.(d) 137.(c) 138.(a) 139.(d)

EXERCISE

1.

(c) Distance between the points =

x

1

2

 –3  5   2  4  =  2   2  =1,3    

2

– x 2   y1 – y 2 

4. 5=

2

=

25 = (4)² + (A – 3)² 25 = 16 + A² + 9 – 6A 9 – 9 = A² – 6A  A² – 6A= 0

A(4,2

=

25 = C² + 16 C² = 9 C = 3

3.

)

D(8,14)

B (8

1 [4(2–14)+8(14 –2) +8(2 – 2)] 2

=

1 [4×(–12)+8 × 12 + 8 ×0) 2

=

1 [–48 + 96] = 24 sq. units 2

1 [x (y – y3)+x2(y3 – y1) +x3(y1 – y2)] 2 1 2

1 [4(14–10)+8(10 –2) +4(2 – 14)] 2

=

1 [4 ×4 + 8 ×8 – 4 ×12] 2

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1 [x (y – y3)+x2(y3 – y1) +x3(y1 – y2)] 2 1 2

=

, 2)

=

(c) Co-ordinates of Mid-points  x1  x 2 y1  y 2   , of AB =   2   2

0)

ERna

,1 C (4

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2.

Area of Δ ABD

(d) Area of Triangle ADC

2

8 – 4  A – 3

 A (A – 6) = 0 A = (6,0) (c) (5)² = (C – 0)² + (4 – 0)²

= 16 sq. units

Hence Area of Qadrilateral ABCD = Area of [ Δ ABD + Δ ADC] = 16 + 24 = 40 Sq. units

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C(10,6)

(c) D

9.

(b) –6x + 9y = 36

A Y

O is the mid point of diagonal AC and BD Co-ordinate

X'

 4  10 2  6   , of O =   2   2

x y + =1 8 7 Length of Intercepts made at x axis at x = 8 y axis at y = 7

(c)

x y – =1 5 5



P(2,8) and Q(6,9)

8 6 B 4 2

.

X'

2 4 6 8 10 12 –2 –4 –5 –6 C –8 –10

–10 –8 –6 –4 –2

Y'

Area of Δ OAB 1 × 8 ×3 2 = 12 Sq. units

=

X



x y + =1 3 6



x y + =1 6 12

Y'

OA = 5 OB = 10/3 OC = –5 (B ut d istance is always positive) Area of Δ ABC = area of Δ OAB

Y

8

12.

.

x y + =1 3 –8

3 A 0 X –1 1 2 3 4 –2 –3 –4 –5 –6 –7 –8 B

.A

1 80 160 ×4× = sq. units 2 13 13 (b) 4x + 2y = 12 8x + 4y = 48 =

=1

10

x y – =1 3 8

X'

x 3y + 5 10

Y

wwM wa. th Les B aryn

9– 8 1 Slope m = = =.25 6– 2 4 (a) 8x –3y = 24

Area of Δ ABC

2x + 3y = 10 =



X'

+ area of Δ OAC =

1 10 1 ×5× + ×5×5 2 3 2

=

50  .25  25 25 + = + 6  2  3 2

=

50  75 125 = sq.units. 6 6

Sol.11(a) On x-axis; y = 0 4x + 3y = 12 = x = (3,0) .......(i) 5x + 7y = 35 x = (7,0) ......(ii) 

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×5

20x+15y = 60 20x+28y = 140 – – – –13y = – 80 80 y=  13

1 ×6×4=12 sq. units 2 (b) 2x –2y = 10

(x2,y2)

y2 – y1 Slope of line m = x – x 2 1

8.

Y'

10.

(7,0)

Y'

Area of Δ OAB =

4

X

4x +3y = 12 4×5x+7y=35×4

Q

(x1,y1)

0

–6 –5 –4 –3 –2 –1

R Enak

7.

B

(3,0)

5×

= (7,4) Point O is also situated on line x –3y + k =0 7 – 3 × 4 + k =0 7 – 12 + k = 0 k=5 (b) 7x + 8y–56 = 0 7x + 8y = 56

P

X'

4 3 2 1

B

geisnh eeYa ridna gv.i Sni

A(4,2)

   6.

7 C  13  6 5 4 3 2 1 A B X 1 234 5 6 7

x y + =1 –6 4

O

 80   x, 

Y

r

5.

Y

(0, 12) D 12 11 10 9 8 7 (0,6) A 6 5 4 3 2 1B

(6,0) C

1 2 3 4 56 7 8 9 –1 (3,0) –2 –3 –4 –5 –6 –7 –8 –9 –10 –11 –12

X

Y'

Area of quadrilateral = Area of Δ OCD – Area of Δ OAB =

1 1 ×6 ×12 – ×3 ×6 2 2

= 27 sq. units

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16.

(a) x + y = 2 3x + 4y = 24 Y D

C X' –1 –2 –3 –4 –5 –6 –7 –8

8 7 6 5 4 3 2 1

17.

(8,0) B X 1A2 3 4 5 6 7 8 9

18.

Y'

  19.

B

y=

X

2

20.

O

D

A

2

2

A(2,0) D

4  16 = 20 = 2 5 Area of Rhombus = Base × height

Y'

=

1 AE(AB + CD) 2

1 × 2 (2 + 4) = 6 sq. unit. 2 15. (a) 5x + 20y = 11 and 2x + ky = 17 will be intersecting lines or having unique solution

=

21.

5 20   2 k  k8

22.

2

0 – 4  y – 3

36 + (y –5)² = 16 + (y – 3)² 36 + y² – 10y + 25 16 + y² – 6y + 9 4y = 36 y=9  Required point is (0,9). (d) a = 2 , b =3.

 x1  x 2  x 3 y1  y2  y3     ,  3 3    0  8  8 6  12  0    ,   3 3   16    ,6   3  (a) Vetices of triangle are (3, – 5)(–7,4) and (h,k) and centroid is (2,–12)

 3 – 7  h –5  4  k   , (2,–12)    3 3  

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2

=5

2

2

=4

2

2

=3

7 – 3  4 – 7

b=

7 – 3  4 – 4

c=

3 – 3  7 – 4

ax1  bx 2  cx 3 ay1  by2  cy 3    ,   a b c  a b c  5  3  4 3  3  7 5  4  4  7  3  4  ,  =  543 5 43 







= (4,5)

25.

x y  =1 a b Given, a + b = 18 ........(i) Pasing throug (4,4)

(d) Let a line

4 4 + =1 a b  ab = 4(a + b) =72 (a – b)² = (a + b)² – 4ab = (18)² – 4 × 72 = 324 – 288 = 36 a – b =  6 .....(ii) On solving equation (i) and (ii) (a,b) =(12,6) or (6,12)

It means

x y + =1 2 3  3x + 2y = 6 (a) Centroid a triangle=

On solving. (h,k)  (10,–35)

2

a=

Incentre I 



Point D wil be (1, –2) ABCD is a trapezium. AB = 2, CD = CE + DE + 2 + 2 =4 height = AE = 3 – 1 = 2 Area of trapezium ABCD

2

4 – 2  4 – 0

2

0 – 6  y – 5 2

X

2

AB =

x – 2  5 – 3

2

C

O

x y + =1 a b

.

Y

h=4

4 – 2  3 – 3

wwM wa. th Les aBryn

E

2 y=3

(d)

B(4,4)

ERna

x=1 x=3

. X'

=     

2

(x –2)² = 0 x=2 (a)Let the required point be P (0,y).  PA = PB =

x y  =1 3 –3

C

2 5 × 4 = 8 5 sq. units 24. (c)

=

1 1 × 8 ×6 – ×2 ×2 2 2 = 22 Sq. units (a) x = 1, x = 3, y = 2 and x = y + 3

Y

–5 3 (c) OA = OB = radius

=

=



23.

k =

x y + =1 ......(i)  2 2 x y + =1 .....(ii)  8 6 Area of quadrilateral = Area of Δ OBD – Area of Δ OAC

14.

(c) 4x + ky = 3 and 3x + 2y =7 are perpendicular to each other. Condition for perpendicular : a1a2 + b1b2 = 0 4 × 3 + k × 2 = 0  k=–6 (b) (k + 1)x + ky = 3 and 5x – 2y = 7 are perpendicular for perpendicular : a1a2 + b1b2 = 0  (k + 1) × 5 + k × (– 2) = 0  5k + 5 – 2k = 0

kgei snhe eYari dnag v.iSn ir

13.

x y + =1 12 6  x + 2y = 12 Or,

Line will be

x y + =1 6 12  2x + y = 12

26.

(c)

A (5,–2)

3:1 P(x,y)

B (9,6)

 9  3  1 5 6  3  1 –2   , (x,y)=   3 1 3 1   =(8,4)

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

 θ= 4

28.

y2 – y1 (a) m1 = tan θ = x – x 2 1

=

AB =

BC =

1 (x – 1) 4 4(y – 3) = x – 1 x –4y + 11 = 0 (b)

AC =

2

2

2

2

2

–1 – 4  –1 – 4

3 –1 1 m1 = 2 – (–2) = 2 and

–4 – 3 7 = –2 – 2 4 Let θ be the angle between BA and BC. Then

m2 =

m2 – m1 tan θ = 1  m m 1 2

 –3  5 2  4  =  2 , 2  = (1,3)   (b) The required coordinates of the point which divides AB in the ratio 5 : 3

wwM wa. th Les B aryn x y x y + =1, + =1 6 7 –4 7 x -axis BC = 4 + 6 = 10 unit AO = 7 units

and

34.

1 Area of triangle = ×BC × OA 2

1 ×10 ×7 = 35 units 2 (b) Let A = (4,3), B = (7, –1), C = (9,3)

=

2

7 – 4  –1 – 3

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5 6 – 32 58 – 3 4 = , 5–3 5–3 = (12,14) (b) Let A = (4, –2), B= (5,5) and C = (–2,4) Then a = BC =

7 1 – 4 2 2 7 1 =  = 1  3 4 2

2

2      θ = tan–1  3 

37.

2

–2 – 5  4 – 5

=5 2 b = AC =

2

c = AB =

2

2

5 – 4  5  2

(b) We have 3x + 4y – 12 = 0  3x + 4y = 12 3x 4y + =1  12 12 x y  + =1 4 3 x y + =1 a b Then the required intercepts on the axes are 4 and 3. (c)Let the equation of the cost curve as a straight line be y = mx + c ....(i)

Which is of the form

2

4  2  –2 – 4

= 6 2

=5 2.

18

site sides are equal. Hence ABCD is a parallelogram. 36. (b) Let m1 and m2 be the slope of BA and BC respectively. Then

mx2 – nx1 my2 – ny1 , m –n m –n

Y'

10 ,AD =

 AB = CD and BC = AD i.e., the oppo-

 x1  x 2 y1  y2    , =   2   2

33.

5 5  =  ,  2 2  {substitute the value of (a,b,c} {x1,x2,x3, y1,y2 and y3}. (c) Let A, B, C and D be the vertices of the quadrilateral whose coordinates are (–2,1) (1,0), (4,3) and (1,2) respectively. Now, AB = 10 ,BC = 18 ,DC

=

–1 – 3  –1 – 5

R Enak

12 11 10 9 8 7 6 5 4 3 2 1 (6,0) C

2

35.

2

= 50 AB² + AC² = BC²  Hence by the Pythagoras Theorem, ABC is a right angled triangle. 32. (a) The coordinates of the mid points of AB

0 X' X –6 –5 –4 –3 –2 –1 1 2 3 4 5 6 7 8 9 –1 –2 –3 –4 –5 –6 –7 –8 –9 –10 –11 –12

AB =

ax1  bx 2  cx 3 ay1  by2  cy3    ,   a b c  a b c 

3 – 4  5 – 4

= 52

Y

30.

2

= 2

(y –3) =

(–4, 0) B

2

4 – 9  3  3

= 25 =5  AB = CA = 5 Hence ABC is an isosceles triangle 31. (c)Let A = (4,4), B = (3,5), C = (–1,–1) Then

3 1 = –4 1– 2 –1 –1 1 Slope of line L = m = = –4 4 1 Equation of line L will be

(0, 7) A

2

9 – 7  3  1

20 = 2 5

CA =

=

29.

2

BC = =

and (x1, y1) = (4, – 2), (x2,y2) = (5,5). (x3,y3) = (–2,4)  The coordinates of the incentre of the  ABC are

25 = 5

r

(b) xcos θ + ysin θ =2  ysin θ = – xcos θ + 2  y = – xcot θ + 2cosec θ m1 = – cot θ x – y = 3 y = x – 3 m2 = 1 Both lines are perpendiculars to each other so m1.m2 = – 1 (– cot θ ).(1)= –1  cot θ = 1

geisnh eeYa ridna gv.i Sni

27.

38.

642

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B (0,b) 4 C(8,10) 5 (a,0)

2

2

36  10  6 = 4 units. 169 40. (b) The intersection point of the lines 3x + 4y – 7 = 0 and x – y + 2 = 0 is, 3x + 4y – 7 = 0 ...... (i) x – y + 2 = 0 ........ (ii) x = y – 2 puting in equation (i) 3y – 6 + 4y – 7 =0 13 13 1 y= , x = y –2 = –2 = 7 7 7 equation of line y = mx + c m = slope of line y = 3x + c .... (iii) equation (iii) passing the point

A

O

B(0, 6)

X

5 b  4  0 and =10 54 or a = 18 and b = 18 Hence from (1), the required equation of the line AB is

42.

x y + =1 18 18 or x + y = 18 (d) Distance between two points

=

=

2

2

13 –3 = +c 7 7 c=

16 puting in equation (iii) 7

16 y = 3x + 7 21x – 7y + 16 = 0 41. (a) Let the equation of the line

x y  =1 ....(i) a b Then the coordinates of A and B are respectively (a,0) and (o,b). since C (8, 10) divides AB in the ratio 5 : 4, we have

AB be

 ar( Δ OAB) 

2

1  8  6 = 24 sq. units 2 (b) at y - axis, x = 0  2  0 – 3y = 6  y = –2  Required point (0, – 2) (c) at x-axis, y = 0  4x + 7  0 = 12  x = 3  Required point = (3, 0) (a) The equation of y-axis is x = 0 (b) The given equation is x = 5. Here, y-coordinate is 0. So, the given line is parallel to y-axis

51.

52.

53. 54.

2

64  4 = 68 = 2 17 units 43. (d) The point (6, –3) lies in the fourth quadrant. 44. (b) If x < 0 & y > 0, (x, y) lies in quadrant II 45. (a)????? When x = 2, y = 3  2 + 5 = 6 + 5 = 11 So, (2, 11) lies on y = 3x + 5 46. (b) Clearly, the point of x-axis has ordinate 7 and abscissa 0 So, the point is (7, 0) 47. (c) Clearly, the point is (0, –8). 48. (a) 2x + 7y = 1 ...........(i) 4x + 5y = 11 .........(ii) On solving (i) and (ii), we get x = 4 and y = –1  Required point of intersection = (4, –1) 49. (c) When x = 4  y = 2  4+3 = 11

ERna

=

50.

Rakesh Yadav Readers Publication Pvt. Ltd.

So, (4, 11) lies on y = 2x + 3 but (4, 10) does not lie on it. (d)

1  OA  OB 2



3  5  1 – 3

wwM wa. th Les aBryn

 1 13   – ,    7 7

Clearly, OA = 8 units and OB = 6 units

x 2 – x1  y2 – y1

Let (x1,y1) = (–5,3) and (x2,y2) = (3,1)  Required distance =

A(8, 0) X

O

5  0  4 a =8 54

12  3 – 5  –2  6

12  –5

Y

Y

kgei snhe eYari dnag v.iSn ir

39.

Where x = number of units of a good produced and y = cost of x units in rupees. Given, when x = 50, y = 320 and when x = 80, y = 380 from (1) 320 = 50m + c ...(2) 380 = 80m + c ....(3) Subtracting (2) from (3), we get m=2 Substituting m = 2 in equation (2), we get c = 220  From (1) y = 2x + 220 When x = 110, y = 2 × 110 + 220 = 440 Hence, the required cost of producing 110 units is ` 440. (b) Length of the perpendicular from the point (3, – 2) to the straight line 12x – 5y + 6 = 0 is

Y

X’

O 1234 5

X

Y’

55.

(b) The given equation is x = – 3 Here y-coordinate is 0. So,the given line is parallel to y-axis Y

X’

–4 –3 –2 –1 O

X

Y’

643

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(a) The given equation is y = 2, here x-coordinate is 0. So, the given line is parallel to x-axis.

A

X

Y

4 x 3 which passes through origin as(0, 0) satisfy it. (b) Taking the points in the order given, it is easily seen that they are represented P, Q, R, S. Then the shape shows rectangle (PQRS) having sides. PS = QR = 3 - (-2) = 5 PQ = SR = 4 - (-3) = 7

63.

 11  (c) Point 3,  will satisfy the  2 given equation. 11 2  a  3  5  6  3a or a  2 2

2

59.

Q(–3, 3)

5 10  a  3 – 5  3a  10 or a  2 3

(d)

X

P(4, 3)

B(0, 6)

r (

O

Clearly, OP 



66.

X

O

c ,0) a

P

c , a

OQ 

X

c b

1 ar(OPQ) OPOQ 2

1 c c c2    2 a b 2ab (b) (–3 , 7) lie s in sec ond quadrant and (4, 6) lies in first quadrant. Y

R(–3, –2)

Y

c b

=

R Enak



Q

Y

 5 (a) Point 3, 2  will satisfy the   given equation.

58.

y

(c) 3y + 4x = 0 or y = –

62.

Y’



c b When y = 0  ax = c c or x  a

X

O(o, o)

Y

X

O

57.

(4, 3)

(b) The equation is - ax + by = c (given) When x = 0  by = c or

B (–4, –3)

3 2 1

X’

65.

Y

geisnh eeYa ridna gv.i Sni

56.

S(4, –2)

(–3, 7)

(4, 6)

Y

wwM wa. th Les B aryn

X

X

O

A(8, 0)

Y

Required distance = AB 2

2

OA  OB

60.

64. (d) The graph of x = 2 will be a line parallel to y-axis at a distance of 2 units to its right. Similarly, the graph of x = –2 will be a line parallel to y-axis at a distance of 2 units to its left. and y = 2 and y = – 2 will be the lines parallel to x-axis at a distance of 2 units to its right and 2 units to its left respectively. Thus ABCD is a square having each side of 4 units.

 82  62  10 units (b)

Y

B

P(7, 24)

X O A Y Clearly, OA = 7, AP = 24  OP² = OA² + AP² = 7² + 24² = 625

61.

 OP  625  25 (a) This graph shows AB is a straight line which passes through the origin.

a b c 1  1  1 a b c 2 2 2

 X

X

68.

Rakesh Yadav Readers Publication Pvt. Ltd.

D(2, –2) Y

X

(d) The given equations are of the form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0.These will represent coincident lines it they have infinitely many solutions.The condition for which is

A(2, 2)

C(–2, –2)

O Y

67.

Y

B(–2, 2)

X

X

1 2 6   3 k 18

k6 (c) a1x + b1y + c 1 = 0 and a2x + b 2 y + c 2 =0 will have no solution if

a b c 1  1 1 a b c 2 2 2

644

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

69.

(a)

74.

Y

(–, +)

X

ab > 0 Hence, a and b are both positive or negative.  If (a, b) lies in first quadrant, (–b, –a) will lie in third quadrant. (a) 5x + 7y = 35

75.

x y   1.........(i) or 7 5 and 4x + 3y = 12 or

h

Y

X

E

O

C(4, 2)

D

X

M

(0, 4)

O

C (3, 0)

Y

and AD = AM – DM = 4 – 2 = 2 units

B (7, 0)

 ar(  ABC)

 Base of Δ ABC = BC = 7 – 3 = 4 unit let height (OA) = h



1  2  7  7sq.unit 2

79.(d)

Y

A(4, 11)

80 from(i) and (ii) we get y = 13 B(–2, 3)

76.

Y

(b) 3x – 6y = 12 .........(i) 3x – y = 3 ..........(ii) (i) – (ii) we get, – 5y = 9

Here, AB = 4, BC = 3

2

21 22 916 = 5

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or y = –

C

E

X

D

1 80 160  4  sq.unit 2 13 13

X

=

(a) Draw AM  x-axis meeting BC at D. Now, BC = BE + EC = 3 + 4 = 7 units

78.

B(–3, 2)

 ar Δ ABC

C(2, –2)

y = x y  x

(0, 5)

A(–1, 2)

X

77.

A(2, 4)

A

a b 1  1  k  1  k 3  a b 9 3 2 2 73. (a) Y

2

27  2.7sq .unit 10 (c) Consider any point in the shaded portion let (3, –3) i.e. x = 3, y = – 3  –3 < 3 i.e. y