Navneet Maths Digest Std 8th

August 30, 2017 | Author: Altamash | Category: Circle, Rectangle, Fraction (Mathematics), Perpendicular, Euclidean Geometry
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This is the navneet maths digest for std 8th students....

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BASED ON THE TEXTBOOK

NA[/NEET HIETTf,EffiET!$$

DIGEST

NAVNEET MATHEMATICS DIGEST : STANDARD

CONTENTS

Squares and Square Roots 1. Squares and Square Roots 2. Inational and Real Numbers

5

3. Parallel Lines 4. Quadrilaterals

17

to 7)

:

You have studied in previous standard to find the square root by the

division method

:

and Proportion

3I 35

48

...

11. Equations in One Variable

84

*

9I

lffil

tffil Cubes_ and Cube Roots

13. Indices

+l

289

I

+4

27

+7

34

:

4

1

85

+5

000

2025 16

0'425 425

Ant. The square root of 289 is 17.

Ans. The square root of 2025 is 45.

(t)

(4) 646416

6400e

804

253

101

105

14. Construction'of Quadrilaterals 15. The Arc of a Circle

116

16. Joint Bar Graphs

123

17. Compound Interest

t28

18. Polynomials

138

19. Discount and Commission

142

20. Volume and Surface Area 21. Division of Polynomials

150

r62

22. Factors of Polynomials

173

*

180

Miscellaneous Exercise-2 (Textbook page 170)

Q. Find the square root by the division method (2) 202s (1) 28e

62 69

Miscellaneous Exerbise-1 (Textbookpage g9)

(Textbook page 1)

55

10. Identities

12.

square root

I

24

5. The Circle 6. Area 7. The Circumference and Area of a Circle 8. Statistics

9. Variation

[. Revision-Square,

t4

(Textbook pages

VIII

108

Anl, 'l'ho

square root

of

64(l0t) is 253.

Ans. The squarc root of 646416 is 804.

I x 9:81 .'. 9 is the square root of 81. ( - 9) x (-9):81 .'. (-9)isthe squarerootof ,',

tltc square root of 81 is 9

or -9.

81.

NAVNEET MATIIEMATICS DIGEST : STANDARD YIIT

SQUARES AND SQUARE ROOTS

(1) Every positive number has two square roots. (2) These squre roots are opposite numbers of each other.

(4) The positive square root of,625.

Q, 2. Write in symbols : Ane. (1) The positive sqqare root of a00:

(Textbook page 2)

Q. Write the square roots of each of the following numbers (1) e Ans. The square root of 9 is 3 or

(2)

-

(3) 100

3:

of 25 is 5 or

-5.

'_glTg

19_91

$yg:

of 100 is t0 or _ -19-gl

Q, 3. lVrite the values of Anc. (1) 10.

(4) ts6 Ans. The square root of 196 is t4 or _14. (s) 324

4ry:

fr:

'

576 is 24

:

,r/-,

-

(D (5)

\E:s

/roo:

and

- j64- -

8.

(3) -

1

tE4

(3) The negative square root of 625.

-

3; 5,

-

5 and

The gquare root of a number that is not a porfect squar.e : A number which is not rquare root is 20 is

a perfect square also has a square root. e.g., the

,/n. (Textbook page 4)

Q,

l. The sides forming the right angle of a right-angled triangle are 7 cm and 5 cm. Find the length of its hypotenuse. Let AABC be a right-angled + trlangle, in which LB:?O, /(AB):5 r* end ,(BC):7 cm. F Bolutlon :

a. 1. Write the following numbers in words : (3) - j6rs @{@ Q) +!l !11 T_"_ pgll_t-1y_: square root of 324. p) fhe negarive square root of 324.

lx: -s

ro.

nttilibers are called non-perfect square numbers.

utU

(Textbook page 2)

g \F

:

13 respectively

- J-,

is read as the posirive square root of sixty_four. is read as the negative square root of sixty_four.

t , ,[1.

2, 3, 5, 20, 27, etc., are not the perfect squares of any integers. Such

Writing: The symbol is used to write the square root. meaning of this symbol is positive square root. For. negative square the symbol ' is used. Reading :

, _J4_:

The square of an integer is called a perfect square number. 13,

or _24.

3. Writing and reading a squarO root

121

9,f,, the numbers 9, 25 and 169 are the squares of 3,

lgtg:1-919!324 is tB or _ t8.

Ans. The square root of

-O- -3

- .Jnt: - 11 t *tt-.t tqt"r"r ""a * (4)

s76

-.JA JA:S

of 81 :

(4) The positive square root of

r!:

4T:

square root

(3) The negative square root of

3.

2s

$l|:

(6)

(2) The negative

:

uffi. .Ai'

I

rl

By Pythagoras' theoiem,

ll(Ac)1'z: u(AB)12 + t/(BC)],

: (5)' + Q)2 :25 * 49 :74 ,', /(AC) : uE4 cn.

Anc. The length of the hypotenuse is J-l+

"

.

U

NAVNEET MATTIEMATICS DIGEST

3

STANDARD

VIII

SQUARES AND SQUARE ROOTS

Q. 2. Find the length of the diagonal of a rectangle of length breadth 7 cm. Solution : Let E ABCD be a rectangle. (BC) :7 cm and /(AB) : 12 cm. Each angle of a rectangle js a right angle. .'. in the figure, AABC is a righrangled

12 cm and

'l'he approximate value of the square root of a non-perfect square number:

7,

ln the above example, we have calculated the square root of 30 up to lirur decimal places.

a

t-

../30:

triangle. By Pythagoras' theorem,

u(Aqlr:

:io})z + e)2 :144 *

.'.

/(AC)

: uEi

49

lhrrt digit and keep the remaining digits as

:193

tu;rproximate value

"

"",493

cm.

30.00000000

-25 5 00 - 416

1087 7

542

7

44 In the above example,

"^

0

219084

02

:

up to one decimal place will be 5.5.

by division

fr .l

of

denote that further there

will

be

infinite number of digits. up to four decimal places.

t6 fr

t2 6 l',)"42

t2 h | )44

:

up to three decimal places

6t 03900 3756

ll2

Ans. ./io:3.1622... Approximate values

0100 (r

:

3.1622 9

rl

s.4772...

third, second and first decimal places

10.00000000

It frl

the value of the square root. This process is never endi In 5.4772..., the dots on the ri

,,,m:5.4772is

tlt l0 I

we continue the process

6629

02

2

Hqurre root up to the

As 30 is not a perfect square, wo

07I

47

.f,0

;llncc by division method. Then write the approximate value of each 8cm

will get more decimal places, i

084 7 609

+7

*50:

.f

'l'lro approximate value of

up to two decimal places.

Q, li'lnd the square roots of the following numbers to the fourth decimar

Ans. The length of the diagonal of the square is .",42g cm.

30 is a non-perfect square number. Let us find 5.47 7 2

nET:5.48

1.

(Textbook page 6)

... /(AC) : ,EZS

6. The square root of a non-perfect square number

'f'lrus, we get,

tr o

B

:64 * 64:128

up to three decimal places.



l/(AC)1'z: t1(AB)l' + t/(BC)l' (8)2 + (8)2

.f

thop 7 and increase the previous digit by

Q. 3. Find the Iength of the diagonal of a square with side g cm. Solution : Let n ABCD be a square whose side is 8 cm. Each angle of a square is a right angle. .'. in the figure, AABC is a right-angled triangle. By Pythagoras' theorem,

of

it is. So, .160:5.477 isthe

T'he digit 7 in the third decimal placb is greater than 5 or 5. Hence, we

Ans. The length of the diagonal of the rectangle is

:

5.4772.

l{ere, the digit2 in the fourth decimal place is less than 5 or 5. We drop

+ t/(AB)12

u(BC)12

9

0t4400 t2644 0175600 126484

049tt6

"/to:ltez

up to two decimal places

u'Eo:ne up to one decimal place

u/n:s.z

:

:

:

10

NAVNEET MATHEMATICS DIGEST

(2) 13s

STANDARD

VIII

SQUARES AI\D SQUARE ROOTS

11.6189

Isroooooooo

I +1

035

+1 226

+6

23228

14 00

up to two decimal places

23 23 69 8

(3) 777

+2 47

up to one decimal place

207 900

JBs:11.6

377

+7

-329

548 67

+7

2229

4

557 487 7

557494

22

76

04012400 3902409 0109991

:

up to two decimal places

126484

:

4427 r29 048

142

+2

:

.4ooo:11.62 up to one decintal place

:

/ooo:31.6

72.9931 5328.OOOOOOOO

.7

+7

:

./tooo :31.623

r2644 0r756

3?D437-49

,l,ns.

o4

t44 00 I30 4L t

013 5900 13 1301 0 459900 437 9 49

:

.f3x

:72.9931...

Approximate values

284

ffi:27.8747-.

up to two decimal places

38969 0263 r00

55744

:

(5) 5328

An*

Approximate values

37 56

02t95100

'nn:2ir.B7s

048 00 43 84 04 1600

+8

6t

+1

Approximate values : up to three decimal places

-4

+l

Ans. .r/tooo :31.6227.-

2

232t

in.ooo-ooooo

9

JBs:1t.62

27.87 47 |

2

+3 61 :

27

2

up to three decimal places

209 t32r 0 t 1627 9

9

3r.6

tooo.oooooooo

3

tt.6t9

185824 2207 600

8

23237

./35:

u.6189...

JBs:

004400

1

Ans.

2t 1356

232r

(4) rooo Approximate values : up to three decimal places

-1

2t

55

T

1459861

:

:

up to three decimal places

:

!Br8:72.ss3 up to two decimal places

:

!&r8:72.ss up to one decimal place

:

J$rB:73.0.

Jm:27.88 up to one decimal place

$n =27.e

:

tl. The square root of a decimal fraction by the division method

:

Itlx. Find the square root of 249.3241by the division method. | 5 .1 .7 9 part of the i if,irst, the digits in the integer iFirst, (249) paired are inumber o.ff from the right, I , 4 9 .3 Z 4 I jThen, the digits in the decimal part (ot 3241) rl -l 25 149 i are also paired off, but starting from the left. i Now, find the square root as per the

f5

-r25

I

4

-r

2t 634

I

jcompleted, place a decimal point after the obtained till tllt.lh* then (i.e., after. after 15)

H++iquotient

Ans.

NAYNEET MATHEMATICS DIGEST : STANDARD

VIII

SQUARES AND SQUARE ROOTS

(3) 34.1s8

(Textbook pageT)

Q. 1. Find

(t)

the square root of the following numbers by the division method

s6.2s

Q) tst.zs

+5 108

+8

1

tt64

29 Ans. n/56.25:7.5.

(3) 49.s616

,tttr. r,ry5Lrr:12.3. (4) 443.s236

7.04

2t.o6

2 -L1

41

+1

4

n.i2TZ

-4

o4 02 0 00 2

2 5236 00

49.5616:7.04.

Q. 2. Flnd the approximate value of the square roots of the following numbers up to the second place :

(1) s9.03

(2) 3.41s8

7.683

+7 146 +6 r528 +8

5

9.0 3 0 0 0 0

-49

10 87

1536

2

+8

6

r27 0 r2224

00

3.415800 -1 2

-224 01

00

46089 66 J5c:o3:7.683... Ans. yte.o3 :7.68 correcr up to two decimal places.

1.8 4 8

I +1

1456 36 36

+4

11588

34.r5S000

-25

0915 864 05180 4656

052400 46736 05664

,', /5als8:5.844... Ans. .f,4rSS :5.84 correct up to two decimal places.

2

+2 46

+6

^

320

-27 6 528 044 50 42 24 +8 5364 02 2600 2 r456 +4 53682 ot 14400 2 r 07364 53684 0 07036

41

42

7

+4

I t684

l

26.842 720.500000

:

5

r 2.3

(4) 720.s

29504

tr54158:1.848...

i two

decimal places.

rrr-

Irrational and Real Numbers rttl 2 ir IrrationalandRea

IRRATIONAL AND REAL NUMBERS (Textbook pages 8 ro t2)

1. Revision : Rational numbers : Ifp is any integer and q is any non-zero integer, then 4 is called

(Textbook page 10) a

Q. Write

the following numbers in the non-terminating recurring

form

:

Ans.

rational number.

The decimal form of a rational number is obtained by dividing its numerator by the denominator. (i) The decimal form of ! is O.O. (ii) The decimal form of I is O.OO... and that of is 0.090909... fr In case (i), the process of division comes to an end. The decimal fraction, so obtained, is calred a terminating decimal fraction.

ln case (ii), the process is unending. In this-case either a aigi, o, a group of digits is repeated. The decimar fraction, so obtained, is called a

0.16

ttlll

i

7.439 i 10.505

i

0.058

'I

i

1.06 i 0.0002

I

I

I

t-

-l

10.6050i 0.0580

i

1.060

I

I I

I

4. Non-terminating recurring decimal forms and rational numbers Every nutmber

non-terminating recuning decimal fraction. ln case (ii), 3 : 0.66... is written as :0.6 and

:

in the non-terminating recurring decimal form is a

rational number.

52

*:0.090909... is written I

il

as

fr:0.09.

[The recurring digits are marked with a line drawn above them. ] (Textbook page 9)

a. chssify the following

decimal fractions as terminating and nonterminating recurring decimals : (t) 0.777... Q) 0.777 (3) 4.7152 (4) 4.71n (t e.16s16s (6) e.16s (7) O.s2sss (s) 0.s2s (s) 72.136. Ans. Terminating decimals : (2) 0.777 (3) 4.7182 (5) 9.165165 (7) 0.s2888. Non-terminating recurring decimals

non-terminating recurring decimal frbction.

3.75:3.750 Every rational number can be written in the non-terminating recurring

iI

:

Numbers whose decimal.form is non-terminating and non-recurring are

calted irrational numbers.

rf;:2.230067977...i.e., and non-rectrrring. So,

.f

the decimal form of

..,6 i, non-tenninating

is an irrational number.

6, tlquare roots of numbers that are not perfect squares

:

The square roo[s ofnumbers that are not perfect squeres are irratiorutl numbers.

The terminating decimal fraction 3.75, if written as 3.750, 3.7500, etc., its value does not.change. writing zeroes on the right side of a terminating decimal it becomes a

l

Irrational numbers

:

(1) 0.777... (4) 4.7182 (6) 9.165 (8) 0.s28 (s) 72.136. 3. writing a terminating decimal in the form of a non-terminating

form.

t,

(Textbook page 12)

Q, Clussify the following numbers into two groups and label each group correctly

:

l\ 4.en

(2) 0.3104s693...

@\E

@trB

(s) 10.0s

(6) 0.1010010001...

Ans. Rational numbers : G) a9n @ J4g (5) 10.05 lrrtrtional numbers : (2) 0.31045693... (3) aEi 6> 0.1010010001...

! Rcul

numbers : The collection together is called real numbers.

of rational and irrational numbers

NAVNEET MAI'HEMATICS DIGEST : STANDARD VTII

r.

Parallel LineS lS -t llParallellrnes

(Textbook pages 13 to 22)

ttt

(Textbook page 12)

Q. Make

l.

a table

with columns for rational numbers, irrational numbers and real numbers and write the following numbers in their proper

places in the table

:

(1) 1.57 Q) v5

o\F

(8)

(3) 4.10547194:.. (4)

4.8

(5)

0.73s (O lE

Rational numbers (4) 4.8

(s) 0.73s

@ d *t

Jzs

(8)

set squares.

2. Lines parallel to the sarne line In the figure, line / ll lnne m and line n ll line m.

v4%

:

Using the ruler and set squares, we find that line

,/\%

I

ll line n.

Lines parallel to the same line are parallel to each other.

Ans.

(r) r.s1

Drawing parallel lines : You have studied how to draw a line parallel to a given line using

Irrational numbers

ReaI numbers

Q).,fr

All the given numbers

(3) 4.10547r94...

are real numbers

0lo (9) 3.819023... (10) 6.10203040...

3..Lines perpendicular to the same line : In the figure,Ttne m r line / and line a r line /. Using the set squares and a ruler, check whether lines m and n are parallel to each other.

intercept : (1) In the figure, transversal n intersects line'/ and line m in two distinct points P and Q. Segment PQ is called the intercept formed by

4, The

lines

/ and m on transversal

n.

(2) Intercepts made by three parallel lines on I

l

a transversal

:

In the figure, line /, line m and line n are parallel to one another.

The intercept cut off by lines 'l' and m on transversil p is seg CD. The inteicept cut off by lines ru and n on tranSversal p is seg DE. The intercept cut off by lines I and n on transversal p is seg CE. (Textbook page 16)

a. ln

each of the following figures, name the intercepts, the lines that form them and the transversal on which they are formed :

NAVNEET MATIIEMA.TICS DIGEST : STANDARD

18

VIII

PARALTEL LINES

Ans.

Fig. (1) : Lines g and ft make intercept seg DE on the transversal4. Lines ft and I make intercept seg EF on the transversal 4. I.ines g and i make intercept seg DF on the transversal q.

t and w make intercepts : seg HD on the transvers al p and IC on the transversal s. Lines w and ft make intercepts : seg DF on thetansversalp and

Fig. (2) : Lines seg

seg CR on the transversal s. Lines r and ft make intercepts : seg rrF on the transversal p and seg IR on the fransversal s. Lines p and s make intercepts : seg HI on line r; seg DC on line and seg FR on line ft.

rig. (3) : Lines s and p make intercepts seg AI on the transversal y.

: seg RD on the transversal r and

p and ft make intercepts : seg DO on the transversal / seg IN on the transversal v. Lines s and ft make intercepts seg RO on the transversal I and seg AN on the transversal v. Lines / and y make intercepts : seg AR on line s; seg ID on line p, Lines

il I

I

l. In the figureo line ft | line /

ll tine rz. Their transversals, line c and line d, cut them at points X, Y, Z and P, Q, R respectively.

If t(xY)

:5, l(YZ):3,

/(PQ) :5.5,

flnd r(QR). Solution z Line k ll line 1 ll line m. Line c and line d are theilr transversals. .'. by the property of three parallel lines and their transversals,

/(xY): /(PQ) (QR) /(YZ)

",53

5.5

/(QR)

... (Substituting the given values)

.'. 5 x /(QR) :5.5;

3

.'.

/(QR):Y:

1.1

x

3

:3.3.

Ans. /(QR) :3.3.

and seg NO on line fr.

I

I

(Textbook page 19 & 20)

5. The prope4ies of parallel lines and their transversals with respect to intercepts

:

(1) If the intercepts formed by three parallel lines"on any one transversal are congruent, the intercepts they forrn on any other transversal are also congruent. Line / ll bne m ll line n and linesp and, q arethe transversals.

In the figure, if intercept seg AB seg

BC, then intercept segDE

=

=

intercept

intercept seg EF.

(2) The ratio of the lengths of tho intercepts made by three parallel lines on one transversal and the ratio of the lengths of the corresponding intercepts made by the same lines on any other transversal are equal. Line.r ll line y ll line z and lines p and q are the transversals.

/(cD) _ /(LM) In the fisure. -'l(DE) " /(MN)'

Q. 2. In the figure, Iine a ll line D ll line c. l.lne d and line e are their transversals lntersecting them in points P, Q, R and polnts Lo M, No respectively. Point Q is the mldpoint of seg PR. tf ,(QR) :7.2 andr(LM) :6.2,find,/(PQ) and /(MN). Solution : Point Q is the midpoint of seg pR

t*?11 l(eR) (Given) ,'. /(QR):7.2 ,.. /(Pe):

...

.'. I(PQ):7.2. 'Line a jl line b ll line c. Line d and line e are thek transversals. ,'. by the property of three parallel lines and their transversals, intercept LM:intercept MN ... t'.' l(PO: /(QR) ... From (l)l .'. /(LM): /(MN)

t(LM):6.2 ... (Given) ,: /(MN):6.2 Ans. /(PQ) :7.2; /(MN) :6.2.

20

NAVNEET MATIilMATICS DTGEST : STANDARD VrrI

(1) Dividing a line segment into a given number of equal parts

PARATLEL LINES

(Textbook page 22) :

Q, 1. Divide seg LM of length 9 cm into 5 congruent parts. equal parts.

Ans.

Construction : (1) Draw a 6 cm long seg pe. (2) At P, draw a ray PM making an acute angle of some suitable measure. (3) At Q, draw a ray QN on the opposite side of seg PQ making an acute angle of the same measure. (4) On the ray PM starting from p, mark off 4 congruent segments pp1, p1p2,

PrP. and P.Po with the help of of

Seg

LA, seg AB, seg BC,

neg CD and seg

DM are the five

congruent parts of seg-LM. p

a

compass. On the ray QN also, starting ,lil

from Q mark off 4 congruent segments eer, erer, ezer and er each of, which is congruent to seg ppr. (s) Draw seg PQo, seg PrQr, seg prer, seg pre, and seg poe. Let seg pre, seg PrQ, and seg P.Q, ihtersect seg pe in points A, B and

{1.

Seg PA, seg

Dtviae seg CD of length 6.4 cm into 3 congruent parts.

Ans. Sog CA, seg

AB

and

rog BD are the three gongruent parts of seg CD.

respectively.

(6)

l,

AB, seg BC and seg Ce are the four congruent parts

seg PQ.

Ex. Divide 5.5 cm long seg Xy in the ratio /(XL) : l(Ly):2 2 3. Ans, As per the above construction, divide seg Xy into 5 (Z+3 congruent parts. Take point

/Qo-):t(LY):2:3.

L on seg Xy

such that

io of division

:l

l$

::g-T:_11

ttber of congruent parts of ,legment

Q,

{. Dlvide

2:3 1:5 l:2 3:4 5:3 5

6

3

7

seg ST of length 10 cm in the ratio 2 : 3. Ans. [Seg ST is to be divided in the ratio 2 : 3. ,', seg ST will have to be divided into 2 * 3 :5 congruent parts.l

8

NAVNEET MATHEMATICS DIGEST : STANDARD

VIII

PARALLEL LINES

Q, 7. Draw a line segment of length 7 cm. Mark l(PR) : t(RO:4 t, L. Anr. /(PR) : /(RQ) :4: I.

.',

l(Yl):2'

3

Q. 5. Divide 7 cm long line segment AB in the ratio 3 : t Ans.Ratio3:2.

.'.

total divisions

:3 * 2: 5

4s

Point C divides seg AB in the ratro 3 : 2. .'. /(AC) : /(CB) :3 :2.

a. 6. Divide seg PQ of length 6 crn in the ratio 1 : 2. Ans.Ratiol:2. .'. total divisions : I -12:3. Pl

,'t \ o\R

\ .,. Point R divides seg PQ in the ratio I :2. .'. /(PR) : /(RQ): 1 :2.

:4 * | :5

lloint R divides seg PQ in the ratio 4 .', /(PR) : /(RQ):4: L

Point P divides seg ST in the ratio 2 : 3.

/(SP) :

total divisions

:

1.

a point R on

it such that

tr

QUADRILATERALS

The types of quadrilaterals named on the basis of their angles and

s

are shown below. (Textbook pages 23 & 24)

Recall the properties of quadrilaterals and

fill in the blanks

(l) mISTR:90' :

(L) Parallelogram : A quadrilateral with opposite sides parallel is called a parallelogram. Properties of parallelogram

____

----- -- -- ----::--

--

--

(2) /(PT) : /(TR) (3) nLSPQ:mlSRQ

(l)

:

,-(1) The opposite sides of a parallelogram are congruent. (W'The opposite angles of a parallelogram are congruent. (3) The diagonals of a parallelogram bisect each other.

- -'..{-'-

l) /(AB) = /(AD) (2) /(BC):/(DC) (3) rn IAMD:90' (4) ! ABCD is a kite. (

1. Rc

HC

.r side DF

(2) /(HF):l(CD) (3)

'(cM):t(DM)

-----

(2) Rhombus : A quadrilateral with all four

Side

sides

: /(EU) (1) '(PN) (2) nlutN:90' (3) r(TN) -r(TP) : /(TU) : /(TE)

congruent is called a rhombus.

Propertiep-of rhombus : (*Fh6" diagonals of rhombus bisect each ot[rer at right angles.

diagonal of a rhombus is Q)dacn -

the

perpendicular bisector of the other.

(U#dtbposite

angles of a rhombus are congruent.

(3) Rectangle : A quadrilateral with all angles right angles is called a rectangle. Properties of a rectangle : (1) The opposite sides of a rectangle are

:

Flnd the length of diagonal QS of the square PQRS, if the length of dlrgonal PR is I cm. Ans. The diagonals of a square are congruent. .'. /(PR): /(QR) ,', they are of equal length. l(PR):8 cm ... (Given)

,', /(Qs):8

squar€.

l,

:

diagonals of a square are congruent. diagonals of a square bisect each other.

{t.Pr!9t-lsMrof a square is the perpendicular

hoblems based on the properties-of quadrilaterals

l,

(4) Square : A quadrilateral with all the sides congruent and every angle a right angle is called a

,(fr'lhe

X

(Textbook page 25)

.(J),.The diagonals of a rectangle bisect each other.

i(l)-The

PE

0)

L congruent. gffine diagonals of a rectangle are congruent.

Properties of a square

N

bisector ofthe ot

ll'ln

cm.

the square ABCD, /(AB)

:

4.5 cm, what are the lengths 0f the

0thcr sides of the square? Anl. All the sides of a square are congruent. ," /(AB): l(BC): /(CD): /(DA). l(AB) :4.5 cm ... (Given) ,', /(BC) : /(CD) : l(DA) :4.5 cm.

NAVNEET MATIIEMATICS DIGEST : STANDARD VIII

a. 3. The diagonals seg DF and seg BG of the square DEFG intersect each other in point M. If ,(DM) -7 cm, find t(EG).

Fhnilarly, seg PQ

/tPg;

...

#

... (D-M-F) ... [From (1) and (2)] ... (3)

The diagonals of a square are congruent.

.'. /(EG): /(DF):14 cm .'. /(EG): L4 cm. Q. 4. The

...

[From (3)]

XZ and,yW are the diagonals of the square XYZW.M is their point of intersection. segments

Find zlXMY. Ans. LXMY is the angle at the point of intersection of the diagonals seg XZ and seg yW. The diagonals of a square are perpendi-cular bisectors of each other.

.'. zLXNIY:90o. Q. 5. In the square HDFC, if t(HD :5 cIn, find J(CD). Ans. Seg IIF and seg CD are the diagonals of square HDFC.

/(CD): /(HF) ". /(HF;:5 ... "- cm. .'. /(CD):5 (2) RECTANGLE

,(PS)

:9

/(sR):7

cm.

'fhe diagonals AC and BD of rectangle ABCD intersect each other in point O. If n/ CAB:2S', find zLDAC and nt/ LCD. Folutlon: Each angle of a rectangle is a rlght angle.

o

,', mLons:90". Hrl DAB

:

nLDAC + nLCAB

,', 90: zLDAC+25 ... (Given : nLCAB:25") ,' , aLDAC: 90 - 25 :65".

The opposite sides of a rectangle are parallel. ,', $eg AB ll seg DC. AC is the transversal. LACD ..- (Alternate angles) .., /-CAB

=

m/ cAB:25'

,',

mLACD

:25"

:

65o;

...

(Given)

zLACD

:

25o.

J, The diagonals AC and BD of rectangle ABCD intersect each other ln the point K. If /(AK) :3.5 cm, then /(KC) : ?, /(AC) : ? 'l'ltc point K is the point of intersection of diagonals AC and BD. ,', /(AK): /(KC). ... (Given) /{AK;:3.5

cm

:

PQRS, l(PQ)

SR .'. /(PQ): /(SR)

(Given)

fiolution : The diagonals of a rectangle bisect each other.

(Given)

/(KC):3.5

(Textbook page 26)

a. 1. In rectangle

seg

t,

Ann mLolC

The diagonals of a square are congruent.

=

Ans. /(QR):9 cm;

(1)

(2)

:7 .-

,', /(SR):'7 cm

Ans. The diagonals of a square bisect each other.

.'. /(DM): /(FM) /(DM):7 cm ... (Given) .'. /(FM) :7 cm l(or;-/(DM)+/(MF) :(7 + 7) cm: L4 cm .'. /(DF):14 cm

QUADRILATERALS

:7

cm,

cm

"' : zl(KC):2 x 3.5 cm:7 cm. /{AC; Anr. /(KC) :3.5 cm; /(AC) :7 cm.

cm. f ind ,(QR) and t(SR).

Solution : The opposite sides of a rectangle are

4.

congruent.

rr,ctangle XYZW intersect each other ln the point M. If l(XZ):8 cm, then

.'. seg PS = seg QR .'. /(PS): /(QR) /(PS1:9 sm ... (Given) .'. /(QR):9 cm

'fhe diagonals XZ and WY of

llnd /fiM) and /(YM). llolution : The diagonals of a rectangle nre congruent and bisect each other.

M

NAVNEET MATTIEMATICS DIGEST : STANDARD

VIII

.'. /(xM) :I t6z)

:

:j :4

x

B

cm ... [Given : t(XZ):8 cm]

cm

t(XZ) ... (The diagonals of a rectangle.) t(XZ):8 cm ... (Given) .'. /(YW):8 cm /(YM) : j lgWl :1 * 8 cm :4 cm Ans. /(XM) :4 cm; /(YM):4 cm. /(YW)

mLL:mLM:mLN:90o.

nlp. What of quadrilateral is n LMNP ? Solution : The sum of the measures of the angles of a quadrilateral 360'. .'. mLL * mLM * mLN * mLp: 360o .'. 90 + 90 + 90 *mLP:360 ... (Substituting the given values) .'. 270 * mLP :360

Q. 5. In n LMNP,

Find

.'. mLP:360*270 .'. mLP:90"

Now, each angle of

tr LMNP is a right angle.

QUADRILATERALS

l,

ll'

nt/ Ql'S in the rhombus PQRS is 65o, find nLQRS.

Ittlttllorr : 'ilre opposite angles of a rhombus are congruent. / Ql,s : LQRS lrl Ql']S: lnIQRS

mi t;l's : 65' ... (Given) . , n/ QRS:65' Ann, m/ QRS:65o.

{, 'l'lrc diagonals

AC and BD of rhombus ABCD intersect each other

ln llrc prrint O. Find mLAOD and,mLBOC. Holutlon : The diagonals of a rhombus are perpendicular bisectors of F*tclt rttlter.

'l'lte point O is the point of intersection of diagonals AC and BD.

,', ttt/ AOD: nLBOC:90". : 90o; wLBOC:

Anx, rrrl.AOD

90o.

KING, mLK:70o and LI:110". Find the nr[Huros of the other angles of the rhombus KING.

l, ln the rhombus

.'. n LMNP is a rectangle.

llolulkrn : The opposite angles of a rhombus

!ns. mLP:90o; n LMNP is a rectangle.

HIT r'ongruent.

(3) RHOMBUS

:

'/N.rLKandLG=LI lll N:mLK:70o and ttt/

(Texfbook page 27)

a. 1. If

Anx.

ttt

the length of one side of a rhombus is 7.5 cm, find the lengths

the remaining sides. Solution : All the sides of a rhombus a.re congruent. The length of one side of a rhombus is given to be 7.5 cm.. Ans. The length of each of the remaining sides of the rhombus is 7.5

Q. 2. The diagonals segXZ and seg YW of rhombus XyZW in each other in the point P. If /(XP) :8 cmo find l(XZ). Solution : The diagonals of a rhombus bisect each other. The point P is the point of intersection of the diagonals. .'. P is the midpoint of diagonalXZ.

.'. t(xz):2t(xP)

:2x8cm

:16 cm Ans. /(XZ): L6 cm.

...

[Given :

/(XP):8 pml

C:mLI=I10" / N : 70' and mLG:

IIAIIAI,I,ELOGRAM

110'.

70"

I 109

: ii::l

(Textbook page 28)

l,

'l'hc diagonals LN and MT of parallelogram LMNT intersect each rtllttr in the point O. If /(MO):5 crrr /(LN):6 cmo find /(OT) and

,tNO). tiolulion : The diagonals of a paralleIoplr'irnr bisect each other.

'l'lrr' point O is the point of intersection ul' llrc diagonals of parallelogram

I MN'f.

'

/(or):

/(Mo)

/(Mo):5 cm ... /(oT):5 cm

(Given)

NAVNEET MATI{EMATICS DIGEST : STANDARD

Similarly,

/(NO) :

/(NO):ry

Ans. /(OT)

:5

The CifCle

... (L-o-N) ... [Given:l(LN):6cm] /(NO;:3

cm; /(NO)

:3

(Textbook paee 29)

l.

cm.

(4) The length of

l.

parallelogram are congruent.

.'. LS = tQ .'. mLS : mLQ .'. mLS:130o ... [Given : mLQ:130o] . mLP -l mLQ: 180 ... (The pair of interior angles are supple .'. mLP * 130:180 ... [Given : mLQ:130"] .'. mLP:180-130 .'. mLP:50" Now, rt lR : mLP ... (The opposite angles of a .'. mLR:50" ... lmLP:50"1 Lns. mLP = 50o1 mLR:S}o; mLS: 130o.

AB is twice that of seg OD.

Look at the adjoining figure. Write the

Look at the adjoining figure and write ilhether the following statements are true or

l.

hhc: Anr. (t)

Seg TS is not a chord.

(2) Seg KM is a chord. (3) Seg CK is a radius.

t2.te ru i: lg!19i"-ej:.':

Q. 3. The measures of the opposite angle of a parallelogram

-x)o

seg

chord and the diameter. Ans. The centre of the circle : point C; Radius : seg CD, seg CM, seg CL; ehord : seg RT and seg LM; Diameter : seg LM.

.'

(50

:

ngmes of the centre of the circle, the radius, the

The opposite sides of a parallelogram are parallel to each other. .'. side PS ll side QR. Side PQ is the transversal.

-2)'and

Fill in the blanks

the circle with centre O shown alongside, ( t) Seg OD is a radius. (2) Seg AB is a diameter. (3) Seg PQ is a chord.

"*.

Q. 2. In parallelogram PQRS, mLQ:130'. Find the measures of other angles of n PQRS. p Solution : The opposite angles of a

(3x

(Textbook pases 29 to 32)

/(OL)

/(No)+/(OL):/(LN) ..' 2/(NO):6 cm .'1

VIII

(False)

(True) (True)

.!TSi)

Thc distance between the centre of a circle and a chord :

respectively. tr'ind the measure of each angle

ln the figure, seg OM r chord AB. Thc length of perpendicular OM means the dlntnnce of centre O from the chord AB.

the parallelogram.

Solution : The opposite angles of a parallelogram are congruent. .'. they are of equal measures.

.'. (3x-2)":(50-x)' .'. 3x-2:50-x .'. 3x*x:50+2

3x-2:3 x 13 -2:39-2:37. .' . (3x - 2)o :37o.

I

Now, the adjacent angles of a parallelogram are supplementary an Let the supplementary angle of 37" be y".

Theny*37:180 .'. y-180-37

.'.

y:I43

.'.

)o:1

Ans. The measures of the angles of the parallelogram : 37", 143", 143".

|

l'he property of the perpendicular drawn from the centre of a circle to a chord : The perpendicular drawn from the centre of a ('lrcle to its chord bisects the chord. ln the figure, seg PT r chord LM. ,', /(LT): /(TM)'

NAVNEET MATIIEMATICS DIGEST : STANDARD

VIII

of the distance between the centre and congruent chords of the circle : In a circle, congruent chords are equidistant from

(2) The property

the centre of the circle. In the figure, chord AB = chord PQ. Seg OM r chord AB and seg ON r chord PQ.

.'. /(OM):

/(ON).

(3) The property of the angles made at the centre of the circle by congruent chords of that circle : Congruent chords of the same circle form congruent angles at the centre of the circle. In the figure, chord AB = chord CD. These chords subtend /AOB and ICOD respectively at the centre O. ... lAOB = LCOD, (Textbook page 32)

Q. 1. In

a circle, chord MN = chord RT. Chord RT is at a 6 cm from the centre. Find the distance of the chord MN from centre.

Solution : The distance of chord RT from the centre of the circle is 6 The congruent chords of a circle are equidistant from the centre of circle. Chord MN = chord RT ... (Given) .'. the distance of chord MN from the centre is 6 cm. Ans. Chord MN is at a distance of 6 cm from the centre.

Q. 2. In the figure, seg OM r chord AB, /(AM) : 1.5 cm. Find the lengths of seg BM and seg AB.

l. tn a circle with centre P, chord AB = il/ APB :40o. Find measure of ICPD. lolutlnn t zLAPB:40o

chord CD

.'. l(BM):1.5

cm

AB ry chord

,', /(tPD=IAPB.

m/ CPD: ruLAPB:40" ... [Given : .rnLAPB:40"] Anr. The measure of /-CPD is 40o.

i,

'l'he radius of a circle is 5 cm. The distance of a chord from the Cintrt ls 4 cm. Find the length of the chord. Folutkln : Let P be the centre of the circle of

l,

tndlun .5 cm. The tlistance of the chord AB from the centre is

4 etn, i.e., /(PM)

- 4 "*.

is a right-angled triangle.

APAM ,', hy Pythagoras' theorem, 2 [,(AM) l'z + t/(PM)] : I\(PA)12 ... (Substituting the values) ,', [/(AM)]2 [email protected])2:(5)2 .'. [/(AM)]z - 9 ,', l/(AM)12 :25 -16

,', l(AM):3

cm.

The porpendicular from the centre to its chord bisects the chord.

,', /(AB):21(AM) :2 x3 cm:6

cm

Anr. 'fhe length of the required chord is 6 cm. 'l'hc radius of a circle is L3 cm. The length of a chord is 10 cm. Find Che dlstance of the chord from the centre. lolutkrn : Let P be the centre of the circle of rnrlius t3 cm and chord AB of length 10 cm' $og PM r chord AB. Tlte perpendicular from the centre of a circle to

l,

x

Ans. /(BM)

1.5

:

cm:3

cm.

1.5 cm; /(AB)

,, /(AM): j llnn; ,', /(AM) =|xl0 cm:5 cm.

:3

t-

10

cm --'t

Itt righrangled APAM, by Pythagoras' theorem, l/(AM)12 +

,', ,',

/(AB) :2/(AM)

and

CD. at the centre. angles congruent form of circle chords ?hF congruent and chord

llx clrord bisects the chord.

Solution : The perpendicular from the centre of a circle to its chord bisects the chord. Seg OM r chord AB. .'. /(AM): /(BM) /(AM): 1.5 cm ... (Given)

:2

THE CIRCLE

(5)2

t/eM)1'z: t/eA)l'

+ t/eM)12

:Qr2

:144

: \844:

t2 cm. 25 .'. l(PM) l/(PM)1'? :169 12 cm. is the centre from chord of the Anr. The distance

-

cm. Mathematics Digest : Std.

Vm lE059il

NAVNEET MATHEMATICS DIGEST : STANDARD VIII

(Textbook pages 33 to 46)

r0u

e. 6. A chord of a circle

is 30 cm long. Its distance from the Find the radius of the circle. sm. Solution : Let P be the centre of a circle. Chord AB is of length 30 cm. Seg PM r chord AB. /(PM):8 cm.

The perpendicular from the centre of 1s

its chord

('nleulute the area of the rectangle given its length and breadth lll t.2 cm, 2.5 cm (2) 2.1m, 1.5 m (3) 3.5 m, L.2 m.

l,

a circle

bisects the chord.

... /(AM):jl1an)

llt t-5,2 cm, b:25 cm 'f

:13

fght-angled APAM, by Pythagoras' theorem,

tt(?A)L'z: U(AM)12 + U(PM)1'z : (15)2 + (8)2 :225 * 64:289

... /(PA).,: JrSr:17

46.

Ixb :5.2 x

'lte nrca of a rectangle:

:jx30cm:15cm.

1n

:

2.5

... ...

(Formula) (Substituting the values)

sq cm.

Anr. 'l'he area of the rectangle:13 sq cm. 2,1 m, b:1.5 m 'f'lte urca of a rectangle: I x

{t) , -

cm

b

:2.1x I.5

The radius of the circle is 17 cm.

:3.15

:3.15

,: .1.5 m, b:1.2 m 'Ihe area of a rectangle: I x b :35

x

:4.2

sq m.

Ann. The area of the rectangle

(Formula) (Substituting the values)

sq m.

Ann. The area of the rectangle

(i)

... ...

1.2

sq m.

... ...

(Formula) (Substituting the values)

:4.2 sqm.

l,

'l'hc slde of a square is given. Calculate its area. (l) 25 cm (2) 2.8 m (3) 7.2 cm (4) 13'5 m.

(lf Sitlc:25

cm 't'lu: area of a square

: (side)2 :(25)2 :625

... ...

(Formula) (Substituting the value)

sq cm.

Ans. The area of the square:625 sq cm.

lJt Sitlc:2.8

m

'l'hc area of a square

: :

(side)2 (2.8)2

:7.84

... ...

(Formula) (Substituting the value)

sq m.

Ans. The area of the square :7.84 sq m.

NAVNEET MATHEMATICS DIGEST : STANDARD

(3) Side : 7.2 cm The area of a square: (side)2

:

... ...

(7.2)2

AREA

VIII

(Formula) (Substituting the value

'l'hc area of parallelograln

51.84 sq cm.

Ans. The area of the square:51.84 sq cm.

(4) Side: 13.5 m The area of a square: (side)2

:

:

3. The area of a parallelogram is 56 sq cm and its height is 7 cm. What is its base ? llolution : The area of a parallelogrilm: 56 sq cm, height: 7 cm, base: ?

... ...

(13.5)2

(Formula)

,'.

56

.'

56 "-r 7

:

: base x 7 : base ,'. base :

base x

height ... (Formula) ... (Substituting the values)

8 cm

Ans. The base:8 cm.

(Substituting the value

4. What is the area of a parallelogram having base

192.25 sq m.

I

Ans. The area of the square:182.25 sq m.

L3 cm and height

cm?

ilolution: Base:13 cm, height:5 cm, area:? 2. The formula for the area of a parallelogram The area of a parallelogram base x height In the figure, n ABCD is a parallelogram. Seg BC is the base and seg AE is the corresponding height.

The area of parallelogram

:

:65

:

.'.

A(

I

ABCD)

:

Am.

The area of a parallelogram is 390 sq cm. If its height is 26 cmo what is its base ? Folutlon: The area of a parallelogram:39O sg cm, height:26 cm, hurc

(Textbook page 34)

sq cm.

Ans. The area of parallelogram:84 sq cm.

is its height ? Solution : The area

height:

is 26 sq cm.

:

?

The area of a parallelogram

0. 1. What is the area of a parallelogram whose base is 12 cm height 7 cm? Solution: Base:12 cm, height:7 cm, area:? The area of parallelogram : base x height ... (Formula) :12 x7 ... (Substituting the val

Q. 2. The area of a parallelogram

sq cm.

The area of parallelogram:65 sq cm.

!,

i

/(BC) x /(AE).

:84

: base x height .. ' (Formula) :13x5 ... (Substituting the values)

,', .190:

its base is 6.5 cm,

of a parallelogram :26 sq cm, base :

base x

base x 26

390

: ;;:base

"'

base:

height ... (Formula) ... (Substituting the values)

15 cm

Attx. The base: 15 cm.

Irytgl:fe-E3lggj 'f'hr, rrrca

If

:

of a triangle

:j

x base x height

'fhu l,.ou of u tight'utg fhe rrrea of a right-angled triangle

: ] x the product of the lengths of the sides fonning the right angle.

6.5

?

The area of parallelograrn: base x height .'. 26:6.5 x height

)6 height .'. height :4 cm . . -: 6.) Ans. The height:4 cm.

... (Formula) ... (Substituting

(Textbook page 37)

the va

l,

A t:crtain triangular plot has a base 20 rn and a height 30 m. What ltn trea ?

h Itrlullrrn :

:20

m, height:30 m, area:? 'l'lte ut'ca of a triangle : ] x base x height ... (Formula) Base

NAVNEET MATIMMATICS DIGEST : STANDARD

:\x20x30 :300

...

VIII

AREA

(Substituting the

Aros of an equilateral triangle

:

sq m

Ans. The area of the triangular plot:300 sq m.

'l'|rc area

of an equilateral

triangle:f tria")'

Q. 2. What is the area of a triangle whose base lg.2 cm and 7.5 cm? Solution : Base : 78.2 cm, height :7.5 cm, arca: ? The area of a triangle : j x base x height ... (Formula)

:lxIB.2x7.5

Ans. The

:68.25 sq cm area of the triangle :68.25

...

(Textbook page 39)

l. If the side of an equilateral

(Substituting the val

triangle is 12 cm, what is its area? :12 cm, area: ?

Solution : The side of an equilateral triangle 't'hc area of an equilateral

tr

triangle:f triO"l2 ... (Formula)

sq cm.

E

v _ 12xlZ _ ^/-1 + ,.. (Substituting

Q. 3. The sides of a right-angled triangre forming the right angle 16 cm and 8 cm. What is its area ? solution : The sides forming the right angle are 16 cm and g cm. The area of a right-angled triangle : j x the product of the lengths

:jx16x8 :64

...

sides forming the right angle. (Substituting the values)

:64

:36rfi Ans. The area of the equilateral

height:

sq cm.

125 sq cffi,

base

has an area

rr.6

sq cm.

:225t/3

'l'he urea

rf its height is 2.9 cm, what

Solution : The area of the triangle : 11.6 cm, height : 2.9 cm, base The area of a triangle : j x base x height ... (Formula)

.'. 11.6: I x base x 2.9 ... (Substituting 11.6 x 2 : ZS : base .'. base: 8 cm.

_...

Aroa of a rhombus

base ?

Ans. The base of the triangle:8 cm.

sq cm.

-4 :{(side)2 ... (Formula) t; :V,'"30x30

Ann. The area of the equilateral

triangle:j x base x height

.'. 125:|x25 xheight 125 x2 :. 25 : height .'. height: 10 cm Ans. The height of the triangle : 1.0 cm. Q. 5. A triangle

triangle:36\F

Thc area of an equilateral triangle

:25

?

The area of a

sqcm

4

of the triangle :

the value)

What will be the area of an equilateral triangle of side 3Q cm? Folution : The side of an equilateral triangle:30 cm, area:?

Q. 4. A certain triangle has an area 125 sq cm. rf its base is 25 cm, is its height ? Solution : The area

Lz

^

t.

sq cm

Ans. The area of the given right-angled triangle

A

the v

(Substituting the value)

sq cm

triangle:225\fr

sq cm.

:

of a rhombur

:1 r product of the lengths of the diagonals.

l.

'fhe diagonals of a rhombus are 84 cm and 42 cmking. What is the lren of the rhombus ?

tlolution : The length of the diagonals are 84 cm and 42 crn, are?: ? :1 product of the lengths of the diagonals " ... (Formula) : ! x 84 x 42 ... (Substituting the values)

'l'lre urea of a rhombur

:

1764 sq cm

Anr. The area of the rhombus

:1764

sq cm.

NAVNEET MATHEMATICS DIGEST : STANDARD

Vm

a. 2. The area of a rhombus is 1280 sq cm. If

one of its 64 cm long, what is the length of the other diagonal ? Solution : The area of the rhombus : l28O sq cm, the length diagonal : 64 cm, the length of the other diagonal : x (Say) The area of a rhombur:

...

1280

"

1280

:I

**

product of the lengths of the di (

...

x 64 x x

x2 64 :x

(Substituting the v

"'x-40cm

Ans. The length of the other diagonal:40 cm.

Q. 3. The lengths of the diagonals of a rhombus are

12 cm and 18 What is the area of the rhombus ? Solution : The diagonals are of lengths 12 cm and 18 crn, are?: The area of a rhombu, : j t product of the lengths of the di

:!x12x18 :108

:

432 sq cm. If one of the diagonals length of 24 cm, find the length of the other diagonal. Solution : The area of the rhombus :432 sq cm, the length of cm, the length of the other diagonal

The area of a rhombur

:1

...

(Substituting the v

one of its diagonals is

long, what is the length of the other diagonal ? Solution: The area of the rhombus:702 sg cm, the length of

.'. 702:I

cm, the length of the other diagonal

x 54 x x

:26 cm.

:

rcg l'M (or seg RT) is the height a ttf lltc ltrrpczium. 'l'he nrcu ol' a trapezium : 1 x the sum of the lengths of the parallel sides x height 'flte nrcu ol trapezium PQRS : j x t/@S) + /(QR)l x l(PM). ..' tl(PM): l(RT)l tl

x

(Textbook paee 42)

Eolullorr : The lengths of the parallel sides of the trapezium are ltt:H,7 ctn and bz:5.3 cm. h:4.5 cm, zrl?:?

lltr. rrttl of a lrapezium tlro sum of the lengths of theparallel sides

. (1t.7+5.3)x4.5 " 14 x 4.5:3I.5 sq cm

...

x height ... (Formula)

(Substituting the values)

Altr. 'l'lrc area of the trapezium:31.5 sq cm.

l,

Q. 5. The area of a rhombus is 702 sq cm. If

:i

ttl'n trapezium

"

Ans. The length of the other diagonal:36 cm.

The area of a rhombur

'lre lcngths of the other diagonal

Itt llrc ligure. n PQRS is a trapezium !n wlrich side PS ll side QR and reg l'M t side QR. seg RT r line PS. f lltfyl1 lor /(RT)] is the distance between llte lrlrullcl sides of the trapezium.

:

(

:54

At-en

x (Say) x product of the lengths of the di

.'. 432:| x 24 x x :36 cm x-*432x2 24 "'

diagonal

I'

'f

e, l, 'l'ltr lcngths of the parallel sides of a trapezium are 8.7 cm and l,J em. lf the perpendicular distance between them is 4.5 cm' what is llr urert'!

L08 sq cm.

Q. 4. The area of a rhombus is diagonal :24

Ailr,

(Substituting the

sq cm

Ans. The area of the rhombus

702x2 54 -:26cm

t

(

...

4t

AREA

*

:

.

-r (Say)

product of the lengths of the di

...

'f'hc area of a trapezium is 262.5 sq cm and the perpendicular rlirllnct between its parallel sides is 15 cm. What is the sum of its pnnrllel sides ? t*rlullon : A (trapezium):262.5 sq cm, h:15 cm. A { tr rrlrczium I : I the sum of the lengths of parallel sides x height " ... (Formula) :12 x the sum of the lengths of parallel sides x 15 162.-5

(Substituting the v

'

rltc sum of the lengths of parallel

(Substituting the values)

sides:'e#:35

Atn. 'l'hc sum of the lengths of parallel

sides

:35

cm.

cm

NAVNEET MATIIEMATICS DIGEST : STANDARD

VIII

0. 3. If

the area of a trapezium with parallel sides of lengths 30 cm 23 cm respectively is 265 sq cm, what is its height ? Solution : A (trapezium) : 265 sq cm, the lengths of parallel sides 30 and 23 cm, height: ? A (trapeziuml :1 x the sum of the lengths of parallel sides x

265:iG0+23)xheight

height

'#:

...

... height:

I'l['

triangle:

rcnrilrcrirneter (s) of the

'l'he nrerr ttl' the triangle

: :

+39 +

22

120 56: _:60

cm

60(60

- 2s)(60 - 39X60 - s6)

60x35x21x4

12x5 x5 x7 x7 x3 x4

:^/12x12x5x5x7x7 :15x5x7:42Asqcm Anr. 'l'hc

Q. 4. The area of a trapezium is 84.5 sq cm and its height is 6.5 one of its parallel sides is 15.2 cm long, what is the lengtli of the

area

of the triangle :420 sq cm.

l,

side ?

Solution : A (trapezium) :84.5 sq cm, height:6.5 cm, length of one the parallel sides: I5.2 cm,length of the other parallel side:,r A (trapezium):1* the sum of the lengths of parallel sides x hei

llow nuch will it cost to have a triangular field weeded at the rate of lk 2 por square metre, if the length of the field's sides are 11 m,

0l

ltr und 60 m ? Hglullrln : Here, a:

ttts

tl

m,

b:61 m and c :60

renriperimeter (s) of the triangular

84.5:IxQ5.Z*x)x6.5

:L5.2+x .'.26:I5.2+x .':15.2-tx:26 6.5 .'. x:26-I5.2 .'. .x:10.8

field:

a

m.

* b* c 2

:uu: 1l +612+60: 132 z

84.5 x 2

'l'lr$ rrlcrl o1'the triangular

Ans. The length of the other parallel side: 10.8 cm.

field: :

s(s-a)(s-bXs-c) 66(66

-

1

1X66

/66x55x5x6

-

61)(66

-

60)

:"/6x11x11x5x5x6 :6 x 11 x 5:330 sq m

:

c.

l'lrr, t'rst

.l'

weeding

:

f:"riffi : Rs 660

Aru. 'l'ho expenditure is Rs 660.

a*b*c 2

A(AABC)

2

s(s-aXs -b)(s-c)

(Substituting the

Ans. The height of the trapezium: L0 cm.

In AABC, /(BC) : a. I(AC): b, I(AB): The perimeter of AABC : a * b I c The semiperimeter (s) of AABC

a*b+c 25

1o cm

8. Another formula for the area of a triangle

43

AREA

:./s(s -

a)(s

-

bXs

-

This formula for finding the area of a triangle, when the lengths sides are given, is known as Heron's formula.

ll'the lengths of the sides of a triangular plot of land are 20 m,2l nt ntul 13 m, what is its area? Flolttllort : Here, a :20 m, b :2I m, c: 13 m

t.

c)

of

f

d a. 1. If the sides of a triangle

'lrc scruiperimeter (s) of the

a*b*c triangle: z

_20+21 +

(Textbook pages 45 & 46)

22

are 25 cm, 39 cm and 56 cmo what is

area of this triangle? Solution : Here, a :25 cm, b :39 cm and c : 56 cm.

l'lrc ru't:t of the triangular

field:

13

:54 :27

^

s(s-aXs-b)(s-c) 27(27 -20)(27 -21)(27

-13)

NAVNEET MATIIEMATICS DIGEST : STANDARD VItr

A(APQS)

9x3x7x3x2x7x2 9x9x7x7x2x2

plot:

b

, ../s(s = axs

-

b)(s

-

c)

A(APQS) + (ASQR) (180 + 390) se

The area of the quadrilateral

"(i)"(t) " (t)

plot:570

sq m.

sq m.

rcmiPerimeter (s) of AGHF

e+b

?4+so+sz 136 ,T* r:68m

{AGHF) :"u/s(s -a)(s -bXs-c)

x!- *P- rP2222

3 xP-

68(68

:,fi *E*t:{r, Ans. The area of the equilateral triangle

68x34x18x16 2x34x34x 18 x 16 34x34x36x16

:fO'.

is the map of a plot of land. Using the gi

area.

- 34X68 - s0)(68 - s2)

t ttilru

aomiperimeter (s),

of aEHF

tr

A(AEHF): 66,''

Solution : The area of the quadrilateral plot PQRS : zraa of right-angled -APQS * area of ASQR. In right-angled APQS, PQ:9 m and PS :40 m.

:'lY r20 :- 20+52+48 ---bu 11

a,

!

tn:570

;if +, '

measures, find its

13 rn.

G Urlng the measures given in the {ilrcr ffnd the area of n EFGH. lOlutlon : Area of ! EFGH rA(AGHF) + A(AEffi') L AGnr, l(GH) :a:34 ttt, : {EF) = b:50 m, /(HF) s/- J/ 1aY

-fu

a. 5. In the figure,F PQnS

: :

Tbr area of plot PQRS

lU,

_3p

:

ST:

:)x6Ox13:390sqm

whose side is p. Solution : Here, a: p, b: p, c : p The semiperimeter (s) of equilateral triangle

The area of the equilateral triangle

ASQR, base QR:60 rn, height

A(ASQR):jbasexheight

126 sq m.

Q. 4. Use Heron's formula to find the area of an equilateral

_a*b*c _p*p*p 222

:]x/(PQ)xl(PS)

:tx9x40:180sqm

:9 x7 x2:126 sqm Ans. The area of the triangular

: ! x theproduct of the lengths of the sides forming the right angle

s(s-aXs-bXs-c) 60(60

-

60 x 40

20X60

-

s2X60

x8 x12

5x12x40x8x12

-

48)

m

NAVNEET MATHEMATICS DIGEST : STANDARD VIII

:40 x 12:480 tl

: :

ll

sq m

A(AGHF) + A(AEHF) (816 + 480) Se m: 1296 sq m Ans. The area of tr EFGH: 1296 sq m. A(

EFGH)

(2) By Pythagoras' theorem, t/(Aql'z: t/(Bc)12 + tl(AB)l'

*

:

.'. /(AC): l3 cm (3) In AABC, /(BC) : a:5 cm, /(AC) : The semiperimeter (s) of AABC

b

A-^*--T*-*----*

H(l rirl crn

t erlret'livcly. 'l'lro rrrcu ol' a rectangle: I x b :5 x 4:20 sq cm 'l'lrt. nrcu ol'the coloured portion: A(AABC) A(rectangle)

: Anr, 'l'lrc

area

i lAll('l)

is a parallelogram.

llnee All :20 169

:

13 cm, /(AB)

: c :12

:*#

cm, height EF

* l(l x 15:300 sq cm. Itr AliAB, base AB :20 Itr.lgltt IiF : ,r,r,,,

'rca

cm

A(AABC), by Heron's formula

:^/I5x10x2x3

:

.,600

:30

-

20) sq cm:60 sq cm

:

15 cm.

base x height

A

cm,

: i :T ; Tgjl' : :

sq cm

A( n ABCD) (300

-

-

A(aEAB)

150) Se cm: 150 sq cm

-

dl lrr tlrc given figure, the shaded region isT A(JNli.

(3) 30 sq cm.

Q. 8. In

each of the figures given, find the area of the coloured

'l'lrc base NE of this triangle is 18 cm. 'l'hc height of this triangle : the breadth of the

Solution

:

rrr'lirrtgle.

(1) Each side of the given triangle is 8 cm.

Ir:

.'.

the given triangle is an equilateral triangle. The area of an equilateral

triangle

e

S

:4,rro"r,:.frx8x8 4',---''

\ta

I

4

a

sq cm

Ans. The area of the coloured portion

F

Arrr. 'l'he area of the shaded region: 150 sq cm.

sq cm

Ans. (1) 30 sq cm (2) 13 cm

:I6.,fr

20cm

15 cm.

of AEAB

s(s-a)(s-bXs-c)

-

:

'l'lre lrr:a of the shadei r"gion 13X15

-

(80

of the coloured portion:60 sq cm.

'l'lte uret of a parallelogram

:tl#!:+:r5 1s(l5 -s)(15

cm,

'l'lre length and the breadth of the rectangle are 5 cm and 4

Jl

144:

AC:20

-1,:o"s

: j x the product of the t B 5 c* angle

+ (0D2 :25

lll'

huse

.- 8 cm.

=_]ltnscxhcight

|

(5)2

nltlt',

lrclglrt

The area of a righ-angled triangle lengths of the sides forming a right

:

lrr

'l'lte nrcrr ol' a triangle

Q. 7. (1) From the measures given in the figure, find the + area of AABC. l\ (2) Usingfothagoras'theorem,findthelengthofsideAC. I \ (3) Find the area of AABC using Heron,s formula. _ I \ Solution : (1) In AABC, mLB:90". I- I \ .'. AABC is a right-angled triangle

:lx5x12:30sqcm

47

AREA

:rcrQ

sq cm.

r

o cl

N

12 cm.

rtc rtrea or A QNE

: i :T : H-j1., sq cm

Ans. The area of the shaded region:108 sq cm. 8cm

18cm

I'III{

The Circumference and Area of a Circle (Textbook pages 47 to

1. Circurnference of a circle

ktfutltrn : I ltorrnula : c :2nrl

lll

:

The relation between circumference and diameter. The ratio of the circumference of a circle to its diameter is

circumference

-n diameter If the circumference, diameter and radius of a circle are denoted by letters c, d and r respectively, then we have,

"'

:2x22x8:352cm

lll

ffi

.'. , - n x2r:2nr

d:6.3 m Circumference (c):

r.-7.7

o'5:

:

352 cm.

cm

: znr :2 xLl x73 :2 x22 x 1.1 :48.4 cm ('ircumference : 48.4 cm. Aru. I

lore, r

-

2.8 m

: znr :2 x! x2.8 :2 x 22 x 0.4: 17.6 m

Atu. ('ircumference =17.6 m. Ftonr the given circumference, find radius and diameter of the

lll

d:35

x

Hete,

Clrele

Here, cm Circumferen ce (c): vf,

l;'*:22

('ircumf'erence

C'lrt'rttrrlbrcnce

l,

(1) 3.5 cm (2) 6.3 m (3) 0.14 m. Solution: [Formula : c:nd.f

Ans. circumfe.J

lrt

diameters

below:

Ant.

Cln'ttrttl'cren ce

(rextbook page 49)

Q. 1. Find the circumference of the circles from their

(l)

(ll

c:nd

Now, diameter:2 x radius. i.e., d:2r [The value of n is raken to be ! or 3.I4]

:Znr :2x21-x56

a

This constant number is denoted by the Greek letter n (pie).

A:n

tlcre,r-56cm C'lrctttrtl'e rcrr ce

number.

c

CIRCUMFERENCE AND AREA OF A CIRCLE

:

(2)

l9tl cm

616

cm

(3) 72.6 m.

Eolutkrn': llore. r.': 198 cm C'll'curnt'erence: nd

lll 11 cm

(2) Here,

: t' t63 22-:

nj

:? * 6.3:22 x 0.9: 19.8 m t.r

: v4 :?, O.l4:22 x 0.02:0.44 m Ans. Circumference : 0.44 m. Circumference (c)

Q. 2. Find the circumference of the circres from their radii given (1) 56 cm (2) 7.7 cm (3) 2.8 m.

198

x7 :9x7:63cm

22

31.5 cm

Ans. Diameter:63 cml radius :31.5 cm.

Ans. Circumference: l9.g m. (3) Here, d:0.14 m

d:

1e6:?| x d

I llcrc, c :616

cm

('ilcrrmference: nd

," 616:?"d

, tl 196 :98 '' ;: ,

.'. d:-

616

_

cm

Ans. Diameler:196 cml radius I l) llcre, c

xj :28x7:196cm

:72.6 m

('ircumference: nd

:98

cm.

NAVNEET MATIIEMATICS DIGEST : STANDARD

d:

... 72.6:2; x d

,:t:d

23.1

2

72.6 x 7

n

,I,IItI: (,IR('TIMI,'I':RENCE AND AREA OF

VIII

:

23.1

m; radius

:

_

51

__r--

l r:x Wrlle llte propcr values in the blanks in the following table

l,

:11.55m

Ans. Diameter

l

:3.3x7:23]m

A CIRCLE

:

Rudlus

lrl

11.55 m.

Q. 4. What is the cost of fencing a circular place of radius 7.7 m, three rounds of wire, if the cost of the wire is Rs 50 per metre solution : To find the length of wire fencing around the circular p we have to find the circumference of the circular place. The radius of the circular place :7.'7 m Circumferen ce

:Znr

Hnlttlkrrr

lltlllrc. t 42cm. cl-)t' 2x42cm:84cm r - 2nr = 2 x + x 42 :2 x 22 x 6 :264 cm Art rr nr' :T x 42 x 42:22 x 6 x 42:5544 sq cm

:2xTx7.j :2x22 x l.l:48.4m

The length of the wire required for three rounds of fencing : 3 x circumference

:3

x

48.4

m:145.2

Arn. l)iameter:84 cml circumference: 264 cm; area:5544 sq cm.

m

The cost is Rs 50 per metre .'. the cost of wtre 145.2 m in length:Rs 50 Ans. The cost of wire : Rs 7260.

x 145.2:Rs

r/: 9.8 m rl 9.8 , : :-:4.9 m 22 9.8 :22 x I.4 :30.8 m t, - nd :?, :T x 4.9 x 4.9:22 x 0.7 x 4.9:75.46 sqm Arcrr - nr'

l,ll I lclc, 7260

Q. 5. The bus has wheels of diameter 0.7 m. How many times must wheel of the bus rotate for covering the distance of 22 km two towns ? solution : The distance covered by the wheel of the bus in 1 rotat : circumference of the wheel : Ttd:! x 0.7 :2.2 m The distance between the two

towns: Z2km:22 x

1000

Atu.

tlt

wheel:

:

distance circumference

22000

ZZ

f

:10000

Area of a circle

:

lf r is the radius of a circle, then area of a circle

:4.9 m; circumference:30.8 m; area = 75.46 sq m.

c:44m t' 2nr .'.44:2x'lxr llerc.

.l

)

lr:re, area:616 sq cm /\rea: nr2 .' . fir2 : 616 f

- 616 x7 : 196 ' rz: 22 d :2r:2 x 14 cm:28

Ans. The wheel will rotate 10,000 times.

2.

Radius

:

nr2.

44 x7 ' r:-:7

2 x22

m

il'2r:2x7:14m Arca: nr' :T x7 x7 :22 x7 : 154 sq m Ans. Radius :7 mi diameter:14 m; area:154 sq m.

m:

To cover a distance of 22000 m, the rotations by the

:

.'. 2| x 12 :616

r:14

cm

cm

c:nd:Tr28:22x4:88cm Ans. Radius : 14 cm; diameter:28

cm; circumference = 88 cm.

n) (6) ao:

(7) a-.

Q. 3. Of which numbers are the foltowing numbers cubes? (t) 40e6 (2) 4et3 (3) ss32 (4) _21n. Solutions

:4x4x256 :4x4x4x64 :4x4x4x4x16

:4x4x4x4x4x4 . 4096 43 x 43 :[email protected][s : __

[email protected]

Q. 2. Simplify: (l) 5a x 53

6) 3-4

:

(1) 4096:4x1024

..

-

E

I

Q)

(2) 78 -73

('

47o

(:). " (?)"

G)

(;)"

(4) (rr x 4)2 (5)

",

Ans.

(1)

54

x

53:

54+3

-

(;)'

oo)

[(;)']'

I -e-n--3o

57

16a

Ans. The number 4096 is the cube of the number 16.

(2\ 4913:I7 x l7 x

(3)

17

:4 x 1458 :4 x2 x729 :2x2x2x9x8l :2x2x2x9x9x9 :. stzz --z-t* gr : 12 9t' :

5832

><

l. Fractional Index : 49J means square root of

lg.

Ans. The number 5832 is the cube of the number

1g.

Let us first find the cube root of 2197,

:13 x 13 x 13

_2197:(_13)3 .'.2t97:t33 Ans. The number (- 2lg7) is the cube of the number ( l3). -

49.

Sqyare root of 49 is written as _t

(4) -2197. 2197

:(;I'':(;)''

(s)

643 means cube root of

t

[email protected] or 491.

64. 1

Cube root of 64 is written

u"

{U

or

641.

ln general, the nth root of a number a is shown by the index 1 n

(23)s

NAVI{EET MATIIEMATICS DIGEST : STANDARD VIII

Ans.

(Textbook page 99)

111*1

Q. 1. Write the meaning of the followins -o index numbers : 1 "t 1 I 1 1 1 (1) 362 @) n G) nt G) 20i 6) 2s64 (6) ri (D t0;

(1)

184

x 18s: 18a

:

Ans.

5+4

18 20

1

1

(5)xt Lxto-x7

1_1 s-3 :73 s :7 Is 2

9

:1820

r11 (6) (xY)t -*i

1_ I to

to-1

-

nt5

_1 11 (4) (9 x 5)o :90 , 56

11 t 11 (3) (4t)n :1412" s - 4rc

(1) The meaning of 36t : Square rootof36. -l (2) The meaning of 271 : Cube root of 27.

!

!

(2) 73 -75

5

*Yi

3

_ - 70 :y7O -L

11

t t

Q. -t 2. Evaluate:

(t)

42 . I1"

125' (6) 10003 Q) 22st Ans.

I

Q)

la -1

(S) 5123

-,rt:2

(a) 15r-

(t

(6) loooi = (7) zWi (s) 51?5:

-L

11L

(15) (13 -15)7 :137 +L57.

,fiooo: lo

:1B:15

if :1

(9)

100t: .u/i6: t0

3. Rules of indiees -.-----^_--_-

-1

(4)

Q) qt (2) t25i = {6:5 (3) l?:.16:1

1111 (9) p2 x p;:pt+t:p 11L (11) (7 x 9)s: 7E x9E 1111 (13) xE i x-l:xE-i 3 11*8 _- 88 _-88

_1 tt 11 (5) 1002 (9) @2 (tu) 646

_1

(3)

{fr,:g

&ir:

(10) 646

rffi:t : ffi:2

:

You have studied the rules of indices in Std. VII. (Textbook page

Q. Simplify: 11rl (1) 18a x lgs (6)

(tt)

I

(rv)'

_7s e) ' 't 7i t

gi;i

(zy 11 (7 x 9); (12)

(3)

_1 _t

(r42)e

-1 -l (E) (ms1t

l1

(mny es) J8 : r11

lffi) l1

(4) (9 x 5)6 (5) x7 -i 11

(9)

pi x pl

I

(Io) mi x

1r (14) x2

+r2

(15) (13+15)

11

L

0 glrz:rt" z:rE

(8)

11

L

@tf :ma"i:mt lllt2

(10)

*1 * *1 :*1*

(t2)

(mn)n

(14)

1 I 11 x2+x2:.xz 2-xo:l

_t_ll

:rnv

3

:mj

x nP

Construction of Quadrilaterals

CONSTRUCTION OF QUADRILATERALS

(Textbook pages 101 to

I

To construct a quadrilateral, the measures of five essential

should be given.

Construction

I

:

To construct a quadrilateral, given the lengths of four sides and diagonal.

IDEFG in which ,(DE):3.5 cm, /(EF)=Z.S t(FG) : 1.7 cm, r(DG) =2.5 cm, ,(DF,):3.5 cm. [Analysis : Suppose n DEFG is the required D 2.5 cm quadrilateral. Here, the lengths of the three sides of ADEF are known. So, by the SSS construction E method, we can construct ADEF. The lengths of :

Q; Construct

the side DG and FG are known. So, we can locate

2. One side and one diagonal of

point G.l Steps

of Construction

:

.

(1) Draw a rough figure of

!

2.5 cm Rough figure

Ans.

D

DEFG show_

(u

ing the given measurements. (2) Draw seg EF of length 2.5 cm.

(3) Complete ADEF using the given measures of side ED and diagonal DF. (4) Now,' complete ADFG using. the meaurements of sides DG and FG.

s

q'/

E

Draw quadrilaterals of the following measures : Q. 1. In n DEFG, /(DE) :5 cm, /(EF) :6 cm, r(FG) cm, /(DF) :9 cm.

Rough figure 108

:7 cm, /(DG) :

a rhombus ABCD are 4.5 cm and 5.5

cm respectively. Explanation : n ABCD is a rhombus. Therefore, all its four sides are congruent.

.'. l(AB): /(BC): l(CD): /(DA):4.5 AC:5.5 cm.

Diagonal

P

Ans.

cm.

110

NAVNEET MATIIEMA,TICS DIGEST : STANDARD

a. 3. ln parallelogram ,(PQ) :3.5

vItr

CONSTRUCTION'OX' QUADRILATEMI"S

PQRS,

cm, ,(QR)

:4.5

(Textbook page 105)

cm, ,(PR)

:6.5

cm. Expl4nation : The opposite sides of a parallelogram are congruent. .'. /(PO: /GS):3.5 cm and /(QR): /(PS):4.5 cm. /(PR):6.5 cm

Draw quadrilaterals of the following measures : 1. In n DEFG, t(EF) :3'5 cm, t(DE) :5'5 cm, l(DF) ,(DG)

:5

cm, [email protected])

:8

:

z's

cm, t(LD)

:5

cm.

Ans.

D

/"F

,U)

5.5 cm Rough figure

4.5 cm Rough figure

' 4.5 cm [Note : ParallelogramPQRs can a]so be constructed by constructing A and APSR with common base seg pR.l Construction 2

:

To construct a quadrilateral, given the lengths of three sides and diagonals.

Q.

Construct

,(AC)

:4

!

ABCD suct that /(AB)

cm,, t(BD) :4.5

Steps of Construction

:

:3

cr& l(BC)

:

Z.S cm,

: 3.5

In n GOLD' /(OD):16 ,(GD):4 cm, /(GL) :6 cltt,

cm.

2.

-s

(1) Draw a rough figure of n ABCD showing grven measurements.

the i

(2) Draw seg BC of length 2.5 cm. (3) With B as cdntre and radius AB :3 cm draw

B 2.5cm

an arc.

(4) With C as centre and radius CA draw another arc intersecting the arc at the point A. (5) Complete

t(CD)

:4

cm,

previous

ABCD using the lengths of

seg BD and seg CD.

(6) Draw seg AD.

n ABCD is the required quadrilateral.

Rough figure

n:t:

C

cm, /(OL)

,'l :7.5

crlr

ttz

NAVNEET MATHEMATICS DIGEST : S.IANDARD VIII

Q. 3.

In n PLAY,

,(PA):8.5

-

cm,

/(PL)

:4.5

t(Ly):7.5

A

cmo

,(LA)

:6.5

cm, /(py)

113

CONSTRUCTION OF QUADRILATERALS

:

.i!49n,::::--,:

S.S

:,r:.r.,riiist:rri:lr,r,,,rrr,ri,rr,,f.!i :

,rffilUSgIW$4;

cm.

(Textbook page 106)

Draw quadrilaterals of the following measures

Ans.

:

l. In IDEFG, /(DE):4.S cnr /(EF):6.5 cmi mLD:65", mLE:100o, mLF:60o.

rl

4.5 cm

4.5 cm

Construction 3

:

To construct a quadrilaterar, given the rengths of two adjacent and the meaures of three angles.

Q.2.

In IMTSN, t(MT):4 cf,r ,(TS):5 cm, mLM:50o,

mLT:ll0"rmLS:70". NS

Q. Construct IABCD in which /(AB):5.4 cm, /(BC) :3.4 m LA : 70, m LB: 60", m LC : llo". Steps of Construction

:

(1) Draw a rough figure

of n ABCD

showing the given measurements. (2) Draw seg AB of length 5.4 cm. (3) At the point B, draw ray BN making

M4cmT Rough figure

ZABN:60o. (4) With B

as a centre and 4cm

radius 3.4 cm draw an arc intersecting ray BN

a. 3.

in point C.

mLY:95", mLZ:85".

(5) At the point A, draw

ray AM

making

IBAM:70o. (6) At the point C, draw

IBCD: n 4PCD

In IXYZW, I(XY):S.S ch: t(XW):3 cD mLX:70o,

A

110o intersecting ray

5.4 cm

AM in point D.

is the required quadrilateral.

Explanation : From the rough figure, measure of LW should be given. (Angles at the end of the given segment XW) In n XYZW, mLx + mLY + mLZ + mLW : 360o. .'. 70 + 95 + 85 + mLW:360.

85"

X

95'

70" 5.5 crn

Rough figure

ltf

NAVITTEET.

MATTEMATICS DIGEST

.'. 250.1-mLW =360 .'. mLW:360-250

:.

STANDARD

VIII

.'. mLW:110o.

'-Ans.

CONSTRUCTION OF QUADRII"ATERALS

Draw quadrilaterals of the follorqng measures i 1. In ITYRE, /(TY):3.5 cm, /(YB)-g crtrr /(RE1:4.5 Cilt

B

z

mLY:60o, mLR:120".

5cm Construction 4

In tl ABCD, /(AB) = 6.5 cil,

mLB:75, mLC:85o. Steps of Construcjion

,(BC)

:

(1) Draw ."g nC of tenj*r 4.5

(2) ltthe

5cm

Rough figure

:

To con$truct a quadrilateral, given the measures of three of its and the two included angles.

Q.

R

i

cm.'

points B and C, draw rays BM'

and CN on the same side

of

BC

:4.5 cm, t(CD):4

2.

In IKING, /(KI)=3.f cmo t(IN):5.5

mLN:75", rnLI:115o.

cm, /(NG)=S

cm,

1:

il

is b'/

making angles of 75o and 85" respect-

ively.

(3) With

B

as centre and radius

Ans.

6.5 cm, draw an arc intersecting ray BM in the point A

5.5 cm

(4) With C as centre and radius 4 cm, draw an arc intersecting ray CN in the point D. (5) Draw seg AD.

n

3. In trRDTc, ,(RD):6 cm, mLD :100", mLT: 90o. Ans.

ABCD is the required

quadrilateral.

4;5 cm

,/(DT)-4

em,

/(TG):l

cnr

THE ARC OF A CIRCLE

Semicircle : Seg RT is the diameter of the circle with centre O. The diameter RT divides

Q. Write the names of the arcs formed by the points A, B and C on the circle. Ans. Arc AXB, arc BYC, arc CZA.

2. Intercepted arc

(Textbook page 109) :

In the figure, the end point N of the arc NTS lies on the side ON of ISON, and the end point S lies on the side OS.

.'. I-SON intercepts arc NTS. Similarly,

/NDS

the circle into two arcs of equal measures. Arc RXT and arc RYT are called semicircles, or semicircular arcs.

intercepts arc NTS.

In the figure, seg SD and seg TN are diameters. Name those minor arcs, major arcs and semicircular arcs whose end points are also the end points of these diameters x Ans. Minor arcs : arc SYN, arc NMD, arcDZT and arc TXS.

Major arcs : arc SMT, arc TYD, arc DXN, (Textbook pages 108 & 109)

0. I. In the figure, name the angles which

ILMT

and

ZLNT

Semicircular arcs : arc SZD, mc SYD, arc T)OI and arc TZN. Central angle :

intercept the arc LXT. Ans.

arc

NZS.

intercept the arc LXT.

Q. 2. Observe the figure and name the

angles

which intercept arcs and name the arc that each angle intercepts.

Ans. LAOB intercepts arc.AXB. LBOC intercepts arc BYC. LCOA intercepts arc CZA. 3. Minor arc and mojor arc

:

Chord AB, not a diameter, of a circle with centre O, divides the circle into two parts arc AXB and arc AYB.

Arc AXB is the minor arc and arc AYB is the major arc. 116

An angle having the eentre of the circle as its vertex is called a central angle. In the figure, IPOQ is the central angle. Measure of an arc : (1) The measure of a minor arc : The measure of the minor arc is equal to the.measure of its central angle. In the figure, m(arc AXB) : zLAOB. zLAOB: 100o. .'. m(arc AXB) : 100'. (2) The measure of a major arc : The measure of a major arc : 360o - the measure of the corresponding minor arc. In the figure, m(arc AYB) : 360" -m(arc AXB):360 - 100:260 .'. m(wc AYB) :269". (3) The measure of a semicircular arc : A The measure of a semicircular arc is 180" In'the figure, arc AXB and arc AYB are semicircular arcs. .'. m(arc AXB) : 180' and ru(arc AYB) : 180'.

NAVNEET MATHEMATICS DIGEST : STANDARD

VIII

The measure of a minor arc is equal to the measure of the corresponding central angle.

(Textbook page 111)

Q. 1. If the measure of a central angle is 120o, find the measure of

.'. m(minor arc PXT):ryLPOT: l00o ... tFror6-(1)l .'. m(minor arc TYN) : nLTON: 80o ... lFrom (4)] .'. ru(minor arc NZC) : zICON: 100o ... tFrom (3)l .'. z(minor arc CWP) :wLPOC:8}" ... [From (2)]

corresponding minor arc and that of the major arc this minor arc. Solution : The measure of a minor arc is'equal to the measure of corresponding central angle. The measure of the central angle

is

:350

.'. m(minor arc):129o. rz (majOr arc

PZT): 360o -

)

m (major

m (major arc CXN)

:360

160o.

(corresponding minor arc)

- 160:200o rn

Ans. The measure of the central angle : 160oi the measure of the maior arc :200'.

Q. 3. In the figure, seg PN and seg TC are the diameters of the circle. Write the measures of the arcs formed by the endpoints of these diameters and explain your answer. Solution : Seg CT is the diameter of the circle. mLpot:100' ... (Given) ... (1) MLPOT +MLPOC:180O ... (Angles in a linear pair) -tmLPoc:1800 100" .'. ... (Given : nLPOT:100')

.'. mLpoc: 180 .'. ruLPOC:80" nLCON:nLPOT

.'. zLCON:100o

I

mtTON:nLPOC .'. zLTON:80o

1oo

...(2) ... (Vertically opposite angles) ... lFrom (1)] ... (Vertically opposite angles) ... [From (2)]

:::3_;,.''":ril*ff,fri"'arc

ryN)

:

360o

-

m(corresponding minor arc NZC)

:360 - 100 ... lFrom (7)] :260o. Ans. Minor arcs : (1) m(minor arc PXT).= l00o (2) zz(minor arc TYN) :80" (3) m(minor arc NZC) : 100o (l) m(minor arc CWP) :80o Major arcs : (1) n(major arc PYC) : 280o (2) m(major arcPZT) :269o (3) m(major arc TWN) :280o (4) m(major arc CXN) =260".

corresponding central angle.

360'

[From (5)]

:280o.

major arc. Solution : The measure of a minor arc is equal to the measure of

:

rz(corresponding minor arc PXT)

:260". arc rwN)

measures of the corresponding central angle and the

arc )

minor arc CWP)

- 80 ... lFrom (8)]

:360_ 100 ...

Q. 2. The measure of the minor arc of a circle is 160.. Hence, find

m(major

... (s) ... (6) ... (7) ... (8)

:280o.

m(major arc) :360o

n(minor arc): 166o ... (Given) .'. measure of the corresponding central angle :

* z(corresponding

Now, m (major arc PYC) :360o

120o ... (Given)

- m(corresponding minor arc) :360 - 120:240" Ans. rn (minor arc) :120"; m(major arc):249o

ttg

TIIE ARC OF A CIRCLE

l,

The inscribed angle : In the figure, O is the cenfe of the circle. M and S, the end points of arc MTS, are on the sides of IMTS and the vertex T of IMTS is on the arc MTS.IMTS is an

angle inscribed in the arc MTS.

p.trffi

(rextbook page

u2)

1. Observe the given figure and filt in the blanks : LPM.L is inscribed in the arc and LTNL is inscribed in arc ......... . Ans. Z PML is ipscribed in the arc PML (or arc PTL) and I-TNL is inscribed in arc TNL (or arc TPL or arc TML).

12fr

NAVNEET MATflDMATICS DTGEST : STANDARD

yIIr

Q' 2. Draw a circle. show an arc LMN on this circre. rnscribe lL . it. How many more angles can you inscribe in this arc ? Ans. The figure shows arc LMN on the circle. Infinite angles can be inscribed in the

TIIE ARC OF A CIRCLE

.'.

mLApB

:|

m(rc AXB)

:jx180'

...

[From (1)]

.'. zLAPB:90"

...

(2)

...

(3)

arc LMN.

(2) Arc PYB is the intercepted arc by inscribed IPAB.

6. Relation between the inscribed angle and intercepted arc : The measure of the inscribed angle is harf the measure of the intercepted by it. In the figure, I-APB is inscribed in arc ApB.

zLAPB:55o.

The inscribed LA9B intercepts arc AXB. IAOB is the corresponding central angle arc AXB.

.'. zLAOB:m(arcAXB) mLAOg:110o ... m(arc AXB):116o. aLAPB:l

l-ApB

ffil$j| (Textbook page 114)

Solution

angle is 120.. Find the measure of the

.-. mLABp:|[email protected] AZp) .'. 50o :| m(nc AZp) ... [From (3)] .'. 50" x2:m(arc AZP) .'. m(nc AZP) : 196'. (2) m(arc PYB) = $go Ans. (1) mLAPn:90" AZP;:199".

:

The measure of the central angle : the measure of the intercepted The measure of the central angle : l20o ... (Given) .'. the measure of the arc it intercepts :120" Ans. The measure of the intercepted arc:120.

;

The cyclic quadrilateral The vertices B, S, N and

:

L of n BSNL

are on the

circle. Such a quadrilateral is called a cyclic quadrilateral.

Q. 2. In the figure, arc AXB is a semicircle. zLPAB:4O". Find the values of (l) mLApB Q) m(arc PYB) (3) m(arc AZp). Solution

(3) In AAPB, %LPAB + %LAPB -t mLABp: 180o .'. 40 + 90 * wLABP: 180 ... lGiven and from (2)]

Arc AZP is the intercepted arc by inscribed LABP. is half the

[email protected] AXB).

Q. 1. The measure of a central it intercepts.

40o)

.'. 130 *%LABP:180 .'. zLABP: 180 - 130 .'. mLABP:50o.

of

This explains that the measure of the inscribed of intercepted arc AXB.

... mLpAB:)[email protected]) .'. 40o : ! [email protected] PYB) ... (Given : nLPAB: .'. 40o x2:m(arcPYB) .'. m(arc PYB) :39'.

[A quadrilateral is

said to be cyclic,

if all its four

vertices lie on a circle.l A

(1) Arc AXB,is a semicircular arc. The measure of a semicircular arc is 1g0?.. (1) The measure of an inscribed angle is equal to half the measure of the intercepted ara.

The opposite angles of a cyclic quadrilateral are supplementary.

.'. in cyclic

quadrilateral BSNL,

mLB + mLN

:

180o and

mLS + mLL

:

180".

(3) m(arc

NAVNEET MATIIEMATICS DIGEST : STANDARD

VIII

Joint Bar Graph tbook page

0. t- In the figure, m LB :85" and m LC Find, mLA and mLD.

:

ll5) A joint bar graph makes comparison of two kinds of information easier. You have studied this in previous standard.

lOSo .

Solution : The opposite angles of a cyclic quadri_ lateral are supplementary.

.'. mLA*mLC: lg0o .'. mLA* 105:190 ... .'. mLA: lgo - 105 .'. mLA:75o Similarly, mLD -f mLB

:

t, (Given :

":L::

y_: y_111:llg_v_h"' to draw such a graph.

Drawing a joint bar graph

:

Q. The number of male and female workers on a work in four villages A, B, Co D under Employment guarantee scheme is given below. Draw a joint bar graph indicating the data.

mLC:105o)

a

.',:,::.$1L:

tQ

D

Male

t0

60

80

70

Female

50

55

50

40

VillA'gq:

lgoo

.'. mLD * 85 : 180 ... [Given mLB:g5"] .'. mLD: 180 - 85 .'. mLD:95" Ans. rn LL : 7S"i m LD : 95". Q' 2. The measures of the opposite

(Textbook pages 116 to 120)

l,

.,.,

angres of a cyric quadrilateral

and 3x. What are the measures of these angles of the quardrila

Solution:

The opposite angles of a cyclic

quadrilateral

supplementary.

.'. r+3x:180o .'. 4x:180o

.

lg0" 4

x:45o 3x:3 x 45o: ,'.'

135"

Ans. The measures

of the opposite angles of the given c

quadrilateral are 45" and l35o respectively.

Steps

(l)

L

for drawing the graph:

Draw X-axis and Y-axis on the graph paper. t23

I24

_

NAVNEET MATHEMATICS DIGEST : STANDARD

VIII

(2) Take the villages on the X-axis and the number of workers on Y-axis.

(3) Take a scale, I cm: 10 workers on the y-axis. (4) Mark villages A, B, C, D on the X-axis keeping equal di between them.

(5) Show males and females in each of the four villages using a j cm from each of the joint bar of 1 cm width. (6) Show the heights of each bar as per the scale taken for the y(7) Draw joint bar graph in this way for each of the villages. (8) To show the two bars in the joint bar graph separately, colour shade) the female's bar.

di3B,d!,,1 lrextnook

pages

tt9 &

120)

0. 1. Given below are the number of girls and boys in Stds V to of a certain school. Draw a joint bar graph from this data.

sl&d

,,11:;:.:lL:.\{:

,.YT$T

,40

30

45

40

Girls

20

30

t5

15

Take the scale

t

?Xlr

Boyi

Ans.

I

VT

I cm:

10 boys/girls on the y-axis.

.IOINT BAR GRAPH

2. Children of four divisions of Std. VIII of a school, planted saplings as shown below. Show this in a joint bar graph. Divi,F,,,,|.p4,

h

Teak

30

Mango

25

Take the scale

Ans.

,,lC'.

ri,D

25

25

20

25

10

5

I cfil:5

saplings on the Y-axis.

126

NAVNEETMATHEMATICSDIGEST:STANDARDvm

JOINT BAR GRAPH

Q. 3. The number of men and women working under the Guarantee scheme in four villages is given berow. Draw lraph to depict this. Villago,r

Pimpa!gaop

Lirnbgaon,,

a joint

Women

150

200

100

50

Men

200

100

200

350

Take the scale 1

cm:50

persons (women/men) on the y_axis.

t27

4. The number of Iiterate and illiterate women in four villages given below. Show this information in a joint bar graph. Viila$e,i

r.,$onake'r

Literate

500

550

600

450

Illiterate

150

150

200

100

.,,.,.,,Neiiarirl

ln this example can we take a scale of 1 cm: 10 women on the Y*axis? 'fhink about it and choose a proper scale. Ans.

Ans.

l[the scale of 1 cm :

10 wqmen is taken, then the bar of maximum height 60 cm (representing 600 women) has to be drawn. This not possible on a graph paper, we use generally.

COMPOUI{D INTEREST

tr

(4) Here, P: Rs 1,25,000; R: 12 p.c.p.a.; N :2 years PxRxN 125000x12x2 :Rs

1.. Revision :

: PxRxN 100 : Here, I simple interest; p: (i)

(I)

(ii) Amount: Principal *

30000

100

A=p* I

R: rate; N: A:P*I

principal;

Interest

number

:

:Rs

Rs (125000 + 30000)

155000

lry: liry9 ilg::l!l):T: 19'm9i "lryT ltl R: 1.:-lls9: (5) Here, P: Rs 1,50,000; R: 10 p.c.p.a.; N :3 years=

of

PxRxN

150000x10x3 :Rs 45000 100

,r8,ffiffisii (Textbook page 121)

Q. L. USe the formulae to find the simple interest and amount

A : P * I : Rs (150000 +45000) :Rs 195000 Ans. Simple interest (I) : Rs 45,000; amount (A) :

:

Rs L'95'000.

Q. 2. If a certain sum of money earns an interest of Rs 3'630 in 3 years at the rate of 11 p.c.p.a., what was the principal? Solution : Here ,

I:3630;

:L#9

Ans. Principal

(1) Here, P=Rs 9000;

PxRxN

R:10

p.c.p.a.;

9000x10x3 100

A: P * I:

Rs (9000 +

Ans. Simple interest (I)

(2) Here, l_

P:Rs

15,000;

PxRxN 100

N:3

:

Rs 2700

p.c.p.a.;

15000x11x2 : 100

N:2

r

A: P * I:

100

:

Rs L8,300.

lgooo

+ 18000): Rs 68000 Ans. Simple interest (I): Rs 18'000; amount: Rs 68,000. Rs (50000

... p

-19##

:

Rs nooo

Rs 11,000.

: PxRxN '. 13500: 45000xRx3 100

"'

I

PxRxN 100

.'. 10000 -

Px10xN

.'. PxN:100000

100

(1) Let the principal be Rs 25000. 100000 : ' : -:-:^-:= 25000

Then. N

4 years

!/Navneet Mathematics Digest : Std.

VtrI tE059il

-tl

I

,

!

i

x 100 :to R- 13500 45goox3

Q. 4. What sum of money will have to be invested and for how many years, in order to get simple interest of Rs 101000 at the rate of 10 p.c.p.a. ? (Write three pairs of values for P and N.) Solution : Here, I : Rs 10,000; R: 10 p.c.p.a.

years

Rs 3300

-r\r

:U,o-

P:?

Rs L1,700.

(3) Here, P: Rs 50,000; R:9 p.c.p.a.; N = 4 years x R x N-5oooo x.9 x 4:Rs

100

years;

Ans. The rate of interest: 10 p.c.p.a.

:

A : P * I : Rs (15000 + 3300) : Rs 18300 Ans. Simple interest (I) : Rs 3300; amount (A)

. r" r-P

:

3630

100

Rs 2700 amount (A)

R:11

...

11p.c.p.a.,N:3

Q. 3. The simple interest on Rs 45,000 in 3 years is Rs 13,500. What is the rate of interest? Solution : Here, I: Rs 13,500; P: Rs 45,000; N:3 yeras; R: ?

years

2700): Rs 11700

:

R:

'

I}

NAVNEET MATImMATICS DTGEST : STANDARD

VIII

OMPOT]ND INTEREST

(2) Let the principal be Rs 20000.

' N:gg:5 20000

Then,

Solution

10000

Ans. Three pairs of principal and period : (1) Principal :'Rs 251000; period:4 years (2) Principal: Rs 20,000; period:5 years (3) Principal: Rs 10,000; period : 10 years. :

The interest of the first year is added to the principal and the is taken as the principal for the second year. The interest for the year is added to the principal of that year and the amount principal for the third year. Interest charged in this way is compound interest.

Formula for finding the amount by compound interest

(ii)

r(r + r*ai A:

Compound interest

:

Amount calculated at compound

amount

-

principal

:

l0

N:

p.c.p.a.;

2

years

Ans. Amount: Rs 9680; compound interest: Rs 1680.

While calculating simple interest; the principal remains the the entire period.

(i) A:

R:

/ R\N / 10\2I A:Pl1+-l:800011+ \ 100/ 1 100/ / I \2: 8000 /ll\2 : lt t 11 : : 8000(t 8000 * *m/ (r-J " r- 10r Rs 9680 Compound interest - A - P : Rs (9680 - 8000) : Rs 1680

(3) Let the principal be Rs 10000. t,11110: rhen,' N: ro years

2. Compound interest

:

(1) Here, P: Rs 8000;

years

131

A

-

:

i

(2) Here, P : Rs 6400; R : I2.5 p.c.p.a. ; N:2 years R :64oo1t*E\' A: P(' +

/

\ roo/

1

'---/

Q. 1. Use the formula to find the amount and the compound

/

g

/g\2 :6400 *r/1\2:6400 " (',/ I s' 8 :Rs 8100 Compound interest - A - P: Rs (8100 - 6400) : Rs 1700 :6400 ('t

g

Ans. Amount: Rs E100 ; compound interest: Rs 1700.

(3) Here,

P

R:

= Rs 10,000;

/ o:r(1+

R\N :

:

*mi

10 p.c.p.a.;

/

N:2

Compound interest

(r/

-

A

years

10\2

* _,,l 100/ / I \2: 10000 /rr\2 : 10000lt 10000lt

p.

(Textbook page 125)

roo

10000 x

- P: Rs (12100 -

R:

10 p.c.p.a.;

/ R\N:1s000(\1 / A:P(1+ * 100/

:15ooo

(t.r1)':

N:3

15000

:

:

Rs 2100

Rs 2100.

years

10\3

*,)

r5ooo (r11)'

11 11t 11 Rs 19965 t0 " tO tO-: Compound interest - A - P: Rs (19965 :

10-:Rs

f0-

10000)

Ans. Amount: Rs l2rIM; compound interest (4) Here, P: Rs 15,000;

11* ll

*

Ans. Amount: Rs

15000)

191965; compound interest

:

:

Rs 4965

Rs 4965.

12100

NAVNEET MATIIEMATICS DIGEST : STANDARD

(5) Here,

P: Rs 20,000; R:5

p.c.p.a.i N

:

YIII

COMPOT]ND INTEREST

3 years

: l2ooo (i*)' : 12ooo

n :p(r +3). : 2oooo(, *1\' \ roo/ loo/

\'

:2oooo :2oooo

(t . r1)':2oooo

(#)'

Q. 4. Bhakti borrowed Rs 40000 from a women's savings group to buy a sewing machine. If the rate of compound interest is 5 p.c.p.a., what

,'JZO"420" n:Rs 23152.50 - A-

p

= Rs (23152.50- 20000)

Ans. Amount: Rs 23152.50; compound interest (6) Here, P: Rs 16,000;

a

: e(r .#). :

:

16ooo

(t . *)'

: r6ooo .;

:

X:

l}ip.c.p.a.;

l6ooo(t 16000

N:

:

amount will she have to return at the end of 3 years Rs 31

= Rs 3I

3 years

.ff)'

A

- p:

Rs (227g1,25

-

16000)

:

Rs 67g

Ans. lWount: Rs 22781.25; compound interest: Rs 67g1.25.

Q. 2. Ganesh deposited Rs 50,000 in a Nationarised Bank for 2 yea 9 p.c.p.a. compound interest. what amount wil Ganesh get at the of 2 years

?

Solution : Elere, P = Rs 50,000; R :9 p.c.p.a.i

e

: e(r .*)^ :

soooo(r

.

N:

#)'

R:

12ooo(t

.#)':

:4ooo(r *rr-1;':4ooo (t

/2t\3

21 21

years

.*)'

21

(#) :4ooo x';"';"'*:4630.s0

5. Shreya won a cash prize of Rs 1250 for standing first in the Std. VII examination. She deposited that amount in a bank for 3 years at a rate of compound interest of 8 p.c.p.a. What is the amount that she will receive from the bank at th_e end of the term?

P: Rs 1250; R: 8 p.c.p.a.; N:3 years / ):3 R\N / 8- \3 :1250 ll+-l A:Pl / 1+-l:12501 l:1250 z5o 1+ * * (1 (t r*-) X) \ loo/ 27 27 27 : - , /27\t :1250( : I :l25Ox:--:X-:-x 2s 2s 2s 1574.64 Solution : Here,

(Textbook page 128)

Q. 1. Fitl in the blanks :

set,

the amount that Attaf must pay back to the bank at the end of Solution : Here, Rs 12,000; l0 p.c.p.a.; N 2 years

e:r(r .*I:

N:3

Ans. She will get ns fSZ+eC from the bank.

Altaf borrowed Rs 12,000 bank at a compoud rate of interest of l0 p.c.p.a. for 2 years.

P:

p.c.p.a.;

\2s/

2 yens

:50000 (1.0D2:50000 x 1.09 x 1.09:Rs 59405 Ans. Ganesh will get Rs 59,405 at the end of the period. Q. 3. In order to buy a colour TV

n:r(r .*)

P: Rs 4000; R:5

?

Ans. Total amount payable: Rs 4630.50.

(;)'

'; f : *. zz7lt.2s -

Solution : Here,

:4ooo

*

Compound interest

,I: *. t4szo

Ans. Altaf will have to pay Rs l4r520to the bank at the end of the period.

2I

Compound interesr

" ,L '

133

12ooo

(r

2 yea

.*"fr

I ".. Iffffi**rtok "--" ;]",T *6** W"'"f "-*1,_S* '**.j [ ,-}ru* T,e#ffii '" f (1)

3000

5

2

-

(2)

1000

10

2

.*)'

(3)

25,000

T2

J

(4)

64,000

12+

-')

(s)

20,000

10.5

2

i]]1lllliilllll illlll'l

l,r,i;i

W.

ffi

I34

NAVNEET MATIIEMATICS DIGIOT : STANDARD

Solution

(l)

VIII

COMPOI,JI\D INTEREST

:

P: Rs 3000; R:5 p.c.p.a.; N :2 PxRxN 3000x5x2 t: --Ioo-- : Rs 3oo Here,

(4) Here, P:Rs 64,000; p.:l2i:12.5 p.c.p.a.; N:3 years -T- PxRxN- 64000x12.5x3 : Rs 241000 100 100

years

-lbo--:

e

: r(r .

:

3ooo

#I

.

= 3ooo(1

e!\': \20/

:

*)'

--p

Rs (3307.50

Difference between C.I. and S.I.

(2) Here,

P: Rs 1000; R:

.

sooo(r

*)'

*, 33o7.so rtj,.*#: 20" 20

3ooo

Compound interest: A

:

:

-

:2

Rs 307.50

-

300)

:

(5) flere, Rs 7

years

Rs 200

:r(r . #). : looo(t .#)' : rooo(r .*)' : looo (i*)': rooo,.1;,. jj:*, ,r,o

e

Compoundinterest:A-p Rs (1210

-

Difference between C.I. and S.I.

(3) Here, P: Rs 25,000;

,t : PxRxN -100-: a

:

:

Rs (2lO -200)

12 p.c.p.a.;

N:3

:

:2sooo

f3)':25ooo

25000(t

/

Compound interest

-

A

.

:

Rs 10.

P:

Rs 20,000;

R:

10.5 p.c.p.a.;

N:2

Rs 3125.

years

PxRxN 20000x10.5x2 Rs 4200 - 100 : -- 100 -: e : r(r . 20000(r . = *). H)' : 20000 (r.ros), t- :

:20000 x 7.221025:Rs 24420.50 Compound interest - A P: Rs (2M20.50- 20000) : Difference betwben C.I. and S.L

:

Rs (442O.5O

-

Rs 4420.50

4200)

= Rs 220.50

for the same period and the same rate at compound interest, how

P: Rs 60,000; R:9 p.c.p.a.; N :3 years PxRxN : 60000x9x3 : Slmple rnterest (D : --mRs 16200 -* 100

#)'

90(x)

:

25000

(, .

*)'

Rs (35123.20_ 25000)

:

Rs

(t:-rl:320_

Rs 112fr.20.

Amount on the same principal at compound interest,

o:t(l

+

loo/

,

:6r

9000)

(l.09)3 )oo )00 (l'09)3

loo/

\

:60000 x 1.295029:Rs 777OtJ4

Compound.interest

Rs 10123.20 Difference berween C.I. and S.L

Rs 2?t,l2S

much more interest would she have to pay?

years

: Rt 35123'20 ^']25*425*?s 25

- p:

:

Solution : Here,

--100 -:'Rs

*I

sttzs

2.'Sunitabai borrowed Rs 60,000 at 9 p.c.p.a. simple interest for 3 years to have her house repaired. Had she borrowed the same sum

Rs 210

25000x12x3

-r(,r .

\25

'

R:

1000)

(|)' : 64000 ,. ;' ; * ! :*,

Compound interest = A - P: Rs (gII25- 64000) Difference between C.I. and S.L : Rs (27125

I

:

:64ooo

('.*)'

-24000):

Rs (307.50

10 p.c.p.a.; N

,_P*RxN 1000xl0x2 : .100 100

:

3000)

a:r(r .*) : o+ooo(r .#)':64ooo

-

A

- P:

(7770I.74

Difference between C.L and S.I.

:

-

60000)

Rs (17701.74

-

:

Rs

17701.74

16200)

:

,

Rs 1501.74

Ans. Sunitabai would have to pay Rs 1501,74 morc as intorcct.

136

NAVNEET MATIIEMATICS DIGEST: STAITIDARD

VIII

COMPOUND INTEREST

Q. 3. Ambadas took a loan of Rs 96,000 at compound interest bank to drill a well in his field. If the rate is 6.25 p.c.p.a., what a will he have to pay to the bank, if he returns the loan after 2 :

How much money would he have saved had he been able to the same amount at simple interest for the ,u_" n""od ;;;; rate?

Period

(N):2

years.

Today's price of the machine

" l:75oool/ t+ :P{/ 1+ R\N \

r00/

:75000 I/ t

:

\2

\

- 4\2 'l 100/

/24\2

24

I :zsooo I 1 I :75ooo x:2sx12s \ --2s/ \2s) 1

24

Rs 69120

Ans. The value"of the machine today:Rs 691120.

Compound inreresr

A_p

-

:

:

Rs I2375

lot--:Rs

12000

Rs (10g375 _ 96000)

x,l,t N-:-96000 x 6 .25 x 2 Simple interesr (D : P

100

Difference between C.I. and S.I.:Rs (tZZflS_ 12000):Rs 375 Ans. Ambadas will have to pay Rs simple interest : Rs 375.

lr0gr375 to the bank, Savi

Q. 3. A wholesale trader sold Rs 20 crore worth of cloth this year. If the sale of cloth increases at a rate of 2Vo per year, how muchwill the sale be after three years? Solution : The sale of cloth today (P) : Rs 20,00,00,000; Increase in the yearly sale (R) :27oi Period (N):3 years. The sale of cloth after 3 years

R\N / :Pl1+-l:

\

: :

100/

2oooooooo

/ 2\3 / 200000000(1*roo-) :200000000 (t

:2oooooooo * 1, 1 r 1 f1)' s0 50 s0 \sol

x 51 x 51 :Rs 212241600 Ans. The sale of cloth after 3 years :Rs 1600 x 51

lry.--]P

ryg_*l

:{

211221411600.

Q. 4. A car is valued at Rs 4,00,000. If its value falls at 2.5Vo per lear, what wiII its value be aftef 3 years? Solution : The value of the car today (P) : Rs 4,00,000; The rate of depreciation every year (R) :2.5Vo; As there is depreciation in the value, R : -2.5; Period (N):3 years. z -r{r3 R\N:400000(t / / 2.5\3 :400000(t o -

:r(t *r*-)

the viuage wiu be 1,38,e1s.

* * )

*)

/, -1\' \ 40l 39x 39x-39 /39\3 :400000x1I :400000x 40 40 40 \40l : Rs 370743.75 :400000

Q' 2' Two years ago Rajani purchased a machine for Rs 75,000. rf i value falls by 4vo every year, what is the value of the machine today : The price (initial value) of rhe machine (p) : Rs 75,000 _q,olulion Yearly depreciation in its value (R) :4Vo; As there is depreciation in value : _ R 4

l\3

+r-)

Ans. The value of the car after 3 years will be Rs 370743.75.

/

tr

POLYNOMIALS

1. Re'

(3) In the expression .'

Q. Classify the following expression, trinomials:

* _or,o*i,

! -2 m

binomials

(8) st.

-n*5;

the power

- I of the variable ru

is not a polynomial.

Are the following expressions polynomials

(t) zt'tu+a2

-

- -2, i:7m-'.

is negative.

2.

(9) t:

17

(7) 35+x3-

7x.

n3

2^

-Gn2+13 (3) t*,

sl

-*1 @ + ($

n-:

t2*$.

(7) (s) s- sb2 +b4 (s) ia3 -7a+te Ans. (1), (2), (4), (6), (7), (8) and (9) are polynomials. {o)

8y..

Q)

?

(3) and (5) are not polynomiatu.

2. Polynomials: If the index of every variable in an expression is a whore number, then expression is a polynomal. For example, gxz + 5x 9 * irs a polynomial

y'+y -5

!i)

and

(ii)

m2

+m-t +13 a whore

urv ruue^

-_negative

---i------

^

----

_ of m - rt or

to

Ans.(l) rrrru. \r.,

ll

I:

nomials or not

(r)

b2

-2s

: 1

e) yt * t8 _ 5y

Ans.

(l)

In rhe expression

b2

number.

.'.

,r,

b2

-

expressions are

_25,

@

1-2.

m

thepower 2 of the variable b is a whole

25 is a polynomial. 1

fr}: whole :y::j"" number. .'.

yi + rs _ 5y, the power j of the variable y is not 4

-1

y2

+

18

-

5y is not 4.polynomial.

I

7

erq

pqlyngll-xqlql

5

@)

3

a3*r"*l

g

-

Q. 4. Write any five expressions which are not polynomials.

1i x1i I

Q' 1' state, giving reason, whether ttre io'owing

ro/,'l

p-'

3*'+1, J4 ,Zrt13 (2) -, (:) *'+t*t-i

+.sl

Ans. (1)

-- -

tsl

Q. 3. Write any five expressions which are polynomials.

(l)-

are not polynomials.

--' a,l'ilffi vv :j:1,:",ii.":.;:tl (i.e. not a whole number). ""rn*. is nor

[Not" ' L

x

!2 ?t --rl11

(2) y2 *y-7 qe le!

G)

Z&+3

(4) mi _m-2 +7

pely_teqi3-ls.:

3. Classification of Polynomials : (1) Monomial : A polynomial having only one term is called a monomial e.g., 5x;37; - J5*. (2) Binomials : A polynomial having two terms is called a binomial.

e.g.3x*5;

-|f -t;t+n.

(3) Trinomial : A polynomial having three terms is called e.g.,5x3 -3x2 +7; 15 * nt +nr: m2 +7m+ 5. 4. The degree of a polynomial

a trinominl.

:

The highest index of the variable

in a given

polynomial is called the

degree of the polynomial. e.g., the degree of the polynomial

(1) x' +7x2 +5"r-9 is 3. (2) 27:27x" .'. the degree of 2j is 0. The degree of a non-zero constant monomial is zero. The d.egree of 0 is not defined.

I4O

NAVNEET MATHEMATICS DIGEST : STANDARD VIII

Coefficient of a monomial :

of

The coefficient

(3)

29 itself

-

is

-

monomial

(I) l3m2

B Q) -'lx

is

is

29. (Textbook page 134

& 135)

Q. 1. Write the coefficient and the degree of each of

monomials:

(D 23x3 (2) 13

$\-6n5

fcl

fr

(6) -ton2

(s)

(e)

x

(11)

15s3

-

(t1

-!t+

(n) 1Ee6 Q'

1y

48

(14)

(s) a2 ,; (10) 5

-t7p

(1s)

Ans.

(1)

23x3

23

3

(2)

18

18

0

6ms

-6

5

(3)

-

l1

(4)

5"

(s) (6)

a2

l0n2

-

11

-= 5

2 2

Q)

-13 9

9

(8)

4y

4

9'

4

1

2. Write the degree of the following polynomials

:

(1)6Q3-thp Q)8nr-25n*9 (3) x_tt*3xa (4) (5) Sl (6) 2ns n3 I6 -7n2 (7) gas - 6t (S) fr* _xa

39m

\

(e) 0

Ans. The degree of the polynomials are as follows : (1)3 (2)2 (3)4 (4).1 (s)0 (6)s (7) s (s)+ (9)cannot determined.

,

I

I

-10 -13

7q"'

trtr 1. Revision

Q.

DISCOUNT AI{D COMMISSION

tr'ill in the blanks

1.

:

Q. 2. A shopkeeper gves t2% discount on the marked price of a TV If TV set is sold for Rs 7480, what is its marked price ? Solution : Let the marked price of the TV set be Rs x.

(1) A ratio with denominator 100 is called a percentage. 15

(2)

15 per

fr:

(4) g5Eo:2 100

(6)

20Vo

of 25000

cent:lS%o. e)

:

25000,

ffi

:

:*

100

Then, discou

xfr:ZO

Its selling

6Vo

l2Vo on Rs x

price:

:

Rs

x*

::n, 100

3. 25

* discount

marked price

set.

/ 3x\l:Rsl/25x-3x\l:Rs-22x -Rsl -^" \^ 25l-"" \ zs )-"" zs But the selling price is given to be Rs 7480

3000

Q) 3Vo

of 100: 100.:

of 100

(3\ 6To

of I

:.

2bc

:748O

" ^-

7480 x 25

:

Rs 8500

22 25 Ans. The marked price of the TV set: Rs 8500.

3:60. 100

Q. 3. The marked price of a scooter is Rs 47,500. If the selling price is

3

" loo:3'

Rs 1[3,700, what is the percentage of discount given by the shopkeeper ?

Solution : The marked price of the scooter: Rs 475A0. Its selling price : Rs 43700 Discount: marked price - selling price

6 11450:11'-+)U x -:687. @) 6aEo of 1610: 1610 I ffi: 9cO.

(3)

nl

sOOO

Ans. (1) 2 per centof 3000:3000 x 3Vo

64per cent

(5) 5 per cent of400:+OO

Q. 2. Find the values : (l) 2 per cent of (4) 60Eo of 1610.

(2)

: :

marked price - discount Rs (650 - 52): Rs 598 Ans. The selling price of the carpet: Rs 598. The selling price

:

of

:Rs

2. Discount:

Percentage

(47500

of discount:

The difference between the marked price and reduced price price) is called discount.

Formulae : Discount: marked price _ selling price. I Discount: marked price x p"r"ntug" of discount.

-43700):Rs

#ffi

3800 r too

\/

38oo

- 47500 x loo:8 Ans. The shopkeeper gave YVo discount.

_,,

Q. 4. A shopkeeper

0. 1._TT marked price of a carpet is Rs 650. If the shopkeeper

87o discount on the marked price, what is. itS selling price ? solution : The marked price of th" .urpt is Rs 650. Discou nt gvo. discount on the marked price: marked price x rate of discount "'

gives 87o discount on the marked price of a camera.

He sold a camera for Rs 460. What was its marked price? Solution : Let the marked price of the camera be Rs x. 87o discount on the marked price.

.'.

discount:rrA:g:nr3. 100 100 "" 25 ,/

Selling

price:

marked price

-

/ 2x\ :ns (x-

r

discount

/25x -2x\ 23x ( rs):n' x,,l:*t rt

But the selling'price is given to be Rs 460.

144

NAVNEET MATHEMATICS DIGEST : STANDARD

23x

:'

.'.

25:460

VIII

x:+eOxfi:20x25:500

Discount: marked price - selling price

Ans. The marked price of the camera was Rs 500.

Q. 5.

.'.

In the following table, from the information

example, find out the figures to

fill in the blanks

DISCOUNT AND COMMISSION

given in

:

3x

3x:20x-2720

"' 2Ox-3x:2720 .: .'. ^:x-136 .'.

:

Selling price

Rs 30

Rs (250

-

30)

:

dlt"?"nt

:

234 1560

x 1oo:

marked price

-

-234):

,

too

15

.'.

discount : l57o

discount

Rs 1326.

The work of selling goods is often done by a person or organisation other than the manufacturers. The remuneration received by the person or organisation for this service is called a commission. A person who brings buyers and sellers in contact with each other is

Discount: marked price selling price .'. selling price : marked price _ discount

:

Rs 24.

marked price

Rs (1560

i00-:

selling price

discount:

-

x

-

Rs (160 - 136) :

(5) Percenrage of

Solution : (1) Discount: marked price x percentage of discount t2 :23U

Rs 160

marked price : Rs 160

Discount: marked price

.'.

x:

L7x:2720

called a commission agent. The agent takes commission which is

Rs 220.

expressed as a percentage.

(2) Discount: marked pirce x percentage of discount

:4800

.'.

selling price

x

(Textbook page 140)

6

,oo:

Rs 2gg

(3) Percenrage of discounl:

-marKed pnce ;

180 :1g00 x 100:

selling price

:

Solution : Let Sharadadevi paid Rs.r as commission. Commission Rs 3 on price Rs 100. Commission Rs x on price Rs 63,500

169

10

...

discount

_ discount :Rs (1800- 1S0):Rs 1620. marked price

(4) Let the marked price be Rs x. Then ryvo discounr: Rs

;*

#;:

:lTVo

3 100 ' x 63500 " 3 x 63500 . ?__ 100

100 x r

..

x:

x:3

x 63500

lg05

She paid Rs 1905 as commission.

R,

ff

was

Rs 631500. She had to pay a commission of 3Vo. How much money did Sharadadevi get for the buffalo?

iisgglnt-

.

.'.

0. L. Sharadadevi sold a buffalo through an agent. Its price

: "marked price _ discount : Rs (4800 - 28S) : Rs 4512.

air.ount

Sharadadevi got Rs (63500 - 1905): Rs 61595 Ans. Sharadadevi got Rs 61,595.

146

NA\TI{EET MATHEMATICS DIGEST : STANDAR-D

Q. 2. Vishwasrao bought a plot of land worth Rs

_

VIII

2150,000 through

estate agent. If the commission was paid at 2.STorhow much did pay the estate agent?

Solution : Let Vishwasrao paid Rs .r as commission to the estate Rs 2.5 commission on price Rs 100. Rs.r commission on price Rs 2,50,000

. 2.5:_

x

. ,.

&-

100

250000

Rs

After giving commission, Nikita will receive

xx

h:Rs #

R'('-#):*'(+#):*.# 49x

2,5 x 250000

2Zo .'. commission on Rs .r:

Commission is

It is given that

100 x -r :2.5 x 250000

she recieved Rs 17150

:17150

' r-__17150 x 50 17500 49 -r

"' 50 Ans. The selling price of two-wheeler : Rs 17,500.

100

fns.

DISCOUNT AND COMMISSION

.

He paid Rs 6250 to the estate agent.

The middleman through whom grain, vegetables, fruits and.other Solution : Commission : Rs (6630 _ 6500) Let the percentage of commisssion be x. Rs 130 commission on Rs 6500

:

farm produce is sold is also called a commission aggnt. The remuneration he gets is called the commission. It is charged as a percentage.

Rs 130.

(Textbook page 142)

Rs -r commission on Rs 100

Q. 1. A dealer sold tea worth Rs 22,500 for a tea company. rf he got a commission at the rate of l\vo, how much was his actual commission ? solution : 187o commission means Rs lg commission on sale of Rs 100. Let Rs .r commission on sale of Rs 22,500.

l0

:18 Th"rr. '22500 x

.'. x:18

:

", # : *r rr1 .'. amount including commission : R. (,. : #) \ The amount given is Rs 303000 ...

commission

Rs

x 22500 100

.'.

100 x

x:

18

x

225O0

.'. r:4050

Ans. Actual commission : Rs 4050. R*

#

lOIx 303000 x 100 :303000 _ ,c0000 .r': 100 101 - J\ Ans. The original price of the jeep : Rs 3,00,000.

lovo corwnission means Rs 10 commission on sale of Rs r00. Let Rs x commission on sale of Rs 32400.

loo

Th"n. '324ffi:

Solution : Let the selling price of the two_wheeler be Rs

Q. 2. A dealer sold 24 rolls of cloth each of worth Rs 1350. what commission did he get at the rate of ltVo ? Solution : Total sale : Rs 1350 x 24:Rs 32400.

.'. x:10 .r.

lo x

x 32400 100

.'.

100

x.r:

10

x 32400

.'. x:3240

Ans. The dealer got Rs 3240 as commission.

NAVNEET MATHEMATICS DIGEST : STANDARD

vru

a. 3. Use the given information and write the proper figures blank spaces

:

DISCOUNT AND COMMISSION

The amount farmer would get : total sale price - total commission : Rs (8400 -252): Rs 8148.

(3) Total sale price of onion :rate x quintals of onion : Rs 180 x 2l : Rs 3780 Total commission at ZSVo : total sale price x rate of commission

:

Rs 3780 x

)<

7: 100

Rs 94.50

The amount farmer would get : total sale price - total commission : Rs (3780 94.50) : Rs 3685.50.

-

(4) Total

Solution : (1) Total sale price of fenugreek :rate x no. of bundles

:5 ffi x 500: Rs 2000 Total commission at 2Vo : total sale price x rate of commission

:

Rs 2000

3: " r00

Rs 40

The amount farmer would get : total sale price total commission : Rs (2000 40) -: Rs 1960.

-

Q) Total

: :

sale price

of grapes

rate x no. of cartons Rs 70 x l2O:Rs 8400

Total commi ssion at 3Vo : total sale price x rate of commission

:

Rs 8400

x;l:

100

Rs 2S2

sale price

of wheat rate x quintals of wheat Rs 850 x 30: Rs 2Sr500 Total commission at 1.52o : total sale price x rate of commission

: :

:

Rs 25500

* E: 100

Rs 3g2.50

The amount farmer would get : total sale price - total commission : Rs (25500 - 382.50) : Rs 25117.50.

(5) Total

: :

sale price of pomegranate

rate x quintals of pomegranate Rs 1450 x 16 : Rs 23,200 Total commission at 4Vo : total sale price x rate of commission

:

Rs 23200 x

a

,05:

Rs 928

The amount farmer would get : total sale price - total commission : Rs (23200 -g28):Rs 22,272.

trtr 1. Cylinder

VOLUME AND SURf,'ACE AREA

:

.'.2512:3.14x12xIZ.5

A

cylinder has two plane ends. Each plane bnd is circular in shape. Each ofthes" i. u base of the cylinder. The remaining "ui"O surface of a cylinder is curved. This surface ls k rown

curved surface of the cylinder. The area of curved surface is called the curved

a, the

,'.

.'. r:8

upper base and h is the height

2. Volume of a cylinder The volume of a

:

cylinder:

:

,h"

"f

area

curved_ surFace

as the

Solution : Here, Cylinder

";h;;.

h:

Q' 1' A cylinder has a base of radius

5 cnl and height of 21 cm.

V_

what

Solution : Here,

d:

cyrn#3l;[:T

14 cm

r

\/_o

The volume (V) of a cylinder

d

.'.

I 54000 cm:ffi:154

litre

Ans. The tank can hold 154 litres of water.

14cm

: 7 cm; h: l7

132 x7 ." r:2"n:2lcm The volume of a cylinder : nr2lt :! x2t x2t x25

:34650 cu cm Ans. The volume of the cylinder:34r650 cu cm. Q. 6. what is the volume of iron required to make a 70 cm long rod of diameter 2.1cm?

cu cm

Solution : Here, diameter

:

I2.5 c{ni

nr21

litres

: ZrEr ^ 22 tJz:zx ... 1.^ 7 xr

"-

Q. 3. The volume of a cylinder is 2512 cu cm and its height is 12.5 Find the radius of its base. (Take :3.14) n Solution : Here, volume : 2512 cu cm; h: The volurne of a cylinder

cm: I

--

-1:#

:2618

nr2h

:?r35x35x40 :154000 cu cm

Now, 1000 cu 154000 cu

:

cm

Circumference

: nr2h 22 :7x /x7x17

Ans.

:70 cm ... r:!:,0 _"^:35 22

"-.

Q. 5. The circumference of the base of a cylinder is r32 cm and its height is 25 cm. What is the volume of the cylinder? solntion: Here, the circumference of the base :132 cm; h:25 cm; \rY .r.jr

?

22 :7x5x5x2l

T: ::1g

cm; d

The volume of a cylindrical tank

nr2h

its volume ? Solution : Here, r:5 cm; h:21cm; The volume (V) of a cylinder : nr2h

or tne

h:40

of the base x height

nr2 x

cylinder: g cm.

Q. 4. The height of a cylindrical tank is 40 cm and its aiu*"tu, i, io How many litres of water can it hold? (1000 cu cm: 1 litre)

(Textbook page 146)

Ans.

2512

_-:64 3.14 x 12.5

cm Ans. The radius of the base of the

surface area.

r is the radius of lower base.as well

. ::-

7"

r:

?

:2.1

cm;

... r:i:T:1.05

cm;

length h:70 cm, V: ? The volume of the iron required: volume of the rod (i.e. cylinder) : nr2h

:?*1.05x1.05x70

Ans. 242s5 cu cm or iron ,.

*ii13'ljff*

the rod.

152

NAVNEET MATHEMATICS DIGEST : STANDARD

VIII

VOLUME AI\D SURFACE AREA

Q. 7. The radius of the base of a cylindrical tank is 0.4 m and its is 0.8 m. How many litres of oil will the tank hold?

2. A cylinder has a height of L m and the circumference of its base is 176 cm. How many sq cm is its total surface areal Solution : Here, circumference:176 cm; h:1 m : 100 cm; S, : l

1 litfe : 1000 cu cm) Solution : Here, r:0.4m:40 cm, h:0.g m : g0 cm The volume of a cylindrical tank : nr2h

'- (n:3.14,

:3.14 x 40 x 40 x 1000 cu

.'.

cm:

176 x7 .'. 176:2nr .'. 176:2 x T x r .'. r: x22 :28 2 The total surface area of a cylinder :2nr (h * r) :2x 22_ x28(100+28)

g0

:401920 cu cm 1

litre

4olg2ocu cm

-=4or92o :401.92 litres 1000

7

22

-2x-x28xI28 7

Ans. The tank can hold 401.92 litres of oil.

:22528 Q. 8. The radius of the

cylindrical wooden block is 5 cm and volume is 1100 cu cm. How many discs_of radius 5 cm and height 2 can be cut from this block of wood ? Solution: Here, r:5 cm, V:1100 cu cm The volume of the wooden block : nr2h base of a

... 1100:?*5x5xh

- 1100x7 .'. h::=---=-_: zzxJx5

.'.

3. A cylinder has a height of 15 cm and the radius of its base is 5 cm. What is the area of its curved surface? (Take n :3.t4.) Solution : Here, r: 5 cm; h: 15 cm, S" - ? The curved surface area of a cylinder :2nrh

:2x3.I4x5x15:471

:

radius of the wooden disc

height of wooden block the number of disc: height of disc I4

:

5

L

Ans. 7 discs can be prepared.

.'.8448:2x+x28(h+28) .'. 8448 : r76(h + 28)

3. The surface area ofa cylinder : (1) The curved surface area of a cylinder :2nrh (2) The total surface area of a cylinder :2nr (h*r)

8448

.'. h+28:,=-

t76

.'. h+28:48

.'. h:48-28 ), h:20 is g cm and height 35 cm.

sqcm

Ans. The curved surface area of the cylinder:471 sq cm. Q. 4. The total surface area of a cylinder is 8448 sq cm. If the radius of its base is 28 cm, what is its height ? Solution : Here, S, : 3448 sq cm; r :28 cm; h: ? The total surface area of a cylinder :2nr(h * r)

:;:,

Q. 1. The radius of a cyrinder

sq cm

Ans. The total surface area of the cylinder :221528 sq cm.

14 cm

Now, the radius of the wooden block

cm

cm

Ans. The height of the cylinder:2O cm.

what is the

of its curved sur{ace? Solution: Here, r:8 cm; h:35 cm,.S,:? The curved surface area of the cylinder :2nrh:2x+ x 8 x 35 :1760 sq cm Ans. The curved surface area of the cylinder :1760 sq cm.

Q. 5. The radius of the base of a cylindrical column of a building is 25 cm and its height is 3.5 m. It costs Rs 15.50 per sq m to paint this column. What will it cost to paint 10 such columns? Solution : Here, h : 3.5

m; r : 25 cm :2 ^:! 21004

^:

1

-

I54

NAYIYEET MATTftMATICS DIGEST : STAITIDARD

VIII

/

VOLUME AND SURFACE AREA

The curved surface area of the cylindrical column

:2nrh:z

:,f,i.i,tl

*?7 r!4"*!

1. Find the volume of the cone of height 7 cm and Solution : Here, r:9 cm; h:7 cmtY:'l The volume of a cone :! nr2h

2

:t

11

to

-

.'. the curved surface area of Expenditure for painting l0

: :

l0

such

columns:10

x#:55

sq

such columns at Rs 15.50 per sq m

Rs 15.50 x 55

Q' 6'

The totar surface area of a cyrinde r is 2464sq cm. The height the radius of the cyrinder are equat. Find the radius of the base or

cylinder. Solution ; Here, total surface area : 2464 sq cm, h:r The total surface area ofa cylinder :2nr

(h*r)

: x (Say)

.' ?1go:2x+xx(x*x) ...fn:r':*l . 2464x7 :56 x 7

.'.

x:2x7:14

, ... x':

56x7

2 :28x7:4x7x7

The volume of a cone

:

1

r-

x area of the

:1l^nrzh.

base x height

the radius and

_ ... t-2_924x3x7 -7x7 - nrn

:

?

.'. r:7cm

Ans. The radius of the base of the cone :7 cm. base has a radius

:576 .'. h*24 cm :! The volume of a cone nr2h : I"? x7 x7 x24:1232 cu cm h2

:252 -72 :625 -49

Ans. The volume of the cone-shaped figure:1232 cu cm.

cone has one circular flat surface and one

the height of the cone.

height of a cone is 18 cm and its volume is 924 cu cm, find the

radius of its base. Solution : Here. h : 18 cm; V :924 cu crr; r The volume of a cone :! nr2h .'. 924:!"? x12 x18

.'.

4. Cone: curved surface. In the figure, seg OA (ft) is the height, seg AP (r) is radius of the base and seg OP (/) is the slant height of rhe cone. There is a right angle between

t}lLe

of 7 cm and its slant height is 25 cm. Solution: Here, r:7 cm:, I:25 cm;Y:? 12+12:h2 ... (Formula) .'. 252:72 +h2 ... (Substituting the values)

Ans. The radius of the base of the cylinder: 14 cm.

A

cu cm

Ans. The volume of the cone - 594 cu cm. =1"?

3. Find the volume of a cone-shaped figure, if its

. Z"n:xx2x 2x2

a base radius of 9 cm.

x9 x9 x7 :594 2. If

Rs 852.50

Ans. The expenditure for painting: Rs g52.50.

.'.

(rexttook page t5o)

,I I I I I I I I I I

4. The volume of a cone is

462 cu cm and it height is 9 cm. Find the radius of the base of the cone. Solution: Here, Y:462cu cm; h:9 cm; r:? The volume of a cone :! nrtl,

I 22 .'.462::X-=-xr'x9 3 7 .'. r2:'7 x7 .'. r:7

462x3x7 .'. 22x9 :rzI cm

Ans. The radius of the base of the cone :7 cm. Q. 5. The volume of a cone is 9856 cu cm.If the diameter of its base is 28 cm, what is its height and its slant height? Solution : Here, V:9856 cu cm; d:28 cm

156

NAVNEET

tUArfrnUlrrCs

d28 ... r:r:T:14

cm;

The volume of a cone

DIGEST : STANDARD VIII

h:?, r:?

:

:22 x 16:352

9856

nrzh

x3x7

71;10;10:h "' h-48 cm Now, /2 : 12 + h2 ... (Formula) : r42 + 482_ ... isuu*il;, .'. l-

:196 *2304:2599

3. The radius of the base of a cone is 9 cm and its height is 40 cm. What is its curved surface area? What is its total surface area?

(n:3.74) Solution : Here, r :9 cmi h: 40 cm; S" : ? S, : 12:r2 +h2 ... (Formula)

the values)

50 cm

:4g

Ans. The height

cm,. the slant height -c

:50

:92 +402 cm.

il;;;""a i,, o",rn;; i; f:j1.""::T,:_g'i:.il; "i,."* the volume of the cone. " Find :l.lqj Solution : Here,

r:

(zc

5 cm;

The volume of a cone

Ans. The volume

h:

nrr6: jx:.r+ x 5 x 5 x 12:314 cu of the cone :314 cu cm. surface area of

gl IP 11{ :gTg tu

a

a cone: TErl

: 191_g.t.1 !on"

nr (t

* r)

of a,cone is t0 cm and the radius of the ba I7 cm. ^la::,,:lr.n:i*nr What is its curved surface Solution : Here,

/:

]!lcurveo

10

surrace area or

1600: 1691:

(41)2

4't cm

The total surface

cone:nrl :3.14 x 9 x 41.: 1158.66 sq cm : area of a cone nr (l I r) :3.14 x 9 (41 + 9) :3.14 x 9 x 50 :'141,3 sq cm

Ans. The curved surface area:115E.66 sq cm; the total surface ar€a: 14tr3 sq cm. Q. 4. The height of a cone-shaped tent is 10 m and the radius of its base is 24 m. (i) What is its slant height? (ii) How much canvas is required to make this tent? (n:3.I4) Solution : Here, h: I0 trri r:24 m, I --,? ... (Formula) Q) 12 : 12 + h2

:242 +102 ... (Substituting :576 * 100:676: (26)2

:t

.'. l:26

*:-";':;3;1'.'n "'

... (Substituting the values)

(ii)

the values)

m

The canvas required for the conical tent

: the ctirved surface area of the tent : nrl:3.I4 x 24 x 26: 1959.36 sq m Ans. (i) The slant height :26 m; (ii) The canvas required: 1959.36 sq m.

Sllution : Here, I :9

(i)

." t-

*

?

The curved surface area of a

u."ut---'

cflli l": Z ".;nrl"i" The curved surface area of a cone: Ans.

=

g1

12 cm

:!

5. Surface area of a cone : .--------tr'ormulae : (1) The curved

sq cm

Ans. (i) The curved surface arcii:198 sq cm. (ii) The total surface area:352 sq cm.

! ...9856:I"?xt4x14xh

.

VOLUME AND SURFACE AREA

cm; r The curved surface area

rea

of

:7

cm; S"

:

?; S,

:

?

of the cone: nrl a cone

= ,lri*\'x

:; 22 x 7(9 +7)

9:

198 sq cm

Q. 5. How much metal sheet will be required to make a cone (hollow) of height 4 m and base radius 3 m? (z:3.I4) Solution: Here, h:4m; r:3 m 12:12 +h2 ... (Formula) :32 + 42 ... (Substituting the values)

158

NAYNEET MATIIEMATTCS DIGEST : STANDARD

:

9

+

16

:25 = (5)2 ... /:5

VOLT]ME AND ST]RFACE,AREA

VIII

m

(Textbobk page 154)

The curved surface area of the metallic cone : TErl:3.14 x3 x 5 :47.1 sq m metal sheet wiu be required.

1. The radius of a sphere is 30 cm. What is the volume of thesPh.ere

_1i:: 1l:1 :1-_"_r Q. 6. The slant height of an ice-cream cone is 12 cmand its surface area is 113.04 sq cm. What is the radius of the base ice-cream cone ? (n : 3.14) solution :,Here, 12 cm, curved surface

r:

The curved surface area of a. corro: ftrl

area:

rr3.04 sq cm

?

Solution: Here, r:30 cm, V:? The volume of a sphere :!nf

.1x :

1

3.,14x30 x 30 x 30

13040 cu cm

Ans. The volume of the sphere: 1,13,040 cu cm.

.'. 113.04:3.l4xrx12

. ..;-:

2. The volume of a sphere is 360fi)z cu cm. What is its radius?

113.04

3.14 x tZ -icm Ans. The radius of the base,:3 cm.

Solution : Here, V

The height of a cone-shaped paper hat is 24 cm and the radi the base is 7 cm. How m""n pap"r will be n""n"* ,0.-' ""nrr".*" r--r-ihats?

h:24

cm;

r:7 ci

..

. .J"

,il::

"-

6. Sphere: Fon4ula : The volume of a sphere

:!

nr3

nr3

4

3

-?.7000

r: Jtrw:30

cm

Q. 3. 27 spheres (metallic) of radius 'F' arre melted and a new sphere is formed. What is the radius of this new sphere? Solution : The volume of a sphere with radius 7:f, nf

x7 xzs

rhe paper required for preparin, llo"l .'. the paper required for preparing l0 hats :550 x l0:5500 Ans. 5500 sq cm of paper will be required.

?

Ans. The radius of the sphere:30 cm.

cone:lttl 22

36000 x

I

.

:;

:!

r:

... 36oooz :!nf

...

!'. 12:rz+h2 ... (Formula) . :J2 qc4z (Substituting the values) :49 * 576 :625 :252 .'. l-25 cm The curved surface area of &

360002,

The volume of a sphere

Q' 7'

Solution : Here,

:

.'. sq

the total volume of 27 sphere

s:27 ,! nrt .J

...

(1)

...

(2)

This is the volume of the new sphere.

Let the radius of the new sphere be R Then, the volume of the new sphere :! From (1) and(2),

AA

t

nRt

:27 x

.'. Rt :2713 .'.

R

nR3

,nr3 ... [Cancelli;^r1"from both the sides]

:3r

Ans. The radius of the ngw sphere

:3r.

160

NAYIIEET MATIIEMATICS DIGEST : STANDARD VIII VOLUME AI\D STJRFACE AREA

7. Surface area ofa sphere : ---

Formula: The

' 12^:

:4nr2.

surface area of a sphere

x7 4 x22

616

.'

.r2

:49 .'. r:7

Ans. The radius of the sphere

Q' 1' Find

the surf'ace area of spheres of the foltowing radii

(l) r : I sm (4) r:2.8 cm

(2) r :10.5 cm (3) r: l0 cm fl (5) r:9.g m (6) r: {2 n1.

Solution :

(1) Here,

r:7

:

:

Q. 3. If the surface area of a sphere is 314 sq cm' find its Yolume. (Take

(2) tlere, r: 10.5 cm The surface area of a sphere :4nr2

(3) Here,

r:

10 cm The surface area

ofa

The surface area of a sphere

:4nr2

. 314:4 x 3.l4 x r' .'. r:5 cm

'

:4nrz:4

cm

(4) Here,

r:2.8

Now, the volume of a sphere

: 1386 sq

"r,n"lolJ,j:il:;; :H

sphere

x 3.14 x 10 x 10: I

cm

:

:

:

15lo :!33 cu cm x 3.r+x 5 x 5 5 : " Ans. The volume of the sphere

The surface area of a sphere

Solution : Here,

d18

7:-:--9 '22

9.8 m

:4ftr2

:

+

:

T

cu cm.

:

d:

:3.I4)

18 cm

Cm

The volume of air in the balloon: the volume of the sphere 4a

?x : 1207.36 sq m.

" Ans. The surface area of the sphere

r:42 m The surface area of a sphere

3t4 4 x 3.14

:!"f

area of the balloon? (Take z 9g.56 sq

cm.

r:

^

rZ:-:)5

Q. 4. The diameter of an inflated balloon (spherical) is 18 cm. How many cubic centirnetres of air does it contain? What is the surface

The surface area of a sphere 4nr2 4 x +x 2.g x 2.g Ans. The surface area of the sphere: gi.SO sq

(5) Here,

z:3.14)

Solution : Here, the surface are'?: 314 sq cm.

.'

*""

cm.

=7

Zl+l

cm

Ans. rhe surrace

cm

9.g x 9.g

-anr-

1207.36 sq

:!x3.Iax9x9x9

(6) Here,

:3052.08 cu cm

:4ftr2

The surface aiea of the balloon

:

the surface area of the sphere

:4nr2 4 x.3.14 x 9 x 9 : 1017.36 sq cm Ans. The volume of air in the balloon:3052.08 cu cm;

:

the surface area of the balloon :1017.36 sq cm.

.'.616:4x?xr, 6/I.,lavneet Mathematics Digest : Std.

VIII lE059fl

,E

DIVISION OF POLYNOMIALS

I5m3 2bs

Q. Comptete the fo'or+i.r* .*r"_"oo

oriil*

,oil'btanks

:

(l)

(4)

+b2:2b3

+n2 e) 36x5 +4x3 (S)

Ana

Ans.

(l)

4na

.',43:5x8*3

'.'

15,m3

2b5 : 28xa

--,- -

:5m x 3m2l b2 x 2b3l

-

=7x2

9-v--'.:

+l3m 4M2 -(-104)

5?nt3

-n2:4n2

(3) 20c2 + L5c:!c (5) 40a2 -:(-10c) -

2yx5y-10y,

...

28xa+(-7xz):-&2 - 8yt * (-yt) :8J' Q. 2. Divide:

m2

'.'

+5m:3m2

x

1-

4xz)l

:rl $r1

(3) 20c2:l5c (6) (-33ya)+(-11y).

(2) 52m3 -l3m:4rnz (4) 36x5 - 4x3 :9x2 (6) (-33yo)+(-11y):3J3. .:

-h

Q. 3. Write five different pairs of dividends and divisors such that the quotient in each case is 2r. Solution : The degree ofthe quotient 2; is 1. .'. the degree of the dividend must be 2 or more than 2 whole number.

x9tn2:9m4

.2. Division of polynomials :

Also, the degree of the divisor must be less than that of the dividend by Five different pairs of dividend and divisor :

Ex. 35rs -:7.r2 To divide, we must consider 35-rs : 7x2

35:7 x5and x5:x2xx3 .". 35.ts :7x2 x Sx3 '.

35-15

x

wtichpolynomial

+7x2:5x3

?

4. Dividing a binomial by a monomial

Ex. 1.

(27ms

9m3

-9mt)-3mz

-3m

z*tlfi*t -gr; (1) 21b +7 -3b (2) Z6Oz' , 6p:6p (3) 5a3+a3-S

(4) -30n2 -3n:

, -l0n

t

.''

t"'.'

x 3bl 36p2 :6p x 6pl

5a3:s3y51 t... -30n2:3nx(-l0z)l I

27m5

2lb:7

0

,

-9m3 -9m3 +, 0

:

Explanation : Divide 27ms by 3m2. 27m5 :3m2 x 9m3

.'. quotient 923 Subtract the product 3m2 x9m3 :27ms from the dividend. Divide the remainder -9m3 by 3m2 -9m3 :3m2 x (-3m)

Subtract the product 3m2

from the dividend. Remainder 0.

x

(-3m):

-gm3

l.

164

NAVNEET MATIfiMATICS DIGEST : STANDARD

vrn

DTVISION OF'POLYNOMIALS

(4)

2y

(s)

y2)2y3 +5y2

8a3

0 *5y2

0 -12a2

_5yt

-+ I2a2

0

0. 1. Fitl in the blanks in Ans.

br:i

the division

+5 Remainder:0

-7b3 +3b5 +303 shown

(7')

3bs:3b3 xb2

3rlros -76t 3bs

0 -7b3

.'. 3b5:3b3:h2 -7b3 :ZO3 x 1-!1 -7b3 :3b3 : _ z3

:

b2

(4) 2y3 + (7) 24ma Ans.

(1)

n

Syt;

t'

-16m2;

n

n)n2

(2)

gn (2)

+3

_n2

0 +3 Quotient: z Remainder:3

p'-6;

.

p (5) 8d3 -1h3; taz (g) 24ma _16m2; gmz.

lF= p?

0 -6 Quotient:p Remainder

(3) 3r,

- -6

- tSr; +t

(6) l8,b4

(S)

3m2

wtlz+rrf

r6m2

0

15,

- 15.r - 15x + 0

Quotient:.r - 5 Remainder: 0

+Tt

-2

-to*t

24ma

0

-16m2 + 0

0

+

Quotient:3m2 Remainder:0

Quotient:3m3 -2m Remainder:0

-2

Q. 3. Divide:

(l)

-8p)+4p (2) (9c2 -2lc)-:3c (3) (7x3 *28x)+7x (5) (t2b4 - 7b2\ -" 3b2 (6) (3dt + a3) -+ a3 @) 04 - 5y') * y', 0) $W1 +15p5)+3p4 (8) (8126 *20m4)-.8m3. (12p3

.- Ans.

(3) x-5 Z*)-U' ^) 5X-

p

Remainder:0

Rernainder =.0

-1,6m2 - !6m2

0

:

*3;

Quotient:90'

O -16m2

Q' 2' Divide the binomiat by the monomiar. write the quotient and remainder n2

0

24ma

-l; Remainder:0 ... 3bs _76s : (3b3) x (b2 _ll + fOi.

Quotient

-

o + 13b3 _ l3b3

Quotient:4a-6

-2m

Bm)24m4

=7b3 +

-:-

(l)

3m3

+ r3b3

0

-

Quotient :2y

2b2)t8b4 tgb4

2yu

0 *.r

(6) su'z+|u

4a-6 2a ,$at _tn,

*5

(2) 3c -7

(1) 3p'-2 qpltzpt -sp ]zP' o -8p

3c)9c2

+4 7x)7x3 +28x

-zrc

0

0

x2

7x3

2"'

-8p + Quotient:3p2 Remainder:0

(3)

7*

0 *28x

-2lc -2lc +

28x 0

0

-2

Quotient:k

-7

Remainder:0

=xz +4 Remainder:0 Quotient

NAVNEET MATIffiMA"ICSDIGEST : STANDARD VUI

(4)

y2

-5

v'zlf -iuz

(5) 4b'-4 3b2)72b4

_12b4

0

Quotient:y2 -S Remainder:0

3a'+ t ^i-..a')3a'+a3

i i-

Ja-

i ia3 i-

o +a3

jo

0

Quotient

ro

i

-762

0 _7b2 --r 7b2

- 5v'

(7)

j i

:4h2

IlL : JJ_ -z3r :: \
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