Number Systems

CHAPTER

1 NUMBER SYSTEMS

Points to Remember : 1. Number used for counting 1, 2, 3, 4, ... are known as Natural numbers. 2. All natural numbers together with zero i.e. 0, 1, 2, 3, 4, ..... are known as whole numbers. 3. All natural numbers, zero and negative numbers together i.e. ...., –4, –3, –2, –1, 0, 1, 2, 3, 4, ... are known as Integers. p 2 5 4 , 4. Rational Numbers : Numbers of the form where p, q both are integers and q  0. For e.g. , q 3 7 1 etc. 5. Every rational number have either terminating or repeating (recurring) decimal representation. Terminating Repeating (Recurring) For eg. 2  0.4 , 13  3.25 etc. 5 4

For e.g. 1  0.333.....  0.3 3 15 = 2.142857142857... 7

here, prime factors of denominator are 2 and 5 only.

 2.142857 etc.

6. There are infinitely many rational numbers between any two given rational numbers. p 7. Irrational Numbers : Numbers which cannot be written in the form of q , where p, q are integers and

q  0. For e.g.

2 , 3 , 17 , , 0.202202220......,3 9 etc.

7 8. Real numbers : Collection of both rational and irrational numbers. For e.g.  3, , 0, 2 , 5 ,  etc. 5

9. Every real number is represented by a unique point on the number line. Also, every point on the number line represents a unique real number. 10. For every given positive real number x, we can find

x geometrically..

11. Identities related to square root : Let p, q be positive real numbers. Then, (i)

pq 

p. q

(iii) ( p  q ) ( p  q )  p  q

(ii)

p  q

p

;q  0

q

(iv) ( p  q ) 2  p  2 pq  q

12. Laws of Radicals : Let x, y > 0 be real numbers and p, q be rationals. then (i) xp × xq = xp+q

(ii) x p  x q  x p  q

(iii) ( x p ) q  x pq

(iv) x p . y p  ( xy ) p

MATHEMATICS–IX

NUMBER SYSTEMS

1

ILLUSTRATIVE EXAMPLES Example 1. Find six rational numbers between 3 and 4. —NCERT. Solution. We know that between two rational numbers a and b, such that a < b, there is a rational number a b . 2 1 7 A rational number between 3 and 4 is (3  4)  . 2 2 Now, a rational number between 3 and

A rational number between

1 7  1  6  7  13 7  . is  3     2 2 2 2  4 2

17  1  7  8  15 7  . and 4 is   4    22  2 2  4 2

Also, a rational number between 3 and

1  13  1  12  13  25 13  is  3     2 4  2 4  8 4

A rational number between

1  15  1  15  16  31 15  and 4 is   4    2 4 4   2 4  8

A rational number between

1  31  1  31  32  63 31  and 4 is   4    2 8 8  2  8  16

25 13 7 15 31 63      4 8 4 2 4 8 16 This can be represented on number line as follows :

 3

OR (without using formula) 10 40 We have, 3  3  10  30 and 4  4  10 10 10 10 We need to find six rational numbers between 3 and 4 i.e. 40 31 32 33 34 35 36 30 , , . Ans. and , which are , , , 10 10 10 10 10 10 10 10

Example 2. Find four rational numbers between 1 and 4 . 4 3 Solution.

1 1 3 3 4 4 4 16    and    4 4 3 12 3 3 4 12

 4 rational numbers between 2

3 16 1 4 4 5 6 7 and i.e. and are , , , . 12 12 4 3 12 12 12 12

NUMBER SYSTEMS

MATHEMATICS–IX

Example 3. Express 0.12 in the form of rational number, Solution.

Let x  0.12  0.121212 ..... multiplying both sides by 100, we get 100 x = 12.1212..... Subtracting (1) from (2), 100 x – x = 12.1212..... – 0.1212....  99 x = 12  x 

Example 4. Represent Solution.

p. q

...(1) ...(2)

12 4  Ans. 99 33

3 on the number line.

Let XOX be number line with O as origin. Let OA = 1 unit. Draw AB  OA such that AB = 1 unit. Join OB. Then using Pythagoras theorem, In OAB OB  OA 2  AB2  12  12  2 units Again, draw DC  OB such that BC = 1 unit. Join OC. then, OC  OB 2  BC 2  ( 2 ) 2  (1) 2  3 units . With O as centre and OC as radius, draw on arc, meeting OX at

P. Then OC = OP = 3 units. Example 5. State whether the following statements are true or false. Justify your answers. (i) Every irrational number is a real number. (ii) Every point on the number line is of the form

m , where m is a natural number..

(iii) Every real number is an irrational number —NCERT (i) True, since collection of real numbers is made up of rational and irrational numbers. (ii) False, no negative number can be the square root of any natural number. (iii) False, for example 5 is a real but not irrational. Example 6. Write the following in decimal form and say what kind of decimal expansion each as: 36 1 3 329 1 2 (iii) 4 (vi) (ii) (iv) (v) —NCERT (i) 100 8 13 400 11 11

Solution.

Solution.

36  0.36  Terminating decimal expansion. 100 1 , by long division, we have : (ii) Consider 11 0.090909

(i)

11 1.000000

99 100 99 100 99 1 MATHEMATICS–IX

NUMBER SYSTEMS

3



1  0.090909......  0.09, which is non-terminating and repeating decimal expansion. 11

1 4  8  1 33 (iii) 4   . By long division, we have 8 8 8

4.125 8 33.000

32 10 8 20 16 40 40 0  33  4.125, terminating decimal expansion. 8 (iv) Consider,

3 by long division, we have 13

0.23076923.... 13 3.0000000

26 40 39 100 91 90 78 120 117 30 26 40 39 1 3  0.23076923.......  0.230769, which is non-terminating and repeating decimal expan13 sion.



4

NUMBER SYSTEMS

MATHEMATICS–IX

(v) Consider,

2 , by long division, we have 11

0.181818 11 2.00000

11 90 88 20 11 90 88 20 11 90 88 2



2  0.181818.....  0.18 , which is non-terminating and repeating decimal expansion. 11 329 , by long division, we have 400

(vi) Consider, 0.8225 400 329.0000

3200 900 800 1000 800 2000 2000 0



329  0.8225, which is terminating decimal expansion. 400

Example 7. What can be the maximum number of digits be in the repeating block of digits in the decimal expansion of

MATHEMATICS–IX

1 ? Perform the division to check your answer.. 17

NUMBER SYSTEMS

—NCERT

5

Solution.

0.588235294117647.... 17 1.0000000000000000

85 150 136 140 136 40 34 60 51 90 85 50 34 160 153 70 68 20 17 30 17 130 119 110 102 80 68 120 119 1 Thus,

1  0.588235294117647 7

 The maximum number of digits in the quotient while computing

1 are 15. 17

p Example 8. Look at several examples of rational numbers in the form q (q  0) , where p and q are integers

Solution. 6

with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy? —NCERT Let us consider various such rotational numbers having terminating decimal representation. NUMBER SYSTEMS

MATHEMATICS–IX

1 3 5  0.5 ;  0.75;  0.625 2 4 8 39 11 43  1.56;  0.088;  0.215 etc. 25 125 200 from the examples shown above, it can be easily observe that, ‘‘If the denominator of a rational number in standard form has no prime factors other than 2 or 5 or both, then and the only then it can be represented as a terminating decimal.’’ Example 9. Visualise 3.765 on the number line, using successive magnification. —NCERT Solution. We know that 3.765 lies between 3 and 4. We divide portion of number line between 3 and 4 in 10 equal parts i.e. 3.1, 3.2, ....., 3.9 and then look at the interval [3.7, 3.8] through a magnifying glass and observe that 3.765 lies between 3.7 and 3.8 (see figure).

Now, we imagine that each new intervals [3.1, 3.2], [3.2, 3.3], ...... , [3.9, 4] have been sub-divided into 10 equal parts. As before, we can now visualize through the magnifying glass that 3.765 lies in the interval [3.76, 3.77]. (see figure).

Again, 3.765 lies between 3.76 and 3.77. So, let us focus on this portion of the number line, and imagine to divide it again into 10 equal parts. The first mark represents 3.761, second mark represents 3.762, and so on. So, 3.765 is the 5th mark in these subdivisions. Example 10. Recall,  is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). c . This seem to contradict the fact that  is irrational. How will you resolve this d contradiction? —NCERT Solution. There is no contradiction. Remember that when you measure a length with a scale or any other device, you only get on approximate rational value. So, you may not realise that either c or d is irrational. Example 11. Simplify the following :

That is,  

(i) 3 7  4 7 (iii) ( 5  3 ) 2 MATHEMATICS–IX

(ii) ( 7  3 ) ( 7  3 ) (iv) 8 30  2 5 NUMBER SYSTEMS

7

Solution.

(i) 3  4  ( 7 ) 2  12  7  84 (ii) ( 7 ) 2  ( 3 ) 2  7  3  4 (iii) ( 5 ) 2  ( 3 ) 2  2( 5 )( 3 )  5  3  2 5  3  8  2 15 (iv)

8 30  4 6 2 5

Example 12. Find 5.3 geometrically.. Solution. Draw AB = 5.3 units and extend it to C such that BC = 1 unit. Find mid-point O of AC. With O as centre, and OA as radius, draw a semicircle. Draw

BD  AC, interesting semicircle at D. Then BD =

5.3 units. With B as centre and BD as radius, draw an arc, intersecting AC produced at E. Then, BE  BD  5.3 units.

Example 13. Find value of a and b, where a  b 2 

Solution.

3 2

We have,

3 2

3 2



3 2

3 2

.

3 2



3 2

(3  2 ) 2



3 2



(3) 2  ( 2 ) 2

9  2  6 2 11  6 2  92 7

11 6 2  ab 2  7 7



11 6 and b  7 7 Example 14. Simplify the following : 

a

2

4

1

(i) 3 5  3 5 Solution.

2 4  5

(i) 3 5

1 1  4

(ii) 7 3

3

2 4 5

7

8

2 5

(iv) (32) 2/5

6

4 3 12

2

4 1 (iii) (3 )

 35 1

 7 12

(iii) 34 x(–1)  34 

(iv) (25 )

1

(ii) 7 3  7 4

5

2 5

1 3

4



1 81

 22 

1 2

2



1 4

NUMBER SYSTEMS

MATHEMATICS–IX

PRACTICE EXERCISE 1. Represent each of the following rational numbers on the number line : (i)

2.

3.

3 5

(ii)

7 4

(iii) – 3.6

(i) Find 4 rational numbers between

1 2 and . 5 3

(ii) Find 5 rational numbers between

3 4 and . 4 3

(i) Represent

2 and

(iv) 4.53

3 on the some number line

(ii) Represent

5 on the number line. 4. Without actual division, find which of the following rationals are terminating decimals. 19 6 13 (iii) (ii) 125 35 24 5. Find the decimal expansions of the following :

(i)

(i) –

16 9

(ii)

22 7

6. Find the decimal representation of

(iii)

11 30

(iv)

17 80

(iv)

37 175

2 3 4 5 , , and . 7 7 7 7

7. Express each of the following recurring decimals in the form of a rational number, (ii) 0.123

31 200

1 1 . Deduce from the decimal representation of , without actual 7 7

calculations, the decimal representation of

(i) 0.7

(v)

(iii) 0.45 6

p : q

(iv) 3.456

8. (i) Find three irrational numbers between

2 4 and . 3 5

(ii) Insert three irrational numbers between

2 and

3.

9. Give an example of two irrational numbers whose : (i) difference is a rational number. (ii) difference is an irrational number. (iii) sum is a rational number. (iv) sum is an irrational number. (v) quotient is a rational number. (vi) quotient is an irrational number. (vii) Product is a rational number. (viii) Product is an irrational number. 10. Simplify each of the following : (i) (7  3 ) (7  3 )

(ii) ( 5  3 ) 2

(iii) 12 20  3 5

(iv) (2 3  3 2 ) 2

(v) (4 2  3) (4 2  3)

(vii) 3 20  3 5  2 2  4 18

11. Represent the following on the number line : (i)

(ii)

2.4

(iii)

5.7

(iv) 9.2

6.8

12. Rationalize the denominator : 1

(i)

2

(ii)

2 3

(iii)

3 5

1

2 3

(iv)

2 3

5 3

13. Rationalize the denominator: 3

1

(i)

52 3

(ii)

1

(iii)

6 3 2

1

5 33 5

(iv)

2 3 3 2

14. Simplify the following : (i)

4 3 4 3



4 3

(ii)

4 3

15. If a and b are rational numbers and if

3 2 5 3 2 5

16. If x and y are the rational numbers and

5 1 5 1

3 5 3 5



3 5 3 5

 a  b 5 , find a and b.



5 1 5 1

 x  y 5 , find x and y.

17. Evaluate the following : (i) 53  52

(ii) 58  55

(v) (3 8 )1/2

(vi) 3 –2  4 –2

(iii) (32 ) 2

(iv) (64) 2/3

18. Find the valueof x if : 1 (ii) 32 x 1    9

(i) 23 x 1  1

1 (iii)   6

7 3 x

2 (iv)   3

 6

x 3

3x  2

2   3

2 3 x

3   2

x 1

19. Find whether the product of irrational numbers (5  2 ), (3  7 ), (3  7 ) and (5  2 ) is a rational or irrational number. 1

2  1.414 , find value of

20. (i) Given

2 1

2 3

3  1.732, find value of

(ii)

21. (i) Prove that

.

2 3

2 is not a rational number..

(ii) Prove that

2  3 is an irrational number..

22. Simplify the following : (i) 0.2  0.3  0.4 (ii) 0.42  0.34 23. Simplify the following : (i)

(iv) 4.16 1.25

(iii) 2.13  1.16

2 n 1  2 n

(ii)

2 n  2 n 1

x 1 y y 1 z z 1 x

24. Assuming that x is positive real number and a, b, c are rational numbers, show that :  xb  (i)  c  x 

a

 xc     xa   

b

 xa   xb 

c

 xa (ii)  b x

  1  

2 26. (i) If x  4  15 , find the value of x 

(ii) If x  5  2 6 , find the value of

1 x2

x

27. Simplify the following : 1



2

   

b c

 xc   xa 

   

ca

1

1 . x

(ii)

2 3 5  3 2 5 28. Rationalize the denominator of following : 1

(i)

 xb   xc 

.

1



a b

1 1 1   0 x y z

25. If 2 x  3 y  6  z , show that

(i)

   

3 2 6 3

4 3 6 2



2 3 6 2

2

(ii)

1 2  3



1 3  5

29. Prove that : (i)

1



1 2

1 2 3

1

(ii)



2 3 5



1

 .... 

3 4

1 2 3 5

1

2

8 9



2 2

30. Represent the following on the number line : (i) 13

(ii) 17

(iii) 2 3

(iv) 1  2

PRACTICE TEST M.M : 30

Time : 1 hour

General Instructions : Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each. 1. Find three rational numbers between 2. Represent

3 4 and . 4 3

5 on the number line.

3. Rationalise the denominator : 1 52

4. Find decimal representation of

16 . 45

5. Simplify the following : (i) ( 5  2 ) 2

(ii) (3 2  2 3 )(3 2  2 3 )

6. If a and b are rational numbers and

4 3 4 3

 a  3b , find a and b.

7. Evaluate the following : 7

3

 1  (ii)    32 

 11  4  11  4 (i)      2 2

2 / 5

8. Express 0.246 as a rational number in the simplest form. 9. Represent 10. Simplify :

4.2 on the number line. Also, give step of constructions. 6 2 3



3 2 6 3



4 3 6 2

ANSWERS OF PRACTICE EXERCISE 2. (i)

4 5 6 7 , , , 15 15 15 15

(ii)

4. (ii), (iv) and (v) 6.

10 11 12 13 14 , , , , 12 12 12 12 12

(ii) 3.142857

5. (i)  1.7

(iii) 0.36

(iv) 0.21142856

4 5 1 2 3  0.142857 ,  0.284714,  0.428571,  0.571428,  0.714285 7 7 7 7 7

7 61 137 1711 (ii) (iii) (iv) 9 495 300 495 8. (i) 0.68010010001 ..., 0.69010110111 .... and 0.7101001000...

7. (i)

(ii) 1.501001000 ... and 1.601001000...

(ii) 3 5 and 5

(iii) 3  5 and 3  5

(iv) 4 5 and  2 5

(v)

(vi)

(vii) 2 3 and 3

(viii) 3 2 and 4 3

9. (i) 3  5 and 5  3

10. (i) 46

(ii) 8 – 2 15

12. (i) 2  3 (ii)

1 (3  5 ) 2

20 and 5

(iii) 8

(iv) 30  12 6

(iii) 7  4 3

(iv)

20 and 6

(v) 23

(vi) 3 5  10 2

1 ( 5  3) 2

13. (i)

1 1 1 (5 3  3 5 ) (iv)  1 (2 3  3 2 ) ( 5  2 3 ) (ii) ( 6  3 2 ) (iii) 4 30 7 6

14. (i)

38 (ii)  3 5 13

15. a 

17. (i) 3125 (ii) 125 (iii) 18. (i)

1 3

(ii)

5 8

29 12 ,b  11 11

1 1 (iv) (v) 81 16

(iii)

5 2

(iv)

1 2 (vi) 144

4 5

19. Rational

20. (i) 0.414

23. (i) 2 (ii) 1

26. (i) 62 (ii) 12

28. (i)

16. x = 3, y = 0

(ii) 13.928

7 422 1540 (iii) (iv) 90 231 297 27. (i) 0 (ii) 0

22. (i) 1 (ii)

1 2 ( 2  2  6 ) (ii) (7  3 3  5  2 15 ) 4 11

ANSWERS OF PRACTICE TEST 1.

10 11 12 , , 12 12 12

6.

a

19 8 ,b  13 13

MATHEMATICS–IX

3.

52

7. (i)

4.  0.35

111 11 (ii) –4 8. 450 2

5. (i) 7  2 10 (ii) 6 10. 0

NUMBER SYSTEMS

13

CHAPTER

2 POLYNOMIALS

Points to Remember : 1. A symbol having a fixed numerical value is called a constant. For e.g. 9 , 

7 ,  , 2 etc. 3

2. A symbol which may take different numerical values is known as a variable. We usually denotes variable by x, y, z etc. 3. A combination of constants and variables which are connected by basic mathematical operations, is known as Algebraic Expression. For e.g. x2 – 7x + 2, xy2 – 3 etc. 4. An algebraic expression in which variable have only whole numbers as a power is called a Polynomial. 5. Highest power of the variable is called the degree of the polynomial. For e.g. 7x3 – 9x2 + 7x – 3 is a polynomial in x of degree 3. 6. A polynomial of degree 1 is called a linear polynomial. For e.g. 7x + 3 is a linear polynomial in x. 7. A polynomial of degree 2 is called a Quadratic Polynomial. For e.g. 3y2 – 7y + 11 is a Quadratic polynomial in y. 8. A polynomial of degree 3 is called a Cubic Polynomial. For e.g. 3t3 – 7t2 + t – 3 is a cubic polynomial in t. 9. According to number of terms, a polynomial having one non-zero term is a monomial, a polynomial having two non-zero terms is a bionomial and a polynomial have three non-zero terms is a trinomial. 10. Remainder Theorem : Let f(x) be a polynomial of degree n  1 and let a be any real number. If f(x) is divided by linear polynomial (x – a), then the remainder is f(a). 11. Factor Theorem : Let f(x) be a polynomial of degree n > 1 and a be any real number. (i) If f(a) = 0, then x – a is a factor of f(x). (ii) If (x – a) is a factor of f(x) then f(a) = 0. 12. Algebraic Identities : (i) (x + y)2 = x2 + 2xy + y2 (ii) (x – y)2 = x2 – 2xy + y2 2 2 (iii) x – y = (x – y) (x + y) (iv) (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz 3 3 3 (v) (x + y) = x + y + 3xy (x + y) (vi) (x – y)3 = x3 – y3 – 3xy (x – y) 3 3 2 2 (vii) x – y = (x – y) (x + xy + y ) (viii) x3 + y3 = (x + y) (x2 – xy + y2) 3 3 3 2 (ix) x + y + z – 3xyz = (x + y + z) (x + y2 + z2 – xy – yz – xz) 13. If x + y + z = 0 then, x3 + y3 + z3 = 3xyz.

ILLUSTRATIVE EXAMPLES Example 1. Which of the following expressions are Polynomials? In case of a polynomial, give degree of polynomial. (i) x4 – 3x3 + 7x2 + 3 (iv) x  14

1 x

(ii) (v) 7

3 y3  7 y  6

(iii)

x 3

(vi) x 2 / 3  POLYNOMIALS

2 x

MATHEMATICS–IX

Solution.

(i) is a polynomial of degree 4. (ii) is a polynomial of degree 3. (v) is a polynomial of degree 0. Example 2. Verify whether the following are zeros of the polynomial, indicated against them. 1 3 2 (iii) p(x) = x – 1, x = 1, –1

4 5 (iv) p(x) = (x + 1) (x – 2), x = – 1, 2

(i) p ( x )  3 x  1, x 

(ii) p ( x )  5 x   , x 

(v) p(x) = x2, x = 0

(vi) p(x) = lx + m, x 

(vii) p(x) = 3x2 – 1, x 

1

,

3

Solution.

2

m l

(viii) p(x) = 2x + 1, x 

3

1 2

–NCERT

(i) We have, p(x) = 3x + 1 At x 

1  1    1  , p     3   1  1  1  0 3  3  3 

1 is a zero of p(x). 3 (ii) We have, p(x) = 5x – 



At x 



4 4 4 , p   5     4    0 5 5 5

4 is not a zero of p(x). 5

(iii) We have, p( x)  x 2  1 At x = 1, p(1) = (1)2 – 1 = 1 – 1 = 0 also, at x = – 1, p(–1) = (–1)2 – 1 = 1 – 1 = 0  1, – 1 both are zeros of p(x). (iv) We have, p(x) = (x + 1) (x – 2) At x = – 1, p(–1) = (–1 + 1) (–1 –2) = 0 × (–3) = 0 also, at x = 2, p(2) = (2 + 1) (2 – 2) = 3 × 0 = 0  –1, 2 both are zeros of p(x). (v) We have, p(x) = x2 At x = 0, p(0) = 02 = 0  0 is a zero of p(x). (vi) We have, p(x) = lx + m At, x  

m m m , p   l   m  m  m  0 l  l   l 

m is a zero of p(x). l (vii) We have, p(x) = 3x2 – 1



MATHEMATICS–IX

POLYNOMIALS

15

At, x 

and at, x 



1

2

 1   1  1   1  3   1  1  1  0   3  , p 3 3  3  3

1

is a zero of p(x), but

3 (viii) We have p(x) = 2x + 1

At x 

2

 2   2  4   1  3   1  4  1  3  0   3 , p 3 3  3  3

2

2

is not a zero of p(x).

3

1 1 1 , p   2   1  1  1  2  0 2 2 2

1 is not a zero of p(x). 2 Example 3. Find the zero of the polynomial in each of the following cases : (i) p(x) = x + 5 (ii) p(x) = x – 5 (iii) p(x) = 2x + 5 (iv) p(x) = 3x – 2 (v) p(x) = 3x (vi) p(x) = ax ; a  0 (vii) p(x) = cx + d, c  0, c, d are real numbers. Solution. We know, finding a zero of p(x), is the same as solving the equation p(x) = 0. (i) p(x) = 0  x + 5 = 0  x = – 5  –5 is a zero of p(x). (ii) p(x) = 0  x – 5 = 0  x = 5  5 is a zero of p(x).



(iii) p(x) = 0  2x + 5 = 0  2x = – 5  x 



5 2

5 is a zero of p(x). 2

(iv) p(x) = 0  3x – 2 = 0  3x = 2  x 



2 3

2 is a zero of p(x). 3

(v) p(x) = 0  3x = 0  x = 0  0 is a zero of p(x). (vi) p(x) = 0  ax = 0  x = 0  0 is a zero of p(x).

( Given a  0)

(vii) p(x) = 0  cx + d = 0  cx = – d  x  

 16

—NCERT

d c

d is a zero of p(x). c

POLYNOMIALS

MATHEMATICS–IX

Example 4. Using remainder theorem, find the remainder when p(x) = 2x3 – 5x2 + 9x – 8 is divided by (x – 3). Solution. Remainder = p(3) = 2(3)3 – 5(3)2 + 9 (3) – 8 = 54 – 45 + 27 – 8 = 81 – 53 = 28 Ans. Example 5. Find the remainder when x 3  3x 2  3x  1 is divided by : (ii) x 

(i) x + 1 Solution.

1 2

(iii) x

(iv) x + 

(v) 5 + 2x

—NCERT

(i) x + 1 = 0  x = – 1  by remainder theorem, the required remainder is p (–1). Now, p(x) = x3 + 3x2 + 3x + 1  p(–1) = (–1)3 + 3(–1)2 + 3(–1) + 1 = –1 + 3 – 3 + 1 = 0  Remainder = 0 (ii) x 

1 1 0x . 2 2

1  by remainder theorem, the required remainder is p  . 2 3

Now,

2

1 1 1 1 p      3   3   1 2 2 2 2 

1 3 3 1  6  12  8 27   1   8 4 2 8 8

 Remainder  27 8 (iii) x = 0 by remainder theorem, the required remainder is p(0). Now, p(0) = (0)3 + 3(0)2 + 3(0) + 1 = 1  Remainder = 1 (iv) x +  = 0  x = –   by remainder theorem, the required remainder is p(–). Now, p(–) = (–)3 + 3(–)2 + 3 (–) + 1 = –3 + 32 – 3 + 1  Remainder = –3 + 32 – 3 + 1. (v) 5 + 2x = 0  2x = – 5  x =

5 2

 5 .  By remainder theorem, the required remainder is p   2  3

2

 5  5  5  5  Now, p    3   3  1  2   2   2   2  

MATHEMATICS–IX

125 75 15   1 8 4 2

POLYNOMIALS

17



125  150  60  8 27  8 8

 Remainder  

27 . 8

Example 6. Using Factor theorem, show that (x + 2) is a factor of x4 – x2 – 12. Solution. Let p(x) = x4 – x2 – 12. Now, x + 2 = 0  x = – 2. (x + 2) is a factor of p(x)  p(–2) = 0 Now, p(–2) = (–2)4 – (–2)2 – 12 = 16 – 4 – 12 = 0, which shows that (x + 2) is a factor of p(x). Example 7. Use the factor theorem to determine whether g(x) is a factor of p(x) is each of the following cases: (i) p(x) = 2x3 + x2 – 2x – 1, g(x) = x + 1 (ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2 (iii) p(x) = x3 – 4x2 + x + 6, g(x) = x – 3 —NCERT Solution. (i) In order to prove that g(x) = x + 1 is a factor of p(x) = 2x3 + x2 – 2x – 1, it is sufficient to show that p(–1) = 0. Now, p(–1) = 2 (–1)3 + (–1)2 – 2(–1) – 1 =–2+1+2–1=0  g(x) is a factor of p(x). (ii) In order to prove that g(x) = x + 2 is a factor of p(x) = x3 + 3x2 + 3x + 1, it is sufficient to show that p(–2) = 0. Now, p (–2) = (–2)3 + 3(–2)2 + 3(–2) + 1 = – 8 + 12 – 6 + 1 = – 1  0.  g(x) is not a factor of p(x). (iii) In order to prove that g(x) = x – 3 is a factor of p(x) = x3 – 4x2 + x – 6, it is sufficient to show that p(3) = 0 Now, p(3) = (3)3 – 4(3)2 + 3 – 6 = 27 – 36 + 3 – 6 = – 12  0.  g(x) is not a factor of p(x). Example 8. If x3 + ax2 + bx + 6 has x – 2 as a factor and leaves a remainder 3, when divided by x – 3, find the value of a and b. Solution. Let p(x) = x3 + ax2 + bx + 6 Since, x–2 is a factor of p(x)  p(2) = 0 ( factor theorem)  (2)3 + a(2)2 + b(2) + 6 = 0  2a + b = –7 ...(1) Also, p(x) leaves remainder 3, when divided by x – 3.  p(3) = 3 ( Remainder theorem) 3 2  (3) + a(3) + b(3) + 6 = 3  3a + b = – 10 ...(2) Solving (1) and (2), we get a = – 3, b = – 1 Ans. 18

POLYNOMIALS

MATHEMATICS–IX

Example 9. Factorize the following: (i) 6x2 – 18xy (ii) x3 + 7x2 – x – 7 (iii) 16a2 – 81b2 2 2 (iv) x + 5x – 24 (v) 9x – 22x + 8 Solution. (i) 6x (x – 3y) (ii) x2 (x + 7) – 1 (x + 7) = (x2 – 1) (x + 7) = (x – 1) (x + 1) (x + 7) (iii) (4a)2 – (9b)2 = (4a + 9b) (4a – 9b) (iv) x 2  8 x  3x  24  x( x  8)  3( x  8) = (x – 3) (x + 8) (v) 9x2 – 18x – 4x + 8 = 9x (x – 2) – 4(x – 2) = (x – 2) ( 9x – 4) Example 10. Factorise : (i) 12x2 – 7x + 1 (ii) 2x2 + 7x + 3 (iii) 6x2 + 5x – 6 (iv) 3x2 – x – 4 –NCERT 2 Solution. (i) 12x – 7x + 1 here, p + q = coefficient of x = – 7 pq = coefficient of x2 × constant term = 12 × 1 =12  p + q = – 7 = – 4 – 3, pq = 12 = (–4) × (–3)  12x2 – 7x + 1 = 12x2 – 4x – 3x + 1 = 4x (3x – 1) – 1(3x – 1) = (4x – 1) (3x – 1) Ans. (ii) 2x2 + 7x + 3 = 2x2 + x + 6x + 3 = x (2x + 1) + 3 (2x + 1) = (x + 3) (2x + 1) (iii) 6x2 + 5x – 6 = 6x2 + 9x – 4x – 6 = 3x (2x + 3) – 2(2x + 3) = (3x – 2) (2x + 3) (iv) 3x2 – x – 4 = 3x2 + 3x – 4x – 4 = 3x(x + 1) – 4(x + 1) = (3x – 4) (x + 1) Example 11. Factorize x3 – 3x2 – 9x – 5 using factor theorem. Solution. Let p(x) = x3 – 3x2 – 9x – 5. factors of constant term are ±1 and ±5. Now, p(1) = 13 – 3(1)2 – 9(1) – 5  0 p(–1) = (–1)3 – 3(–1)2 – 9(–1) – 5  0.  (x + 1) is a factor of p(x).  p(x) = (x + 1) (x2 – 4x – 5) = (x + 1) [x2 – 5x + x – 5] = (x + 1) [x(x – 5) + 1 (x – 5)] = (x + 1) (x + 1) (x – 5) Ans. Example 12. Factorise : x3 + 13x2 + 32x + 20 —NCERT 3 2 Solution. Let p(x) = x + 13x + 32x + 20 Now, factors of constant term 20 are ± 1, ± 2, ± 5, ± 10 and ± 20. Now, p(1) = (1)3 + 13(1)2 + 32(1) + 20 = 1 + 13 + 32 + 20 = 66  0 p(–1) = (–1)3 + 13(–1)2 + 32(–1) + 20 = –1 + 13 – 32 + 20 = – 33 + 33 = 0  (x + 1) is a factor of x3 + 13x2 + 3x + 20. Now, divide p(x) by x + 1.

MATHEMATICS–IX

POLYNOMIALS

19

p(x) = (x +1) (x2 + 12x + 20) = (x + 1) [x2 + 2x + 10x + 20] = (x + 1) [x(x + 2) + 10 (x + 2)] = (x + 1) (x + 2) (x + 10) Ans. Example 13. Expand the following : (i) (a – 2b + 3c)2 (ii) (2x + y)3 (iii) (3x – 2y)3 Solution. (i) (a)2 + (–2b)2 + (3c)2 + 2(a) (–2b) + 2(–2b) (3c) + 2 (a) (3c) = a2 + 4b2 + 9c2 – 4ab – 12bc + 6ac (ii) (2x)3 + (y)3 + 3(2x)(y) [2x + y] = 8x3 + y3 + 6xy (2x + y) = 8x3 + y3 + 12x2y + 6xy2 (iii) (3x)3 – (2y)3 – 3(3x) (2y) [3x – 2y] = 27x3 – 8y3 – 18xy (3x – 2y) = 27x3 – 8y3 – 54x2y + 36xy2 Example 14. Factorize the following : (i) x3 + 27y3 (ii) 27a3 – 64b3 (iii) a3 – 8b3 + 64c3 + 24abc [  a3 + b3 = (a + b) (a2 – ab + b2)] Solution. (i) (x)3 + (3y)3 = (x + 3y) (x2 – 3xy + 9y2) 3 3 2 2 (ii) (3a) – (4b) = (3a – 4b) (9a + 12ab + 16b ) [ a3 – b3 = (a – b) (a2 + ab + b2)] (iii) (a)3 + (–2b)3 + (4c)3 – 3 × a × (–2b) × (4c) = (a – 2b + 4c) (a2 + 4b2 + 16c2 + 2ab + 8bc – 4ac) [ a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)] 

1 ( x  y  z ).[( x  y ) 2  ( y  z ) 2  ( z  x) 2 ] 2 LHS = x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)

Example 15. Verify that : x3 + y3 + z3 – 3xyz  Solution.

Consider,



1 (x + y + z) . 2 (x2 + y2 + z2 – xy – yz – zx) 2



1 (x + y + z) (2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx) 2



1 (x + y + z) (x2 + x2 + y2 + y2 + z2 + z2 – 2xy – 2yz – 2zx) 2



1 (x + y + z) [(x2 + y2 – 2xy) + (y2 + z2 – 2yz) + (z2 +x2 – 2zx)] 2

1 (x + y + z) [(x – y)2 + (y – z)2 + (z – x)2] 2 = RHS. Hence verify. Example 16. If x + y + z = 0, show that x3 + y3 + z3 = 3xyz. Solution. We have x + y + z = 0  x+y=–z Cubing both sides, we get (x + y)3 = (–z)3  x3 + y3 + 3xy (x + y) = –z3  x3 + y3 + 3xy (–z) = – z3 ( x + y = – z)

–NCERT



20

POLYNOMIALS

—NCERT

MATHEMATICS–IX

 x3 + y3 + z3 – 3xyz = 0  x3 + y3 + z3 = 3xyz. Hence shown. Example 17. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given : (i) Area : 25a2 – 35a + 12 (ii) 35y2 + 13y – 12 —NCERT Solution. (i) Given, Area of rectangle = 25a2 – 35a + 12 = 25a2 – 15a – 20a + 12 = 5a (5a – 3) – 4 (5a – 3) = (5a – 3) (5a – 4)  Possible length and breadth are (5a – 3) and (5a – 4) units. (ii) Area of given rectangle = 35y2 + 13y – 12 = 35y2 + 28y – 15y – 12 = 7y (5y + 4) – 3 (5y + 4) = (5y + 4) (7y – 3).  Possible length and breadth are (5y + 4) and (7y – 3) units. Example 18. What are the possible expressions for the dimensions of the cuboid whose volumes are given below: (i) Volume : 3x2 – 12x (ii) Volume : 12ky2 + 8 ky – 20 k —NCERT 2 Solution. (i) Volume : 3x – 12x = 3x (x – 4) = 3 × x × (x – 4)  Possible dimensions of cuboid are 3, x and (x – 4) units. (ii) Volume = 12ky2 + 8ky – 20k = 4 k (3y2 + 2y – 5) = 4k (3y2 – 3y + 5y – 5) = 4k [3y (y – 1) + 5 (y – 1)] = 4k (y – 1) (3y + 5)  Possible dimensions of a cuboid are 4k, (y – 1) and (3y + 5) units.

PRACTICE EXERCISE 1. Which of the following expressions are polynomials in one variable? (i) 3x2 – x3 + 7x + 1

4 x (v) 2x4 – 7x3 + 3

(ii) x 

(iii)

2 y2  7 y  2

(iii)

5t 3  7t 2  1

(iv) y 3 / 2  y 2  1 2. Write degree of each the following polynomial: (i) 7x4 – 9x3 + 2x + 4

(ii) 7 – y2 + y3

(iv) 10 (v) 3x2 – 7x + 4 3. Classify each of the following as linear, quadratic and cubic polynomial: (i) 3x3 – 7x (ii) 4 y2 + 3y – 1 2

(iii) 7r

3

(v) y  y (iv) x  3x 3 2 4. Find the value of p(x) = 5x – x + 3x + 4 at (i) x = 0 (ii) x = 2 (iii) x = – 1 5. Find the value of p(0), p(2) and p(–3) where p(x) = x3 – x2 + x – 1. 6. Verify whether the following are zeros of the polynomial, indicated against them: 3 2 (iii) p(x) = x2 + x – 6; x = 3, –2

(i) p( x )  2 x  3; x 

MATHEMATICS–IX

(ii) p(x) = (x + 3) (x – 4) ; x = – 3, 4 (iv) p(x) = x – x3 ; x = 0, 1, –1 POLYNOMIALS

21

7. Find the zero of the polynomial in each of the following : (i) p(x) = x – 4 (ii) p(x) = 3x + 4 (iii) p(x) = 7x (iv) p(x) = rx + s; r  0, r, s are real numbers. 8. Using Remainder Theorem, find the remainder when : (i) 4x3 – 7x2 + 3x – 2 is divided by x – 1 (ii) x3 – 7x2 + 6x + 4 is divided by x – 3 (iii) x3 + 2x2 – x + 3 is divided by x + 3 (iv) 4x3 – 4x2 + x – 2 is divided by 2x + 1 (v) x3 – ax2 + 5x + a is divided by x – a (vi) x3 + ax2 – 6x + 2a is divided by x + a 9. If 35 is the remainder when 2x2 + ax + 7 is divided by x – 4, find the value of a. 10. Without actual division, prove that 2x3 + 13x2 + x – 70 is exactly divisible by x – 2. 11. Show that x – 1 is a factor of x3 – 2x2 – 5x + 6. 12. Find the value of p for which the polynomial x4 – 2x3 + px2 + 2x + 8 is exactly divisible by x + 2. 13. Find values of a and b so that the polynomial x4 + 7x3 + 4x2 + ax + b is exactly divisible by x – 1 and x + 3. 14. The polynomials ax3 – 3x2 + 7 and 2x3 + 7x – 2a are divided by x + 3. If the remainder in each case is same, find the value of a. 15. The polynomials ax3 + 4x2 – 3 and 4x3 + 4x – a when divided by x – 3, leaves the remainder R1 and R2 respectively. Find the value of a if R1 = 3R2. 16. Find the integral zeros of x3 – 3x2 – x + 3. 17. Find the integral zeros of x3 + 4x2 – x – 4. 18. Show that x – 2 is a factor of p(x) = x3 – 6x2 + 15x – 14. 19. Show that x + 4 is a factor of x4 + 3x3 – 4x2 + x + 4. 20. Find value of a for which x + a is a factor of f(x) = x3 + ax2 + 3x + a + 4. 21. Without actual division, prove that 2x4 + x3 – 2x2 + 5x – 6 is exactly divisible by x2 + x – 2. 22. For what value of a is the polynomial 2x3 – ax2 + 8x + a + 4 is exactly divisible by 2x + 1. 23. If x3 + ax2 + bx – 12 has x – 3 as a factor and leaves a remainder 10 when divided by x + 2, find a and b. 24. Factorise the following, using factor theorem : (i) x3 – 6x2 + 11x – 6 (ii) x3 – 6x2 + 3x + 10 (iii) 2x3 – 5x2 – x + 6 (iv) x3 – 3x2 + 3x – 1 (v) x3 + 4x2 – 11x – 30 (vi) 4x3 + 9x2 – 19x – 30 3 2 25. Factorize 6x + 35x – 7x – 6, given x + 6 is one of its factor. 26. Factorize 2x3 + 5x2 – 124x – 63, given x + 9 is one of its factor. 27. Use suitable identity to find the following products : (i) (x + 4) (x + 6) (ii) (x – 3) (x + 8) (iii) (2x + 3) (3x – 2)

 2 5 2 5 (iv)  t    t   2 2 

28. Evaluate the following products without multiplying directly: (i) 87 × 93 (ii) 106 × 94 22

POLYNOMIALS

MATHEMATICS–IX

29. Factorize the following using appropriate identities :

y2 (ii) 9 x 2  6 x  1 144 (iv) 9 x 2  30 xy  25 y 2 (iii) x 4  y 4 30. Factorize by splitting the middle term : (i) x2 – 21x + 108 (ii) 12y2 – y – 6 (iii) t2 – 11t – 42 (v) 8x2 – 2x – 15 (vi) 20x2 + 13x – 84 31. Expand each of the following using suitable identity : (i) 9 x 2 

(i) ( x  2 y  3z ) 2 3

(iii) (3x – 2y)

a  (ii)   b  3c  3  y  (iv)  2 x   2  

(iv) 6z2 + 5z – 6

2

3

32. Evaluate the following, using suitable identity: (i) (101)3 (ii) (99)3 33. Factorize the following: (i) 4x2 + y2 + 9z2 + 4xy – 6yz – 12xz (ii) 64a3 – 144a2b + 108ab2 – 27b3 3 2 2 3 (iii) x + 6x y + 12xy + 8y (iv) 8x3 + 125 (v) 27a3 – 64b3 (vi) 27a3 + 8b3 + c3 – 18abc 34. Without actually calculating the cubes, find the value of each of the following : (i) (–14)3 + (8)3 + (6)3 (ii) (19)3 + (–11)3 + (–8)3 35. If x + y + z = 0, prove that x3 + y3 + z3 = 3xyz. 1 ( x  y  z ) [( x – y ) 2  ( y – z ) 2  ( z – x) 2 ] 2 37. Using factor theorem, show that a – b, b – c and c – a are the factors of a(b2 – c2) + b (c2 – a2) + c (a2 – b2). 38. Factorize the following : (i) 1 – 2ab – a2 – b2 (ii) x4 + 5x2 + 9

36. Prove that: x3 + y3 + z3 – 3xyz 

(iv) 9( x  y ) 2  24( x 2  y 2 )  16( x  y ) 2 (iii) 5 5 x 2  30 x  8 5 (v) x6 – y6 (vi) x6 + y6 39. Factorize the following : (i) x3 (y – z)3 + y3 (z – x)3 + z2(x – y)3 (ii) (a – 4b)3 + (4b – 3c)3 + (3c – a)3 3 3 3 40. Prove that : (a + b) + (b + c) + (c + a) – 3 (a + b) (b + c) (c + a) = 2 (a3 + b3 + c3 – 3abc)

PRACTICE TEST MM : 30

Time : 1 hour

General Instructions : Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each. 1. 2. 3. 4.

Find the remainder when 3x3 – 8x2 + 9x – 10 is divided by x – 3. Find the value of a for which x3 + ax2 – 3x + 14 is exactly divisible by x + 2. Factorize : 64a3 – 27b3 – 144a2b + 108ab2 Evaluate : (103)3 , using suitable identity.

MATHEMATICS–IX

POLYNOMIALS

23

5. Factorize : 3x2 + 13x – 10 x  (ii)   y  3 

2

6. Expand the following : (i) (x – 2y – 3z)

3

7. Without actually calculating the cubes, evaluate the following : (20)3 + (–15)3 + (–5)3 8. Factorize : x3 – 64y3 – 8z3 – 24xyz. 9. Factorize 6x3 + 25x2 + 21x – 10 using factor theorem. 10. Prove that : a3 + b3 + c3 – 3abc =



1 (a  b  c) (a  b) 2  (b  c ) 2  (c  a ) 2 2



ANSWERS OF PRACTICE EXERCISE 1. 3. 4. 6.

(i), (iii) and (v) 2. (i) 4 (ii) 3 (iii) 3 (iv) 0 (v) 2 (i) Binomial (ii) Trinomial (iii) Monomial (iv) Bionomial (v) Bionomial 4, 46, –5 5. –1, 5, –40 (i) yes (ii) yes (iii) no (iv) yes

4 3 8. (i) – 2 (ii) 175

7. (i) 4

(ii) 

5 r (iii) –3 (iv) – 4 (v) 6a

(iii) 0

(iv)

(vi) 8a

9. a = – 1

12. p = – 9

13. a = – 21, b = 9

14. a 

16. 1, –1, 3

17. 1, –1, –4

20. a = 2

22. a 

24. (i) (x – 1) (x – 2) (x – 3) (iv) (x – 1) (x – 1) (x – 1) 25. (x + 6) (3x + 1) (2x – 1)

(ii) (x + 1) (x – 2) (x – 5) (v) (x + 2) (x – 3) (x + 5) 26. (x + 9) (x – 7) (2x – 1)

27. (i) x2 + 10 x + 24

(ii) x2 + 5x – 24

28. (i) 8091

(ii) 9964

11 5

15. a 

109 10

1 23. a = 1, b = – 8 3 (iii) (x + 1) (x – 2) (2x – 3) (vi) (x – 2) (x + 3) (4x + 5)

(iii) 6x2 + 5x – 6 y  y  29. (i)  3 x    3 x   12   12  

4 (iv) t 

25 4

(ii) (3x – 1) (3x – 1)

(iii) ( x  y)( x  y)( x 2  y 2 ) (iv) (3x + 5y) (3x + 5y) 30. (i) (x – 9) (x – 12) (ii) (3y – 2) (4y + 3) (iii) (t + 3) (t – 14) (iv) (2z + 3) (3z – 2) (iv) (2x – 3) (4x + 5) (vi) (4x – 7) (5x + 12) 31. (i) x2 + 4y2 + 9z2 – 4xy – 12yz + 6xz (iii) 27x3 – 8y3 – 54x2y + 36xy2

(ii)

a2 2  b 2  9c 2  ab  6bc  2ac 9 3

(iv) 8 x 3 

y3 3  6 x 2 y  xy 2 8 2

32. (i) 1030301 (ii) 970299 24

POLYNOMIALS

MATHEMATICS–IX

33.

(i) (2x + y – 3z) (2x + y – 3z) (ii) (4a – 3b) (4a – 3b) (4a – 3b) (iv) (2x + 5) (4x2 – 10x + 25) (v) (3a – 4b) (9a2 + 12ab + 16b2) (vi) (3a + 2b + c) (9a2 + 4b2 + c2 – 6ab – 2bc – 3ac) 34. (i) –2016 (ii) 5016 38.

(i) (1 + a + b) (1 – a – b)

(ii) (x2 + x + 3) (x2 – x + 3)

(iii) (x + 2y) (x + 2y) (x + 2y)

(iii) ( 5 x  2) (5 x  4 5 )

(iv) (x + 7y) (x + 7y) (v) (x – y) (x + y) (x2 – xy + y2) (x2 + xy + y2) 2 2 4 2 2 4 (vi) (x + y ) (x – x y + y ) 39. (i) 3xyz (y – z) (z – x) (x – y) (ii) 3 (a – 4b) (4b – 3c) (3c – a)

ANSWERS OF PRACTICE TEST 1. 26 4. 1092727

2. a = – 3 5. (3x – 2) (x + 5)

6. (i) x2 + 4y2 + 9z2 – 4xy + 12yz – 6xz 7. 4500 9. (x + 2) (2x + 5) (3x – 1)

MATHEMATICS–IX

3. (4a – 3b) (4a – 3b) (4a – 3b)

(ii)

x3 x2 y  y3   xy 2 27 3

8. (x – 4y – 2z) (x2 + 16y2 + 4z2 + 4xy – 8yz + 2xz)

POLYNOMIALS

25

CHAPTER

3 CO-ORDINATE GEOMETRY

Points to Remember : 1. Coordinate axes : Two mutually perpendicular lines X´OX and YOY´ known as x-axis and y-axis respectively, constitutes to form a co-ordinate axes system. These axes interests at point O, known as origin.

2. Co-ordinate axes divides the plane into four regions, known as Quadrants. 3. The position of any point in a plane is determined with reference to x-axis and y-axis. 4. The x-coordinate of a point is its perpendicular distance from the y-axis measured along the x-axis. The x-coordinate is known as abscissa. 5. The y-coordinate of a point is its perpendicular distnace from the x-axis measured along the y-axis. The y-coordinate is known as ordinate. 6. Abscissa and ordinate of a point written in the form of ordered pair, (ascissa, ordinate) is known as the co-ordinate of a point. 7. If the point in the plane is given, we can find the ordered pair of its co-ordinate and if the ordered pair of real numbers is given, we can find the point in the plane corresponding to this ordered pair. Quadrant I Sign of 8. Sing Convention : x  coordinate  y  coordinate

26



II

III

IV

–

–





–

–

CO-ORDINATE GEOMETRY

MATHEMATICS–IX

ILLUSTRATIVE EXAMPLES Example 1. Write the answer of each of the following questions: (i) What is the name of the horizontal and the vertical lines drawn to determine the position of any point in the cartesian plane? (ii) What is the name of each part of the plane formed by these two lines? (iii) Write the name of the point where these two lines intersect? —NCERT Solution. (i) Rectangular axes/co-ordinate axes (ii) Quadrant (iii) Origin Example 2. Write the co-ordinate of points A, B, C, D, E and F.

MATHEMATICS–IX

CO-ORDINATE GEOMETRY

27

A (4, 3)

Example 5. Look at the figure given, and write the following : (i) The co-ordinate of A. (ii) The abscissa of point B. (iii) The ordinate of point C. (iv) The co-ordinate of D. (v) The point whose co-ordinate are (–2, 5).

Solution.

28

(i) Co-ordinate of A are (6, –4) (iii) Ordinate of C is –3 (v) Point E have co-ordinate as (–2, 5)

(ii) Abscissa of B is 2. (iv) Co-ordinates of D are (4, 3)

CO-ORDINATE GEOMETRY

MATHEMATICS–IX

Example 6. Plot the points (x, y) given in the following table on the plane choosing suitable units of distance on the axes. x  2 1 0 1 3 y 8 7  1.25 3  1 —NCERT Solution.

Example 7. Plot the points A (–3, –3), B (5,–3), C (5, 2) and D (–3, 2) on the graph paper. Join them in order and name the figure so formed. Also, find its area. Solution.

MATHEMATICS–IX

CO-ORDINATE GEOMETRY

29

ABCD is a rectangle Area of ABCD = AB × BC = 8 × 5 sq. units = 40 sq. units Example 8. Graph the following equations (i) x = –2 (ii) y = 3 (iii) y = x + 2 Solution. (i) x = –2. The given equation can be written as 1.x + 0.y = –2  x is fixed as –2 and y may choose any value. Let us represent following information in a tabular form. x 2 2 2 2 2 y  2 1 0 1 2

(ii) y = 3. Given equation may be written as 0.x + 1.y = 3 y is fixed as 3 and x may choose any value. Let us represent this information in a tabular form. x y

2 3

1 3

0 3

1 3

2 3

(iii) y = x + 2 here, when x = 0, y = 2 ; x = 1, y = 3 ; x = –1, y = 1 etc. Represent this in the tabular form. x 1 0 1 2 y 1 2 3 4

30

CO-ORDINATE GEOMETRY

MATHEMATICS–IX

PRACTICE EXERCISE 1. State the quadrant in which the following points lie. (i) A (3, –4) (ii) B (–5, 11) (iii) C (–10, –15) (iv) D (8, 12) (v) E (–11, 5) (vi) F (–100, –200) (vii) G (10, 50) (viii) H (20, –5) 2. Look at the figure given, and write the following : (i) The co-ordinate of P (ii) The ordinate of Q (iii) The abscissa of R (iv) The point given by (4, –3) (v) The point which is at a distance of 3 units from y-axis (vi) Co-ordinate of point T

MATHEMATICS–IX

CO-ORDINATE GEOMETRY

31

3. Plot the points P (1, 3), Q (3, 7) and R (5, 11). Are these points collinear? 4. Plot the points A (–2, 3), B (8, 3) and C (6, 7). Join them in order. Name the figure obtained. Also find its area. 5. Plot the points A (3, 2), B (11, 8), C (8, 12) and D (0, 6). Join them in order. Name the figure thus obtained. 6. Plot the points P (0, –1), Q (2, 1), R (0, 3), S (–2, 1). Join them in order. Name the figure obtained. 7. Plot the points A (–2, 1), B (1, 1), C (–4, –3) and D (3, –3). Join them in order. Name the figure thus obtained. 8. Plot the points P (7, 3), Q (3, 0), R (0, –4) and S (4, –1). Join them in order. Name the figure thus obtained. 9. Plot the points A (0, 3), B (–4, 1), C (0, –6) and D (4, 1). Join them in order. Name the figure thus obtained. 10. Graph the folloiwng equations : (i) x = 4 (ii) y = –3 (iii) y = x (iv) x + y = 3

PRACTICE TEST M.M : 15

Time : 1/2 hour

General Instructions : All questions carry 3 marks each. 1. Name the quadrant in which the following points lie : (i) A (–7, 9) (ii) B (–10, – 25) (iii) C (7, –9) (iv) D (11, 7) 2. Plot A (–6, 3), B (6, 0) and C (4, 5). Join these points in order. Name the figure thus obtained. 3. Look at the figure and write the following : (i) The co-ordinate of A. (ii) The co-ordinate of B. (iii) The abscissa of C. (iv) The point whose co-ordinates are (–4, 3) Y A

6 5 4 D

3 2 1

–7

–6

–5

–4

–3

–2

0 1

–1

2

3

4

5

6

7

8

X

–1 –2 C

–3 –4 –5

32

B

CO-ORDINATE GEOMETRY

MATHEMATICS–IX

4. Mark the points P (–4, 2), Q (–4, –4), R (3, –4) and S (3, 2) on the graph paper. Join these points in order. Name the figure obtained. Also, find area of the figure obtained. 5. Draw the graph of y = x + 1. Does the point (–7, 6) lie on this line?

ANSWERS OF PRACTICE EXERCISE 1. (i) IVth quadrant (v) IInd qudrant 2. (i) P (–5, 4) (v) S 3. yes

(ii) IInd qudrant (vi) IIIrd quadrant (ii) Q (4, 2) (vi) T (4, –3)

(iii) IIIrd quadrant (vii) Ist quadrant (iii) –6

(iv) Ist quadrant (viii) IVth qudrant (iv) T

4. Triangle, Area = 20 sq. units

MATHEMATICS–IX

CO-ORDINATE GEOMETRY

33

5. Rectangle

34

6. Square

CO-ORDINATE GEOMETRY

MATHEMATICS–IX

9. Kite

10.

(i)

MATHEMATICS–IX

(ii)

CO-ORDINATE GEOMETRY

35

(iii)

(iv)

ANSWERS OF PRACTICE TEST 1. (i) IInd quadrant

(ii) IIIrd qudrant

(iii) IVth quadrant

(iv) Ist quadrant

(ii) B (2, –5)

(iii) –4

(iv) D

2. Triangle

3. (i) A (7, 6)

4. Rectangle, Area = 42 square units 36

CO-ORDINATE GEOMETRY

MATHEMATICS–IX

5. No

MATHEMATICS–IX

CO-ORDINATE GEOMETRY

37

CHAPTER

4 LINEAR EQUATIONS IN TWO VARIABLES

Points to Remember : 1. An equation of the form ax + by + c = 0, where a, b, c are real numbers, such that a and be are not both zero, is known as linear equation in two variables. 2. A linear equation in two variables has infinitely many solutions. 3. The graph of linear equation in two variables is always a straight line. 4. y = 0 is the equation of x-axis and x = 0 is the equation of y-axis. 5. The graph of x = a is a straight line parallel to the y-axis. 6. The graph of y = b is a straight line parallel to the x-axis. 7. The graph of y = kx passes through the origin. 8. Every point on the graph of a linear equation in two variables is a solution of the linear equation. Also, every solution of the linear equation is a point on the graph of the linear equation.

ILLUSTRATIVE EXAMPLES Example 1. Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case. (i) 4x – 7y = 10 Solution.

(ii) x +

y =–5 2

(iii) x = –4y

(iv) 4y – 9 = 0 (v) x = 5 (i) 4x – 7y = 10  4x – 7y – 10 = 0 comparing with ax + by + c = 0, a = 4, b = –7, c = –10 (ii) x 

y y  5  x   5  0 = ax + by + c 2 2

1 ,c=5 2 (iii) x = –4y  x + 4y + 0 = 0 = ax + by + c on comparing, a = 1, b = 4, c = 0 (iv) 4y – 9 = 0  0.x + 4.y – 9 = 0 = ax + by + c on comparing, a = 0, b = 4, c = –9 (v) x = 5  x – 5 = 0,  x + 0.y – 5 = 0 on comparing, a = 1, b = 0, c = –5 Example 2. Which one of the following options is true, and why? y = 3x + 5 has (i) a unique solution, (ii) only two solutions, (iii) infinitely many solutions. —NCERT Solution. Given equation is y = 3x + 5 when x = 0, y = 3(0) + 5 = 5  (0, 5) is one of its solution. on comparing, a = 1, b 

when y = 0, then x  38

5 3

LINEAR EQUATIONS IN TWO VARIABLES

MATHEMATICS–IX

 5  , 0  is another solution.    3  when x = 1, then y = 3(1) + 5 = 8  (1, 8) is also its solution. So, its clear that for every infinitely values we can give to x, we have corresponding value of y.  This equation has infinitely many solutions. Example 3. Bhavya and Anisha have a total of Rs. 100. Express this information in the form of an equation. Solution. Let total amount with Bhavya = Rs x and total amount with Anisha = Rs. y according to the question, x + y = 100, which is required equation. Example 4. Give five integer solutions of the equation 3x + y = 8. Solution. 3x + y = 8  y = 8 – 3x Now, when x = 0, y = 8 – 3 (0) = 8 when x = 1, y = 8 – 3 (1) = 5 when x = 2, y = 8 – 3 (2) = 2 when x = –1, y = 8 – 3 (–1) = 11 when x = –2, y = 8 – 3 (–2) = 14 Solutions can be represented in the tabular form as follows :

x 0 1 2 1  2 y 8 5 2 11 14

Example 5. Write four solutions for each of the following equations : (i) 2x + y = 7 (ii) x + y = 9 (iii) x = 4y —NCERT Solution. (i) 2x + y = 7  y = 7 – 2x For x = 0, y = 7 – 2(0) = 7 For x = 1, y = 7 – 2(1) = 7 – 2 = 5 For x = 2, y = 7 – 2(2) = 7 – 4 = 3  Four solutions of the given equation are (0, 7), (1, 5), (2, 3), (–1, 9). (ii) Given equation x + y = 9  y = 9 – x For x = 0, y = 9 –  (0) = 9 For x = 1, y = 9 –  (1) = 9 –  For x = 2, y = 9 –  (2) = 9 – 2 For x = –1, y = 9 –  (–1) = 9 +   The four solutions of the given equation are (0, 9), (1, 9 – ), (2, 9 – 2), (–1, 9 + ). x (iii) Given equation x = 4y  y  4 0 For, x = 0, y   0 4 4 For x = 4, y   1 4 4  1 For x = – 4, y  4 8 For x = 8, y   2 4  The four solutions of a given equation are (0, 0), (4, 1) (–4, –1) and (8, 2). MATHEMATICS–IX

LINEAR EQUATIONS IN TWO VARIABLES

39

Example 6. Find a if x = 3, y = 1 is a solution of the equation 3x – y = a. Solution.

Given 3x – y = a As x = 3, y = 1 is a solution of this given equation, it must satisfy it.  3 (3) –1 = a  a = 9 – 1  a  8 Ans.

Example 7. Give the equations of two lines passing through (2, 14). How many more such lines are there, and why? —NCERT Solution.

Equations of two lines passing through (2, 14) can be taken as 2x + y = 18 or 3x – y = – 8. Aslo, equations such as 7x – y = 0, 5x + 2y = 38 etc. are also satisfied by the co-ordinates of the point (2, 14). So, any line passing through (2, 14) is an example of a linear equation for which (2, 14) is a solution. Thus, there are infinite number of lines through (2, 14).

Example 8. Draw the graph of the following equations :

Solution.

(i) x = 3

(ii) y = –4

(v) 2x + 5y = 10

(vi) 2x – y = 7

(i) x = 3 As x is constant, y may take any value.

x 3 3 3 3 3 y  2 1 0 1 2

(iii) y = x x y

40

(iii) y = x

(iv) y = –x (ii) y = –4 As y is constant, x may take any value.

x  2 1 0 1 2 y 4 4 4 4 4

(iv) y = –x

2 1 0 1  2 2 1 0 1  2

x  2 1 0 1 2 y 2 1 0 1  2

LINEAR EQUATIONS IN TWO VARIABLES

MATHEMATICS–IX

(v) 2x + 5y = 10 10  2 x y 5

(vi) 2x – y = 7 y = 2x – 7

x 0 5 3 5 y 2 0 0.8 4

MATHEMATICS–IX

x 0 1 4 5 y 7 5 1 3

LINEAR EQUATIONS IN TWO VARIABLES

41

Example 9. From the choices given, choose the equation whose graph is given (i) y = 2x (ii) y = 2x + 1 (iii) x + y = 0

Solution.

(iv) y = 2x – 4

Given points on the graph are (2, 0), (1, –2) (0, –4) and (–1, –6). By trial and error we observe that all these 4 given points satisfy the equation y = 2x – 4, so, it is a graph of y = 2x – 4.

Example 10. The taxi fare in a city is as follows. For the first km., the fare is Rs. 8 and for the subsequent distance it is Rs. 5 per km. Write a linear equation for this information and draw its graph. Solution.

Let total distance covered is x km. Let total fare is Rs. y Since, fare for Ist km is Rs. 8 and for remaining (x–1) km. is Rs. 5 per km.  By given information, we have 8 × 1 + 5 (x – 1) = y  5x – y = –3 or

y  5x  3

Now, equation is y = 5x + 3. Giving different values to x, we get corresponding values of y. Let us represent this in a tabular form. x 1 2 3 4 y 8 13 18 23

42

LINEAR EQUATIONS IN TWO VARIABLES

MATHEMATICS–IX

Example 11. If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also, read from the graph the work done when the distance travelled by the body is : (i) 2 units (ii) 0 units. —NCERT Solution.

Let x be the distance and y be the work done. Then, according to the given problem, we have y = 5x.

( 5 is the constant force).

Let us not draw the graph of this linear equation in two varibles. Required table is :

MATHEMATICS–IX

x 0 1 2 y 0 5 10

LINEAR EQUATIONS IN TWO VARIABLES

43

(i) From the graph, we see that, x = 2 units distance  y = 10 units work done. (ii) From the graph, we see that, x = 0 unit distance  y = 0 unit work done. Example 12. Yamini and Fatima, two students of class IX of a school, together contributed Rs. 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this date. (You may take their contributions as Rs. x and Rs. y). Draw the graph of the same. —NCERT Solution. Let, Yamini contributed Rs. x and Fatima contributed Rs. y. then, according to the given question, we have, x + y = 100 Required table is :

x 20 40 60 y 80 60 40

Every point of the shaded portion including the line x + y = 100, x-axis and y-axis in the first quadrant represent the solution set. 44

LINEAR EQUATIONS IN TWO VARIABLES

MATHEMATICS–IX

Example 13. In countries like USA and Canada, temperature is measured in Fahrenheit, wheres in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius: F

(i) (ii) (iii) (iv) (v)

Solution.

F

9 C  32 . 5 Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for yaxis. If the temperature is 30°C, what is the temperature in Fahrenheit? If the temperature is 95°F, what is the temperature in Celsius? If the temperature is 0°C, what is the temperature in Fahrenheit, and if the temperature is 0°F, what is the temperature in Celsius? Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it. —NCERT 9 C  32 . 5

Required table is :

MATHEMATICS–IX

C 10 20 30 F 50 68 86

LINEAR EQUATIONS IN TWO VARIABLES

45

(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for yaxis. 9 C  32 is shown in the figure. 5 From the graph : C = 30°  F = 86° From the graph : F = 95°  C = 35° If C = 0°  F = 32° and, If F = 0°  C = – 17.8° Yes, clearly from the graph, the temperature which is numerically the same is both Fahrenheit and Celcius is – 40° F = – 40°C.

(i) The graph of the line F  (ii) (iii) (iv) (v)

PRACTICE EXERCISE 1. Which of the following are linear : (i) 3x (x – 2) = 3x2 + 2x – 7 (ii) (x + 1) (x – 2) = 10 2 (iv) 2x 2 + 7x – 3 = 2x (x + 1) (iii) x (x + 1) = –2x + 7x + 2 2. Write each of the following equation in the form ax + by + c = 0 (i) y – 3 = 2 (x – 1) (iii)

(ii)

x  y5 3

x –3 y6 2

(iv) 2 (x – 1) –3 (y + 1) = 1

x y  6 2 3

(iv)

x 1 3y –1  3 2 4 3. Write four solutions for each of the following : (i) 2x – y = 3 (ii) x = 3y 4. Find out which of the following equations have x = 2, y = –1 as a solution :

(iii)

(i) 3x + 2y = 4 (iv) 4x – y = 9

(ii) 4x – y = 8 (v)

x y2 2

(iii) x 

y 3 2

(vi) 7x – 4y = 18

5. Find the value of k if the given point lies on the graph defined by each equation : (i) 3x + ky = 2 ; (1, –1) (ii) y + kx = 8 ; (–3, 2) (iii) 2x – 3y = k ; (–1, 5) (iv) 2x – k = 4y ; (2, 0) 6. Show that x = 1, y = –6 ; x = 2, y = –3 and x = 3, y = 0 are all solution of equation 3x – y = 9. 7. The equation of a graph is given by 2x + y = 10. Indicate which of the following points lies on the graph. (i) (6, –2) (ii) (2, 8) (iii) (2, 6) (iv) (–3, 16) (v) (–6, 18) (vi) (–6, 22) 8. Draw the graph of line 2x + 5y = 13. Is the point (9, –1) lies on the line? 9. Draw the graph of y = –2x + 4. Find co-ordinates of point at which it intersects the axes. 10. Draw the graph of y = 3x + 2 and y = 3x – 1 using the same pair of axes. Are these two lines parallel? 11. Draw the graph of 3x + 2y = 7 and 2x – 3y = 10 using the same pair of axes. Are these two lines perpendicular? 46

LINEAR EQUATIONS IN TWO VARIABLES

MATHEMATICS–IX

12. Draw the graph of 3x – y = 5 and 2x + 3y = 7 on the same axes. What is their point of intersection? 13. Graph the following equations : (i) 2y = –x + 3

(ii)

x 1 y  2 3

(iii) 3x – 4y = 12

(iv) y = |x|

14. Amit invests Rs. 100 at the rate of 5% p.a. simple interest. Assuming rate to be same, find graphically the interest he will earn after 5 years? 15. Unit of temperature measurement, Fahrenheit and celsius are related by relation F  (i) (ii) (iii) (iv)

9 C  32 . 5

Draw graph of linear equation above, using celsius for x-axis and Fahrenheit for y-axis. If the temperature is 35°C, what is the temperature in Fahrenheit? If the temperature is 86°F, what is the temperature in celsius? Is there a temperature which is numerically the same in both units? If yes, find it.

PRACTICE TEST

Time : ½ hour

M.M : 15 General Instructions : All questions carry 3 marks each.

1. Give three integral solutions for equation 5x – y = 9. 2. Find the value of a so that the equation 4x – ay = 7 have (2, –3) as a solution. 3. Give the geometrical representation of x = –3 as an equation. (i) In one variable (ii) In two variables 4. Yamini and Fatima, together contributed Rs. 100 towards PM relief fund. Write a linear equation which this data satisfies. Draw the graph of the same. 5. Draw the graph of 3x + y = 8. Is x = –1, y = 11, a solution of this equation. Also, shade the portion bounded by this line and both the axes.

ANSWERS OF PRACTICE EXERCISE 1. (i) and (iv) 2. (i) 2x – y + 1 = 0

(ii) x – 3y – 15 = 0

(iii) x – 2y – 15 = 0

(iv) 2x–3y–6=0

(v) 3x – 2y – 36 = 0 (vi) 2x – 3y – 9 = 0 3. (i) (0, –3), (1, –1), (2, 1), (–1, –5)

(ii) (3, 1), (6, 2), (9, 3), (12, 4)

4. (i), (iv), (v), (vi) 5. (i) k = 1

(ii) k = –2

(iii) k = –17

(iv) k = 4

7. (i), (iii), (iv), (vi)

MATHEMATICS–IX

LINEAR EQUATIONS IN TWO VARIABLES

47

8. Yes

9. (0, 4) and (2, 0)

48

LINEAR EQUATIONS IN TWO VARIABLES

MATHEMATICS–IX

10. Yes

11. Yes

12. (2, 1)

13. (iv) y = | x |

MATHEMATICS–IX

LINEAR EQUATIONS IN TWO VARIABLES

49

14.

15.

From graph, we observe that after 5 years, he will receive Rs. 25.

From graph, we have : (ii) 95°F (iii) 30°C (iv) Yes, –40°

ANSWERS OF PRACTICE TEST 1. (1, –4), (2, 1) and (3, 6)

2. a  

1 3

--3 3. (i)

50

(ii)

LINEAR EQUATIONS IN TWO VARIABLES

MATHEMATICS–IX

4. x + y = 100

where, x Rs. = contribution by Yamini and y Rs. = contribution by Fatima. 5. Yes

MATHEMATICS–IX

LINEAR EQUATIONS IN TWO VARIABLES

51

CHAPTER

5

J A

INTRODUCTION TO EUCLID’S GEOMETRY Points to Remember : 1. A point, a line and a plane are concepts only and these terms are taken as undefined. 2. Axioms (or Postulates) are assumptions which are self evident truths.

J A

3. Theorems are the statements which are proved, using axioms, previously proved statements and deductive reasoning. 4. Some of Euclid’s axioms were:

(a) Things which are equal to the same thing are equal to one another.

(b) If equals are added (or subtracted) to / from equals, the wholes / remainders are equal. (c) The whole is greater than the part.

(d) Things which are double of the same things are equal to one another. 5. Euclid’s Five Postulates:

B

Postulate 1: A straight line may be drawn from any one point to any other line. Postulate 2: A terminated line can be produced indefinitely.

Postulate 3: A circle can be drawn with any center and any radius. Postulate 4: All right angles are equal to one another.

T I

Postulate 5: If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles. 6. Two equivalent versions of Euclid’s fifth Postulate: (a) Play fair axiom:

“Through a given point, not on the line, one and only one line can be drawn parallel to a given line.”

M A

(b) Two distinct intersecting lines cannot be parallel to the same line.

ILLUSTRATIVE EXAMPLES

Example 1. Prove that an equilateral triangle can be constructed on any given line segment. Solution. Draw a line segment, says PQ. Using Euclid’s postulate 3, draw a circle with center P and radius PQ. Again, we draw another circle with center Q and radius QP. The two circles meet at a point, says R. Join P to R and Q to R. now, in  PQR, PQ = PR and PQ = QR. So, by axiom 4, we get PQ = QR = PR. Hence, PQR is an equilateral triangle. 52

INTRODUCTION TO EUCLID’S GEOMETRY

MATHEMATICS–IX

Example 2. If a point C lies between two points A and B such that AC = BC, then prove that AC = Solution.

1 AB. 2

Since, C lies between AB.  AB = AC + BC ( by addition axiom)  AB = AC + AC ( AC = BC)  AB = 2AC

1 AB. 2 which prove the result. Example 3. Let point C be a mid-point of line segment AB. Prove that every line segment has one and only one mid-point. Solution. Let if possible, D be another mid-point of AB.  AD = DB ....(1) But, it is given that C is the mid-point of AB.  AC = CB ...(2) Subtracting (1) from (2), we get AC – AD = CB – DB  DC = – DC  2DC = 0  DC = 0  C and D must coincides. Thus, every line segment has one and only one mid-point. Example 4. In figure, if AC = BD, then prove that AB = CD. Solution. AC = BD ...(1) ( given ) Also, AC = AB + BC ...(2) ( point B lies between A and C) and, BD = BC + CD ...(3) ( point C lies between B and D) Substituting for AC and BD from (2) and (3) in (1), We get, AB + BC= BC + CD  AB + BC – BC = BC + CD – BC ( subtracting BC both sides)  AB = CD Hence proved. Example 5. Which of the following statements are true and which are false? Give reasons for your answers. (i) Only one line can pass through a single point. (ii) There are infinite number of lines which pass through two distinct points. (iii) A terminate line can be produced indefinitely on both sides. (iv) If two circles are equal, then their radii are equal. (v) In figure, if AB = PQ and PQ = XY then AB = XY.



AC =

—NCERT T Solution.

(i) False, since, through a point infinite number of lines may be drawn. (ii) False, since one and only one line can pass through two distinct points. (iii) True, since a line can be produced infinitely on both the ends. (iv) True, since two circles will be equal only when their radii are the same. (v) True, since AB = PQ, PQ = XY AB = XY. (By transitive property)

MATHEMATICS–IX

INTRODUCTION TO EUCLID’S GEOMETRY

53

PRACTICE EXERCISE 1. 2. 3. 4. 5. 6. 7.

What is the difference between axiom and a theorem? Give any three axioms given by Euclid. What are Euclid’s five postulates? State Play fair’s axiom. Give two equivalent versions of Euclid’s fifth postulate. If P,Q and R are three points on a line, and Q lies between P and R, then prove that PQ + QR = PR. Prove that an equilateral triangle can be constructed on any given segment.

8. Prove that, “Two distinct lines cannot have more than one point in common.”

54

INTRODUCTION TO EUCLID’S GEOMETRY

MATHEMATICS–IX

CHAPTER

6

J A

LINES AND ANGLES Points to Remember :

1. If a ray stands on a line, then the sum of the two adjacent angles so formed is 180° and vice-versa. This property is known as the Linear Pair Axiom. 2. If two lines intersect each other, then the vertically opposite angles are equal. 3. If a transversal intersects two parallel lines, then (a) each pair of corresponding angles is equal. (b) each pair of alternate interior angles is equal. (c) each pair of interior angles on the same side of the transversal is supplementary. 4. If a transversal intersects two line such that, either (a) any one pair of corresponding angles is equal, or (b) any one pair of alternate interior angles is equal, or (c) any one pair of interior angles on the same side of of the transversal is supplementary, then the lines are parallel. 5. Two intersecting lines cannot both be parallel to the same line. 6. Lines which are parallel to a given line are parallel to each other. 7. The sum of three angles of a triangle is 180°. 8. If a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.

T I

B

J A

ILLUSTRATIVE EXAMPLES

Example 1. If two lines intersects, prove that vertically opposite angles are equal. Solution. Given : Two lines AB and CD intersects at O. To prove : 1 = 3 and 2 = 4 Proof : OB stands on line COD. 1 + 2 = 180° ( linear pair) ...(1) Also, OD stands on line AOB 2 + 3 = 180° ( linear pair) ...(2) from (1) and (2) 1 + 2 = 2 + 3 1 = 3 Similarly, 2 = 4 Hence proved

M A

Example 2. The complement of an angle is half of itself. Find the angle and its complement. Solution. Let the given angle be x. Then, its complement is 90° – x. According to given question, 90° – x =

1 x 2



MATHEMATICS–IX

x

x  90 2

3x 2  90  90  x  2 3

LINES AND ANGLES

55

 x = 60° Hence, the given angle is 60° and its complement is 90° – 60° i.e. 30°. Example 3. Two supplementary angles are in the ratio 3 : 2. Find the angles. Solution. Let the angles be 3x and 2x. according to given question, 3x + 2x = 180° ( angles are supplementary)  x

180º  36º 5



5x = 180°



Angles are 3 x  3  36º  108    Ans. and 2 x  2  36º  72  

Example 4. What value of x would make AOB a line in given figure if AOC = (3x –10)° and BOC = (7x + 30)°?

Solution.    

Since, OB and OA are opposite rays BOC + AOC = 180º 7x + 30° + 3x – 10° = 180° 10x + 20° = 180° 10x = 180° – 20° 10x = 160º  x 

160 10

 x  16 Ans.

Example 5. Rays OA, OB, OC, OD and OE have the common initial point O. Show that : AOB + BOC + COD + DOE + EOA = 360° Solution. Let us draw a ray OF oposite to ray OA. B

C 3 4

F D

2 0 5

1 6

A E

so, 1 + 2 + 3 = 180° and 4 + 5 + 6 = 180° ( linear pair axiom) adding (1) and (2), (1 + 2 + 3) + (4 + 5 + 6) = 180° + 180°  1 + 2 + (3+ 4) + 5 + 6 = 360°  AOB + BOC + COD + DOE + EOA = 360° Hence shown. 56

LINES AND ANGLES

...(1) ...(2)

MATHEMATICS–IX

Example 6. Prove that if a transversal intersects two parallel lines, then each pair of interior angle on the same side of the transversal is supplementary. Solution. Given : A transversal l intersects two parallel lines AB and CD at P and Q.

4 + 5 = 180° and 3 + 6 = 180º Proof : Ray QD stands on line l  1 + 4 = 180° ...(1) ( linear pair axiom) also, 1 = 5 ...(2) ( corresponding angles) from (1) and (2) 4 + 5 = 180° Again, ray QC stands on line l  2 + 3 = 180° ...(3) ( linear pair axiom) and, 2 = 6 ...(4) ( corresponding angles) from (3) and (4) 3 + 6 = 180° Hence proved. Example 7. If two parallel lines are intersected by a transversal, prove that the bisectors of the two pairs of interior angles encloses a rectangle. Solution. Given : Two parallel lines AB and CD are intersected by a transversal  at P and R respectively. PQ, RQ, RS and PS are bisectors of APR, PRC, PRD and BPR respectively. To prove :

To prove : PQRS is a rectangle Proof : Since AB || CD and l is a transversal  APR = PRD ( alt. interior angles) 

1 1 APR = PRD  QPR = PRS 2 2

But, these are alternate interior angles.  PQ || RS. Similarly QR || PS.  PQRS is a parallelogram. MATHEMATICS–IX

LINES AND ANGLES

57

Now, ray PR stands on AB  APR + BPR = 180°

( linear pair)

1 1 APR + BPR = 90° 2 2  QPR + SPR = 90°  QPS = 90° Thus, PQRS is a parallelogram, one of whose angle is 90°  PQRS is a rectangle. Hence proved. Example 8. In the given figure lines XY and MN intersect at O. If POY = 90° and a : b = 2 : 3, find c. 

—NCERT Since XY is a line  a + b + 90° = 180° ( linear pair)  a + b = 180° – 90°  a + b = 90° But a : b = 2 : 3. Let a = 2x and b = 3x.  2x + 3x = 90°  5x = 90°  x = 18°  a = 2x = 2 × 18° = 36° and b = 3x = 3 × 18° = 54° Now, OM and ON are opposite rays.  MON is a line. Since ray OX stands on MN  MOX + XON = 180° ( linear pair)  c + b = 180°  c + 54° = 180°  C = 180° – 154° = 126°  c = 126° Ans. Example 9. It is given that XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ZYP, find XYQ and reflex QYP. —NCERT Solution. Since XY is produced to point P,  XP is a straight line. Since, YZ stands on XP,  XYZ + ZYP = 180° ( linear pair)  64° + ZYP =1 80°  ZYP = 180° – 64° = 116° Since ray YQ bisects ZYP

Solution.

116  58 2 Now, XYQ = XYZ + ZYQ  XYQ = 64° + 58° = 122° and reflex QYP = 360° – QYP = 360° – 58° = 302° Ans.



58

QYP  ZYQ 

LINES AND ANGLES

MATHEMATICS–IX

Example 10. In the given figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.

—NCERT

Since, AB || CD and CD || EF  AB || CD || EF. Now, CD || EF and PR is a transversal.  PQD = PRF  180° – y = z  y + z = 180° also, y : z = 3 : 7. Let y = 3a and z = 7a  3a + 7a = 180°  10a = 180°  a = 18°  y = 3 × 18° = 54° and z = 7 × 18° = 126° Now, AB || CD and PQ is a transversal.  x + y = 180° ( consecutive interior angles are supplementary)  x + 54° = 180°  x = 180° – 54° = 126°  x = 126° Ans. Example 11. In the given figure, X = 62°, XYZ = 54°. If YO and ZO are the bisectors of XYZ and XZY respectively of XYZ, find OZY and YOZ. —NCERT Solution.

Solution.

In XYZ, YXZ + XYZ + XZY = 180° ( angle sum property of a triangle)  62° + 54° + XZY = 180°  XZY = 180° – 62° – 54° = 64° Since, YO and ZO are bisectors of XYZ and XZY, 

OYZ 

1 1 XYZ   54  27 2 2

1 1 XZY   64  32 2 2 In OYZ, we have YOZ + OYZ + OZY = 180° ( angle sum property of a triangle)  YOZ + 27° + 32° = 180°  YOZ = 180° – 27° – 32° = 121° Hence, OZY = 32° and YOZ = 121° Ans.

and, OZY 

MATHEMATICS–IX

LINES AND ANGLES

59

Example 12. In the given figure, if PQ  PS, PQ || SR, SQR = 28° and QRT = 65°, then find the values of x and y. —NCERT

QRT = RQS + QSR ( Exterior angle property in SRQ)  65° = 28° + QSR  QSR = 65° – 28° = 37° Since, PQ || SR and the transversal PS intersects then at P and S respectively.  PSR + SPQ = 180° ( sum of consecutive interior angles is 180°)  (PSQ + QSR) + 90° = 180°  y + 37° + 90° = 180°  y = 180° – 37° – 90° = 53° Now, in the right SPQ, we have PQS + PSQ = 90°  x + 53°= 90°  x = 90° – 53° = 37° Hence, x = 37° and y = 53° Ans. Example 13. In each of the following figures, AB || CD, find x.

Solution.

A B 50°

x°

O 25° D

C (i)

Solution.

(i) Through O, Draw EOF || AB || CD. then, 1 + 2 = x Now, EO || AB and BO is the transversal.  1 + 50º = 180º ( Interior angles on the same side of the transversal) A B 50°

E

1 O 2

C

60

F 25° D

LINES AND ANGLES

MATHEMATICS–IX

 1 = 180° – 50° 1 = 130° Again, EO || CD and OD is a transversal  2 + 25° = 180°  2 = 180° – 25° 2 = 155° adding (1) and (2) 1 + 2 = 130° + 155° = 285°

...(1)

...(2)

i.e. x  285º Ans. (ii) Produce AB to intersect CE and F. D

C A

105°

B 125°

G

F x° E

Now, ABE + EBF = 180°

( Linear pair)

 125° + EBF = 180°  EBF = 180° – 125° = 55° Again, CD || FG and CF is a transversal DCF + CFG = 180° ( Interior angles on same side of transversal) 105° + CFG = 180º CFG = 180° – 105° = 75° ( vertically opposite angles)

Also, BFE = CFG = 75°

Now, In BEF, BEF + EBF + BFE = 180° ( angle sum property)  x° + 55° + 75° = 180°  xº + 130° = 180°  x = 180º – 130° 

x  50 Ans.

Example 14. The side QR of PQR is produced to a point S. If the bisectors of PQR and PRS meet at point 1 —NCERT T, then prove that QTR = QPR. 2 Solution. Side QR of PQR is produced to S,  Ext. PRS =P + Q ( Exterior angle sum property of a triangle) 

1 1 1 Ext PRS = P + Q 2 2 2

1 P + 1 ...(1) 2 Again, In QRT., Ext. TRS = T + 1 ( same as above)  2 = T + 1 ...(2)



T

P

2 =

MATHEMATICS–IX

2

1 1 Q

LINES AND ANGLES

2 R

S

61

Equating (1) and (2), we get 1 P + 1 = T + 1 2 1 1  T = P or QTR = QPR 2 2 Hence Proved. Example 15. Bisectors of exterior angles of ABC (obtained by producing sides AB and AC) meet at O. Prove 1 that BOC = 90° – A. 2 Solution. Given : ABC in which bisectors of exterior angles meet at O (as shown) 1 To prove : BOC = 90 – A 2 Proof : In OBC 1 + 2 + 0 = 180º ( angle sum property of a triangle) 1 1  CBD + BCE + O = 180° 2 2  CBD + BCE + 2O = 360°  (180º – B) + (180º – C) + 2O = 360°  B + C = 2O  180 – A = 2O 1  O = (180 – A) 2 1  BOC = 90° – A. Hence proved. 2

T I

B

PRACTICE EXERCISE

1. 2. 3. 4. 5. 6.

J A

J A

If an angle is 24° less than its complement. find its measure. An angle is 40° less than one-third of its supplement. Find the angle and its supplement. Two supplementary angles are in the ratio 11 : 7. Find them. Find the angle whose supplement is four times its complement. Find the measure of an angle if, three times its supplement is 60° more than six times its complement. In the following figure, it is given that 2a – 5b = 10º find a and b.

M A

a

A

b B

7. In the following figures, what value of x will make AOB a straight line?

62

LINES AND ANGLES

MATHEMATICS–IX

8. In thefollowing figure, COA = 90º and AOB is a straight line. Find x and y.

9. Prove that the bisectors of the angles of a linear pair are at right angles. 10. If the bisectors of two adjacent angles form a right angle then prove that their non-common arms are in the same straight line. 11. In given figure, AOB is a line. Ray OD is perpendicular to AB . OC is another ray lying between OA and OD. Prove that DOC =

1 (BOC – AOC) 2 D

90° B

A

12. In the following figure, AB and CD intersects at O and BOE = 70°. Find value of a, b and c.

13. In the given figure 1 : 2 = 5 : 4 and AB || CD. Find all the labelled angles.

7

MATHEMATICS–IX

8

LINES AND ANGLES

63

14. Find missing x in the following diagrams. Given AB || CD.

A

B 140° 85°

E

x° D

C

(v) 15. In the following figure AB || CD. Find the values of a, b and c.

16. In the following figure, show that AB || CD. E

F 50°

A

64

30° 20°

C 160°

LINES AND ANGLES

D

B

MATHEMATICS–IX

17. In the following figure, AB || CD and CD || EF. Also EA  AB. If BEF = 65°, find the value of a, b and c.

x   x  18. The angles of a triangle are   50  ,   60  and (2x – 15)°, find the angles. 2  3 

19. If a transversal cuts two parallel lines and is perpendicular to one of them, show that it will be perpendicular to the other also. 20. If two parallel lines are intersected by a transversal, show that the bisectors of any corresponding angles are parallel. 21. If two lines are intersected by a transversal in such a way that the bisectors of a pair of corresponding angles are parallel, show that the lines are parallel to each other. —NCERT 22. Prove that the bisectors of a pair of alternate angles of two parallel lines are themselves parallel. 23. Prove that if the arms of an angle are, respectively, parallel to the arms of another angle, then the angles either have equal measure or they are supplementary. 24. PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD. —NCERT B

P

Q D

A R

C

S

25. If two straight lines are perpendicular to the same line, prove that they are parallel to each other. 26. One of the angles of a triangle is 55°. Find the remaining two angles, if their diffrence is 35°. 27. In the given figures find x :

MATHEMATICS–IX

LINES AND ANGLES

65

28. Sides QP and RQ of PQR are produced to points S and T respectively. If SPR = 140° and POT = 105°, find PRQ.

29. A square ABCD is surmounted by an equilateral triangle EDC. Find x. E x°

D

C

A

B

30. In the following figure, AB || DC. If x  B y

y 4 and y  z . Find the values of x, y and z. 2 9

x z

C

D A

PRACTICE TEST MM : 30

Time : 1 hour

General Instructions : Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each. 1. What value of x would make AOB a straight line? C

(2x–15)° x°

(3x + 3)° A

66

D

O

LINES AND ANGLES

B

MATHEMATICS–IX

2. If two parallel lines are intersected by a transversal, prove that the bisectors of two interior alternate angles are parallel. 3. Find the value of x°.

4. Find the value of x°, it is given that AB || CD.

5. If two straight lines are perpendicular to the same line, prove that they are parallel to each other. 6. In the given figure, show that AB || CD. A

B 80° C 145°

45° 35°

E

D

F

7. In the following figure, AB || CD, find a and b.

A

E

B

55° b

C

a F

135° G D

8. Angles A, B and C of a triangle satisfy B – A = 30° and C – B = 45°. Find all the angles. 9. Prove that the sum of three angles of a triangle is 180°. MATHEMATICS–IX

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10. In the given figure, the sides AB and AC of ABC are produced to points E and D respectively. If bisectors BO and CO of CBE and BCD respectively meet at O, then prove that : BOC = 90° –

1 BAC 2 A

B

C

E

D

O

ANSWERS OF PRACTICE EXERCISE 1. 33°

2. 15°, 165°

3. 110°, 70°

6. 130°, 50°

7. (i) x = 20° (ii) x = 33° (iii) x = 15°

4. 60°

5. 20°

8. x = 33°, y = 11°

13. 1 = 3 = 5 = 7 = 80°, 2 = 4 = 6 = 8 = 100°

12. a = 22°, b = 44°, c = 92°

14. (i) x = 115° (ii) x = 40° (iii) x = 55° (iv) x = 80°

(v) x = 135°

15. a = 100°, b = 35°, c = 45°

17. a = 25°, b = 115°, c = 115°

18. 65°, 70°, 45°

26. a = 80°, b = 45°

27. (i) 65° (ii) 95°

28. 65°

29. x = 45°

30. 24°, 48°, 108°

ANSWERS OF PRACTICE TEST 1. x = 32°

3. x = 110°

7. a = 55°, b = 80°

8. A = 25°, B = 55°, C = 100°

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4. 290°

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CHAPTER

7 TRIANGLE

Points to Remember : 1. Two figures are congruent, if they are of same shape and same size. 2. If two triangles ABC and XYZ are congruent under the correspondence A  X, B  Y and C  Z, then symbolically, ABC XYZ 3. SAS Congruence Rule : If two sides and the included angle of one triangle are equal to two sides and the included angle of the other triangle, then the two triangles are congruent. 4. ASA Congruence Rule : If two angles and the included side of one triangle are equal to two angles and the side of the other triangle, then the two triangles are congruent. 5. AAS Congruence Rule : If two angles and one side of one triangle are equal to two angles and the corresponding side of the other triangle, then the two triangles are congruent. 6. RHS congruence Rule : If in two right triangles, hypotenuse and one side of a triangle are equal to the hypotenuse and one side of other triangle, then the two triangles are congruent. 7. SSS Congruence rule : If three sides of one triangle are equal to the three sides of anoter triangle, then the two triangles are congruent. 8. Angles opposite to equal sides of a triangle are equal. 9. Sides opposite to equal angles of a triangle are equal. 10. Each angle of an equilateral triangle is 60°. 11. Of all the line segments that can be drawn to a given line from a point not lying on it, the perpendicular line segment is the shortest. 12. In a triangle, angle opposite to the longer side is greater. 13. In a triangle, side opposite the greater angle is longer. 14. Sum of any two sides of a triangle is greater than the third side. 15. Difference between any two sides of a triangle is less than its third side.

ILLUSTRATIVE EXAMPLES Example 1. In the given figure, AB = AC and ACD = 110°, find A. A

110° B

Solution.

Since AB = AC B = C Now ACB = 180° – 110°  ACB = 70°  B = ACB = 70° Now, A B + C = 180º

MATHEMATICS–IX

C

D

(angles opposite to equal sides are equal) ( linear pair)

(angle sum property of a triangle) TRIANGLES

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A + 70° + 70° = 180°



A = 180° – 140°   A  40  Ans.

Example 2. In given figure, AB = CF, EF = BD and AFE = DBC. Prove that AFE CBD. Solution. Given, AB = CF D Add BF both sides, we get AB + BF = CF + BF  AF = CB A Now, In AFE and CBD, F B AF = BC (proved above) AFE = CBD (given) and EF = BD (given) E So, by SAS congruence rule, AFE CBD

C

Example 3. In quadrilateral ABCD, AC = AD and AB bisects A. (see figure). Show that ABC  ABD. What can you say about BC and BD? —NCERT Solution. In ABC and ABD, we have AC = AD (given) CAB = BAD ( AB bisects A) and, AB = AB (common side)  ABC  ABD (SAS congruence condition) also, BC = BD (cpct) Example 4. ABCD is a quadrilateral in which AD = BC and DAB = CBA (see figure). Prove that : (i) ABD  BAC (ii) BD = AC (iii) ABD = BAC Solution. In ABD and BAC, we have AD = BC (given) DAB = CBA (given) AB = AB (common side)  ABD  BAC, (SAS congruence condition) which proves (i) part  BD = AC and ABD = BAC, which proves (ii) and (iii) (cpct)

—NCERT

Example 5. If the bisector of the vertical angle of a triangle bisects the base, prove that the triangle is isosceles. Solution. Given ABC in which AD is the bisector of A which meets BC in D such that BD = DC. To prove : AB = AC Construction : Produce AD to E, such that AD = DE. Join E to C 70

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A Proof : In ABD and ECD BD = DC (given) AD = DE (by construction) 1 2 ADB = EDC (vertically opp. angles)  ABD ECD (by SAS congruence condition)  AB = EC and 1 = 3 (cpct) C B D also, 1 = 2 (AD is angle bisector)  2 = 3  EC = AC (sides opp. to equal angles) 3  AB = AC ( EC = AB) E Hence, ABC is isosceles. Example 6. In a right angled triangle, one acute angle is double the other. Prove that the hypotenuse is double the smallest side. Solution. Given PQR in which Q = 90° and QRP = 2QPR To prove : PR = 2QR Construction : Produce RQ to S such that RQ = QS. Join P to S. Proof : In PSQ and PQR PQ = PQ (common side) PQS = PQR (= 90° each) QS = QR (by construction)  PSQ PQR (SAS congruence condition)  PS = PR and SPQ = RPQ = xº (say) (cpct)  SPR = 2x° = PRQ = PRS Now, In PSR, SPR = PRD = 2x°  PS = SR PR = SR ( PS = PR)  PR = 2QR ( SR = 2QR) Hence proved. Example 7. In the given figure, PQRS is a quadrilateral and T and U are respectively points on PS and RS such that PQ = QR, PQT = RQU and TQS = UQS. Prove that QT = QU. Solution. PQT = RQU (given) ....(1) TQS = UQS (given) .....(2) adding, (1) and (2) we getPQT + TQS = RQU + UQS or PQS = RQS Now, In PQS and RQS P PQ = RQ (given) T PQS = UQS (proved above) QS = QS (common side) 1 S  PQS RQS (SAS congruence condition) Q 2 So, 1 = 2 (cpct) Again, In TQS and UQS U TQS = UQS (given) R 1 = 2 (proved above)

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QS = QS (common side)  TQS UQS (ASA congruence condition) Hence, QT = QU (cpct) Example 8. Line l is the bisector of an  A, and B is any point on . BP and BQ are perpendiculars from B to the arms of A (see figure). —NCERT Show that : (i) APB  AQB (ii) BP = BQ or B is equidistant from the arms of A. Solution. In APB and AQB, we have APB = AQB ( each = 90°) PAB = QAB ( AB bisects PAQ) AB = AB ( common side)  APB  AQB, (AAS congruence condition) which proves (i) part. and, BP = PQ (cpct) which proves (ii) part. Example 9. In the given figure, AC = AE, AB = AD and BAD = EAC. Show that BC = DE.

—NCERT

In ABC and ADE, we have AB = AD (given) BAC = DAE ( BAD = EAC  BAD + DAC = EAC + DAC  BAC = DAE) and, AC = AE (given)  ABC  ADE (SAS congruence condition)  BC = DE (cpct) Hence shown. Example 10. In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see figure). Show that (i) AMC  BMD (ii) DBC is a right angle (iii) DBC  ACB Solution.

1 (iv) CM  .AB 2

—NCERT Solution.

72

(i) In AMC and BMD, we have AM = BM ( M is the mid-point of AB) AMC = BMD (Vertically opp. angles) CM = MD (given)   AMC  BMD (SAS congruence condition) TRIANGLES

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(ii) Now, AMC  BMD  BD = CA and BDM = ACM ...(1) (cpct) Thus, transversal CD cuts CA and BD at C and D respectively such that the alternate angles BDM and ACM are equal.  BD || CA.  CBD + BCA = 180° ( consecutive interior angles on the same side of a transversal are supplementary).  CBD + 90° = 180°  CBD = 90° or, DBC = 90° (iii) Now, in DBC and ACB, we have  BD = CA (from (i)) DBC = ACB ( each = 90°) BC = BC (common side) DBC  ACB (SAS congruence condition) (iv) CD = AB (cpct) 1 1 CD  AB  2 2 1 CM  AB.  Hence proved. 2 Example 11. In the given figure, AB = AC, DB = DC. Prove that ABD = ACD. Solution. Join A to D A In ABD and ACD, we have AB = AC (given) BD = CD (given) AD = AD (common side) D  ABD ACD (SSS congruence condition) C B So, ABD = ACD (cpct) Example 12. In ABC, D is the mid point of base BC. DC and DF are perpendiculars to AB and AC respectively, such that DE = DF. Prove that B = C. Solution. In BED and CFD, we have DE = DF (given) BD = DC (D is mid point of BC) BED = CFD (= 90° each)  BED CFD (RHS congruence condition) So, B = C (cpct) Example 13. In an isosceles triangle ABC, with AB = AC, the bisectors of B and C intersect each other at O. Join A to O. Show that : —NCERT (i) OB = OC (ii) AO bisects A. Solution. (i) In ABC, we have AB = AC  B = C ( angles opposite to equal sides are equal)

1 1 B  C 2 2 OBC = OCB 

1 1 B and OCB  C ) 2 2 OB = OC ( sides opposites to equal angles are equal).

( OB and OC bisect B and C respectively.  OBC   MATHEMATICS–IX

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(ii) Now, In ABO and ACO, we have, AB = AC (given) OBC = OCB (proved above) OB = OC (proved above)  ABO  ACO (SAS congruence condition)  BAO = CAO (cpct)  AO bisects BAC. Hence proved. Example 14. ABC is an isosceles triangle in which altitudes BE and CF are drawn to sides AC and AB respectively (see figure). Show that these altitudes are equal. —NCERT Solution. In ABC, AB = AC  ACB = ABC  ECB = FBC Now, In BEC and CFB, we have BEC = CFB (each = 90°) ECB = FBC (proved above) BC = BC (common side)  BEF  CFB (AAS congruence condition)  BE = CF (cpct) Hence shown. Example 15. ABC and DBC are two isosceles triangles on the same base BC (see figure). Show that ABD = ACD. —NCERT Solution. In ABC, we have AB = AC  ABC = ACB ...(1) ( angles opposite to equal sides are equal) Also, In BCD, we have BD = CD  DBC = DCB ...(2) ( same reason as above) Adding (1) and (2), we get, ABC + DBC = ACB + DCB  ABD = ACD. Hence shown. Example 16. ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that BCD is a right angle. —NCERT Solution. In ABC, we have AB = AC  ACB = ABC ...(1) ( angles opp. to equal sides are equal). Now, AB = AD (given)  AD = AC ( AB = AC). Thus, In ADC, we have AD = AC  ACD = ADC ...(2) Adding (1) and (2), we get ACB + ACD = ABC + ADC  BCD = ABC + BDC ( ADC = BDC) BCD + BCD = ABC + BDC + BCD ( adding BCD both sides) 2BCD = 180° ( angle sum property) BCD = 90° Hence, BCD is a right angle. 74

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Example 17. ABC and DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. (see figure). If AD is extended to intersect BC at P, show that : —NCERT (i) ABD  ACD (ii) ABP  ACP (iii) AP bisects A as well as D. (iv) AP is the perpendicular bisector of BC. Solution. (i) In ABD and DBC, we have AB = AC (given) BD = DC (given) AD = AD (common side)  ABD  DBC (SSS congruence condition) (ii) In ABP and ACP, we have AB = AC (given) BAP = CAP ( ABD  ACD  BAD = DAC  BAP = PAC) AP = AP (common side)  ABP  ACP (SAS congruence condition) (iii) Since ABD  ACD  BAD = DAC  AD bisects A  AP bisects A ...(1) In BDP and CDP, we have BD = CD (given) BP = PC ( ABP  ACP  BP = PC) DP = DP (Common side)  BDP  CDP (SSS congruence condition)  BDP = PDC  DP bisects D  AP bisects D ...(2) from (1) and (2), we get AP bisects A as well as D. (iv) Since AP stands on BC.  APB + APC = 180° ( linear pair) But, APB = APC 180  90  APB = APC  2 Also, BP = PC (proved above)  AP is perpendicular bisector of BC. Example 18. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of PQR (see figure). Show that : —NCERT (i) ABM  PQN (ii) ABC  PQR

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Solution.

Since, BC = QR

1 1 BC  QR  BM  QN ...(1) 2 2 Now, In ABM and PQN, we have AB = PQ (given) BM = QN (from (1)) AM = PN (given)  ABM  PQN, (SSS congruence condition) which proves (i) part.  B = Q (cpct) ...(2) Now, In ABC and PQR, we have AB = PQ (given) B = Q (from (2)) BC = QR (given)  ABC  PQR (SAS congruence condition) which proves (ii) part. Example 19. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles. —NCERT Solution. In BCF and CBE, we have BFC = CEB ( each = 90°) hyp. BC = hyp. BC (common side) FC = EB  BCF  CBE ( RHS congruence condition)  FBC = ECB (cpct) Now, In ABC ABC = ACB ( FBC = ECB)  AB = AC ABC is an isosceles triangle. Example 20. In the given figure, the line segment joining the mid points M and N of opposite sides AB and DC of quadrilateral ABCD is perpendicular to both these sides. Prove that the other two sides of quadrilateral are equal. Solution. Join M to D and M to C. In MND and MNC M DN = CN (given) A B MND = MNC (each = 90°) MN = MN (common side)  MND MNC (RHS congruence condition)  DM = CM and DMN = CMN ....(1) Now, AMN = BMN (= 90° each) ....(2) Subtracting (1) from (2), we get D C N AMN – DMN = BMN – CMN  AMD = BMC Now, In AMD and BMC, we have AM = BM (given)



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MD = MC (proved above) AMD = BMC (proved above)  AMD BMC (SAS congruence condition) So, AD = BC (c.p.c.t.) i.e. other two sides of the quadrilateral are equal. Example 21. In figure, B < A and C < D. Show that AD < BC.

—NCERT

Since, B < A and C < D  AO < BO and OD < OC ( side opp. to greater angle is larger) Adding these results, we get AO + OD < BO + OC  AD < BC. Hence shown. Example 22. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that A > C and B > D. —NCERT Solution. Join A to C and B to D. Since, AB is the smallest side of quadrilateral ABCD.  In ABC, we have BC > AB  8 > 3 ...(1) ( angle opp. to longer side is greater). Since, CD is the longest side of quadrilateral ABCD,  In ACD, we have CD > AD  7 > 4 ...(2) ( angle opp. to longer side is greater). Adding (1) and (2), 8 + 7 > 3 + 4  A > C Again, In ABD, we have AD > AB ( AB is the shortest side)  1 > 6 ...(3) In BCD, we have CD > BC ( CD is the longest side)  2 > 5 ...(4) Adding (3) and (4), we get 1 + 2 > 5 + 6  B > D Thus, A > C and B > D. Hence shown. Example 23. In the given figure, PR > PQ and PS bisects QPR. Prove that PSR > PSQ. —NCERT Solution.

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In PQR, we have PR > PQ (given)  PQR > PRQ ( Angle opposite to larger side is greater)  PQR + 1 > PRQ + 1 (adding 1 both sides)  PQR + 1 > PRQ + 2 ...(1) ( PS is the bisector of P  1 = 2) Now, In PQS and PSR, we have PQR + 1 + PSQ = 180° and PRQ + 2 + PSR = 180° PQR + 1 = 180° – PSQ and PRQ + 2 = 180° – PSR  180° – PSQ > 180° – PSR (Using (1))  – PSQ > – PSR  PSQ < PSR i.e. PSR > PSQ. Hence proved. Example 24. In the given figure, ABC is a triangle and D is any point in its interior. Show that : BD + DC < AB + AC Solution. Extend BD to meet AC in E. A In ABE, AB + AE > BE ( sum of any two sides is greater than the third side)  AB + AE > BD + DE ....(1) E In CDE, DE + EC > DC ....(2) D from (1) and (2), we get AB + AE + DE + EC > BD + DE + DC B C  AB + (AE + EC) > BD + DC  AB + AC > BD + DC  BD + DC > AB + AC Hence shown. Example 25. In the given figure PQRS is a qudrilateral in which diagonals PR and QS intersect in O. Show that (i) PQ + QR + RS + SP > PR + QS (ii) PQ + QR + RS + SP < 2 (PR + QS) Solution. In PSR, we have

Solution.

PS + SR > PR

....(1)

In PQR, PQ + QR > PR

R

S

( sum of any two sides is greater than the third side) ....(2)

adding (1) and (2), we get

O

PQ + QR + PS + SR > 2PR

....(3)

In PSQ, PS + PQ > SQ

....(4)

In RSQ, RS + RQ > SQ

....(5)

P

Q

adding (4) and (5), we get PQ + RQ + RS + SP > 2SQ

....(6)

adding (3) and (6), we get 2 (PQ + QR + RS + SP) > 2 (PR + SQ)  PQ + QR + RS + SP > PR + SQ which proves (i) part 78

....(7)

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Again, In OPQ, OP + OQ > PQ In OQR, OR + OQ > RQ In OSR, OS + OR > SR

....(8) ....(9) ....(10)

In OPS, OP + OS > PS

....(11)

adding (8), (9), (10), (11), we get 2 (OP + OQ + OR + OS) > PQ + QR + RS + SP or

2 (PQ + QS) > PQ + QR + RS + SP

or

PQ + QR + RS + SP < 2 (PR + QS)

which proves (ii) part.

PRACTICE EXERCISE 1. In PQR, P = 80° and PQ = PR. Find Q and R. 2. Prove that the measure of each angle of an equilateral triangle is 60°. 3. In ABC, BAC = 50° and AB = AC. If ACD = x°, find the value of x.

(Ans. 50°, 50°) (Ans. x = 115°)

A 50°

x°

B

C

D

4. If the altitude from one vertex of a triangle bisects the opposite side, prove that the triangle is isosceles. 5. Prove that the perpendiculars drawn from the vertices of equal angles of an isosceles triangle to the opposite sides are equal. 6. Prove that medians of an equilateral triangle are equal. 7. If S is the mid-point of the hypotenuse PR of a right-angled PQR. Prove that QS =

1 PR 2

8. In the given figure, it is given that AE = AD and CE = BD. Prove that ABE ACD. A

E

D O C

B

9. In the given figure, D and E are the points on the base BC of ABC such that BD = CE, AD = AE and ADE = AED. Prove that ABE ACD. A

B

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D

E

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79

10. Equilateral triangles ABD and ACE are drawn on the sides of a ABC. Prove that CD = BE. 11. BD is the bisector of ABC. P is any point on BD. Prove that the perpendiculars drawn from P to AB and BC are equal. 12. In the given figure, AD is the median and BE and CF are drawn perpendicular on AD and AD produced from B and C respectively. Prove that BE = CF.

13. In an isosceles ABC, in which AB = AC, the bisector of B and C meet at I. Show that (i) BI = CI (ii) AI is the bisector of A 14. If the altitudes AD, BE and CF of a ABC are equal, prove that ABC is equilateral. 15. Prove that diagonals of a rhombus bisect each other at right angle. 16. A point O is taken inside a rhombus ABCD such that its distances from the angular points A and C are equal. Show that BO and DO are in one and the same straight line. 17. In the given figure, m || n and E is the mid-point of AB. Prove that E is also the mid-point of any line segment CD having its end points at m and n respectively.

18. In the given figure, C is the mid-point of AB. If ACD = BCE and CBD = CAE, prove that DC = EC.

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19. In the given figure, EDC is an equilateral triangle and DABC is a square. Prove that : (i) EA = EB (ii) EBC = 15° E

D

C

A

B

20. In the given figure, CE  AB and DF  AB, and CE = DF. Prove that OC = OD. C

F

A

B

E

O

D

21. Prove that in a right triangle, the hypotenuse is the longest side. —NCERT 22. In the given figure, D is a point on the side of ABC such that AD = AC. Prove that AB > AD. A

B

C

D

23. Prove that the sum of three altitudes of a triangle is less than the sum of the three sides of a triangle. 24. AD is a median of ABC. Prove that AB + AC > 2AD. 25. In the given figure, D is a point on side AB of ABC and E is a point such that BD = DE. Prove that : AC + BC > AE A

E

D

B

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PRACTICE TEST M.M : 30

Time : 1 hour

General Instructions : Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each. 1. Prove that the angles opposite to the equal sides of a triangle are equal. 2. In ABC, D is the mid-point of BC. If DL  AB and DM  AC such that DL = DM, prove that AB = AC. 3. In the given figure, CPD = BPD and AD is the bisector of BAC, prove that CAP BAP. C P

A

D

B

4. In the given figure, ABCD is a square. E and F are the mid-points of the sides AD and BC respectively. Prove that BE = DF. D

C

F E

A

B

5. In the given figure AB and CD are respectively the smallest and the longest sides of a quadrilateral ABCD. Prove that A > C and B > D.

6. ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. Show that BCD = 90° D

A

B

82

C

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7. Show that of all the line segments drawn from a given point not on it, the perpendicular line segment is the shortest. 8. AD and BE are respectively altitudes of an isosceles triangle ABC with AC = BC. Prove that AE = BD. 9. ABC is a right angled triangle at C. M is the mid-point of AB. Prove that CM =

1 AB 2

10. Prove that “Two triangles are congruent if any two angles and the included side of one triangle is equal to any two angles and the included side of the other triangle.”

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CHAPTER

8 QUADRILATERALS

Points to Remember : 1. The sum of the angles of a quadrilateral is 360°. 2. A diagonal of a parallelogram divides it into two congruent triangles. 3. In a Parallelogram: (i) opposite sides are equal (ii) opposite angles are equal (iii) diagonals bisect each other 4. A quadrilateral is a parallelogram, if (i) opposite sides are equal, or (ii) opposite angles are equal, or (iii) diagonals bisect each other, or (iv) a pair of opposite sides is equal and parallel. 5. Diagonals of a rhombus bisect each other at a right angle and vice-versa. 6. Diagonals of a rectangle bisect each other and are equal, and vice-versa. 7. Diagonals of a square bisect each other at right angles and are equal, and vice-versa. 8. Mid-point Theorem : The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is half of it. 9. A line through the mid-point of a side of a triangle parallel to another side bisects the third side. 10. The quadrilateral formed by joining the mid-points of hte sides of a quadrilateral, in order, is a parallelogram.

ILLUSTRATIVE EXAMPLES Example 1. Four angles of a quadrilateral are in the ratio 1 : 2 : 3 : 4. Find them. Solution. Let the four angles be x, 2x, 3x and 4x. Since, sum of four angles of a quadrilateral is 360°, x + 2x + 3x + 4x = 360°  10x = 360°, or x = 36°  four angles are 36°, 2 × 36°, 3 × 36°, 4 × 36° i.e. 36°, 72°, 108°, 144° Ans. Example 2. In the given figure, PQRS is a trapezium in which PQ || SR. If P = 60° and Q = 75°, find S and R.

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P + S = 180° and Q + R = 180° ( interior angles on the same side of transversal are supplementarly)  60° + S = 180° and 75° + R = 180°  S = 180 – 60° and R = 180° – 75°  S = 120° and R = 105° Ans. Example 3. Show that the diagnomals of a rhombus are perpendicular to each other. Solution. Let ABCD be a given Rhombus. here, AB = BC = CD = DA Now, In AOD and COD, OA = OC ( diagonals of parallelogram bisect each other) OD = OD (common side) AD = CD Solution.

 ΔAOD  ΔCOD (SSS congruence condition)  AOD = COD (c.p.c.t.) But, AOD + COD = 180° (linear pair)  2AOD = 180°  AOD = 90° So, the diagonals of a rhombus are perpendicular to each other. Example 4. If the diagonals of a parallelogram are equal, then show that it is a rectangle. —NCERT Solution. Given : A parallelogram ABCD in which AC = BD. To prove : ABCD is a rectangle. Proof : In ABC and DCB, we have AB = DC (opp. sides of parallelogram) BC = BC (common side) AC = DB (given)  ABC  DCB (SSS congruence condition)  ABC = DCB ...(1) (cpct) But AB || DC and BC cuts them.  ACB + DCB = 180° (sum of consecutive interior angles is 180°)  2ABC = 180°  ABC = 90° Thus ABC = DCB = 90°  ABCD is a parallelogram one of whose angle is 90°. Hence, ABCD is a rectangle. Example 5. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. —NCERT Solution. Given : A quadrilateral ABCD in which the diagonals AC and BD intersect at O such that AO = OC, BO = OD and AC  BD. To prove : ABCD is a rhombus. Proof : In AOD and COB, we have AO = OC (given) OD = OB given) AOD = COB (vertically opp. angles)  AOD  COB (SAS congruence condition)  OAD = OCB ...(i) (cpct) MATHEMATICS–IX

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Now, line AC intersects AD and BC at A and C respectively such that AOD = OCB (proved in (i)) i.e. alternate interior angles are equal.  AD || BC. Similarly, AB || CD Hence, ABCD is a parallelogram. Again, In AOD and COD, we have OA = OC (given) AOD = COD (each = 90°) OD = OD (common side)  AOD  COD (SAS congruence condition)  AD = CD ...(2) (cpct) Now, ABCD is a parallelogram,  AB = CD and AD = BC (opp. sides of a parallelogram are equal)  AB = CD = AD = BC ( using (2)) Hence, quadrilateral ABCD is a rhombus. Example 6. Show that the diagonals of a square are equal and bisect each other at right angles. —NCERT Solution. Given : A square ABCD. To prove : AC = BD, AC  BD and OA = OC, OB = OD. Proof : Since ABCD is square,  AB || DC and AD || BC. Now, AB || DC and transversal AC intersects them at A and C respectively. BAC = DCA ( alternate interior angles are equal)  BAO = DCO ...(1) Again AB || DC and BD intersects them at B and D respectively.  ABD = CDB ( alternate interior angles are equal)  ABO = CDO ...(2) Now, In AOB and COD, we have BAO = DCO (from (1)) AB = CD (opp. sides of a parallelogram are equal) ABO = CDO (from (2))  AOB  COD (ASA congruence condition)  OA = OC and OB = OD (cpct) Hence, the diagonals bisect each other. Again, In ADB and BCA, we have AD = BC (sides of a square are equal) BAD = ABC (each 90°) AB = AB (common side)  ADB  BCA (SAS congruence condition)  AC = BD (cpct) Hence, diagonals are equal. Now, In AOB and AOD, we have, OB = OD (diagonals of a parallelogram bisect each other) AB = AD ( sides of a square are equal) AO = AO (common side)  AOB  AOD (SSS congruence condition)  AOB AOD (cpct) 86

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but, AOB AOD = 180°  AOB AOD =

180 = 90° 2

 AO  BD  AC  BD. Hence, diagonals intersect at right angles. Hence proved. Example 7. Show that if the diagonals of a quadrilateral one equal and bisect each other at righ angles, then it is a square. —NCERT Solution.

Given : A quadrilateral ABCD in which the diagonals AC = BD, AO = OC, BO = OD and AC  BD. To prove : Quadrilateral ABCD is a square. Proof : In AOD and COB, we have AO = OC (given) OD = OB (given) AOD = COB (vertically opp. angles)  AOD  COB (SAS congruence condition)  OAD = OCB ...(1) (cpct) Now, line AC intersects AD and BC at A and C respectively such that OAD = OCB, i.e. alternate interior angles are equal.  AD || BC Similarly, AB || CD. Hence, ABCD is a parallelogram. Now, In AOB and AOD, we have AO = AO (common side) AOB = AOD (each = 90°) OB = OD ( diagonals of a parallelogram bisect each other)  AOB  AOD (SAS congruence condition)  AB = AD (cpct) But, AB = CD and AD = BC (opposite sides of a parallelogram are equal)  AB = BC = CD = AD ...(2) Now, In ABD and BAC, we have AB = AB (common side) AD = BC (opp. sides of parallelogram are equal) BD = AC (given)  ABD  BAC (SSS congruence condition)  DAB = CBA (cpct) But, DAB + CBA = 180°  DAB CBA =

180 = 90° 2

Thus, ABCD is a parallelogram whose all the sides are equal and one of the angle is 90°.  ABCD is a square. Hence proved. MATHEMATICS–IX

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Example 8. Prove that bisectors of a parallelogram form a rectangle. Solution. Given : A paralleogram ABCD in which AR, BR, CP and DP are the bisectors of A, B, C and D respectively forming quadrilateral PQRS.

To prove : PQRS is a rectangle. Proof : DCB + ABC = 180° (co-interior angles of parallelogram are supplementary) 

1 1 DCB  ABC  90 2 2

 1  2  90 ...(1) Also, In CQB, 1 + 2 + CQB = 180° ...(2) from (1) and (2), we get CQB = 180° – 90° = 90°  RQP = 90° ( CQB = RQP, vertically opp. angles) Similarly, it can be shown, QRP = RSP = SPQ = 90° So, Quadrilateral PQRS is a rectangle. Example 9. In the given figure, ABCD is a parallelogram in which P and Q are the mid points of AB and CD respectively. If AQ and DP intersects at S and PC and BQ intersects at R, show that the quadrilateral PRQS is a parallelogram.

Solution.

AB || DC  AP || QC 1 1 AB  DC  AP  QC 2 2  AP || QC and AP = QC  APCQ is a parallelogram. So, AQ || PC or SQ || PR. Similarly, it can be easily shown that PS || QR. Thus, In quadrilateral PRQS, SQ || PR and PS || RQ. So, PRQS is a parallelogram.

Also, AB = DC 

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Example 10. ABCD is a parallelogram and line segments AP and CQ bisects the A and C respectively. Show that AP || CQ. Solution.

here, A = C ( opp. angles of a parallelogram are equal) 1 1 A  C 2 2  1 = 2 ...(1) Now, AB || DC and CQ is a transversal  2 = 3 ...(2) From (1) and (2), 1 = 3 Thus, transversal AB intersects AP and CQ at A and Q, such that 1 = 3 i.e. corresponding angles are equal. Hence, AP || CQ. Example 11. Diagonals AC of a parallelogram ABCD bisects A (see figure). Show that : (i) it bisects C also. (ii) ABCD is a rhombus.



Solution.

(i) Given : A parallelogram ABCD, in which diagonal AC bisects A. To prove : (i) AC bisects C i.e. 3 = 4 (ii) ABCD is a rhombus Proof : Since ABCD is a parallelogram,  AB || DC. Now, AB || DC and AC intersects them  1 = 3 ...(1) ( alternate interior angles) Again, AD || BC and AC intersects them.  2 = 4 ...(2) ( alternate interior angles) but, 1 = 2 ...(3) (given) from (1), (2) and (3), we get 3 = 4 Hence, AC bisects C. (ii) To prove : ABCD is a rhombus. from (i) part, We have 1 = 2 = 3 = 4 Now, in ABC, 1 = 4  AB = BC (sides opp. to equal angles in a triangle are equal) Similarly, In ADC, we have 2 = 3  AD = DC. Also, ABCD is a parallelogram.  AB = CD and AD = BC combining these, we get AB = BC = CD = DA. Hence, ABCD is a rhombus.

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Example 12. ABCD is a rhombus. Show that diagonal AC bisects A as well as C and diagonal BD bisects B as well as D. —NCERT Solution. Given : A rhombus ABCD To prove : (i) Diagonal AC bisects A as well as C. (ii) Diagonal BD bisects B as well as D. Proof : ADC, AD = DC (sides of a rhombus are equal)  DAC = DCA ...(1) (angles opp. to equal sides of a triangle are equal) Now, AB || DC and AC intersects them  BCA = DAC ...(2) (alternate angles) from (1) and (2), we get DCA = BCA ...(3)  AC bisects C. In ABC, AB = BC (sides of a rhombus are equal) from (3) and (4), we get BAC = DAC  AC bisects A. Hence, diagonal AC bisects A as well as C. Similarly, diagonal BD bisects B as well as D. Example 13. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show that : (i) APD  CQB (ii) AP = CQ (iii) AQB  CPD (iv) AQ = CP (v) APCQ is a parallelogram. —NCERT Solution. Given : ABCD is a parallelogram P and Q are points on the diagonal BD such that DP = BQ. To prove : (i) APD  CQB (ii) AP = CQ (iii) AQB  CPD (iv) AQ = CP (v) APCQ is a parallelogram. Construction : Join A to C to meet BD in O. Proof : We know that the diagonals of parallelogram bisect each other. Now, AC and BD bisect each other at O.  OB = OD But BQ = DP (given)  OB – BQ = OD – DP  OQ = OP Thus, in quadrilateral APCQ, diagonals AC and PQ are such that OQ = OP and OA = OC i.e., the diagonals AC and PQ bisects each other. Hence, APCQ is a parallelogram, which proves (v) part. (i) Now, In APD and CQB, we have 90

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AD = CB (opp. sides of a a parallelogram ABCD) AP = CQ (opp. sides of a a parallelogram APCQ) DP = BQ (given) APD  CQB (SSS congruence condition) (ii) AP = CQ (cpct) (iii) In AQB and CPD, we have AB = CD (opp. sides of a a parallelogram ABCD) AQ = CP (opp. sides of a a parallelogram APCQ) BQ = DP (given) AQB  CPD (SSS congruence condition) (iv) Since, AQB  CPD  AQ = CP (cpct) Example 14. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD respectively. (see figure). Show that : (i) APB  CQD (ii) AP = CQ —NCERT

Solution.

(i) Since, ABCD is a parallelogram,  DC || AB. Now, DC || AB and transversal BD intersects them at B and D respectively. ABD = BDC (alt. int. angles) Now, In APB and CQD, we have ABP = QDC ( ABD = BDC) APB = CQD (each = 90°) AB = CD (opp. sides of a parallelogram) APB  CQD (AAS congruence condition) (ii) Since, APB  CQD  AP = CQ (cpct)

Example 15. ABCD is a trapezium in which AB || CD and AD = BC (see figure). Show that : (i) A = B (ii) C = D (iii) ABC  BAD (iv) diagonal AC = diagonal BD Solution.

—NCERT Given : ABCD is a trapezium, in which AB || CD and AD = BC. To prove : (i) A = B (ii) C = D (iii) ABC  BAD (iv) diagonal AC = diagonal BD Construction : Produce AB and draw a line CE || AD. Proof : (i) Since AD || CE and transversal AE cuts them at A and E respectively.  A + E = 180° ...(1)

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Since AB || CD and AD || CE,  AECD is a parallelogram.  AD = CE  BC = CE (AD = BC (given)) Thus, In BCE, we have BC = CE CBE = CEB  180° –B = E  B + E = 180° ...(2) from (1) and (2), we get, A + E = B + E  A = B (ii) Since A = B  BAD = ABD  180° – BAD = 180° – ABD  ADB = BCD  D = C i.e. C = D (iii) In ABC and BAD, we have BC = AD (given) AB = AB (common) A = B (proved above)  ABC  BAD (SAS Congruence condition) (iv) Since, ABC  BAD  AC = BD (cpct) Hence proved. Example 16. Show that the quadrilateral formed by joining the mid-points of the adjacent sides of a quadrilateral is a parallelogram. Solution. Given : A quadrilateral ABCD, in whcih P, Q, R, S are the mid-points of AB, BC, CD and DA respectively. To prove : Quadrilteral PQRS is a parallelogram. Construction : Join A to C. Proof : In ABC, P and Q are mid-points of AB and BC respectively. 1 AC ( mid-point theorem) 2 Again, In DAC, R and S are mid-points of sides CD and AD respectively.

 PQ || AC and PQ 

1 AC [ mid-point theorem] 2 Now, PQ || AC and SR || AC  PQ || SR

 SR || AC and SR 

1 AC  SR  PQ  SR 2  PQ || SR and PQ = SR Hence, PQRS is a parallelogram.

Again, PQ 

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Example 17. ABCD is a rhombus and P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. —NCERT Solution. Given : ABCD is a rhombus in which P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined to obtain a quadrilateral PQRS. To prove : PQRS is a rectangle. Construction : Join A to C. Proof : In ABC, P and Q are the mid-points of AB and BC respectively. 1 AC ...(1) 2 ( mid-point theorem) Similarly, In ADC, R and S are the mid-points of CD and DA respectively.

 PQ || AC and PQ =

1 AC 2 ( mid-point theorem) from (1) and (2), we get PQ || SR and PQ = SR  PQRS is a parallelogram. Now, AB = BC

 SR || AC and SR =

...(2)

(sides of rhombus are equal)

1 1 AB = BC  PB = BQ  3 = 4 ( angles opp. to equal sides of a triangle are equal) 2 2 Now, In APS and CQR, we have AP = CQ (half of equal sides AB and BC) AS = CR (half of equal sides AD and CD) PS = QR (opp. sides of parallelogram PQRS)  APS  CQR (SSS congruence condition)  1 = 2 (cpct) Now, 1 + SPQ + 3 = 180° ( linear pair)  1 + SPQ + 3 = 2 + PQR + 4 but, 1 = 2 and 3 = 4 (proved above)  SPQ = PQR ...(3) Now, SP || RQ and PQ intersects them,  SPQ + PQR = 180° ...(4) from (3) and (4), we get 2SPQ = 180°  SPQ = 90° Thus, PQRS is a parallelogram whose one angle is 90°. Hence, PQRS is a rectangle. Example 18. ABCD is a rectangle and P, Q, R, S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. —NCERT



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Solution.

Given : ABCD is a rectangle, in which P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined to obtain a quadrilateral PQRS. To prove : PQRS is a rhombus. Construction : Join A to C. Proof : In ABC, P and Q are the mid-points of sides AB and BC. 1 AC ...(1) ( mid-point theorem) 2 Similarly, In ACD, R and S are the mid-points of sides CD and DA.



PQ || AC and PQ =

1 AC ...(2) 2 from (1) and (2), we get PQ || SR and PQ = SR  PQRS is a parallelogram. Now, AD = BC (opp. sides of a rectangle ABCD)



RS || AC and RS =

( mid-point theorem)

1 1 AD  BC  AS  BQ 2 2 Now, In APS and BPQ, we have AP = BP ( P is the mid-point of AB) PAS = PBQ (each = 90°) AS = BQ (proved above)  APS  BPQ (SAS congruence condition)  PS = PQ (cpct) Thus, PQRS is a parallelogram in which adjacent sides are equal.  PQRS is a rhombus. Example 19. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see figure). Show that F is the mid-point of BC. —NCERT Solution. Given : A trapezium ABCD, in which AD || BC. E is the mid-point of AD and EF || AB.



To prove : F is the mid-point of BC. Construction : Join B to D. Let it intersect EF in G. Proof : In DAB, E is the mid-point of AD (given) EG || AB ( EF || AB)  By converse of mid-point theorem, G is the mid-point of DB. Now, In BCD, G is the mid-point of BD ( proved above) GF || DC ( AB || DC, EF || AB  DC || EF)  By converse of mid-point theorem, F is the mid point of BC. 94

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Example 20. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that : (i) D is the mid-point of AC

(ii) MD  AC

(iii) CM  MA 

1 AB 2

—NCERT Solution.

Given : ABC is right angled at C, M is the mid-point of hyp. AB . MD || BC. To prove : (i) D is the mid-point of AC (ii) MD  AC 1 AB 2 Proof : (i) In ABC, M is the mid-point of AB and MD || BC.  D is the mid-point of AC i.e. AD = DC ...(1) (ii) Since MD || BC  ADM = ACB (corresponding angles)  ADM = 90° ( ACB = 90°, MD || BC) But, ADM + CDM = 180° ( linear pair)  90° + CDM = 180°  CDM = 90° Thus, ADM = CDM = 90° ...(2)   MD AC. (iii) Now, In AMD and CMD, we have AD = CD (from (1)) ADM = CDM (from (2)) MD = MD (common side)  AMD  CMD (SAS congruence condition)   MA MC (cpct)

(iii) CM  MA 

Also, Hence,

MA 

CM = MA 

1 AB, 2

Since M is the mid-point of AC.

1 AB. 2

PRACTICE EXERCISE 1. In the given figure, PQRS is a rectangle whose diagonals PR and QS intersect at O. If OPQ = 32°, find OQR. (Ans. : 58°)

2. Prove that, if in a parallelogram diagonals are equal and perpendicular to each other, then it is a squqre. 3. In the given figure, PQRS is a parallelogram and X and Y are the points on the diagonal QS such that SX = QY. Prove that PYRS is a parallelogram. MATHEMATICS–IX

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95

4. Let ABC and DEF be two triangles drawn in such a way that AB || DE ; AB = DE; BC || EF and BC = EF. Show that AC || DF and AC = DF.

5. In a parallelogram ABCD, the bisectors of consecutive angles A and B intersect at O. Prove that AOB = 90°. 6. In the given figure, ABCD is a parallelogram and P is the mid-point of AD. A line through D, drawn parallel to PB, meets AB produced at Q and BC at R, prove that : (i) AQ = 2DC (ii) DQ = 2DR

7. In a parallelogram ABCD, if A = (4x + 20)° and B = (3x – 15)°, find the value of x and the measure of ecah angle of a parallelogram. (Ans. x = 25 and angles are 60°, 120°, 60°, 120°) 8. A ABC is given. If lines are drawn through A, B, C parallel respectively to the sides BC, CA and AB forming XYZ, as shown. Show that BC 

1 YZ. 2

9. In a parallelogram ABCD, the bisector of A also bisects the side BC at E. Show that AD = 2AB. 10. If a transversal cuts two parallel lines, prove that the bisectors of the interior angles form a rectangle. 96

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11. ABC is a triangle right angled at B and D is the mid-point of AC. DE is drawn perpendicular to BC. Prove that BD 

1 AC. 2

12. If P, Q and R are respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC, prove that PQR is also equilateral triangle. 13. ABCD is a parallelogram in which E and F are mid-points of the sides AB and CD respectively. Prove that the line segments CE and AF trisect the diagonal BD. 14. Let ABCD be a trapezium in which AB || DC and let E be the midpoint of AD. Let F be a point on BC such that EF || AB. Prove that: (i) F is the mid-point of BC 1 (AB  DC) 2 Prove that the line segment joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides and is equal to half their difference. Show that the quadrilateral formed by joining the midpoints of pairs of adjacent sides of a rhombus is a rectangle. Show that the quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a rectangle is a rhombus. Show that the line segments joining the mid-points of opposite sides of a quadrilateral bisect each other. Show that the four triangles formed by joining the mid-points of the three sides of a triangle are congruent to each other. In ABC, AD is the median, and E is the mid-point of AD . BE produced meets AC in F. Prove that

(ii) EF 

15. 16. 17. 18. 19. 20.

AF 

1 AC . 3

21. The diagonals of a quadrilateral are perpendicular. Prove that the quadrilateral, formed by joining the mid-points of its sides is a rectangle. 22. D, E and F are respectively the mid-points of the sides BC, AC and AB of an isosceles triangle ABC, in which AB = AC. Prove that AD is perpendicular to EF and is bisected by it. 23. In the given figure, ABC is a triangle and through A, B, C lines are drawn parallel to BC, CA and AB respectively intersecting at P, Q and R. Prove that perimeter of PQR is double the perimeter of ABC.

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24. ABC is any triangle. D is a point of AB such that AD 

1 AB and E is a point on AC such that 4

1 1 AC. Prove that DE  BC. 4 4 25. BX and CY are perpendiculars to a line passing through the vertex A of a triangle ABC. If Z is the midpoint of BC. Prove that XZ = YZ. AE 

PRACTICE TEST MM : 30

Time : 1 hour

General Instructions : Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each. 1. In parallelogram ABCD, AB = 10 cm and BC = 8 cm. If B = 120°, find : (i) DA (ii) C (iii) D 2. ABCD is a parallelogram and AB is produced to X so that AB = BX. Prove that DX and BC bisect each other. 3. Two adjacent angles of a parallelogram are (2x + 15)° and (3x – 25)°. Find the value of x and also the measure of the angles. 4. ABCD is a trapezium, in which AB || DC. X and Y are respectively mid-points of AD and BC. If AB = 13 cm and CD = 9 cm, find XY. 5. In a parallelogram, show that the angle bisectors of two adjacent angles intersect at right angle. 6. ABCD is a parallelogram whose diagonals intersect each other at O. Through O, XY is drawn as shown. Prove that OX = OY.

7. PQRS is a parallelogram. PO and QO are respectively the angle bisectors to P and Q. Line AOB is drawn parallel to PQ. Prove that : (i) PA = QB (ii) AO = OB

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8. Prove that in a right angled triangle, the median bisecting the hypotenuse is half of the hypotenuse. 9. ABC is a triangle right angled at C, A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that : (i) D is the mid-point of AC. (ii) MD  AC. 1 AB. 2 10. Prove that the quadrilateral formed by joining mid-points of the sides of a square is again a square.

(iii) CM  MA 

ANSWERS OF PRACTICE EXERCISE 1. (i) 8 cm (ii) 60° (iii) 120° 3. x = 38, angles are 89°, 91°, 89°, 91° 4. 11 sq. units

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CHAPTER

9

AREAS OF PARALLELOGRAMS AND TRIANGLES Points to Remember : 1. Two congruent figures must have equal areas. However, two figures having equal areas need not to be congruent. 2. Two figures are said to be on the same base and between the same parallels, if they have a common base and the vertices (or the vertex) opposite to the common base of each figure lie on a line parallel to the base. 3. Parallelograms on the same base and between the same parallels are equal in area. 4. Area of parallelogram = Base × corresponding height. 5. Parallelograms on the same base (or equal bases) and having equal areas lie between the same parallels. 6. Two triangles on the same base (or equal bases) and between the same parallels are equal in area. 7. Two triangles having the same base (or equal bases) and equal areas lie betwen the same parallels. 8. Area of Triangle 

1 × Base × corresponding height. 2

9. Area of a Rhombus 

1 × product of diagonals. 2

1 × (sum of the parallel sides) × (distance between them). 2 11. A median of a triangle divides it into two triangles of equal area. 12. The diagonals of a parallelogram divides it into four triangles of equal area.

10. Area of a Trapezium 

ILLUSTRATIVE EXAMPLES Example 1. In the given figure, ABCD is a parallelogram, AB = 12 cm, altitude DF = 7.2 cm and DE = 6 cm, find the perimeter of parallelogram ABCD. Solution. Area of parallelogram ABCD = AB × DE ...(1) also, Area of parallelogram ABCD = BC × DF ...(2) From (1) and (2), we get AB × DE = BC × DF  12 × 6 = BC × 7.2  

12  6  10 cm 7.2 Perimeter of parallelogram ABCD = 2 (AB + BC) = 2 (12 + 10) cm = 2 (22) cm = 44 cm Ans. BC 

Example 2. P, Q, R, S are respectively, the mid-points of sides AB, BC, CD and DA of parallelogram ABCD. Show that the quadrilateral PQRS is a parallelogram and its area is half the area of the parallelogram ABCD. 100

AREAS OF PARALLELOGRAMS AND TRIANGLES MATHEMATICS–IX

Solution.

Join A to C and Q to S

Now, In ABC, P and Q are respectively mid-points of AB and BC.  PQ || AC and PQ 

1 AC 2

Similarly, SR || AC and SR 

1 AC 2

...(1) ( mid-point theorem) ...(2)

From (1) and (2), we get PQ || SR and PQ = SR So, PQRS is a parallelogram. Now, SPQ and parallelogram ABQS stand on same base SQ and between same parallels AB and QS,  ar (PQS) 

1 ar (parallelogram ABQS) 2

...(3)

1 ...(4) ar (parallelogram SQCD) 2 adding (3) and (4), we get, 1 ar (PQS) + ar (SRQ)  [ar (parallelogram ABQS)  ar (parallelogram SQCD)] 2 1  ar (parallelogram PQRS)  ar (parallelogram ABCD) 2 which proves the result. Example 3. P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC). —NCERT Solution. APB and parallelogram ABCD stand on the same base AB and lie between the same parallels AB and DC.

Similarly, ar (SRQ) 

1 ar(ABCD) ...(1) 2 Similarly, BQC and parallelogram stand on the same base BC and lie between the same parallels BC and AD.

 ar (APB) =

1 ar (ABCD) ...(2) 2 from (1) and (2), we get ar (APB) = ar (BQC). Hence proved. Example 4. A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field seperately. How should she do it? —(NCERT)

 ar (BQC) 

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Solution.

Clearly, the field parallelogram PQRS is divided into 3 parts. Each part is in shape of a triangle. Since APQ and parallelogram PQRS stand on the same base PQ and lie between the same parallels PQ and SR. 1 . ar (PQRS) ...(1) 2 Clearly, ar (APS) + ar (AQR) = ar (PQRS) – ar (APQ)

 ar (APQ) =

= ar (PQRS) –

1 ar (PQRS) ( using (1)) 2

1 ar (PQRS) ...(2) 2 From (1) and (2), we get ar (APS) + ar (AQR) = ar (APQ) Thus, the farmer should sow wheat and pulses either as [(APS and AQR) or APQ] or as [APQ or (APS and AQR)]. Example 5. Show that a median of a triangle divides it into two triangles of equal area. —NCERT Solution. Let ABC be a given triangle and AD is a median. Draw AE  BC. Since, D is mid point of BC, so we have, BD = DC ...(1)

=

Now, ar (ABD) 

1  BD  AE 2

and, ar (ADC) 

1  CD  AE 2

...(2)

1  BD  AE ...(3) 2 ( BD = DC) From (2) and (3), we get ar (ABD) = ar (ADC) which proves the desired result. Example 6. E is any point on median AD of a ABC. Show that ar (ABE) = ar (ACE) Solution. Given : AD is a median of ABC and E is any point on AD. To prove : ar (ABE) = ar (ACE) Proof :  AD is the median of ABC  ar (ABD) = ar (ACD) ...(1) also, ED is the median of EBC,  ar (BED) = ar (CED) ...(2) Subtracting (2) from (1), we get ar (ABD) – ar (BED) = ar (ACD) – ar (CED)  ar (ABE) = ar (ACE). Hence shown. 

Example 7. In a ABC, E is the mid-point of median AD. Show that ar (BED) = Solution.

—NCERT T

Given : ABC in which E is the mid-point of median AD. To prove : ar (BED) 

102

1 ar (ABC). 4

—NCERT

1 . ar (ABC) 4

AREAS OF PARALLELOGRAMS AND TRIANGLES MATHEMATICS–IX

Proof : Since AD is a median of ABC and median divides a triangle into two triangles of equal area.  ar (ABD) = ar (ADC)  ar (ABD) =

1 ar (ABC) 2

...(1)

In ABD, BE is a median, 

ar (BED) = ar (BAE)

 ar (BED) =

1 ar (ABD) 2



1 1  ar (ABC) 2 2

=

1 ar (ABC) 4

( using (1))

Hence shown. Example 8. If the medians of a ABC intersect at G, show that

Solution.

ar (AGB) = ar (AGC) = ar (BGC)  1 ar (ABC) 3 We know that a median of a triangle divides it into two triangles of equal areas. In ABC, AD is the median. 

ar (ABD) = ar (ACD)

Again, In GBC, ar (GBD) = ar (GCD)

...(1) ...(2)

Subtracting (2) from (1), ar (ABD) – ar (GBD) = ar (ACD) – ar (GCD)  ar (ABG) = ar (AGC)

...(3)

Similarly, ar (AGB) = ar (BGC)

...(4)

From (3) and (4), we get, ar (ABG) = ar (AGC) = ar (BGC) Now, ar (ABC) = ar (ABG) + ar (AGC) + ar (BGC) = 3 ar (ABG) , ar (ABC) 

1 ar (ABC) 3

Hence, ar(AGB) = ar (AGC) = ar (BGC)  1 ar ( ABC) 3 Example 9. PQRS and ABRS are parallelograms and X is any point on the side BR. Show that : (i) ar (parallelogram PQRS) = ar (parallelogram ABRS) (ii) ar (AXS) =

1 ar (parallelogram PQRS) 2

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Solution.

Since, parallelogram PQRS and ABRS stand on same base SR and between same parallels SR and PB,  ar (parallelogram PQRS) = ar (parallelogram ABRS) 1 ar (parallelogram ABRS) 2 ( ASX and parallelogram ABRS stand on same base AS and between parallel lines AS and BR) But, ar (parallelogram ABRS) = ar (parallelogram PQRS)

Again, ar (ASX) =

 ar (ASX) 

1 ar (parallelogram PQRS) 2

Hence proved. Example 10. In ABC, D, E and F are mid-points of the sides BC, CA and AB respectively. Prove that : (i) BDEF is a parallelogram 1 (ii) ar (DEF)  ar (ABC) 4 1 (iii) ar (BDEF)  ar (ABC). 2 Solution. (i) FE || BC or FE || BD and DE || BF ( mid point theorem)  BDEF is a parallelogram. (ii) Since a diagonal of a parallelogram divides it into two triangles equal in area.  ar (BDF) = ar (DEF) Similarly, ar (DCE) = ar (DEF)

and ar (DEF) = ar (AFE) Combining all these, we get ar (BDF) = ar (DEF) = ar (DCE) = ar (AFE) But, ar (ABC) = ar (BDF) + ar (DEF) + ar (DCE) + ar (AFE) = 4 ar (DEF) 1 . ar (ABC) 4 (iii) Again, 4ar (DEF) = ar (ABC) 1  2 . ar (DEF) = ar (ABC) 2 1  ar (BDEF) = . ar (ABC) 2 Hence proved.

 ar (DEF) 

104

AREAS OF PARALLELOGRAMS AND TRIANGLES MATHEMATICS–IX

Example 11. In the given figure, ABC and ABD are two triangles on the same base AB. If line segment CD is bisected by AB at O, show that ar (ABC) = ar (ABD). —NCERT

Solution.

J A

J A

Given : ABC and ABD are two triangles on the same base AB. A line segment CD is bisected by AB at O. i.e. OC = OD. To prove : ar (ABC) = ar (ABD) Proof : In ACD, we have OC = OD (given)  AO is a median.  ar (AOC) = ar (AOD) ...(1) ( median divides a triangle in two triangles of equal areas). Similarly, In BCD, BO is the median.  ar (BOC) = ar (BOD) ...(2) adding (1) and (2), we get ar (AOC) + ar (BOC) = ar (AOD) + ar (BOD)  ar (ABC) = ar (ABD). Example 12. In the given figure, diagonals AC and BD of a quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that : (i) ar (DOC) = ar (AOB) (ii) ar (DCB) = ar (ACB) (iii) DA || CB or ABCD is a parallelogram. —NCERT Solution.

T I

B

(i) Draw DN  AC and BM  AC. In DON and BOM DNO = BMO (each = 90°) DON = BOM (vert. opp. angles) OD = OB (given)  DON  BOM (AAS congruence condition)  ar (DON) = ar (BOM) ...(1) ( congruent triangles have equal area) Again, In DCN and BAM DNC = BMA (each = 90°) DC = AB (given) DN = BM (DON  BOM  DN = BM)  DCN  BAM (RHS congruence condition)  ar (DCN) = ar (BAM) ...(2) ( congruent triangles have equal areas) adding (1) and (2), we get ar (DON) + ar (DCN) = ar (BOM) + ar (BAM)  ar (DOC = ar (AOB) (ii) Since, ar (DOC) = ar (AOB)  ar (DOC) + ar (BOC) = ar (AOB) + ar (BOC)

M A

MATHEMATICS–IX AREAS OF PARALLELOGRAMS AND TRIANGLES

105

 ar (DCB) = ar (ACB) (iii) DCB and ACB have equal areas and have the same base. So, these triangles lie between the same parallels.  DA || CB. i.e. ABCD is a parallelogram. Example 13. A point O inside a rectangle ABCD is joined to the vertices. Prove that : ar (AOD) + ar (BOC) = ar (AOB) + as (COD) Solution. Draw POQ || AD and ROS || AB. Since, POQ || AD and DC cuts them. ADC = PQC = 90° ( corr. angles) i.e. OQ  CD. Similarly, OR  AD, OP  AB and OS  BC. Consider, ar (AOD) + ar (BOC) 

1 1  AD  OR   BC  OS 2 2



1  AD  (OR  OS) 2



1 1  AD  RS   AD  AB 2 2

( AD  BC)

1 ar (rect. ABCD) 2 Again, ar (AOB) + ar (COD) 



1 1  AB  OP   CD  OQ 2 2



1  AB  (OP  OQ) 2



1 1  AB  PQ   AB  AD 2 2

...(1)

( AB = CD)

1 ar (rect. ABCD) ...(2) 2 From (1) and (2), we get ar (AOD) + ar (BOC) = ar (AOB) + ar (COD) Example 14. The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see figure). Show that ar (ABCD) = ar (PBQR). 

Solution.

106

Join A to C and P and Q. Since AC and PQ are diagonals of parallelogram ABCD and parallelogram BPQR respectively. AREAS OF PARALLELOGRAMS AND TRIANGLES MATHEMATICS–IX

 ar (ABC) =

1 ar (ABCD) 2

...(1)

1 ar (BPRQ) ...(2) 2 Now, ACQ and AQP are in the same base AQ and CP.  ar (ACQ) = ar (AQP)  ar (ACQ) – ar (ABQ) = ar (AQP) – ar (ABQ) ( Subtracting ar (ABQ) from both sides)  ar (ABC) = ar (BPQ)

and, ar (PBQ) =

1 1 ar (ABCD)  ar(BPRQ) ( using (1) and (2)) 2 2  ar (ABCD) = ar (BPRQ) Hence proved. Example 15. In the given figure, PQ is a line parallel to side BC of ABC. If BX || CA and CY || BA meet the line PQ produced in X and Y respectively, show that : ar (ABX) = ar (ACY).



Solution.

Parallelogram XBCQ and ABX stand on the same base BX and between the same parallels BX and CA, we have 1 ar (parallelogram XBCQ) ...(1) 2 Also, parallelogram BCYP and ACY stand on same base CY and between the same parallels CY and BA, we have

ar (ABX) 

1 ar (parallelogram BCYP) ...(2) 2 But, parallelogram XBCQ and parallelogram BCYP stand on same base BC and between same parallels BC and XY,  ar (parallelogram XBCQ) = ar (parallelogram BCYP) ...(3) from (1), (2) and (3), we get, ar (ABX) = ar (ACY) Hence proved. Example 16. In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that (i) ar (ACB) = ar (ACF) (ii) ar (AEDF) = ar (ABCDE) —NCERT Solution. (i) Since ACD and ACF are on the same base AC and betweeen the same parallels AC and BF,  ar (ACB) = ar (ACF) (ii) Now, ar (ACB) = ar (ACF) adding ar (ACDE) both sides, we get

ar (ACY) 

MATHEMATICS–IX AREAS OF PARALLELOGRAMS AND TRIANGLES

107

ar (ACB) + ar (ACDE) = ar (ACF) + ar (ACDE)  ar (AEDF) = ar (ABCDE) Hence proved. Example 17. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium. —NCERT Solution. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that, ar (AOD) = ar (BOC) ...(1) adding ar (ODC) on both sides, we get ar (AOD) + ar (ODC) = ar (BOC) + ar (ODC)  ar (ADC) = ar (BDC) 1 1  DC  AL   DC  BM 2 2  AL = BM  AB || DC. (distance between two parallel lines is same) Hence ABCD is a trapezium. Example 18. In the given figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums. Solution. ar (BDP) = ar (ARC) ...(1) (given) and ar (DPC) = ar (DRC) ...(2) (given) subtracting (2) from (1), we get ar (BDP) – ar (DPC) = ar (ARC) – ar (DRC)  ar (BDC) = ar (ADC)  DC || AB Hence, ABCD is a trapezium. ar (DRC) = ar (DPC) (given) on subtracting ar (DLC) from both sides, we get ar (DRC) – ar (DLC) = ar (DPC) – ar (DLC)  ar (DLR) = ar (CLP) on adding ar (RLP) to both sides, we get ar (DLR) + ar (RLP) = ar (CLP) + ar (RLP)  ar (DRP) = ar (CRP)  RP || DC Hence, DCPR is a trapezium.



Example 19. Let P, Q, R and S be respectively the mid-points of the sides AB, BC, CD and DA of quadrilateral ABCD. Show that PQRS is a parallelogram such that ar (parallelogram PQRS)  Solution.

1 ar (quad. 2

ABCD). Join A to C and A to R. In ABC, P and Q are mid-points of AB and BC respectively. 1 AC. 2 In DAC, S and R are mid-points of AD and DC respectively.

 PQ || AC and PQ 

1 AC. 2 Thus, PQ || SR and PQ = SR.

 SR || AC and SR 

108

AREAS OF PARALLELOGRAMS AND TRIANGLES MATHEMATICS–IX

 PQRS is a parallelogram. Now, median AR divides ACD into the triangles of equal area. 1 . ar (ACD) 2 median RS divides ARD into two triangles of equal area.

 ar (ARD) =

 ar (DSR) =

1 ar (ARD) 2

...(2)

from (1) and (2), we get, ar (DSR) = Similarly, ar (BQP) =

...(1)

1 1 1 1 ar (ARD) = ( ar (ACD)) = ar (ACD) 2 2 4 2

1 ar (ABC). 4

 ar (DSR) + ar (BQP) =

1 [ar (ACD) + ar (ABC)] 4

 ar (DSR) + ar (BQP) =

1 ar (quad. ABCD) 4

Similarly, ar (CRQ) + ar (ASP) =

1 ar (quad. ABCD) 4

...(3) ...(4)

Adding (3) and (4), we get 1 ar (quad. ABCD) 2 But, ar (DSR) + ar (BQP) + ar (CRQ) + ar (ASP) + ar (parallelogram ABCD) = ar (quad ABCD) Subtracting (5) from (6), we get

ar (DSR) + ar (BQP) + ar (CRQ) + ar (ASP) =

...(5) ...(6)

1 ar (quad. ABCD). Hence proved. 2 Example 20. Prove that, of all the parallelograms of given sides, the parallelogram which is a rectangle has the greatest area. Solution. In parallelogram ABCD of sides a and b, let h be the height corresponding to the base a.

ar (|| PQRS) =

Now, In DAE, h L2. Now, L1 – L2 = 400 cm2 – 360 cm2 = 40 cm2  The cubical box has larger lateral surface area and is greater by 40 cm2. (ii) Total surface area of the cubical box (S1) = 6 (edge)2 = 6 × (10)2 cm2 = 600 cm2 Total surface area of the cuboidal box (S2) = 2 (lb + lh + bh) = 2 (12.5 × 10 + 10 × 8 + 8 × 12.5) cm2 = 2 (125 + 80 + 100) cm2 = 610 cm2 Clearly, S2 > S1.  S2 – S1 = 610 cm2 – 600 cm2 = 10 cm2 Thus, the cuboidal box has greater surface area and is greater by 10 cm2. Example 2. Two cubes each of 15 cm edge are joined end to end. Find the surface area of the resulting cuboid. Solution. here, l = length of resulting cuboid = 15 cm + 15 cm = 30 cm b = breadth of resulting cuboid = 15 cm h = height of resulting cuboid = 15 cm 2

SURFACE AREAS AND VOLUMES

MATHEMATICS–IX

 Surface area of resulting cuboid = 2 (lb + bh + lh) = 2 (30 × 15 + 15 × 15 + 30 × 15) cm2 = 2 (450 + 225 + 450) cm2 = 2 (1125) cm2 = 2250 cm2 Ans. Example 3. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine: (i) The area of the sheet required for making the box. (ii) The cost of sheet for it, if a sheet measuring 1 m2 costs Rs. 20. –NCERT Solution. We have, Length; l = 1.5 m, Breadth, b = 1.25 m and Depth = Height, h = 0.65 m (i) Since the plastic box is open at the top,  Plastic sheet required for making such box = 2 (l + b) × h + lb = 2 (1.5 + 1.25) × 0.65 m2 + 1.5 × 1.25 m2 = 2 × 2.75 × 0.65 m2 + 1.875 m2 = 3.575 m2 + 1.875 m2 = 5.45 m2 (ii) Cost of 1 m2 of sheet = Rs. 20  Total cost of 5.45 m2 of sheet = Rs. 5.45 × 20 = Rs. 109 Ans. Example 4. Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m × 3 m? –NCERT Solution. here, l = 4m, b = 3m, h = 2.5 m. Since there is no tarpaulin for the floor.  Tarpaulin required = [2 (l + b) × h + lb] = [2 (4 + 3) × 2.5 + 4 × 3] m2 = (2 × 7 × 2.5 + 12) m2 = (35 + 12) m2 = 47 m2 Ans. Example 5. The sum of length, breadth and height of a cuboid is 21 cm and the length of its diagonal is 12 cm. Find the surface area of the cuboid. Solution. Let the length, breadth and height of the cuboid be l cm, b cm and h cm respectively. Then, l + b + h = 21 ...(1) Now, diagonal = 12 cm 

l 2  b 2  h 2  12

 l2 + b2 + h2 = 144 ...(2) Now, l + b + h = 21 Squaring both sides, we get (l + b + h)2 = (21)2  l2 + b2 + h2 + 2lb + 2bh + 2lh = 441  144 + 2 (lb + bh + lh) = 441  2 (lb + bh + lh) = 441 – 144 = 297  Surface area of the cuboid is 297 cm2 Ans. MATHEMATICS–IX

( using (2))

SURFACE AREAS AND VOLUMES

3

Example 6. Aggarwal sweets stall was placing an order for making card board boxes for packing their sweets. Two size of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. 5% of the total surface area is required extra, for all the overlaps. If the cost of cardboard is Rs. 5 for 1000 cm2, find the cost of cardboard required for supplying 300 boxes of each kind. Solution. Surface area of Ist box = 2 (25 × 20 + 20 × 5 + 25 × 5) cm2 = 2 (500 + 100 + 125) cm2 = 1450 cm2 Surface area of IInd box = 2 (15 × 12 + 12 × 5 + 15 × 5) cm2 = 2 (180 + 60 + 75) cm2 = 630 cm2 Total combined surface area = 1450 cm2 + 630 cm2 = 2080 cm2 Area of overlaps = 5% of 2080 cm2  5  2080 cm 2  104 cm 2 100  Total surface area of 2 boxes (one of each kind) = (2080 + 104) cm2 = 2184 cm2  Surface area of 300 boxes of each kind = 300 × 2184 cm2 = 655200 cm2 Now, cost of cardboard for 1000 cm2 = Rs. 5 

Cost of cardboard for 1 cm 3  Rs.

5 1000

 Cost of cardboard for 655200 cm2  Rs.

5  655200 1000

= Rs. 3276 Ans. Example 7. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m2. –NCERT Solution. Diameter of cylindrical pillar = 50 cm 50 cm  25 cm  0.25 m 2 also, height (h) = 3.5 m

 radius (r) 

Now, curved surface = 2rh  2  22  0.25  3.5 m 2 = 5.5 m2 7 Cost of painting 1 m2 = Rs. 12.50  Cost of painting 5.5 m2 = Rs. 12.50 × 5.5 = Rs. 68.75 Ans. Example 8. In the given figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade. –NCERT

4

SURFACE AREAS AND VOLUMES

MATHEMATICS–IX

Solution.

20 cm  10 cm 2 and, height (h) = 30 cm + 2 × 2.5 cm = 35 cm (for margin)  Cloth required for covering the lampshade = Its curved surface area = 2  rh

Here, radius (r ) 

 2

22  10  35 cm 2 = 2200 cm2 Ans. 7

Example 9. It is required to make a closed cylindrical tank of height 120 cm and base diameter 140 cm from a metal sheet. How many square metres of the sheet is required for the same? Solution. We have, diameter of base = 140 cm. 140 cm  70 cm 2 and, height of cylinder = 120 cm.  Total surface area of required tank = 2  r (r + h)



radius of base 

 2

22  70 (70  120) cm 2 7

83600 2 m  8.36 m 2 Ans. 10000 Example 10. The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition? –NCERT Solution. Cardboard required by each competitor = curved surface area of one penholder + base area = 2rh + r2, where r = 3 cm, h = 10.5 cm  440  190 cm 2  83600 cm 2 

 22   22   2   3 10.5    (3) 2  cm 2 7  7  

= (198 + 28.28) cm2 = 226.28 cm2 (approx)  Cardboard required for 35 competitors = 35 × 226.28 cm2 = 7920 cm2 (approx) Ans. Example 11. The diameter of a roller is 84 cm and its length is 100 cm. It takes 300 complete revolution to move once over to level a playground. Find the area of the playground in m2. Solution.

Radius of roller (r ) 

84 cm  42 cm 2

Length of the cylindrical roller (h) = 100 cm. Area moved by the roller in one revolution  2  rh  2 

MATHEMATICS–IX

22  42  100 cm 2 7

SURFACE AREAS AND VOLUMES

5



Area moved in 300 revolutions

22  42  100  300 cm 2 7 = 44 × 180000 cm2 = 44 × 18 m2 = 792 m2 Ans. Example 12. Find (i) the curved surface area of a cylindrical petrol storage tank that is 4.2 m in diameter and  2

4.5 m high. (ii) how much steel was actually used if making the closed tank? Solution.

1 of the steel actually used was wasted in 12 —NCERT

4.2 m  2.1 m , h  4.5 m 2 (i) Curved surface area = 2  r h

Here, r 

22  2.1 4.5 m 2 7 = 59.4 m2 (ii) Total surface area of closed tank  2

 2 r h  2 r 2  2r (r  h) 22  2.1 ( 2.1  4.5) m 2 7 = 87.12 m2  2

Let the total sheet used for making the cylindrical tank be x m2. Given, wastage  according to given question, x 

x 2 m . 12

x  87.12 12

87.12  12 11 x  87.12  x   95.04 m 2 11 12  Steel used for making closed tank including wastage = 95.04 m2 Ans. Example 13. Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find (i) radius of the base and (ii) total surface area of the cone. —NCERT Solution. (i) Here,  r l = 308 cm2, l = 14 cm.



22 308 14  r  308  r   7 cm. 7 22  2 and, (ii) Total surface area = r (r + l)



22  7 (7  14) cm 2 7 = 22 × 21 cm2 = 462 cm2 Example 14. How many meters of cloth, 5 m wide, will be required to make a conical tent, the radius of whose base is 7 m and height is 24 m? Solution. Radius of the tent, r = 7 m height of the tent, h = 24 m 

 6

slant height, l  r 2  h 2  7 2  24 2 m  625 m  25 m SURFACE AREAS AND VOLUMES

MATHEMATICS–IX



curved surface area =  r l

22  7  25 m 2  550 m 2 7 i.e., area of the cloth = 550 m2 

Now, length of cloth required  area  550 m  110 m width 5  length of cloth required = 110 m. Example 15. A conical tent is 10 m high and the radius of its base is 24 m. Find : (i) slant height of the tent. (ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is Rs. 70. Solution. (i) here, r = 24 m, h = 10 m Let l be the slant height of the cone. then,

–NCERT

l2 = h2 + r2  l  h 2  r 2  24 2  10 2  576  100  676  26 m.

(ii) Canvas required to make the conical tent = curved surface of the cone  rl 

22  24  26 cm 2 7

Now, Rate of canvas for 1 m2 = Rs. 70  Total cost of canvas  Rs. 22  24  26  70 = Rs. 137280 Ans. 7 Example 16. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps. –NCERT Solution. here, radius of cap (r) = 7 cm height of cap (h) = 24 cm Let l be the slant height. then, l  h 2  r 2  24 2  7 2  576  49  625  25 cm Sheet required for one cap = curved surface of the cone = rl 22  7  25 cm 2  550 cm 2 7  Sheet required for 10 such caps = 10 × 550 cm2 = 5500 cm2. Ans. Example 17. A bus stop in barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs. 12 per m2, what will be the cost 

of painting all these cones? (Use  = 3.14 and 1.04  1.02 ). Solution.

here, radius (r ) 

–NCERT

40 cm  20 cm  0.2 m and height (h) = 1 m 2

slant height (l )  r 2  h 2  0.04  1  1.04  1.02 m MATHEMATICS–IX

SURFACE AREAS AND VOLUMES

7

Now, curved surface of 1 cone =  rl 

2  0.2  1.02 m 2 7

 Curved surface of 50 such cones  50  22  0.2  1.02 m 2 7 Now, cost of painting 1 m2 = Rs. 12  Total cost of painting  Rs. 12  50 

22  0.2  1.02 = Rs. 384.68 (approx) Ans. 7

Example 18. A corn cob (see figure), shaped somewhat like a cone, has the radius of its broadest end as 2.1 cm and length as 20 cm. If each 1 cm2 of the surface of the cob carries an average of four grains, find how many grains you would find on entire cob? —(NCERT)

Solution.

We have, r = 2.1 cm, h = 20 cm let, slant height be l cm. then, l  r 2  h 2  (2.1) 2  (20) 2 cm  4.41  400 cm



 404.41 cm  20.11 cm curved surface area of corn cob =  r l

22  2.1  20.11 cm 2 7 = 132.726 cm2 Now, number of grains on 1 cm2 = 4  number of grains on 132.726 cm2 = 4 × 132.726 = 530.904  531 Hence, total number of grains on the corn cob = 531 Ans. Example 19. The surface area of a sphere is 154 cm2. Find its radius. Solution. Let the radius of the sphere be r cm. then, 4  r2 = 154 (given) 



r2 

 r  8

—NCERT

154 154  7 49   4 4  22 4

49 7 cm  cm 4 2

radius of the sphere 

7 cm. 2

SURFACE AREAS AND VOLUMES

MATHEMATICS–IX

Example 20. The radius of a spherical baloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases. —NCERT Solution. Let S1 and S2 be the total surface area in two cases of r = 7 cm and R = 14 cm.  S1 = 4 r2 = 4 (7)2 cm2 and S2 = 4 R2 = 4 (14)2 cm2 

Required ratio 

S1 4π  7  7 1   i.e. 1 : 4 Ans. S 2 4π  14 14 4

Example 21. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas. –NCERT Solution.

Let the diameter of earth be R and that of the moon will be The radii of moon and earth are

R 4

R R and respectively.. 8 2 2

R 1 4  8 Ratio of their surface area     64  1  4  1 i .e. 1 : 16 Ans. 2 1 64 1 16 R 4   4 2 Example 22. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of bowl. –NCERT Solution. Inner radius, r = 5 cm Thickness of Steel = 0.25 cm  Outer radius, R = (r + 0.25) cm = (5 + 0.25) cm = 5.25 cm 22  Outer curved surface  2 r 2   2   5.25  5.25  cm 2 7  

= 173.25 cm2 Example 23. The internal and external diameters of a hollow hemispherical vessel are 20 cm and 28 cm respectively. Find the cost of painting the vessel all over at 15 paisa per cm2. Solution. Outer radius of vessel, R = 14 cm Inner radius of vessel, r = 10 cm Area of the outer surface = 2  R2 = 2  × (14)2 cm2 = 392  cm2 Area of the inner surface = 2  r2 = 2  × (10)2 cm2 = 200  cm2 Area of the ring at the top =  (R2 – r2) =  (142 – 102) cm2 =  (14 – 10) (14 + 10) cm2 = 96  cm2  Total area to be painted =3 92  cm2 + 200  cm2 + 96  cm2 = 688 cm2 Now, cost of painting 1 cm2 = 15 paisa  15 Rs. 100 MATHEMATICS–IX

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9

15  688 π Rs 100 15 22   688  Rs 100 7 = 324.34 Rs. Ans. Example 24. A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1 m3 = 1000 l). –NCERT Solution. Here, l = 6m, b = 5m and h = 4.5 m  Volume of the tank = lbh = (6 × 5 × 4.5) m3 = 135 m3  The tank can hold = 135 × 1000 litres = 135000 litres of water. ( 1 m3 = 1000 litres) Example 25. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m. —NCERT Solution. Given capacity of a cuboidal tank 50000 3  50000 l  m  50 m 3 1000 Let the breadth of cuboidal tank be b m. according to given question, we have 2.5 × b × 10 = 50 50  25 b  50  b  b2 25  breadth of the tank is 2m. Ans. Example 26. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute? —NCERT Solution. Volume of water that flows in 1 hour (60 minutes) = volume of water of a cuboid whose dimensions are 3 m, 40 m and 2000 m. ( 2 km = 2000 m) = 3 × 40 × 2000 m3  Volume of water that flows in 1 minute

 Cost of painting 688  cm 2 

3  40  2000 3 m  4000 m 3 Ans. 60 Example 27. Three cubes whose edges are 3 cm, 4 cm and 5 cm respectively are melted and recast into a single cube. find the surface area of the new cube. Solution. Let x cm be the edge of new cube. Then, volume of the new cube = sum of the volumes of three cubes.  x3 = 33 + 43 + 53 = 27 + 64 + 125 = 216 = (6)3  x = 6 cm.  Edge of the new cube is 6 cm. and, surface area of the new cube = 6 (6)2 cm2 = 216 cm2 Ans. Example 28. A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20 m × 15 m × 6 m. For how many days will the water of this tank last? –NCERT Solution. Here l = 20 m, b = 15 m, and h = 6 m  Capacity of the tank = lbh = (20 × 15 × 6) m3 = 1800 m3 Water requirement per person per day = 150 litres 

10

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MATHEMATICS–IX

Water required for 4000 person per day = (4000 × 150) l  4000 150  3 3   m  600 m  1000 

Number of days the water will last 

Capacity of tank Total water required per day



1800  30 600

Thus, the water will last for 30 days Ans. Example 29. A godown measures 60 m × 25 m × 10 m. Find the maximum number of wooden crates each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown. –NCERT 3 3 Solution. Volume of the godown = (60 × 25 × 10) m = 15000 m Volume of 1 crate = (1.5 × 1.25 × 0.5) m3 = 0.9375 m3 Number of crate that can be stored in the godown 

Volume of the godown Volume of 1 crate



15000  16000 Ans. 0.9375

Example 30. If the lateral surface of cylinder is 94.2 cm2 and its height is 5 cm, then find (i) radius of its base (ii) its volume (use  = 3.14) . –NCERT Solution. (i) Let r be the radius of the base and h be the height of the cylinder. Then, Lateral surface = 94.2 cm2  2rh = 94.2  2 × 3.14 × r × 5 = 94.2 

r

94.2 3 2  3.14  5

Thus, the radius of its base = 3 cm. (ii) Volume of the cylinder = r2h = (3.14 × 32 × 5) cm3 = 141.3 cm3 Ans. Example 31. It costs Rs. 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs. 20 per m2, Find : (i) inner curved surface area of the vessel (ii) radius of the base (iii) capacity of the vessel –NCERT Solution.

(i) inner curved surface area of the vessel 

Total cost of painting Rate of painting

 2200  2 2   m  110 m 20  

MATHEMATICS–IX

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11

(ii) Let r be the radius of the base and h be the height of the cylindrical vessel.  2rh = 110 22  r  10  110 7



2



r

110  7 7   1.75 2  22  10 4

Thus, the radius of the base = 1.75 m  22 7 7  (iii) Capacity of the vessel = r2h      10  m 3 7 4 4  

= 96.25 m3 Example 32. The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it? –NCERT Solution. Capacity of a closed cylindrical vessel = 15.4 litres 1  3  3  15.4   m  0.0154 m 1000  

Let r be the radius of the base and h be the height of the vessel. Then, Volume = r2h = r2 × 1 = r2 ( h = 1m)  r2 = 0.0154 

22 2  r  0.0154 7

 r2 



0.0154  7  0.0049 22

r  0.0049  0.07

Thus, the radius of the base of vessel = 0.07 m. Metal sheet needed to make the vessel = Total surface area of the vessel = 2rh + 2r2 = 2r (h + r)  2

22  0.07  (1  0.07) m 2 7

= 44 × 0.01 × 1.07 m2 = 0.4708 m2 Example 33. A patient in a hospital in given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients? –NCERT Solution. Diameter of the cylindrical bowl = 7 cm 7 cm 2 Height of serving bowl = 4 cm.

 Radius 

12

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MATHEMATICS–IX

 Soup served to 1 patient = Volume of the bowl = r2h  22 7 7       4  cm 3 = 154 cm3 7 2 2   Soup served to 250 patients = (250 × 154) cm3 = 38500 cm3 = 38.5 l. Ans.

Example 34. The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 gm. —NCERT Solution.

here, inner radius (r ) 

24 cm  12 cm 2

28 cm  14 cm 2 h = length of the pipe = 35 cm.  volume of wood used in making the pipe

and outer radius (R) =

  R 2h   r 2h   h (R 2  r 2 ) 

22  35  [14 2  12 2 ] cm 3 7



22  35  (14  12) (14  12) cm 3 7

22  35  2  26 cm 3  5720 cm 3 7 Now, 1 cm3 of wood = 0.6 g  5720 cm3 of wood = 0.6 × 5720 g = 3432.0 g = 3.432 kg. Ans. Example 35. A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 10 cm, find the weight of the whole pencil if the specific gravity of the wood is 0.7 gm/cm3 and that of the graphite is 2.1 gm/cm3. —NCERT Solution. For graphite cylindrical rod : 

radius (r) of graphite cylinder  1  1 cm  1 cm 2 10 20 and, length of graphite rod (h) = 10 cm.  volume of graphite cylindrical rod =  r2 h 2

22  1      10 cm 3 7  20 

 Weight of graphite used for pencil = volume × specific gravity 

22 1 1    10  2.1 gm 7 20 20

( 1 cm3  2.1 gm)

= 0.165 gm. MATHEMATICS–IX

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13

Again, for pencil including graphite rod, we have, 7 7 mm  cm 2 20 and, length of pencil (h) = 10 cm

radius of pencil (R) 



volume of pencil =  R 2 h 2

22  7      10 cm 3 7  20 

 volume of wood used for pencil

  R2 h   r 2 h 

22 7 7 22 1 1 1    10 cm 3     cm 3 7 20 20 7 20 20 10



11 1   48 cm 3 7 20

11 1   48  0.7 gm ( 1 cm3  0.7 gm) 7 20 = 2.64 gm.  Total weight = 2.64 gm + 0.165 gm = 2.805 gm Ans. Example 36. A well of diameter 3m is dug 14 m deep. The earth taken out of it has been spread evenly all around it to a width of 4 m to form an embankment. Find the height of the embankment.

 weight of wood 

Solution.

3 m , height (h) = 14 m 2  Volume of the earth taken out of the well

Radius of the well (r ) 

22 3 3    14 m 3  99 m 3 7 2 2 Outer radius of the embankment   r2 h 

11 3 m4 m  m 2 2  Area of embankment = outer area – inner area R

  R 2  r 2 

2 2 22  11   3   22  11 3   11 3                7  2   2   7  2 2   2 2   

22  7  4  88 m 2 7  Height of the embankment 



14

Volume 99 9  m  m  1.125 m Ans. Area 88 8

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MATHEMATICS–IX

Example 37. The height of a cone is 15 cm. If its volume is 1570 cm3. Find the radius of the base (Use  = 3.14). –NCERT Solution. Here h = 15 cm and volume = 1570 cm3 Let the radius of the base of the cone be r cm. Now, Volume = 1570 cm3   

1 2 r h  1570 3 1  3.14  r 2  15  1570 3 1570 r2   100 3.14  5

 r  100  10 Thus, the radius of the base of cone is 10 cm Ans. Example 38. The radius and height of a right circular cone are in the ratio of 5 : 12. If its volume is 2512 cm3, find the slant height and radius of the base of the cone. (use  = 3.14) Solution. Let the radius of cone be 5x and height be 12 x.  Volume of the cone  1 r 2 h 3 according to question, we get 1  3.14  (5 x ) 2  (12 x)  2512 3



1 314   25 x 2  12 x  2512 3 100

2512 8 x  2 314  radius of base = 5x = 5 × 2 cm = 10 cm and, height = 12 x = 12 × 2 cm = 24 cm



x3 

 slant height  r 2  h 2  10 2  24 2  100  576  676  26  radius of cone = 10 cm and slant height = 26 cm Ans. Example 39. The volume of a right circular cone is 9856 cm3. If the diameter of the base is 28 cm, find (i) height of the cone (ii) slant height of the cone (iii) curved surface area of the cone —NCERT Solution. Given, volume of cone = 9856 cm3 and, radius of base (r) 

28 cm  14 cm 2

(i) we know, volume of cone  1 r 2 h 3 1 22  14  14  h  9856  3 7 MATHEMATICS–IX

SURFACE AREAS AND VOLUMES

15



h

3  9856 cm  48 cm. 22  2 14

(ii) Slant height of a cone (l)

 h2  r 2

 (48) 2  (14) 2 cm

 2304  196 cm  2500 cm  50 cm (iii) Curved surface area of a cone =rl 22  14  50 cm 2  2200 cm 2 . 7 Example 40. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required. –NCERT 

Solution.

Diameter of the base of the cone = 10.5 m 10.5  radius  r    m  5.25 m  2  Height of the cone = 3m

 Volume of the cone 

 1 22  3 1 2 r h     5.25  5.25  3  m = 86.625 m3 3 7 3  

To find the slant height l We have, l 2  h 2  r 2  32  (5.25)2 = 9 + 27.5625 = 36.5625  l  36.5625  6.0467 m (approx.)

Canvas required to protect wheat from rain = Curved surface area  22   rl    5.25  6.0467  m 2 = 99.77 m2 (approx)  7 

Example 41. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained? Solution.

The solid obtained is a cone with r = 5 cm and h = 12 cm. 

Volume  1 r 2 h 3 

1  3.14  5  5  12 cm 3 3

 100 3.14 cm3 = 314 cm3 Ans. 16

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MATHEMATICS–IX

Example 42. If the surface area of a sphere is 616 cm2, find its volume. Solution. Let r be the radius of sphere. then, 4  r2 = 616 616 616  7   49 4 4  22



r2 



r  49 cm  7 cm

 volume of sphere  4  r 3  4  22  7  7  7 cm 3 3 3 7 = 1437.3 cm3 Ans. Example 43. The diameter of the moon is approximately one-fourth the diameter of the earth. What fraction of the volume of the earth is the volume of the moon? –NCERT Solution. Let the radius of earth be R. then, radius of moon 

R . 4

 volume of earth 

4  R3 3

and, volume of moon 

4 R   . 3 4

3

4  R3 volume of earth 64 3    volume of moon 4  R  3 1   3 4

i.e., volume of earth is 64 times the volume of the moon. 1 times that of earth. 64 Example 44. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm3? –NCERT Solution. Diameter of the ball = 4.2 cm

i.e., volume of moon is

4.2  Radius    cm  2.1 cm  2 

Volume of the ball  4 r 3 3 

4 22   2.1  2.1  2.1 cm 3  38.808 cm 3 3 7

Now, Density of metal = 8.9 gm per cm3  Mass of the ball = 38.808 × 8.9 g = 345.3912 g = 345.4 g (approx) Ans. MATHEMATICS–IX

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17

Example 45. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1m, then find the volume of the iron used to make the tank. –NCERT Solution. Let R cm and r cm be respectively the external and internal radii of the hemispherical vessel, then R = 1.01 m and r = 1 m ( as thickness = 1cm = 0.01 m) Volume of iron used = External volume – Internal volume 

2 3 2 3 2 R  r   ( R 3  r 3 ) 3 3 3



2 22   [(1.01)3 – (1)3 ] m 3 3 7

44 44  (1.030301  1) m 3   0.030301 m 3 21 21 = 0.06348 m3 (approx) Example 46. A dome of a building is in the form of hemi-sphere. From inside, it was white-washed at the cost of Rs. 498.96. If the cost of white-washing is Rs. 2 per square meter, find: (i) the inside surface area of the dome (ii) the volume of the air inside the dome. –NCERT Solution. Let r be the inner radius of the hemispherical dome. Then, inside surface area of the hemisphere = 2  r2. Since, at the rate of Rs. 2 per square metre, the total cost of white-wash is Rs. 498.96, 

 surface area of hemisphere 

498.96 2 m  249.48 m 2 2

according to question, 2r 2  249.48 

r2 

249.48  7  39.69 2  22



r  39.69 m  6.3 m (i) Inside surface are of the dome = 2  r2 = 249.48 m2 (ii) Inside volume of the dome

(calculated above)

2 3 2 22 r    (6.3) 3 m 3 3 3 7 = 523.908 m3 Ans. Example 47. Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface are S. Find the: (i) radius r of the new sphere. (ii) ratio of S and S. –NCERT 

Solution.

18

4 (i) Volume of 27 solid spheres of radius r  27  r 3 3 4 3 Volume of the new sphere of radius r  r  3 SURFACE AREAS AND VOLUMES

...(1) ...(2)

MATHEMATICS–IX

According to the problem, we have, 4 3 4 r   27  r 3 3 3



r3  27r 3  (3r )3  r   3r

(ii) Required ratio 

S 4r 2 r2 r2 1      1: 9 2 2 2 S  4r 9 (3r ) 9r

Example 48. A wooden bookshelf has external dimensions as follows : Height = 110 cm, Depth = 25 cm, Breadth = 85 cm (see figure). The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 paisa per cm2 and the rate of painting is 10 paisa per cm2, find the total expenses required for polishing and painting the surface of the bookshelf. –NCERT

Solution.

Area to be polished = (110 × 85 + 2 × 85 × 25 + 2 × 25 × 110 + 4 × 75 × 5 + 2 × 110 × 5) cm2 = (9350 + 4250 + 5500 + 1500 + 1100) cm2 = 21700 cm2  cost of polishing @ 20 paisa per cm2 = 21700 

20  Rs. 4340 100

Also, Area to be painted = (6 × 75 × 20 + 2 × 90 × 20 + 75 × 90) cm2 = (9000 + 3600 + 6750) cm2 = 19350 cm2  Cost of painting @ 10 paisa per cm2  19350 

10 Rs.  Rs. 1935 100

 Total expense = Rs. 4340 + Rs. 1935 = Rs. 6275 Ans. Example 49. The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in figure. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paisa per cm2 and black paint costs 5 paisa per cm2. –NCERT MATHEMATICS–IX

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19

Solution.

Clearly, we have to subtract the area of the circle on which sphere is resting while calculating the cost of silver paint. Surface area to be painted silver = 8 (curved surface area of the sphere – area of circle on which sphere is resting) = 8 (4 R2 – r2) where R 

21 cm , r  1.5 cm 2

 441   8π  4   2.25  cm 2  8π (441  2.25) cm 2  8π (438.75) cm 2 4  

 Cost of silver paint @ 25 paisa per cm2 25   22  Rs.  8   438.75    Rs. 2757.86 (approx) 100 7  

Again, surface area to be painted black = 8 × curved surface area of cylinder = 8 × 2rh = 8  2 

22  1.5  7 cm 2  528 cm 2 7

 Cost of black paint @ 5 paisa per cm2  Rs. 528 

5  Rs. 26.40 100

 Total cost of painting = Rs. 2757.86 + Rs. 26.40 = Rs. 2784.26 (approx) Ans. Example 50. The diameter of a sphere is decreased by 25%. By what percent does its curved surface decrease? –NCERT 2

Solution.

d Let d be the diameter of the sphere. Then, its surface area  4   d 2 2 On decreasing its diameter by 25%,

New diameter  d1  d  25% of d  d  25 d  75 d  3 d 100 100 4 20

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MATHEMATICS–IX

2

d   1 3d   New surface area  4  1   4    2 4   2

 4.

2

9d 2 9  . d 2 64 16

 Decrease in surface area  d 2 

9 9 7  d 2  1  d 2  d 2 16 16  16 

 Percentage decrease in surface area 

decrease in surface area  100% original surface area

7 d 2 7  16 2  100%   100% = 43.75% Ans. 16 d

PRACTICE EXERCISE Questions based on Surface area of cuboid and cube 1. Find the surface area of a cuboid 16 m long, 14 m broad and 7 m high. 2. Find the length of the longest pole that can be placed in a room 12 m long, 8 m broad and 9 m high. 3. The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of whitewashing the walls of the room and the ceiling at the rate of Rs. 8.50 per m2. 4. Find the percentage increase in the surface area of a cube when each side is doubled. 5. The paint in a certain container is sufficient to paint an area equal to 9.375 m 2. How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container? —NCERT 6. A small indoor green house (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high. (i) What is the area of the glass? (ii) How much of tape is needed for all the 12 edges? —NCERT 7. Three cubes each of side 6 cm are joined end to end. Find the surface of the resulting cuboid. 8. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is to be open at top. Ignoring the thickness of the plastic sheet, find: (i) area of the sheet required to make the box. (ii) the cost of the sheet for it, if a sheet measuring 1 m2 cost Rs. 22. 9. If the surface area of the cube is 96 cm2, find its edge and length of its diagonal. 10. The dimensions of a rectangular box are in the ratio of 2 : 3 : 4 and the difference between the cost of covering it with sheet of paper at the rate of Rs. 4 and Rs. 4.50 per m2 is Rs. 650. Find the dimensions of the box. 11. Mary wants to decorate her christmas tree. She wants to place the tree on a wooden box covered with coloured paper with picture of Santaclaus on it (see figure). She must know the exact quantity of paper to buy for this purpose. If the box has length, breadth and height as 80 cm, 40 cm and 20 cm respectively, how many square sheets of paper of side 40 cm would she require? —NCERT MATHEMATICS–IX

SURFACE AREAS AND VOLUMES

21

12. The length and breadth of a hall are in the ratio 4 : 3 and its height is 5.5 meters. The cost of decorating its walls (including doors and windows) at Rs. 6.60 per m2 is Rs. 5082. Find the length and breadth of the room. Questions based on Surface area of a cylinder 13. The curved surface area of a right circular cylinder of height 14 cm is 176 cm2. Find the diameter of the base of the cylinder. 14. A metal pipe is 77 cm long. The inner diameter of a cross-section is 4 cm, the outer diameter being 4.4 cm. (see figure). Find its : —NCERT

15.

16. 17. 18. 19.

22

(i) inner curved surface area (ii) outer curved surface area (iii) Total surface area The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find: (i) its inner curved surface area, (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2. An iron pipe 20 cm long has exterior diameter 25 cm. If the thickness of the pipe is 1 cm, find the total surface area of the pipe. A rectangular sheet of paper 88 cm × 50 cm is rolled along its length and a cylinder is formed. Find curved surface area of the cylinder formed. A solid cylinder has total surface area of 462 cm2. Its curved surface area is one-third of its total surface area. Find the radius and height of the cylinder. The total surface area of a hollow metal cylinder, open at both the ends of external radius 8 cm and height 10 cm is 338  cm2. Find thickness of the metal in the cylinder. SURFACE AREAS AND VOLUMES

MATHEMATICS–IX

20. In the given figure, you see the frame of a lamp shade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade. —NCERT

Question based on surface area of a Cone 21. Find the curved surface area of a cone, if its slant height is 50 cm and the diameter of its base is 28 cm. 22. Find the total surface area of a cone, if its slant height is 21 cm and diameter of its base is 24 cm. —NCERT 22   23. The curved surface area of a cone is 4070 cm2 and its radius is 35 cm. What is its slant height?  use π   7  

24. The radius and slant height of a cone are in the ratio 4 : 7. If its curved surface area is 792 cm2, find its 22   radius.  use π   7  

25. The circumference of the base of a 10 m high conical tent is 44 m. Calculate the length of canvas used in making the tent if width of canvas is 2 m.  use π  22  7   26. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. (use  = 3.14) —NCERT 27. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m2? —NCERT 2 28. The curved surface area of a cone of radius 6 cm is 188.4 cm . Find its height. Questions based on surface area of sphere and hemi-sphere 29. The surface area of a sphere is 5544 cm2. Find its radius. 30. A hemi-spherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs. 16 per 100 cm2. —NCERT MATHEMATICS–IX

SURFACE AREAS AND VOLUMES

23

31. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas. —NCERT 32. In the given figure, a right cylinder just encloses a sphere of radius r. Find

(i) Surface area of the sphere (ii) curved surface area of the cylinder (iii) ratio of the areas obtained in (i) and (ii). 33. The internal and external diameters of a hollow hemi-spherical vessel are 20 cm and 28 cm respectively. Find the cost of painting the vessel all over at 15 paisa per cm2. 34. A toy is in the form of a cone mounted on a hemi-sphere. The diameter of the base and the height of the cone are 6 cm and 4 cm respectively. Find the surface area of the toy. (use  = 3.14) Question based on Volume of Cuboid and Cube 35. The total surface area of a cube is 1350 cm2. Find its volume. 36. 500 persons took dip in a rectangular tank which is 80 m long and 50 m broad. What is the rise in the level of water in the tank, if the average displacement of water by a person is 4 m3? 37. Three cubes of a metal with edges 6 cm, 8 cm and 10 cm respectively are melted and formed into a single cube. Find the edge of the new cube formed. Also, find its volume. 38. The volume of a cuboid is 1536 m3. Its length is 16 m, and its breadth and height are in the ratio 3 : 2. Find surface area of the cuboid. 39. A field is 70 m long and 40 m broad. In one corner of the field, a pit which is 10 m long, 8 m broad and 5 m deep, has been dug out. The earth taken out of it is evenly spread over the remaining part of the field. Find the rise in the level of the field. 40. The areas of three adjacent faces of a cuboid are 15 cm2, 40 cm2 and 24 cm2. Find the volume of the cuboid. 41. How many bricks, each measuring 25 cm × 15 cm × 8 cm will be required to build a wall 10 m × 4 dm × 5 m when one-tenth of its volume is occupied by mortar? 42. A rectangular reservoir is 120 m long and 75 m wide. At what speed per hour must water flow into it through a square pipe of 20 cm wide so that the water rises by 2.4 m in 18 hours. Question based on Volume of a Cylinder 43. The radius of a cylinder is 14 cm and its height is 40 cm. Find (i) curved surface area (ii) the total surface area (iii) volume of the cylinder. 44. The total surface area of a cylinder is 462 cm2. Its curved surface is one-third of its total surface area. Find the volume of the cylinder. 45. The curved surface area and the volume of a pillar are 264 m2 and 396 m3 respectively. Find the diameter and the height of the pillar. 24

SURFACE AREAS AND VOLUMES

MATHEMATICS–IX

46. The sum of the height and radius of the base of a solid cylinder is 37 m. If the total surface area of the cylinder is 1628 m2, find its volume. 47. A cylindrical tube, open at both ends, is made up of metal. The internal radius of the tube is 5.2 cm and its length is 25 cm. The thickness of the metal is 8 mm. Calculate the volume of the metal. 48. A soft drink is available in two packs : (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much? —NCERT 49. If the diameter of the cross-section of a wire is decreased by 5%, how much percent will the length be increased so that the volume remains the same? —NCERT 50. Water flows out through a circular pipe, whose internal diameter is 2 cm, at the rate of 70 cm per second into a cylindrical tank, the radius of whose base is 40 cm. By how much time will the level of water rise in half an hour? 51. A rectangular piece of paper is 22 cm long and 12 cm wide. A cylinder is formed by rolling the paper along its length. Find the volume of the cylinder. 52. A well, with inner radius 4m, is dug 14 m deep. The earth taken out of it has been spread evenly all round 22   it to a width of 3m to form an embankement. Find the height of this embankement.  use π   7  

Question based on Volume of a Cone 53. The base radii of two cones of the same height are in the ratio 3 : 4. Find the ratio of their volumes. 22   54. A cone of height 24 cm has curved surface area 550 cm2. Find its volume.  use π   7  

55. The radius and height of a right circular cone are in the ratio of 5 : 12. If its volume is 314 cm3, find the slant height and radius of the base of the cone. (use  = 31.4) 56. Find the slant height and curved surface area of a cone whose volume is 12935 cm3 and the radius of the base is 21 cm. 57. A semi-circular thin sheet of metal of diameter 28 cm is bent and an open conical cup is made. Find the capacity of the cup. 58. Monica has a piece of canvas whose area is 551 m2. She uses it to have a conical tent made with a base radius of 7 m. Assuming that all the stitching margins and the wastage incurred while cutting, amounts to approximately 1 m2, find the volume of the tent that can be made with it. —NCERT 59. Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 9 cm. 60. A right angled triangle of which the sides containing the right angle are 6.3 cm and 10 cm in length, is made to turn round on the longer side. Find the volume of the solid, thus generated. Also, find its curved surface area. Question based on Volume of a Sphere and Hemi-Sphere 61. Find the surface area of a sphere whose volume is 606.375 m3. 62. A solid sphere of radius 3 cm is melted and then cast into smaller spherical balls, each of diameter 0.6 cm. Find the number of small balls thus obtained. 63. The surface areas of two spheres are in the ratio 1 : 4. Find the ratio of their volumes. 64. The diameter of a metallic sphere is 6 cm. It is melted and drawn into a wire having diameter of the crosssection as 2 mm. Find the length of the wire. MATHEMATICS–IX

SURFACE AREAS AND VOLUMES

25

65. The radii of the internal and external surfaces of a metallic spherical shell are 3 cm and 5 cm respectively. It is melted and recast into a solid cylinder of height 10

2 cm. Find the diameter of the base of the 3

cylinder. 66. The largest sphere is carved out of a cube of side 7 cm. Find the volume of the sphere. (use  = 3.14) 67. Determine the ratio of the volume of a cube to that of a sphere which will exactly fit inside the cube. 68. A hemi-spherical bowl is made of steel 0.5 cm thick. The inside radius of the bowl is 4 cm. Find the volume of the steel used in making it. 69. A hemi-spherical bowl of internal radius 9 cm contains a liquid. This liquid is to be filled into cylindrical shaped small bottles of diameter 3 cm and height 4 cm. How many bottles are required to empty the bowl? 70. A hemi-sphere of lead of radius 8 cm is cast into a right circular cone of base radius 6 cm. Find the height of the cone. Miscellaneous Questions 71. The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to its base. If its volume be

1 of the volume of the given cone, at what height above the base is the section made? 27

72. A circus tent consists of a cylindrical base surmounted by a conical roof. The radius of the cylinder is 20 m. The height of the tent is 63 m and that of the cylindrical base is 42 m. Find the volume of air contained in the tent and the area of canvas used for making it. 73. How many litres of water flows out of pipe having an area of cross-section of 5 cm2 in one minute, if the speed of water in the pipe is 30 cm/sec ? 74. A sphere of diameter 6 cm is dropped into a cylindrical vessel partly filled with water. The radius of the vessel is 6 cm. If the sphere is completely submerged in water, find by how much will the surface level of water be raised. 75. An ice-cream cone has a hemispherical top. If the height of the conical portion is 9 cm and base radius 22   2.5 cm, find the volume of ice-cream in the ice-cream cone.  use π   7   76. In the given figure, a solid is made of a cylinder with hemispherical ends. If the entire length of the solid is 108 cm and the diameter of the hemispherical ends is 36 cm, find the cost of polishing the surface of the solid at the rate of 7 paisa per cm2.

77. A spherical copper ball of diameter 9 cm is melted and drawn into a wire, the diameter of whose thickness is 2 mm. Find the length of the wire in meters. 78. The difference between the inside and outside surfaces of a cylindrical water pipe 14 m long is 88 m2. If the volume of pipe be 176 m3. Find the inner and outer radii of the water pipe. 26

SURFACE AREAS AND VOLUMES

MATHEMATICS–IX

79. Water is flowing at the rate of 2.5 km/hr through a circular pipe 20 cm internal diameter, into a circular cistern of diameter 20 m and depth 2.5 m. In how much time will the cistern be filled? 80. A conical vessel of radius 6 cm and height 8 cm is filled with water. A sphere is lowered into the water (see figure), and its size is such that when it touches the sides of the conical vessel, it is just immersed. How much water will remain in the cone after the overflow?

PRACTICE TEST M.M. : 30

Time : 1 hour

General Instructions : Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each. 1. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs. 10 per m2 is Rs. 15000, find the height of the hall. 2. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs. 1250 per m2. 3. A right triangle PQR with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained. 4. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold? 5. A joker’s cap is in the form of a right circular cone of base 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps. 6. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl. 7. A village, having a population of 4000, requires 150 l of water per head per day. It has a tank measuring 20 m × 15 m × 6 m. For how many days will the water of this tank last ? 8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients? 9. The radius and height of a cone are in the ratio 4 : 3. The area of the base is 154 cm2. Find the area of the curved surface. 10. The diameter of a sphere is decreased by 25%. By what percent its curved surface area decrease?

ANSWERS OF PRACTICE EXERCISE 1. 868 cm2

2. 17 m

5. 100

6. (i) 4250 cm2 (ii) 320 cm 7. 360 cm2

8. (i) 5.45 m2 (ii) 119.90 Rs.

9. 4 cm, 4 3 cm

10. 10 m, 15 m, 20 m

12. 40 m, 30 m

13. 4 cm MATHEMATICS–IX

3. Rs. 629

11. 7

14. (i) 968 cm2 (ii) 1064.80 cm2 (iii) 2038.08 cm2 SURFACE AREAS AND VOLUMES

4. 300%

15. (i) 110 m2 (ii) Rs. 4400 27

16. 3168 cm2

17. 440 cm2

18. r = 7 cm, h = 14 cm

19. 3 cm

20. 2200 cm2

21. 2200 cm2

22. 1244.57 cm2

23. 37 cm

24. 12 cm

25. 134.2 m

26. 63 m

27. Rs. 1155

28. 8 cm

29. 21 cm

30. Rs. 27.72

31. 1 : 16

32. (i) 4  r2 (ii) 4r2 (iii) 1 : 1

33. 324.34 Rs

34. 103.62 cm2

35. 3375 cm3

36. 50 cm

37. 12 cm, 1728 cm3

38. 832 cm2

39. 14.7 cm

40. 120 cm3

41. 6000

42. 30 km/hr

43. (i) 3520 cm2 (ii) 4752 cm2 (iii) 24640 cm3

44. 539 cm3

45. 6 m, 14 m

46. 4620 m3

47. 704 cm3

48. cylindrical tin, 85 cm2 49. 10.8%

50. 78.75 cm

51. 462 m3

52. 6.8 m (approx)

53. 9 : 16

54. 1232 cm3

55. 13 cm, 5 cm

56. 35 cm, 2310 cm2

57. 622.3 cm2

58. 1232 m3

59. 190.93 cm3

60. 415.8 cm3, 233.9 cm2

61. 346.5 m2

62. 1000

63. 1 : 8

64. 36 m

65. 7 cm

66. 179.5 cm3

67. 6 : 

68. 56.83 cm3

69. 54

70. 28.44 cm

71. 20 cm

72. 61600 m3, 7102.85 m2

73. 9 l

74. 1 cm

75. 91.66 cm3

76. Rs. 855.36

77. 121.5 m

78. 1.5 m, 2.5 m

79. 10 hours

80. 188.57 cm3

ANSWERS OF PRACTICE TEST 1. 6 m

2. Rs. 68.75

3. 100  cm3

4. 0.303 l (approx.)

5. 5500 cm2

6. 173.25 cm2

7. 3 days

8. 38.5 l

9. 192.5 cm2

10. 43.75%

28

SURFACE AREAS AND VOLUMES

MATHEMATICS–IX

CHAPTER

14 STATISTICS

Points to Remember : 1. Facts or figures, collected with a definite pupose, are called Data. 2. Statistics is the area of study dealing with the collection, presentantion, analysis and interpretation of data. 3. The data collected by the investigator himself with a definite objective in mind are known as Primary data. 4. The data collected by somone else, other than the investigator, are known as Secondary data. 5. Any character which is capable of taking reversal different values is called a variable. 6. Each group into which the raw data are condensed is known as class-interval. Each class is bounded by two figures known as its limits. The figure on the left is lower limit and figure on the right is upper limit. 7. The difference between true upper limit and true lower limit of a class is known as its class-size. 8. Mid-value of a class (or class mark) = 9. 10. 11. 12. 13.

upper limit  lower limit 2

Class size is the difference between any two successive class marks (mid-values). The difference between the maximum value and the minimum value of the variable is known as Range. The count of number of observations in a particular class is known as its Frequency. The data can be presented graphically in the form of bar graphs, histograms and frequency polygons. The three measures of central tendency for an ungrouped data are : (i) Mean : It is found by adding all the values of the observations and dividing it by the total number of observations. It is denoted by x . n

 xi

x1  x2  ........  xn i 1 .  n n For an ungrouped frequency distribution,

So,

x

n

x

f1 x1  f 2 x2  ...... f n x n  f1  f 2  ........ f n

 f i xi i 1 n

 fi i 1

(ii) Median : It is the value of the middle-most observation(s).  n 1   If n is an odd number, then median = value of the   2 

th

observation. th

n n  and, if n is an even number, then median = mean of the values of   and   1  2 2  (iii) Mode : The mode is the most frequently occurring observation. Empirical formula for calculating mode is given by, Mode = 3 (Median) – 2 (Mean)

MATHEMATICS–IX

STATISTICS

th

observations.

173

ILLUSTRATIVE EXAMPLES Example 1. The relative humidity (in %) of a certain city for a month of 30 days was as follows : 98.1 98.6 99.2 90.3 86.5 95.3 92.9 96.3 94.2 95.1 89.2 92.3 97.1 93.5 82.7 95.1 97.2 93.3 95.2 97.3 96.2 92.1 84.9 90.2 95.7 98.3 97.3 96.1 92.1 89.0 (i) Construct a grouped frequency distribution table with classes 84-86, 86-88 etc. (ii) Which month or season do you think this data is about ? (iii) What is the range of this data? —NCERT Solution. (i) Frequency distribution table Relative humidity (in %) Tally Marks Frequency | 84  86 1 86  88

|

1

88  90 90  92

|| ||

2 2

92  94

|||| ||

7

94  96 96  98

|||| | |||| ||

6 7

|||| 98  100 4 Total 30 (ii) Month-June or season-Monsoon (iii) Range = Maximum observation – minimum observation = 99.2 – 84.9 = 14.3 Example 2. The value of  upto 50 decimal places is given below : 3.14159265358979323846264338327950288419716939937510 (i) List the digits from 0 to 9 and make a frequency distribution of the digits after the decimal point. (ii) What are the most and the least frequency occurring digits? —NCERT Solution. (i) Frequency distribution table Digit Tally Marks Frequency 0 || 2 1 2

|||| ||||

5 5

3 4

|||| ||| ||||

8 4

5 6

|||| ||||

5 4

7 8

|||| ||||

4 5

9 Total

|||| |||

8 50

(ii) Most frequency occuring digits are 3 and 9, and least occurring digit is 0. 174

STATISTICS

MATHEMATICS–IX

Example 3. The following table gives the life times of 400 neon lamps: Life time (in hours)

No. of lamps

300 - 400 400 - 500

14 56

500 - 600 600 - 700

60 86

700 - 800

74

800 - 900 900 - 1000

62 48

(i) Represent the given information with the help of a histogram. (ii) How many lamps have a life time of more than 700 hours? —NCERT Solution.

(i)

(ii) No. of lamps having life time more than 700 hours = 74 + 62 + 48 = 184 Example 4. The following two tables give the distribution of students of two sections according to the marks obtained by them : Section A

Section B

Marks

Frequency

Marks

Frequency

0-10 10-20 20-30 30-40 40-50

3 9 17 12 9

0-10 10-20 20-30 30-40 40-50

5 19 15 10 1

Represent the marks of the students of both the sections on the same graph by frequency polygon. MATHEMATICS–IX

STATISTICS

175

Solution.

Required frquency polygon is as follows :

Example 5. Draw histogram of the weekly pocket expenses of 125 students of a school given below :

Solution.

Weekly pocket expenses (in Rs.) 10 - 20 20 - 30

No. of students 10 15

30 - 50 50 - 60 60 - 90

40 25 30

90 - 100

05

Here, we observe that class intervals are unequal, so we will first adjust the frequencies of each class interval. Here, the minimum class size is 10. We know, Adjusted frequency of a class interval Minimum class size   frequency of the class class size  The adjusted frequency of each class interval is given below : Weekly pocket expenses Frequency Adjusted frequency 10 10  10 10 - 20 10 10 10  15  15 20 - 30 15 10 10  40  20 30 - 50 40 20 10  25  25 50 - 60 25 10 10  30  10 60 - 90 30 30 10  5  05 90 - 100 05 10

176

STATISTICS

MATHEMATICS–IX

So, required histogram is given below.

Example 6. In a mathematics test given to 15 students, the following marks (out of 100) are recorded : 41, 48, 39, 46, 52, 54, 62, 40, 96, 52, 98, 40, 42, 52, 60. Find the mean, median and mode of the above marks. 15

 xi Solution.

(i) Mean ( x ) 



i 1

15

41  48  39  46  52  54  56  62  40  96  52  98  40  42  52  60 15

822  54.8 15 (ii) Arranging the data in the ascending order : 

39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98 here, 

n = 15, which is odd.  n 1 median = value of    2 

 15  1     2 

th

observation

th

observation = 8th observation = 52

(iii) Since, 52 occurs most frequently i.e. 3 times, so mode is 52. MATHEMATICS–IX

STATISTICS

177

Example 7. Find the mean salary of 60 workers of a factory from the following table: Salary (in Rs) 3000 4000

No. of workers 16 12

5000 6000 7000

10 8 6

8000 9000

4 3

10000

1

—NCERT

Solution. Salary (in Rs.) xi 3000

No. of workers f i 16

 Mean ( x ) 

f i xi 48000

4000

12

48000

5000 6000

10 8

50000 48000

7000

6

42000

8000 9000

4 3

32000 27000

10000 Total

1  f i  60

10000  f i xi  305000

 f i xi 305000   Rs. 5083.33 Ans.  fi 60

Example 8. The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number. Solution.

Here, n = 5, x  18. Now, x 

 xi   xi  5  18  90 n

So, total of 5 numbers is 90. Let the excluded number be a. Then, total of 4 numbers is 90 – a. Mean of 4 numbers 

90  a 4

90  a  16 ( Given, new mean = 16) 4  90 – a = 64  a = 26  the excluded number is 26. Ans.



178

STATISTICS

MATHEMATICS–IX

Example 9. The median of the observations 11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41 arranged in ascending order is 24. Find x. Solution. Here, n = 10. Since n is even : th th n n  observatio n   1     observation 2  Median   2  2 5th observation  6th observation  24  2 ( x  2)  ( x  4) 2x  6  24    24  24  x  3 2 2  x = 21 Ans. Example 10. Find the mode for the following data : 14, 25, 28, 14, 18, 17, 18, 14, 23, 22, 14, 18. Solution. Arranging the given data in ascending order: 14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28 Since, 14 occurs maximum number of times (4 times),  14 is the required mode.

PRACTICE EXERCISE 1. Construct a frequency table for the following ages (in years) of 30 students using equal class intervals, one of them being 9-12, where 12 is not included. 18, 12, 7, 6, 11, 15, 21, 9, 13, 8, 15, 17, 19, 22, 14, 21, 8, 23, 12, 17, 6, 18, 15, 23, 16, 22, 9, 21, 16, 11 2. The electricity bills (in Rs.) of 40 houses in a locality are given below : 116, 127, 100, 107, 80, 82, 65, 91, 101, 95, 87, 105, 81, 129, 92, 75, 78, 89, 61, 121, 128, 63, 76, 84, 62, 98, 65, 95, 108, 115, 101, 65, 52, 59, 81, 87, 130, 118, 108, 116 Construct a grouped frequency table. 3. For the following data of weekly wages (in Rs.) received by 30 workers in a factory, construct a grouped frequency distribution table. 258, 215, 320, 300, 290, 311, 242, 272, 268, 210, 242, 258, 268, 220, 210, 240, 280, 316, 306, 215, 236, 319, 304, 278, 254, 292, 306, 332, 318, 300 4. Construct a frequency table, with equal class-intervals from the following data on the weekly wages (in Rs.) of 25 labourers working in a factory, taking one of the class-intervals as 460-500 (500 not included). 580, 625, 485, 537, 540, 425, 637, 605, 607, 430, 611, 632, 600, 640, 638, 612, 584, 440, 536, 515, 449, 480, 556, 561, 508 5. Given below are two cumulative frequency distribution tables. Form a frequency distribution table for each of these. (i) (ii) Ages ( in years ) No . of persons Below 10 15 Below 20 28

Marks obtained More than 10 More than 20

No. of Students 0 17

Below 30 Below 40 Below 50

39 60 73

More than 30 More than 40 More than 50

27 39 52

Below 60

80

More than 60

60

MATHEMATICS–IX

STATISTICS

179

6. On a certain day, the temperature in a city was recorded as under : Time 5 a.m. 8 a.m. 11a.m. 3 p.m. 6 p.m. Temperature (in C ) 20 24 26 22 18 Draw a bar graph to represent the above data. 7. Read the bar graph given below and answer the questions given below : (i) What information is given by the bar graph? (ii) In which year was the production maximum? (iii) After which year was there a sudden fall in the production? (iv) Find the ratio between the maximum and minimum production during the given period.

8. The table given below shows the number of blinds in a village : Age group 0  20 20  40

No . of blinds 5 9

40  60 60  80 80  100

10 4 2

Total 30 Draw a histogram to represent the above data. 9. The following table shows the average daily earnings of 40 general stores in a market, during a certain week. Daily earning (in Rs.) No. of stores

180

600 - 650 650 - 700

5 10

700 - 750 750 - 800

2 7

800 - 850

12

850 - 900 Total

4 40

STATISTICS

MATHEMATICS–IX

Draw a histogram to represent the above data. 10. The following table gives the heights of 50 students of a class. Draw a frequency polygon to represent this. Height (in cm) No . of Students 120 -135 1 135 - 150 18 150 -165 165 - 180 180 -195

23 7 1

Total

50

11. In a study of diabetic patients in a village, the following observations were noted. Represent the given data by a frequency polygon. Age (in Years) No. of Patients 10 - 20 20 - 30

2 6

30 - 40 40 - 50

12 20

50 - 60

9

60 - 70 Total

4 53

12. Draw a histogram and a frequency polygon on the same graph to represent the following data : Weight (in cm) No . of Persons 40 - 50 30 50 - 60 25 60 - 70 70 - 80 80 - 90

40 30 10

Total

135

13. Draw a histogram to represent the following frequency distribution. Class Interval Frequency 10 - 15 6

MATHEMATICS–IX

15 - 20 20 - 30

9 10

30 - 50 50 - 80

8 18

STATISTICS

181

14. Draw a histogram for the marks of students given below : Marks No . of Students 0 - 10 8 10 - 30 32 30 - 45 45 - 50 50 - 60

18 10 6

Total

74

15. The runs scored by two teams A and B on the first 120 balls in a cricket match are given below : Number of balls Team A Team B 0 - 12 10 4 12 - 24 12 2 24 - 36 36 - 48

4 20

16 18

48 - 60 60 - 72 72 - 84

10 12 6

8 10 12

84 - 96 96 - 108

8 16

20 12

108 - 120

20

4

Represent the data of both the teams on the same graph with the help of frequency polygons. 16. Find the mean for each of the following sets of numbers : (i) 25, 12, 37, 19, 43, 40, 11 (ii) 6.2, 4.9, 7.1, 2.9, 5.7, 8.3 2 2 2 2 2 2 (iii) 1 , 2 , 3 , 4 , 5 , 6 (iv) 13, 23, 33, 43, 53 17. Calculate the mean ( x ) for each of the following distribution : (i)

x f

2 4 6 8 10 1 3 5 2 6

(ii)

x f

5 10 15 20 25 30 6 3 2 5 1 3

18. The following table shows the number of accidents met by 120 workers in a factory during a month : No. of accidents 0 1 2 3 4 5 No. of workers 36 34 21 25 3 1 Find the average number of accidents per workers. 19. The marks obtained out of 50 by 80 students in a test are given below : Marks 15 20 22 24 25 30 33 38 35 No. of Students 5 8 7 16 12 18 7 3 4 Calculate the average marks. 20. If the mean of the following data is 18.75, find the value of p. xi 10 15 p 25 30 f i 5 10 7 8 2 182

STATISTICS

MATHEMATICS–IX

21. The average of height of 30 boys out of a class of 50, is 160 cm. If the average height of the remaining boys is 165 cm, find the average height of the whole class. 22. The average of six numbers is 30. If the average of first four is 25 and that of last three is 35, find the fourth number. 23. The mean of 100 observations was calculated as 40. It was found later on that one of the observations was misread as 83 instead of 53. Find the corrected mean. 24. The mean of 10 numbers is 18. If 3 is subtracted from every number, what will be the new mean? n

25. If x is the mean of n observations x1, x2, ...., xn, then prove that

 ( xi  x )  0 i.e. the algebraic sum of i 1

deviations from mean is zero. 26. Find the median of following data : (i) 14, 6, 18, 9, 23, 22, 10, 19, 24 (ii) 17, 13, 28, 19, 23, 22, 12, 32 27. The numbers 5, 7, 10, 12, 2x – 8, 2x + 10, 35, 41, 42, 50 are arranged in ascending order. If their median is 25, find the value of x. 28. Find the median of the following observations : 46, 64, 58, 87, 41, 77, 35, 55, 90, 92, 33. If 92 is replaced by 99 and 41 by 43 in the above data, find the new median. 29. Find out the mode of the following data : (i) 14, 28, 19, 25, 14, 31, 17, 14, 12, 27 (ii) 8.3, 8.9, 8.1, 8.7, 8.9, 7.9, 8.7, 8.9, 8.1 30. Given below is the number of pairs of shoes of different sizes sold in a day by the owner of the shop. Size of shoe 1 2 3 4 5 6 7 8 9 No. of pairs sold 2 2 3 4 5 5 6 9 1 What is the modal shoe size?

PRACTICE TEST MM : 15

Time :

1

2

hour

General Instructions : Each question carry 3 marks. 1. Three coins were tossed 30 times simultaneously. Each time the number of heads occuring was noted down as follows: 0 1 2 2 1 2 3 1 3 0 1 3 1 1 2 2 0 1 2 1 3 0 0 1 1 2 3 2 2 0 Prepare a frequency distribution table for this data. 2. A random survey of the number of children of various age groups playing in a park was found as follows: Age ( in years ) 1 - 2 2 - 3 3 - 5 5 - 7 7 - 10 10 - 15 15 -17 No. of children 5 3 6 12 9 10 4 Draw a histogram to represent the data above. MATHEMATICS–IX

STATISTICS

183

3. Find mean ( x ) for the following distribution: x 10 15 20 25 30 35 f 3 2 4 7 3 1 4. The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x. 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95 5. The marks obtained by 80 students in a test are given. Find the modal marks. Marks 4 12 20 28 36 44 No. of students 8 10 15 24 15 8

ANSWERS OF PRACTICE EXERCISE 1.

2.

Class 6 - 9 9 -12 12 - 15 15 -18 18 - 21 21 - 24 Frequency 5 4 4 7 3 7

Class 50 - 60 60 - 70 70 - 80 80 - 90 90 - 100 100 - 110 110 - 120 120 - 130 130 -140 Frequency 2 6 3 8 5 7 4 4 1

3.

Weekly Wages 200 - 220 220 - 240 240 - 260 260 - 280 280 - 300 300 - 320 320 - 340 ( in Rs.) No. of 4 2 6 4 3 9 2 workers

4.

Weekly Wages 420 - 460 460 - 500 500 - 540 540 - 580 580 - 620 620 - 660 ( in Rs.) No. of 4 2 4 3 7 5 workers

5. (i)

(ii)

184

Age (in year ) 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 No. of persons 15 13 11 21 13 7

Marks obtained 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 No. of Students 8 13 12 10 17

STATISTICS

MATHEMATICS–IX

6.

7. (i) The given bar graph shows the production (in million tonnes) of food grains during the period from 2000 to 2004. (ii) 2002 (iii) 2000 (iv) 5 : 2

8.

10.

MATHEMATICS–IX

9.

11.

STATISTICS

185

9

6 5

12.

13. 3 2

0

14.

15.

186

STATISTICS

MATHEMATICS–IX

16. (i) 26.71

(ii) 5.85

(iii) 15.16

(iv) 45

17. (i) 7.05

(ii) 15.25

18. 1.4

19. 26.7 (approx)

21. 162 cm

22. 25

23. 39.7

24. 15

26. (i) 18

(ii) 20.5

27. 12

28. 58, 58

29. (i) 14

(ii) 8.9

30. 8

20. 20

ANSWERS OF PRACTICE TEST 3. 22

MATHEMATICS–IX

4. 62

5. 28

STATISTICS

187

CHAPTER

15 PROBABILITY

Points to Remember : 1. An activity which gives a result is called an experiment. 2. An experiment which can be repeated a number of times under the same set of conditions, and the outcomes are not predictable is called a Random Experiment. 3. Performing an experiment is called a trial. 4. Any outcome of an experiment is known as an event. 5. In n trials of a random experiments, if an event E happens m times, then the probability of happening of E is given by, P(E ) 

Number of outcomes favour to E m  Total number of possible outcomes n

6. For any event E, which is associated to an experiment, we have 0  P ( E )  1. 7. If E1, E2, E3, ....., En are n elemantary events associated to a random experiment, then P(E1) + P(E2) + P(E3) + ........... + P(En) = 1

ILLUSTRATIVE EXAMPLES Example 1. In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit the boundary. —NCERT Solution. Since, she hits a boundary 6 times, it means she has missed 30 – 6 = 24 times.  Required probability = P (she did not hit the boundary) Number of time she did not hit the boundary  Total number of balls she plays 24 4   0.8 Ans. 30 5 Example 2. Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes: 

Outcome Frequency

Solution.

3 heads 2 heads 1 heads no heads 23

72

77

28

If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up. —NCERT Out of total of 200 outcomes, 2 heads come for 72 times.  P (2 heads coming up)  72  9 Ans. 200 25

188

PROBABILITY

MATHEMATICS–IX

Example 3. An organisation selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below: Monthly income (in Rs.)

0

Less than 7000   7000-10000 10000-13000 13000-16000 16000 or more

Solution.

Vehicles per family 1 2

10 0 1 2 1

160 305 535 469 579

Above 2

25 27 29 59 82

0 2 1 25 88

Suppose a family is chosen. Find the probability that the family chosen is : (i) earning Rs. 10000-13000 per month and owning exactly 2 vehicles. (ii) earning Rs. 16000 or more per month and owning exactly 1 vehicle. (iii) earning less than Rs. 7000 per month and does not own any vehicle. (iv) earning Rs. 13000-16000 per month and owning more than 2 vehicles. (v) owning not more than 1 vehicle. —NCERT The total number of families = 2400 (i) Number of families earning Rs. 10000 – 13000 per month and owning exactly 2 vehicles = 29  Required probability  29 2400 (ii) Number of families earning Rs. 16000 or more per month and owning exactly 1 vehicle = 579 Required probability  579  193 2400 800 (iii) Number of families earning less than Rs. 7000 per month and does not own any vehicle = 10 

Required probability  10  1 2400 240 (iv) Number of families earning Rs. 13000 – 16000 per month are owning more than 2 vehicles = 25.



Required probability  25  1 2400 96 (v) Number of families owning not more than 1 vehicle = families having no vehicle + families having 1 vehicle = (10 + 0 + 1 + 2 + 1) + (160 + 305 + 535 + 469 + 579) = 2162 

Required probability  2162  1031 2400 1200 Example 4. A die is throw 300 times and the outcomes are noted in as given below : 

Outcome

1

2

3

4

5

6

Frequency 70 45 47 30 35 48 If a die is thrown at random, find the probability of getting. (i) 1 (ii) 2 (iii) 3 (iv) 4 MATHEMATICS–IX

PROBABILITY

(v) 5

(vi) 6 189

Solution.

Total number of trials = 300 (i) P (getting 1)  70  7  0.23 300 30 (ii) P (getting 2)  45  0.15 300 (iii) P (getting 3)  47  0.156  0.16 300 (iv) P (getting 4)  30  0.1 300 (v) P (getting 5)  35  0.116 300

48  0.16 300 Example 5. The table given below shows the marks obtained by 90 students of a class in a test with maximum marks 100.

(iv) P (getting 6) 

Marks 0  15 15  30 30  45 45  60 60  75 Above 75 No. of students 7 14 18 25 17 9

Solution.

A student of the class is selected at random. Find the probability that student gets : (i) less than 30% marks (ii) 60 or more marks (iii) marks between 45 and 75. (iv) distinction Total number of students = 90 (i) P (getting less than 30% marks)  7  14  21  0.233 90 90 (ii) P (getting 60 or more marks)  17  9  26  0.288 90 90 (iii) P (getting marks between 45 and 75)  25  17  42  0.466 90 90

(iv) P (getting distinction)  9  0.1 90 Example 6. A tyre manufacturing company kept a record of the distance covered before a type to be replaced. Following table shows the result of 1000 cases. Distance (in km) less than 400 400  900 900  1400 more than 1400 No. of tyres 200 335 375 90 If you buy a tyre of this company, what is the probability that : (i) it will need to be replaced before it has covered 400 km. (ii) it will last more than 900 km? 190

PROBABILITY

MATHEMATICS–IX

Solution.

(iii) it will need to be replaced? (iv) it will not need to be replaced at all? (v) it will need to be replaced after it has covered somewhere between 400 km and 1400 km? We have, total number of trials = 1000 (i) The number of tyres that needs to be replaced before it has covered 400 km = 200.  Required probability  200  0.2 1000 (ii) The number of tyres that last more than 900 km = 375 + 90 = 465  Required probability  465  0.465 1000 (iii) Since, all the tyres we have considered to be replaced. Required probability  1000  1 1000 (iv) The number of tyres that do not need to be replaced at all = 0



0 0 1000 (v) The number of tyres which require replacement after covering somewhere between 400 km and 1400 km is 335 + 375 = 710.



Required probability 

 Required probability  710  0.71 1000 Example 7. A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is: (i) White (ii) not black (iii) Red or white Solution. Total number of balls = 3 + 5 + 4 = 12. (i) P (white ball)  4  1 12 3 (ii) P (not black) = 1 – P (black)  1 

5 12  5 7   12 12 12

3 4 7 .  12 12 Example 8. Cards marked with the numbers 2, 3,...., 101 are placed in a box and mixed thoroughly. One card is drawn from this box. Find the probability that the number on the card is: (i) an even number (ii) a number less than 14 (iii) a perfect square number (iv) a prime number less than 20. Solution. Total numbers = 100 (i) Total number of even numbers = 50

(iii) P (red or white) 

 P (an even number)  50  1 100 2 (ii) P (a number less than 14)  P ( 2 , 3,....,13) 

MATHEMATICS–IX

12 3  100 25

PROBABILITY

191

(iii) P (a perfect square number) = P (4, 9, 16, ....., 100)  9 100 (iv) P (a prime number less than 20) = P (2, 3, 5, 7, 11, 13, 17, 19) 8 2  100 25 Example 9. Find the probability that a leap year, selected at random will have 53 sundays. Solution. A leap year has 366 days. 52 weeks = 52 × 7 = 364 days Days left = 366 – 364 = 2 52 weeks means there will be 52 sundays. From remaining 2 days, the possibilities are as follows : (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday) and (Saturday, Sunday). So, from 7 possibilities, we have 2 cases in which we have 53rd sunday. 

2 Ans. 7 Example 10. A bag contains 10 white balls and x black balls. If the probability of drawing a black ball is double that of a white ball, find x. Solution. Total number of balls = 10 + x.

 Required probability 

P (black ball) 

x 10  x

10 10  x according to given question, P (black ball) = 2 P (white ball)

and, P (white ball) 

x 2  10  10  x 10  x  x = 20 Ans. ( 10 + x  0)



PRACTICE EXERCISE 1. A number is chosen from 1 to 20. Find the probability that the number chosen is : (i) a prime number (ii) a composite number (iii) a square number (iv) an odd number (v) an even number (vi) number between 7 and 14 2. A bag contains 9 red and 6 blue balls. Find the probability that a ball drawn from a bag at random is (i) Red ball (ii) blue ball 3. In a sample of 500 items, 120 are found to be defective. Find the probability that the item selected at random is (i) defective (ii) non-defective 4. In a school of 1800 students, there are 875 girls. Find the probability that a student chosen at random is (i) a boy (ii) a girl 5. In a cricket match, a batsman hit a boundary 12 times out 45 balls he plays. Find the probability that he did not hit a boundary. 192

PROBABILITY

MATHEMATICS–IX

6. A coin is tossed 700 times and we get head : 385 times; tail : 315 times. When a coin is tossed at random, what is the probability of getting : (i) a head? (ii) a tail? 7. Two coins are tossed 600 times and we get two heads : 138 times, one head : 192 times ; no head : 270 times. When two coins are tossed at random, what is the probability of getting : (i) 2 heads? (ii) 1 head? (iii) no head? 8. Three coins are tossed 250 times and we get: 3 heads : 46 times; 2 heads : 56 times; 1 head : 70 times; 0 head : 78 times. When three coins are tossed at random, what is the probability of getting : (i) 3 heads ? (ii) 2 heads? (iii) atleast 2 heads? (iv) atmost 2 heads? 9. A die is thrown 300 times and the outcomes are noted as given below : Outcomes 1 2 3 4 5 6 Frequencies 58 75 52 39 42 34 When a die is thrown at random, what is the probability of getting a: (i) 4 (ii) 6 (iii) number less than 3 (iv) number which is prime 10. In a survey of 350 ladies, it was found that 235 like coffee, while rest of them dislike it. Find the probability that a lady chosen at random: (i) likes coffee (ii) dislikes coffee. 11. On one page of a telephone directory, there are 200 phone numbers. The frequency distribution of their units digit is given below : Units digit 0 1 2 3 4 5 6 7 8 9 Frequency 21 20 21 21 22 25 22 17 15 16 One of the numbers is chosen at random from the page. What is the probability that the units digit of the chosen number is : (i) 5 (ii) 8 (iii) an even number (iv) an odd number 12. The blood groups of 30 students of class IX are recorded as follows : A, B, O, AB, O, A, O, O, B, A, O, A, B, O, O, A, O, B, A, B, O, A, B, AB, O, O, A, A, O, AB A student is selected at random from the class for blood donation. Find the probability that the blood group of the student chosen is: (i) A (ii) B (iii) AB (iv) O 13. Following are the ages (in years) of 350 patients, getting medical treatment in a hospital. Age (in years) 10  20 20  30 30  40 40  50 50  60 60  70 No. of Patients 85 50 55 75 55 30 One of the patients is selected at random. Find the probability that his age is : (i) 40 years or more but less than 50 years. (ii) 30 years or less than it. (iii) less than 10 years. (iv) 50 years or more. MATHEMATICS–IX

PROBABILITY

193

14. Given below is the frequency distribution of wages (in Rs.) of 40 workers in a certain factory: Wages (in Rs) 100  120 120  140 140  160 160  180 180  200 200  220 220  240 No. of workers 4 3 6 5 9 8 5 A workers is selected at random. Find the probability that his wages are: (i) less than Rs. 160. (ii) atleast Rs. 180. (iii) more than or equal to Rs. 140 but less than Rs. 200. (iv) more than Rs. 200. 15. The following table gives the life time of 500 neon lamps: Life time 300  400 400  500 500  600 600  700 700  800 800  900 900  1000 (in hours) No. of 24 66 75 96 89 72 78 Lamps

A lamp is selected at random. Find the probability that the life time of the selected lamp is: (i) less than 400 hours (ii) atleast 800 times (iii) atmost 600 hours (iv) between 500 hours to 900 hours.

PRACTICE TEST MM : 15

Time :

1

2

hour

General Instructions : Each Questions carry 3 marks. 1. Two coins are tossed simultaneously 350 times with the following frequencies of different outcomes: Two heads : 105 times; One head : 125 times; no head : 120 times. Find the probability of getting : (i) 2 heads (ii) atleast one head 2. Following table shows the marks scored by 80 students in a mathematics test of 100 marks. Marks 0  20 20  30 30  40 40  50 50  60 60  70 70  100 No. of students 5 8 10 18 15 18 6 Find the probability that a student obtained: (i) less than 40% marks. (ii) 60 or more marks. 3. 1200 families with 2 children were selected at random and the following data were recorded; No. of girls in a family 0 1 2 No. of families 255 515 430 194

PROBABILITY

MATHEMATICS–IX

If a family is chosen at random, compute the probability that it has : (i) no girl (ii) 2 girls (iii) at most one girl (iv) at least one girl 4. The table given below shows the ages of 90 teachers in a school. Age (in years) 18  29 30  39 40  49 50  59 No. of teachers 8 32 40 10 A teacher from this school is chosen at random. What is the probability that the selected teacher is : (i) 40 or more than 40 years old ? (ii) age less than 40 years ? 5. An insurance company selected 2000 drivers at random, in a particular city to find out a relationship between age and accidents. The data obtained are given in the following table : Age of drivers (in years)

Accidents in one year 1 2

0

18-29 30-50 above 50

440 505 360

160 125 45

110 60 35

3

over 3

61 22 15

35 18 9

Find the probabilities of the following events for a driven chosen at random from the city: (i) being 18-29 years of age and exactly 2 accidents in one year. (ii) being 30-50 years of age and having one or more accidents in a year. (iii) having no accidents in one year.

ANSWERS OF PRACTICE EXERCISE 1. (i)

2 5

(ii)

11 20

(iii)

2. (i)

3 5

(ii)

2 5

3. (i)

4. (i)

37 72

(ii)

35 72

5.

6. (i)

11 20

(ii)

9 20

7. (i)

8. (i)

23 125

(ii)

28 125

(iii)

9. (i)

13 100

(ii)

17 150

(iii)

10. (i)

47 70

(ii)

23 70

MATHEMATICS–IX

1 5

(iv)

3 10

6 25

(ii)

19 25

23 100

(ii)

32 100

51 250

(iv)

102 125

133 300

(iv)

169 300

11 15

PROBABILITY

(iii)

9 20

195

11. (i)

1 8

(ii)

3 40

(iii)

101 200

(iv)

99 200

12. (i)

3 10

(ii)

1 5

(iii)

1 10

(iv)

2 5

13. (i)

3 14

(ii)

27 70

(iii) 0

(iv)

17 70

14. (i)

13 40

(ii)

11 20

(iii)

1 2

(iv)

13 40

15. (i)

6 125

(ii)

3 10

(iii)

33 100

(iv)

83 125

ANSWERS OF PRACTICE TEST 1. (i)

3 10

(ii)

23 35

2. (i)

33 80

(ii)

3 10

3. (i)

17 80

(ii)

43 120

4. (i)

5 9

(ii)

4 9

5. (i)

11 200

(ii)

9 80

196

(iii)

77 120

(iii)

261 400

(iv)

63 80

PROBABILITY

MATHEMATICS–IX