Navneet Maths Digest Std 8th

BASED ON THE TEXTBOOK

NA[/NEETHIETTf,EffiET!$$

DIGEST

CONTENTS

NAVNEET MATHEMATICS DIGEST : STANDARD

VIII

Squares and Square Roots1. Squares and Square Roots 2. Inational and Real Numbers5

(Textbook pages

I

to 7)

[. Revision-Square,division method:

square root

:

t417

3. Parallel Lines 4. Quadrilaterals 5. The Circle 6. Area 7. The Circumference and Area of a Circle 8. Statistics

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

24

3I35

(Textbook page 1)

48

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

:

...

55 62 69 84

9. Variation10. Identities

and Proportion

+l

I

2891

4

11. Equations in One Variable

*

+7(t)

27

+4 +585

202516

Miscellaneous Exerbise-1 (Textbookpage g9)

9I

34

000

0'425 425

tffil12.Cubes_ and Cube Roots

lffil

Ant. The square root of 289 is 17.6400e

Ans. The square root of 2025 is 45.

(4) 646416

101

253

804

13. Indices 14. Construction'of Quadrilaterals 15. The Arc of a Circle 16. Joint Bar Graphs 17. Compound Interest 18. Polynomials 19. Discount and Commission

105 108 116

123

t28138

142150

Anl, 'l'ho

square root

of

Ans. The squarc root of646416 is 804.

20. Volume and Surface Area 21. Division of Polynomials

64(l0t) is 253.

r62173180

22. Factors of Polynomials

*

Miscellaneous Exercise-2 (Textbook page 170)

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

81.

or -9.

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.(Textbook page 2)

(4) The positive square root of,625.

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

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

(2) The negative

square root

of 81 :121

uffi. .Ai'

-

3.

(3) The negative square root of (4) The positive square root of

, _J4_:

(2)

2s

t , ,[1.(3) -

$l|:4T:

r!: 3:

(3) 100

'_glTg

19_91

of 25 is 5 or

-5.10.

Q, 3. lVrite the values ofAnc. (1)

:

$yg:

of 100 is t0 or _ -19-gl

(4) ts6Ans. The square root of 196 is t4 or _14. (s) 324

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

-O- -3

(D(5)

\E:sro.

lx: -s

/roo:

'

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

4ry:

fr:

lgtg:1-919!324 is tB or _ t8.576 is 24

9,f,, the numbers 9, 25 and 169 are the squares of 3,13,

-

3; 5,

(6)

s76

-

-

5 and

13 respectively

Ans. The square root of

or _24.:

2, 3, 5, 20, 27, etc., are not the perfect squares of any integers. Suchnttilibers are called non-perfect square numbers.

3. Writing and reading a squarO root

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.

,r/-,

1

The gquare root of a number that is not a porfect squar.e : A number which is notrquare root is 20 isa perfect square also has a square root. e.g., the

- J-,utU

,/n.(Textbook page 4)

Reading :

-.JA JA:S

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

and

- j64- -

Q,

8.

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. FBolutlon : By Pythagoras' theoiem,rlI

(Textbook page 2)

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.

g \F

tE4

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

(3) The negative square root of 625.

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

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

U

NAVNEET MATTIEMATICS DIGEST

3

STANDARD

VIII12 cm and7,


9

Q. 2. Find the length of the diagonal of a rectangle of lengthbreadth 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 righrangledtriangle. By Pythagoras' theorem,

'l'he approximate value of the square root of a non-perfect square number:ln the above example, we have calculated the square root of 30 up to lirur decimal places.

t-

a

../30:

5.4772.

l{ere, the digit2 in the fourth decimal place is less than 5 or 5. We drop49

u(Aqlr:.'./(AC)

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

u(BC)12

+ t/(AB)12

: uEi

:193"",493 cm.

lhrrt digit and keep the remaining digits as tu;rproximate value

it is. So, .160:5.477 isthe

of

.f

up to three decimal places.

Ans. The length of the diagonal of the rectangle is

"

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

thop 7 and increase the previous digit by 'f'lrus, we get,

1.

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

nET:5.48

up to two decimal places.

'l'lro approximate value of

.f,0

up to one decimal place will be 5.5.(Textbook page 6)

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

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

:

B

;llncc by division method. Then write the approximate value of each8cm

(8)2 + (8)2

:64 * 64:128

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

... /(AC) : ,EZS:

Hqurre root up to the

third, second and first decimal placesAns. ./io:3.1622...Approximate values:

:

"^

tlt l0I

3.1622

6. The square root of a non-perfect square number30 is a non-perfect square number. Let us find 5.47 7 2

10.000000009

.f

by division

It frl

0100(r

up to three decimal places up to two decimal places

:

30.00000000

rlfr .l

+7

1087477

-25 5 00 - 416

*50:

s.4772...

6t 039003756

"/to:ltezu'Eo:ne

:

As 30 is not a perfect square, wo

will get more decimal places, iwe continue the process

t6fr

084 7 609

of

ll2

07I7

5422

6629

02

44In the above example,

219084

0

the value of the square root. This process is never endi In 5.4772..., the dots on the ridenote that further there

t26 l',)"42

0t4400 t26440175600 126484

up to one decimal place

:

u/n:s.z

t2h | )44

will

be

049tt6

02

infinite number of digits.up to four decimal places.

,,,m:5.4772is

10

NAVNEET MATHEMATICS DIGEST

T

STANDARD

VIII

SQUARES AI\D SQUARE ROOTS

(2) 13s

11.6189

(4) rooo Ans.

3r.69

2

27Ans. .r/tooo :31.6227.-

I+1

Isroooooooo-1035

./35:

u.6189...:

3

tooo.oooooooo

2t

Approximate values : up to three decimal places

+3 61

+1

2t14 00

JBs:

tt.6t9:

+l

6t37 56

Approximate values

:

up to three decimal places

:

+6

226

up to two decimal places

1356

222

./tooo :31.623up to two decimal places:

232r1

004400

JBs:1t.62up to one decimal place:

r2644 0r756

232t207 9008

1264844427 r29 048

.4ooo:11.62up to one decintal place:

2322823 23 699

JBs:11.6

185824 2207 600209 t32r 0 t 1627 927.87 47 |

3?D437(5) 5328

+1

/ooo:31.6

72.99315328.OOOOOOOO

23237(3) 7772

8

+7An*

.7

+2ffi:27.8747-.:

142

-49o4

,l,ns.

.f3x

:72.9931...::

284

Approximate values

in.ooo-ooooo

+7+855

+2 47

-4

377

Approximate values : up to three decimal places

54867

-329

+7557444

048 00 43 84 04 1600

up to two decimal places

'nn:2ir.B7s

013 5900 13 1301 0 459900 437 9 49

t44 00 I30 4L t

up to three decimal places

!Br8:72.ss3up to two decimal places:

!&r8:72.ssup to one decimal place:

02t95100

:

1459861

J$rB:73.0.:

38969 0263 r0022297 76

Jm:27.88up to one decimal place:

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

$n =27.e

557 487

557494

04012400 3902409 0109991

Itlx. Find the square root of 249.3241by the division method. | 5 .7 9 iFirst, the digits in the iinteger part of the .1 if,irst, I , 4 9 .3 Z 4 I inumber (249) are paired o.ff from the right, 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

-r252t

4 634

I

-r

I

H++iquotient

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

NAYNEET MATHEMATICS DIGEST : STANDARD

VIII


(Textbook pageT)

(3) 34.1s8:

(4) 720.s

Q. 1. Find

(t)

s6.2s

the square root of the following numbers by the division method

Q) tst.zsr 2.3

5

+5

34.r5S000

2

26.842 720.500000^

1

+829

108

-25

+4 +4

tt64

0915 864 05180 4656

Ans. n/56.25:7.5.

I t68411588

l(3) 49.s616

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

052400 46736 05664

7.042 -L1 4

n.i2TZ41

2t.o6

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

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

+6

+2 46

320

+1

41

-4

o4 02 0 002

4249.5616:7.04.

i two

decimal places.

2 5236 00

Q. 2. Flnd the approximate value of the square roots of the following(1) s9.03numbers up to the second place :

7.683

(2) 3.41s81.8 4 8

+7 146 +6 r528 +8153666

7

5

9.0 3 0 0 0 0

-49

10 87

I +1

6

r27 0 r2224 4608900

+836 36

2

3.415800 -12

-22401

00

1456

29504

J5c:o3:7.683... Ans. yte.o3 :7.68 correcr up totwo decimal places.

tr54158:1.848...

Irrational and Real r 2 ir IrrationalandRea Numbers ttl

rrr-

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 calledrational number.

(Textbook page 10)a

Q. WriteAns.

the following numbers in the non-terminating recurring

form

:

non-terminating recuning decimal fraction. ln case (ii), 3 : 0.66... is written as :0.6 and52

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

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.

0.16

ttlll

i

7.439 i 10.505'I -l

iI I I

0.058

iI I

1.06 i 0.00021.060

t-

10.6050i 0.0580

iI I

4. Non-terminating recurring decimal forms and rational numbersEvery nutmber rational number.

:

in the non-terminating recurring decimal form is a

*:0.090909... is writtenI

as

il

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

fr:0.09.

t,

Irrational numbers

:

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

calted irrational numbers.

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:

a. chssify the following

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

the decimal form of

..,6 i, non-tenninating

.f

is an irrational number.:

6, tlquare roots of numbers that are not perfect squaresnumbers.

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

(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-terminatingThe 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

(Textbook page 12)

Q, Clussify the following numbers into two groups and label each groupcorrectly:

l\ 4.en

(2) 0.3104s693...

@\E

@trB

(s) 10.0s

(6) 0.1010010001...

non-terminating recurring decimal frbction.

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

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

form.

! Rcul

iI

l

numbers : The collection together is called real numbers.

of rational and irrational numbers

NAVNEET MAI'HEMATICS DIGEST : STANDARD VTII

(Textbook page 12)

Parallel LineS lS llParallellrnes -tttt

r.

(Textbook pages 13 to 22)

l.

Q. Make

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

a table

Drawing parallel lines : You have studied how to draw a line parallel to a given line usingset squares.:

places in the table

o\FAns.

(1) 1.57 Q) v5(8)

(3) 4.10547194:.. (4)

4.8

(5)

0.73s (O lE

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

,/\%Irrational numbersReaI numbers

Using the ruler and set squares, we find that line

I

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

ll line n.

Rational numbers

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.4, The

(r) r.s1(s) 0.73s

(4) 4.8

Q).,fr(3) 4.10547r94...

All the given numbersare real numbers

@d *t

Jzs

(8)

v4%

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

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 bylines

/ and m on transversal:

n.

(2) Intercepts made by three parallel lines onI

a transversal

l

In the figure, line /, line m and line n are parallel

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)

to one another.

a. ln

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

18

NAVNEET MATIIEMA.TICS DIGEST : STANDARD

VIII

PARALTEL LINES (Textbook page 19 & 20)

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. Fig. (2) : Linesseg

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

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

l. In the figureo line ft | line /

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.

If t(xY)

:5, l(YZ):3,

/(PQ) :5.5,

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

: seg RD on the transversal r and

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

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,

ilII

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,and seg NO on line fr.

",53

5.5

/(QR)

... (Substituting the given values)3

.'. 5 x /(QR) :5.5;Ans. /(QR) :3.3.

.'.

/(QR):Y:

1.1

x

3

:3.3.

I I

5. The prope4ies of parallel lines and their transversals with respectto intercepts:

(1) If the intercepts formed by three parallellines"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 ABseg

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

=

intercept

BC, then intercept segDE

(2) The ratio of the lengths of tho intercepts made by three parallel lines on one transversal and the ratio of the lengths of thecorresponding 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.

=

intercept seg EF.

.'. 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)

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

...

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

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

20

NAVNEET MATIilMATICS DTGEST : STANDARD VrrI

PARATLEL LINES

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

(Textbook page 22):

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

Seg

LA, seg AB, seg BC,DM are the five

neg CD and seg

congruent parts of seg-LM.p

PrP. and P.Po with the help of of

a

compass. On the ray QN also, startingil ,l

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 andrespectively.

{1.

l,

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.

(6)

Seg PA, segseg PQ.

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

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+3congruent parts. Take point

L on seg Xy

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

such that

io of division

:l

l$

::g-T:_11

2:3 1:5 l:2 3:4 5:35

ttber of congruent parts of,legment

6

3

7

8

Q,

{. Dlvide

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


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.

a point R on

it such that

.',

total divisions

:4 * | :5

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

/(SP) :

l(Yl):2'

3

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

:

1.

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.

tr1. RcThe types of quadrilaterals named on the basis of their angles and are shown below. (Textbook pages 23 & 24)s

QUADRILATERALS

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

(l) mISTR:90':

Recall the properties of quadrilaterals and

fill in the blanks

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

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

(l)

Side

HC

.r side DF

NX:PE:

____

- -'..{-'-

,-(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.----- -- -- ----::---

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

--

-----

'(cM):t(DM)

(2) Rhombus : A quadrilateral with all fourcongruent is called a rhombus.

sides

Q)dacn diagonal of a rhombus is perpendicular bisector of the other.

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

: /(EU) (1) '(PN) (2) nlutN:90' (3) r(TN) -r(TP) : /(TU) : /(TE)0)

(U#dtbposite

angles of a rhombus are congruent.

hoblems based on the properties-of quadrilaterals

(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

(Textbook page 25)

L congruent. gffine diagonals of a rectangle are congruent..(J),.The diagonals of a rectangle bisect each other.

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

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)

l,

,', /(Qs):8

cm.4.5 cm, what are the lengths 0f the

Properties of a square

:

l,bisector ofthe ot

ll'ln

the square ABCD, /(AB)

i(l)-The,(fr'lhe

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

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

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

QUADRILATERALS

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).Ans. The diagonals of a square bisect each other.

Fhnilarly, seg PQ

/tPg;

:7 .-

=

seg

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

(Given)

,', /(SR):'7 cmAns. /(QR):9 cm;

.'. /(DM): /(FM) /(DM):7 cm ... (Given) .'. /(FM) :7 cm l(or;-/(DM)+/(MF) :(7 + 7) cm: L4 cm .'. /(DF):14 cmThe diagonals of a square are congruent.

/(sR):7

...

(2)

(1)

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

#

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

...

[From (3)]

'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.

t,

o

,', mLons:90".Hrl DAB

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.

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

:

nLDAC + nLCAB

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

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

m/ cAB:25'

=

.'. 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.The diagonals of a square are congruent.

,',

mLACD

:25"

...

(Given)

Ann mLolC

:

65o;

zLACD

:

25o.

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

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

(Given)

cm

(Textbook page 26)

a. 1. In rectangle,(PS)

:9

PQRS, l(PQ)

:7

cm,

"' /{AC; : zl(KC):2 x 3.5 cm:7 cm. Anr. /(KC) :3.5 cm; /(AC) :7 cm.4.

/(KC):3.5

cm

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

Solution : The opposite sides of a rectangle arecongruent.

'fhe diagonals XZ and WY ofM

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

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

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

NAVNEET MATTIEMATICS DIGEST : STANDARD

VIII

QUADRILATERALS

.'. /(xM) :I t6z)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)

:

:j :4

x

B

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

l,

ll'

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

cm

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.

Q. 5. In n LMNP,

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 :360Find

mLL:mLM:mLN:90o.

{, 'l'lrc diagonalsF*tclt rttlter.

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

Anx, rrrl.AOD

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

90o.

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

Now, each angle of

tr LMNP is a right angle.

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.!ns. mLP:90o; n LMNP is a rectangle.(3) RHOMBUS:

llolulkrn : The opposite angles of a rhombusHIT r'ongruent.

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

(Texfbook page 27)

a. 1. If

Anx.

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

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

110'.

70"

I 109

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

IIAIIAI,I,ELOGRAM

ii::l

(Textbook page 28)

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.

'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

l,

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

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

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

:2x8cm

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

...

[Given :

/(XP):8 pml

I MN'f.

'

/(or):

/(Mo)(Given)

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

NAVNEET MATI{EMATICS DIGEST : STANDARD

VIII

Similarly,

/(NO) :

The CifCle

(Textbook pases 29 to 32) (Textbook paee 29)

/(OL)

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

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

/(NO):ry:5

l.

Fill in the blanks

:

Ans. /(OT)

cm; /(NO)

:3

"*.

cm.

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

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.

(4) The length of

seg

AB is twice that of seg OD.

.'. LS = tQ .'. mLS : mLQ .'. mLS:130o ... [Given : mLQ:130o]The opposite sides of a parallelogram are parallel to each other. .'. side PS ll side QR. Side PQ is the transversal.

l.

Look at the adjoining figure. Write the

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

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.

. 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..'

l.

Look at the adjoining figure and write ilhether the following statements are true orSeg TS is not a chord.

hhc: Anr. (t)

(False)

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

(True) (True)

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

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

.!TSi)

-2)'and

(50

the parallelogram.

-x)o

respectively. tr'ind the measure of each angle

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

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

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

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

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

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

|

Theny*37:180 .'. y-180-37143".

.'.

y:I43

.'.

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

)o:1

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

THE CIRCLE

(2) The property

of the distance between the centre and congruent chords of the circle : In a circle, congruent chords are equidistant fromthe centre of the circle. In the figure, chord AB = chord PQ. Seg OM r chord AB and seg ON r chord PQ.

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

chord CD

and

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

AB ry chord

.'. /(OM):

/(ON).

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

,', /(tPD=IAPB.

i,

(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)

'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)

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.

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

is a right-angled triangle.

- 4 "*.

,', l(AM):3

cm.

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

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 ofcircle. 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.

,', /(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,

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

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) /(AB) :2/(AM)

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

,, /(AM): j llnn; ,', /(AM) =|xl0 cm:5 cm.,', ,',(5)2

t-

10

cm --'t

.'. l(BM):1.5

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

cm

+ t/eM)12

:Qr2-

:2

x

Ans. /(BM)

:

1.5

cm:3

cm.

1.5 cm; /(AB)

:3

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

:144

: \844:

Vm lE059il

NAVNEET MATHEMATICS DIGEST : STANDARD VIII

r0u

(Textbook pages 33 to 46)

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 of1s

('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

its chord

bisects the chord.

... /(AM):jl1an)1n

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

'lte nrca of a rectangle:

:jx30cm:15cm.

Ixb :5.2 x

2.5

... ...

(Formula) (Substituting the values)

fght-angled APAM, by Pythagoras' theorem,

:13{t) , 2,1 m, b:1.5 m 'f'lte urca of a rectangle: I x

sq cm.

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

Anr. 'l'he area of the rectangle:13 sq cm.

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

cm

46.

The radius of the circle is 17 cm.

:2.1x I.5:3.15sq m.

b

... ...


Ann. The area of the rectangle

:3.15

sq m.

(i)

,: .1.5 m, b:1.2 m 'Ihe area of a rectangle: I x b :35x1.2

... ...


:4.2

sq m.

Ann. The area of the rectangle

:4.2 sqm.

'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.

l,

(lf Sitlc:25

cm 't'lu: area of a square

: (side)2 :(25)2:625sq cm.

... ...

(Formula)(Substituting the value)

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

lJt Sitlc:2.8

m

'l'hc area of a square

: :

(side)2(2.8)2sq m.

... ...

(Formula)(Substituting the value)

:7.84

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


VIII

AREA

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

:

(7.2)2

... ...

(Formula)(Substituting the value

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: ?'l'hc area of parallelograln

51.84 sq cm.

:

base x

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

,'.

56 56 "-r7

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

.'... ...(Formula)(Substituting the value

: base x 7 : base ,'. base :

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

8 cm

:

Ans. The base:8 cm.

:

(13.5)2 192.25 sq m.

4. What is the area of a parallelogram having base

L3 cm and height

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

I

cm?

ilolution: Base:13 cm, height:5 cm, area:? 2. The formula for the area of a parallelogramThe 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

: base x height .. ' (Formula) :13x5 ... (Substituting the values):65sq cm.

:

Am.

The area of parallelogram:65 sq cm.

.'.

A(

I

ABCD)

:

i

/(BC) x /(AE).(Textbook page 34)

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

:

?

The area of a parallelogram

:

base x

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

,', .190:390

base x 26

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

: ;;:base

Attx. The base: 15 cm.

"'

base:

15 cm

:84Q. 2. The area of a parallelogramis its height ? Solution : The area

sq cm.

Irytgl:fe-E3lggj'f'hr, rrrca

Ans. The area of parallelogram:84 sq cm.is 26 sq cm.

of a triangle

:j

x base x height

If

its base is 6.5 cm,6.5

of a parallelogram :26 sq cm, base : ... (Formula) ... (Substituting

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

: ] x the product of the lengths of thesides fonning the right angle.(Textbook page 37)

height:

?

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

the va

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

l,

h Itrlullrrn :

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

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

:20

NAVNEET MATIMMATICS DIGEST : STANDARD

VIII

AREA

:\x20x30:300sq m

...

(Substituting the

Aros of an equilateral triangle'l'|rc area

:

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

of an equilateral

triangle:f tria")'(Textbook page 39)

Q. 2. What is the area of a triangle whose base lg.2 cm and7.5 cm? Solution : Base : 78.2 cm, height :7.5 cm, arca: ? The area of a triangle : j x base x height ... (Formula)

l. If the side of an equilateral't'hc area of an equilateral

:lxIB.2x7.5

Ans. The

:68.25 sq cm area of the triangle :68.25

...sq cm.

(Substituting the val

Solution : The side of an equilateral triangle

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

triangle:f triO"l2 ... (Formula)v _ ^/-1 _ 12xlZ + ,.. (SubstitutingA ^ Lz

trE

Q. 3. The sides of a right-angled triangre forming the right angle16 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

:36rfiAns. The area of the equilateral

the value)

sqcmsq cm.

triangle:36\F

:jx16x8:64sq cm

...

sides forming the right angle. (Substituting the values)

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

t.

Ans. The area of the given right-angled triangleis its height ? Solution : The area

:64

sq cm.

Q. 4. A certain triangle has an area 125 sq cm. rf its base is 25 cm,

-4 :{(side)2 ... (Formula) t; :V,'"30x304

of the triangle :

height:

125 sq cffi,

base

:25Ann. The area of the equilateral Aroa of a rhombus'l'he urea:

:225t/3

_...

(Substituting the value) sq cm.

?

sq cm

The area of a

triangle:j x base x height

triangle:225\fr

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

of a rhombur

:1 r product of the lengths of the diagonals.

has an area

rr.6

sq cm.

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)

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

l.

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

the v

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)

:

Ans. The base of the triangle:8 cm.

1764 sq cm

Anr. The area of the rhombus

:1764

sq cm.


Vm

AREA

4t

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:

tAilr,'f

702x2 54 -:26cm'lre lcngths of the other diagonal:

:26 cm.

**

product of the lengths of the di(

I'

At-en

ttl'n trapezium

...

1280 1280

:I

x 64 x x

...

(Substituting the v

"

x2 64 :x

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

"'x-40cm

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

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)ltl

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.

:!x12x18:108Ans. The area of the rhombussq cm

(

...

(Substituting the

x

(Textbook paee 42)

:

L08 sq cm.

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'!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?:?

Q. 4. The area of a rhombus isdiagonal :24

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

lltr. rrttl of a lrapezium

The area of a rhombur

:1

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

:

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

"

tlro sum of the lengths of theparallel sides

x height ... (Formula)

...

(Substituting the values)

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

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

...

(Substituting the v

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

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

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 ofdiagonal

'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

l,

:54

cm, the length of the other diagonal

The area of a rhombur

:i

:

-r (Say)

.

(Substituting the values) cm

*

product of the lengths of the di

'

rltc sum of the lengths of parallel

sides:'e#:35sides

.'. 702:I

x 54 x x

...

(Substituting the v

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

:35

cm.


VIII

AREA

43

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

I'l['

rcnrilrcrirneter (s) of the

triangle:

a*b+c2

25

+39 +

22

56 _:60 : 120

cm

'l'he nrerr ttl' the triangle

: :

s(s-aXs -b)(s-c)60(60

265:iG0+23)xheight

...1o cm

(Substituting the

60x35x21x4

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

'#:

height

... height:

12x5 x5 x7 x7 x3 x4

Ans. The height of the trapezium: L0 cm.

:^/12x12x5x5x7x7:15x5x7:42AsqcmAnr. 'l'hcarea

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

of the triangle :420 sq cm.

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,

l,

0l

ltr und 60 m ? Hglullrln : Here, a:

tl

m,

b:61 m and c :60field:a

m.

ttts

renriperimeter (s) of the triangular

* b* c2

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:10.8Ans. The length of the other parallel side: 10.8 cm. 8. Another formula for the area of a triangle In AABC, /(BC) : a. I(AC): b, I(AB): The perimeter of AABC : a * b I c The semiperimeter (s) of AABCc.:

84.5 x 2

: 1l +61 +60: 132 :uu2 z'l'lr$ rrlcrl o1'the triangular

field: :

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

/66x55x5x6l'lrr, t'rst

-

1

1X66

-

61)(66

-

60)

:"/6x11x11x5x5x6 :6 x 11 x 5:330 sq m.l'weeding

:

f:"riffi : Rs 660

a*b*c2

Aru. 'l'ho expenditure is Rs 660.

A(AABC)

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

:./s(s -

a)(s

-

bXs

-

c)

of

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.

f

'lrc scruiperimeter (s) of the

a*b*c triangle: z

da. 1. If the sides of a triangle

(Textbook pages 45 & 46)

_20+21 +

13

are 25 cm, 39 cm and 56 cmo what is l'lrc ru't:t of the triangular

22

:54 :27

^

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

field:

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

-13)

NAVNEET MATIIEMATICS DIGEST : STANDARD VItr

A(APQS)

: ! x theproduct of the lengths of the sides forming the right angle:]x/(PQ)xl(PS)

:9 x7 x2:126 sqmAns. The area of the triangular

9x3x7x3x2x7x2 9x9x7x7x2x2126 sq m.

:tx9x40:180sqmbASQR, base QR:60 rn, height

ST:

13 rn.

plot:

A(ASQR):jbasexheightTbr area of plot PQRS

:)x6Ox13:390sqm

Q. 4. Use Heron's formula to find the area of an equilateralwhose side is p. Solution : Here, a: p, b: p, c : p The semiperimeter (s) of equilateral triangle

: :

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

tn:570

sq m.

lU,

The area of the quadrilateral

plot:570

sq m.

_a*b*c _p*p*p 222

_3p

,:../s(s = axs

The area of the equilateral triangle

-

b)(s

-

c)

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

"(i)"(t) " (t)3 xP-

;if +,'

-fu

rcmiPerimeter (s) of AGHF

e+b

?4+so+sz 136 ,T* r:68m68(68

x!- *P2222 rP-

{AGHF) :"u/s(s -a)(s -bXs-c) 68x34x18x16 2x34x34x 18 x 16 34x34x36x16t ttilru

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

- 34X68 - s0)(68 - s2)

:fO'.

a. 5. In the figure,F PQnSmeasures, find its

is the map of a plot of land. Using the gia ,

area.

aomiperimeter (s),

of aEHF

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

tr

A(AEHF):

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

!

66,''

-

20X60

-

s2X60

-

48)

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.

60 x 40

x8 x12

5x12x40x8x12

NAVNEET MATHEMATICS DIGEST : STANDARD VIII

AREA

47

:40 x 12:480A(

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

tl

: :

sq m

ll

lrr

nltlt',lll'

huse

AC:20

cm,

lrclglrt

.- 8 cm.

'l'lte nrcrr ol' a triangle

=_]ltnscxhcight

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". II \ .'. AABC is a right-angled triangle|

-1,:o"s:H(l rirl crn

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

'l'lre length and the breadth of the rectangle are 5 cm and 4t 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)

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

:lx5x12:30sqcm

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

:Anr, 'l'lrcarea

(80

of the coloured portion:60 sq cm.

-

20) sq cm:60 sq cm

-

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

:

(5)2

+ (0D2 :25

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

*

144:b

llnee All :20169

i lAll('l)

is a parallelogram.cm, height EF

:

15 cm.

'l'lte uret of a parallelogram

:

base x height

:

13 cm, /(AB)

: c :12

:*#

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

cm,

A

20cm

F

15 cm.

:tl#!:+:r5A(AABC), by Heron's formula

cm

'rca

of AEAB

: i :T ; Tgjl': :(300

sq cm

'l'lre lrr:a of the shadei r"gion13X15

A( n ABCD)

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

:^/I5x10x2x3

1s(l5 -s)(15

-

-

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

-

150) Se cm: 150 sq cm

-

A(aEAB)

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

:

.,600

:30

sq cm

dl lrr tlrc given figure, the shaded region isT(3) 30 sq cm.A(JNli.'l'lrc base NE of this triangle is 18 cm. 'l'hc height of this triangle : the breadth of therrr'lirrtgle.o cl

Q. 8. InSolution

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

(1) Each side of the given triangle is 8 cm.the given triangle is an equilateral triangle. The area of an equilateral

N

18cm

.'.

Ir:Se\tar

12 cm.

triangle

rtc rtrea or A QNE

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

Ia8cm

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

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

:I6.,fr

sq cm

Ans. The area of the coloured portion

:rcrQ

sq cm.

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

I'III{

CIRCUMFERENCE AND AREA OF A CIRCLE

1. Circurnference of a circle

:

llla

ktfutltrn : I ltorrnula : c :2nrl

tlcre,r-56cmC'lrctttrtl'e rcrr ce

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

:Znr:2x21-x56

:2x22x8:352cmAnt.('ircumf'erence

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

:

352 cm.

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

circumference

(ll

Hete,

Cln'ttrttl'cren ce

A:n

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

"'

c:nd.'. , - n x2r:2nr

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

r.-7.7

cm

lrt

lore, r

-

2.8 m

C'lrt'rttrrlbrcnce

lllbelow:

ffi

(rextbook page 49)diameters

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

Q. 1. Find the circumference of the circles from their(1) 3.5 cm (2) 6.3 m (3) 0.14 m. Solution: [Formula : c:nd.fHere, cm Circumferen ce (c): vf,

Atu. ('ircumference =17.6 m.

l,

Ftonr the given circumference, find radius and diameter of the:

Clrele

(l)

d:35

lll

l9tl cm

(2)

616

cm

(3) 72.6 m.

Ans. circumfe.J (2) Here,

l;'*:22nj

x

o'5:

lll11 cm

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

1e6:?| x d

d:

198

d:6.3 m Circumference (c):

:? * 6.3:22 x 0.9: 19.8 mt.r

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

Ans. Diameter:63 cml radius :31.5 cm.

: t' t63 22-:

22

x7 :9x7:63cm

31.5 cm

I llcrc, c :616

cm

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

('ilcrrmference: nd

," 616:?"d

.'. d:cm

616

_

xj :28x7:196cm

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

, tl 196 :98 '' ;: ,I l) llcre, c

Ans. Diameler:196 cml radius

:98

cm.

:72.6 m

('ircumference: nd


VIII

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

A CIRCLE

51

... 72.6:2; x d

d:

72.6 x 7

,:t:d

n :

:3.3x7:23]m l,

l

_

__r-:

l r:xWrlle llte propcr values in the blanks in the following tableRudlus

23.1

2

:11.55m

Ans. Diameter

:

23.1

m; radius

11.55 m.

lrl

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 pwe have to find the circumference of the circular place. The radius of the circular place :7.'7 mCircumferen ce

:Znr

Hnlttlkrrr

:

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

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

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

:3

x

48.4

m:145.2

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

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 kmtwo towns ? solution : The distance covered by the wheel of the bus in 1 rotat : circumference of the wheel : Ttd:! x 0.7 :2.2 mThe distance between the two

r/: 9.8 m rl :-:4.9 m 9.8 ,: 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'Atu.Radius

:4.9 m; circumference:30.8 m; area = 75.46 sq m.44 ' r:-:7 x7

tlt

c:44m t' 2nr .'.44:2x'lxrllerc.

2 x22

m

towns: Z2km:22 xdistance

1000

m:

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

wheel:

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.f

circumference

.l

:2.Area of a circle:

22000

)

ZZ

:10000

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

.'. 2| x 12 :616

Ans. The wheel will rotate 10,000 times.

' rz: 616 x7 : 196 22 d :2r:2 x 14 cm:28 :nr2.

r:14cm

cm

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

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

cm; circumference = 88 cm.

}:2-53.1

tzl5x-4:2 .!x .'.5x-4:2x2x .'. 5x - 4:4x .'. 5x-4x:4.', X:4Ans.

(3)

3x+1:-l

.'. 2x: -3 (3x+ 1) .'. 2x: -9x -3 .'. 2x +9x: -3

i-:-l

3

x:4.

.'. 11x - -3" &J

lt3

Ans. x

86

NAVNEET MATTTEMATICS DIGEST : STANDARD VItr

EQUATIONS

tI\

ONE VARIABLE

-7 (4}-:2 3ct.'. 5a-7:2x3a .'. 5a -7 :6a

5a

5m-3 _ 8 (5) _: 2mq9(5m-3):8(2m)

(6) *2 :2 .4x

-n -3

x-4

+75 -=3m:

114\

4 x-L n: zx - r

(15)

3v*53

5

-7:6a-5a- l:A

45m-27:I6m 45m-l6m:27I

." a= -7 Ans. a: -7.

.'

. 29m:27 .'. 7n: *29i

27iiI

+2:2(x 4x *2:?-x 4x-2x: 2x: - IO4x

=til) = - 3(m+7) 5m= -3m-21

-21-lo

= -313l ,T t-T'31

.'. 3(3Y+5):5(3-7Y) .'. 9y * 15: 15 - 35Y .'. 9Y + 35Y: 15 - 1t .'. 44Y:gAns. Y :6.

-7y: t

.t.

y:

o

_10

27i Ans. lrr: 29i -.(8)x*l -:; 6x- I+:)

i

-------j---------lI

Al:: 1:

- l:

(7)x- - :3|

2x

+3

3x (91 15 :4 " bc+1

.'.2x+3:3(x-L) .'.2x+3:3x-3.'. be -3x: *3-3

-X: -6 .'. ir:6 Ans. x:6.

: 6(x - l) .'. 5x*5:6x-6 .'. 5x-6x: -6-5 -x: -ll.'.5(x1)

.'.3x*5:.'.3x*5:8x.'.3x

-8x:28235

- 5x:23'y:--

.'. x:11Ans. x - 11.

jt:;:l

.'. 3(r8-2Y):40-4) *9:15r-10 .'. 136 - 51Y: 5Y +2 .'. 54-6Y:4Y-16 -6Y-4Y: -16-54 - l5x: - 10-9 -5lY-5Y:2-1,36i r*19 - 10Y: -79 -56Y: -134 134 67 19 . .._-7 " t- 56-28 "'loy:70 ,, y-t 3 Ans. y =7. ffi 19 Ans. y : tg.I

+l 5 =29

(17)

8-3vsv

1

i l) : 5(3x -2) .'. 17(8-3Y):1(5Y+2)i

+z: n

4 18 (18) -2v :j

y_+

i

I

I

III

Ans.

23 x: -=-. 5l

Problems:(Textbook page 8E)

(l0l 5x*4 :2 ' '2x+l.'.5x*4:2(2x*I) .'. 5.r* 4:4x*2 .'.5x-4x:2-4.', X: -2Ans. x

(tt\ llm-l2m-13

lIm-l:2(2m*3\, .'.2(x-t2):71, -:2 Ilm- I:4m * 6 .'.2x+4:7xlIm- 4m:6 + .'. 2x-'lx: -21

Poorva is older than Durva by 6 years. The ratio of Poorva's age to Dgrva's age three years hence, will be 4 : 3. What is Poorva's age

Hlryr&lutlon : Let Durva'spresent age be .r years. ?orva is older than Durva by 6 years. ,', Poorya's present age is (x*6) years Thrce years hence'Eurya's age will be

7m:7 .'. m: IAns.

-5x: -25.'. 5x:25.'. .f:5Ans.

- -2.

m:1.

lgorva's

* 3) Years age will be (x * 6 + 3) :(.x

(x

*

9) years.

r:5.

ThE ratio of Poorva's age to Durya's age

will

be 4 : 3.

" r+3

,

x+9 _43

.'.

3(x+9):4(x+3) .'. 3x*27:4x*12

NAVNEET MATHEMATICS DIGEST : STANDARD VITI.'

EQUATIONS IN ONE VARIABLE

.

3x

- 4x: -x:-15

12

-27

.'. x=15

Poorva's present

age: (x * 6) years

:

(15 + 6) years

:2I

years.

Itlvc years agolltt' tge of the younger brother was (x - 5) years (x + 4 - 5) : (x - 1) years' nrrrl the age of the elder brother waslirrrrn the given conditi.'

Ans. Poorva's age today is 21 years.

on,=:5" I X- I

Q. 2. The difference between two natural numbers is72.lf we getdividing the greater number by the smaller number, find the Solution : Let the smaller of the two numbers be x. The difference between the numbers is 72.

.'.

the greater number is

x*72.x 172 j:4

years' Ans. The present age of the younger brother is 15 Q,

- 5) :5(x - 1) ,',7x-35:5x-5 ,', 7x-5x: -5 +35 ,',2x:30 .'. x:15,7(x

From the given condition,.'

l.

'fne numerator of a fraction is 3 less than its denominator' If the

. x *72:4x

.'. 4x:x*72.'. x:24

.'.4x-x:72 .'.3x:72 .'. x*72:24+72:96

Ans. The required numbers ne 24 and 96.

Q. 3. The ratio of the difference obtained by subtracting

2

from

numeratoristripledandthedenominatorisincreasedby2'thevalue fraction ? of the fraction obtained is j. wtrat was the original be x' f{olution : Let the numerator of the original fraction 'l'ltc numerator is smaller than the denominator by 3' ,'. the denominator: (x + 3) x

number, to the sum of twice the same number and 5 is 13 : 15. is the number ? Solution : Let the required number be x.

,'.

the original fraction is

Now, the new numerator is three times rtnd the denominator is increased by 2'

"

+:.

x:x

3x5

The number obtained by subtractingnumber

The number obtained by adding 5 to twice the number From the given conditi

:3x -

2 from

three

times

,',f

the new denominator

:

x'13 1 2 :

*

2.

:2x

-15.

irrrrn the given condition,,'

-

on,=:: 2x+5

15

.'. 15(3x -2): l3(2x + 5) .'. 45x - 30 : 26x -165 .' . 45x -26x:65 + 30

.'.6x:x-l 5 .'. 6x-x:5 rrrrd.x*3:l*3:4Ans. The original fraction is1

.

3x x

2:

3x1 -: ,.'

I(x -15)

. 5x:5

.'.

x:

I

;

.'. l9x:95

.'.

x-T:t t9

Ans. The required number is 5.

Q. 4. The difference between the ages of two brothers today is 4 Five years ago, the ratio of their ages was 5 : 7. Find the age ofyounger brother. Solution : Let the present age of the younger brother be x years. Then, the present age of the elder brother : (-r * 4) years.

(r. The denominator of a fraction is greater than its numerator by 3' lf 3 is subtracted frorn the numerator and 2 added to the denominator, the value of the fraction we get is f. fina the original fraction. Solution : Let the numerator of the original fraction be -r' 'l'hen, its denominator is x * 3.

.'.

the original fraction

it

#.

3 is subtracted from the numerator.

NAVNEET MATHEMATICS DIGEST: STANDARD

VIII

.'.

the new numrator is -r 3. 2 is added to the denominator. .'. ..the new denominator is x * 3 * 2 x * 5.

-

:

Write the values :

From the given condition,

+:: 'x15

$,

,/ffi

(2)

'$*

.'. 5(r - 3) : 1(-r * .'. 5.r - 15 :.r * 5 .'. 5x - x:5 + 15

5

5)

l, Find the suare roots by division method (l) lose (2\ te36 (3) 10316.

and-r*3:5*3:8

.'.4x:2O

.'.

x:5(z',)

444

(3)1

t041081620

TeT6 336 336 000

Ans. The original fraction is f . I

9

Q. 7. Sandpep has Rs 50 more thal,flre amount with Gayatri. If of themis given Rs 15, the ratio,between their amounts will be How much money did Gayatri have to begin with?Solution.: Let Gayatri have Rs.x.

189 189

+4 -16

+1

-1

000 ,.r/toss: rs.

+4*rs.

84

+0

J6o-:u

+4

204208

008 00 816 816 000

Then, Sandeep has Rs (x * 50). If Rs 15 are given to both, Gayatri would have Rs (x Sandeep would have Rs (-r * 50 + 15): (x + 55)

+ 15) and

From the given conditiorr,

'19: :: '.x*15 I

!.

.'. l(r + 65) :3(x * ;. x*65:3x*45.'. x-3x:45-65

Find the approximate values of the square roots of the following numbers up to two decimal Places : (l) et.32t6 (2) 7.6se.

15)

(t)99

9.5 5 6

(2)2

.'.

?-x:2O

*7-x: -20

9r.32160081 10 32 9 25 01 o7

2.7 67 7.6T 9-o o 6

.'. .r:10

+z

Ans. Gayatri had Rs 10.

l6

9525 119100

+7 546 +6+7

47

-4

3 6s3276

-3 29 0 3690041400 38689o27

5527

9l

L2

004464t6:9.556...

1t4636

5534

tl

lnr. Approximate value of .FtlztOup to two decimal91

Jrn

!/7 56, :2.767...Ans. Approximate value of $-SSS up to two deciinal Places :2.77.

92


VIII

MISCELLANEOUS EXERCISE -

I

a. 4. Classify the following numbersnumbers :

as rational numbers

or i

(1)

2.1s

Ans. Rationd numbers : (1) 2.18 \F% Irrational numbers : (\ ,f tO @) 4.gtl3t61g,.,

g ,F g,F.x

s

(4) 4.812316s9...

7. In the parallelogram ABCD, m LA :(8r + 8)", rnLC : (9x - 7)". Find the measure of each angle of this paralleloiram. Solution : The opposite angles of a parallelogram are congruent.

Q. 5. DivideAns.seg

seg

AB of length 7.5 cm into 6 congruent parts.

AC, seg CD, seg DE, seg EF, seg FG and seg GB are the 6 congruent pa.rts of seg AB.

,'. + 8)' : (9x -7)" .'. 8x* 8:9x-7 .'. 8.x-9x: -7 -8 - x: - 15 .'. x:'1,5 ,'. mLA: (8x * 8)' : (8 x 15 * 8)': (120 * 8)o :(8x

.'. LA= LC

.'. mLA:mLC

... (1)

123".

From (1),

mLC:128".-.'. mLB:180- 128 .'. mLB:52" ... (Opfosite angles of a parallelogram.)

The adjacent angles of a parallelograrn are supplementary. .'. mLA+ mLB: l80o

.'. 128*mLB:180Also,

mLD:mLB

Ans.

.'. mLD:52" mLA:128"; mLB:52"; mLC:128"i mLD:52.PQRS, ,(PQ) :6 cm, /(PS) = 9 cm, find the tengths of seg QR and seg SR. Solution : The opposite sides of a parallelogram are congruent. .'. ses PQ = seg SR .'. l(PQ): l(SR) /(PQ) : 6 cm . .. ( Given) .'. /(SR): 6 cm Similarly, seg PS = seg QR .'. /(PS): l(QR) l(PS):9 cm ... (Given) .'. /(QR):9 cm

8.

If in the parallelogram

Ans. /(QR):9 cmi /(SR;:6

"-.

Q. 6. Draw seg PQ of length 9 cm and divide it in the ratio 3 z 2.Ans.

Point T divides seg PQ in the ratio 3 : 2. i.e., /(PT) : /(TQ) :3 :2,

P. ./

A

Q. 9. The radius of a circle is 17 cm. Find the length of a chord which is at a distance of I cm from the centre of the circle. Solution: In the figure, P is the centre of a circle with radius PQ: 17 cm. Seg PM t chord QR. l(PM):8 cm. In right-angled APQM, by futhagoras' theorem, t(QM)1'? + t/(PM)12: tl(PQ)1'z

g9'l*"

.'. tl(QM)l'*82:I'72 .'. t/(QM)]'z : r72 -82 :289 -64:225 .'. /(QM): 15 cm.The perpendicular from the centre to its chord bisects the chord. . ' /(QR):2/(QM) :2 x 15:30 cm Ans. The length of the chord:30 cm.

f'NAVNEET MATIIEMATICS DIGEST : STANDARD

VIII13. Segscg

0. 10. If in a circle with centre O, Chord pe chord AB = lzLAOB:70o, find the measure of lpOe.

MISCELLANEOUS EXERCISE -

1

AB jl seg CD,

centre.ChordPQ

Solution : In a circle, congruent chords subtend congruent angles ata

BM

r

seg CD,

= chordAB ... (Given) Chord PQ subtends lpoe at the centre O..'. IPOQ = LAOB ... mLyOq: aLAOB ruLAOB:70" ... (Given)Chord AB subtends ZAOB aat the centre O.

/(AB):4 cm, /(BM):3 cm, /(CD): 12 cm. or-.--..--.--.-.--,, l'ind the area of n ABCD. "Solution: In IABCD, seg AB ll seg CD ...

(Given)

..

!ABCDisatrapezium.

:. mLPOq:7}"

Ans. mLpOe:70o.

Q' 1r' rn a circre with centre p, a chorddistance

of rength 16 cm is at a of 8 cm from the centre. Find the radius of this circle. Solution : Let chord AB of length 16 cmbe at a

of the parallel side AB and CD are 4 cm and lZ cm lospectively. 'l'he distance between the parallel sides, /(BM) :3 cm. Area of a trapezium : I x the sum of the lengths of parallel sides x height ... (Formula).- j x t/(AB) + /(CD)l x /(BM) :i"@*12)x3

'f'he lengths

of 8 cm from the centre p of a circle. Seg PMr chord AB. /(AB):16 cm, /(pM):g sm.

:1rx16x3:24sqcmAns. The area of n ABCD

...

(Subsritutingrhevalues)

:24 sqcm.

.'. /(AM) :jl1an; :) x t6:8 cm .'. in right-anged ApAM, by pythagoras, U(PA)lr: [/(AM)]2 + t/1pM)1, : g2 * g2 :64 * 64: l2gAns. The radius of the circle

The perpendicular from the centre of a circle to its chord bisects the chord.

q.

14.

If

triangles and /(HC)theorem,

|l HDFC.triangles.

AHDC and ADFC are equilateral :12 cmtfind the area ofH

Solution : AHDC and ADFC are equilateralSide CD is common to both the

: nM

triangles.

/(HC)

:

12 cm.

cm.15 cm long, what

Q. 12. If the sides of AABC are lg cm, 15 cm and

.'. both the equilateral triangles .'. A(AHDC) : A(ADFC)

are having side 12 cm in length.

and/(AB):c:15cm. .'.

its area? Solution : Let in AABC /(BC):the semiperimerer (s.y: '

'fhe area of an equilateral triangle

:{./:t :T

Ora"l,

...

(Formula)sq cm

a:lg

cm, /(CA)

:b:15

cm 'l'he area

r;

x t2 x t2:36"",6

A(AABC) by Heron's formula s(s-a)(s -b)(s-c)24(24

a*b*c2_

18+ 15+ 15

2

-:T:z+cm

-48 _ nn

of

I

HDFC

:

Ans. The area of

G6^,[3 +36\/E) sq cm D HDFC :72.,/a sq cm.

:

A(AHDC) + A(ADFC)

:72.,fr

sq cm.

24x6x9x94 x 6 x 6 x 9 x 9 :2Ans.

-

18)(24_

lt(Z

15)

A(AABC):108

x6 x 9:

Q. 15. rf the circumference of a circle is 88 cm, find its radius and area. Solution : The circumference of the circle:8g cm 'fhe circumference of a ctrcle :2nr ... (Formula)10g sq cm

sq cm.

.'.88:2x!xr88x7

...

(Substituting the values)

". 2x22-'

.'. r-14cm

NAVI\EET MATI{EMATICS EIGEST ! STAIIDARD VIII

MISCELLANEOUS'EXERCISE -

I

The area of a

circle:nr2 :? x

Ans. The radius

:

x 14 :22 x 2 x 14: 616 sq cm14

... (Formula) ... (Substituting

Solutionrhe val

:i;.,?..,

Xl

14 cm; The

area:616 sq cm.

2025

8 12

160 3001E0

Q. 16. The diameter of a greater circle is 126 cm and that of acircle is

3035

64

ll2

cm. Find the difference between their (i)

(ii)

areas.

L40780

Solution : The diametet @) of the greater circle

:

126 cm.

N:30the sum of N780 xxf _ -/$ 30

.'.

radius

Navneet Maths Digest Std 8th

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