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 Exercise2 (Textbook page 170)
Q. Find the square root by the division method (2) 202s (1) 28e
62 69
Miscellaneous Exerbise1 (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
[. RevisionSquare,
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 _ 19gl
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 rightangled triangle are 7 cm and 5 cm. Find the length of its hypotenuse. Let AABC be a rightangled + 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_t1y_: square root of 324. p) fhe negarive square root of 324.
lx: s
ro.
nttilibers are called nonperfect 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:1919!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 Jl+
"
.
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 nonperfect 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 nonperfect square number. Let us find 5.47 7 2
nET:5.48
1.
(Textbook page 6)
... /(AC) : ,EZS
6. The square root of a nonperfect 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 rightangled 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?D43749
,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.oooooooo
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 nonzero integer, then 4 is called
(Textbook page 10) a
Q. Write
the following numbers in the nonterminating 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 thiscase 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. Nonterminating recurring decimal forms and rational numbers Every nutmber
nonterminating recuning decimal fraction. ln case (ii), 3 : 0.66... is written as :0.6 and
:
in the nonterminating 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. Nonterminating recurring decimals
nonterminating recurring decimal frbction.
3.75:3.750 Every rational number can be written in the nonterminating recurring
iI
:
Numbers whose decimal.form is nonterminating and nonrecurring are
calted irrational numbers.
rf;:2.230067977...i.e., and nonrectrrring. So,
.f
the decimal form of
..,6 i, nontenninating
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 nonterminating
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 segLM. 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$
::gT:_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.
Propertiepof 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!9tlsMrof a square is the perpendicular
hoblems based on the propertiesof 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;
...
#
... (DMF) ... [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 perpendicular 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
... (LoN) ... [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:180130 .'. 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.
.'. (3x2)":(50x)' .'. 3x2:50x .'. 3x*x:50+2
3x2:3 x 13 2:392:37. .' . (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 .'. y18037
.'.
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 rightangled 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 t5,2 cm, b:25 cm 'f
:13
fghtangled 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:feE3lggj 'f'hr, rrrca
If
:
of a triangle
:j
x base x height
'fhu l,.ou of u tight'utg fhe rrrea of a rightangled 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 rightangled 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 rightangled 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 rightangled 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
"'x40cm
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
Aten
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 .':15.2tx:26 6.5 .'. x:26I5.2 .'. .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(sa)(sbXsc) 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(saXs b)(sc)
(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(saXsb)(sc) 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 bXsc)
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 rightangled APQS * area of ASQR. In rightangled 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(saXsbXsc) 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(sa)(sbXsc)

:
'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 righangled 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 rightangled triangle
:lx5x12:30sqcm
47
AREA
:rcrQ
sq cm.
r
o cl
N
12 cm.
rtc rtrea or A QNE
: i :T : Hj1., 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 :2x21x56
a
This constant number is denoted by the Greek letter n (pie).
A:n
tlcre,r56cm 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) 34
:
(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 en3o
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 zt* 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 Lxtox7
1_1 s3 :73 s :7 Is 2
9
:1820
r11 (6) (xY)t *i
1_ I to
to1

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 xl:xEi 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 2xo: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$4;
cm.
(Textbook page 106)
Draw quadrilaterals of the following measures
Ans.
:
l. In IDEFG, /(DE):4.S cnr /(EF):6.5 cmi mLD:65", mLE:100o, mLF:60o.
rl
4.5 cm
4.5 cm
Construction 3
:
To construct a quadrilaterar, given the rengths of two adjacent and the meaures of three angles.
Q.2.
In IMTSN, t(MT):4 cf,r ,(TS):5 cm, mLM:50o,
mLT:ll0"rmLS:70". NS
Q. Construct IABCD in which /(AB):5.4 cm, /(BC) :3.4 m LA : 70, m LB: 60", m LC : llo". Steps of Construction
:
(1) Draw a rough figure
of n ABCD
showing the given measurements. (2) Draw seg AB of length 5.4 cm. (3) At the point B, draw ray BN making
M4cmT Rough figure
ZABN:60o. (4) With B
as a centre and 4cm
radius 3.4 cm draw an arc intersecting ray BN
a. 3.
in point C.
mLY:95", mLZ:85".
(5) At the point A, draw
ray AM
making
IBAM:70o. (6) At the point C, draw
IBCD: n 4PCD
In IXYZW, I(XY):S.S ch: t(XW):3 cD mLX:70o,
A
110o intersecting ray
5.4 cm
AM in point D.
is the required quadrilateral.
Explanation : From the rough figure, measure of LW should be given. (Angles at the end of the given segment XW) In n XYZW, mLx + mLY + mLZ + mLW : 360o. .'. 70 + 95 + 85 + mLW:360.
85"
X
95'
70" 5.5 crn
Rough figure
ltf
NAVITTEET.
MATTEMATICS DIGEST
.'. 250.1mLW =360 .'. mLW:360250
:.
STANDARD
VIII
.'. mLW:110o.
'Ans.
CONSTRUCTION OF QUADRII"ATERALS
Draw quadrilaterals of the follorqng measures i 1. In ITYRE, /(TY):3.5 cm, /(YB)g crtrr /(RE1:4.5 Cilt
B
z
mLY:60o, mLR:120".
5cm Construction 4
In tl ABCD, /(AB) = 6.5 cil,
mLB:75, mLC:85o. Steps of Construcjion
,(BC)
:
(1) Draw ."g nC of tenj*r 4.5
(2) ltthe
5cm
Rough figure
:
To con$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.
.'. ISON 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 ITNL 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, IAPB 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
lApB
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
.',:,::.$1L:
tQ
D
Male
t0
60
80
70
Female
50
55
50
40
VillA'gq:
lgoo
.'. mLD * 85 : 180 ... [Given mLB:g5"] .'. mLD: 180  85 .'. mLD:95" Ans. rn LL : 7S"i m LD : 95". Q' 2. The measures of the opposite
(Textbook pages 116 to 120)
l,
.,.,
angres of a cyric quadrilateral
and 3x. What are the measures of these angles of the quardrila
Solution:
The opposite angles of a cyclic
quadrilateral
supplementary.
.'. r+3x:180o .'. 4x:180o
.
lg0" 4
x:45o 3x:3 x 45o: ,'.'
135"
Ans. The measures
of the opposite angles of the given c
quadrilateral are 45" and l35o respectively.
Steps
(l)
L
for drawing the graph:
Draw Xaxis and Yaxis on the graph paper. t23
I24
_
NAVNEET MATHEMATICS DIGEST : STANDARD
VIII
(2) Take the villages on the Xaxis and the number of workers on Yaxis.
(3) Take a scale, I cm: 10 workers on the yaxis. (4) Mark villages A, B, C, D on the Xaxis keeping equal di between them.
(5) Show males and females in each of the four villages using a j cm from each of the joint bar of 1 cm width. (6) Show the heights of each bar as per the scale taken for the y(7) Draw joint bar graph in this way for each of the villages. (8) To show the two bars in the joint bar graph separately, colour shade) the female's bar.
di3B,d!,,1 lrextnook
pages
tt9 &
120)
0. 1. Given below are the number of girls and boys in Stds V to of a certain school. Draw a joint bar graph from this data.
sl&d
,,11:;:.:lL:.\{:
,.YT$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 yaxis.
.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 Yaxis.
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 N5oooo 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" rP
:
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/ (rJ " 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 *rr1;':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:Ap 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 39x39 /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
+mt +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,
@
12.
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
pqlyngllxqlql
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*ti
+.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 *y7 qe le!
G)
Z&+3
(4) mi _m2 +7
pely_teqi3ls.:
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"r9 is 3. (2) 27:27x" .'. the degree of 2j is 0. The degree of a nonzero 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)6Q3thp Q)8nr25n*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/25x3x\l:Rs22x 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:20x2720
"' 2Ox3x:2720 .: .'. ^:x136 .'.
:
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 : STANDARD
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 twowheeler : 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
2xx28xI28 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:4828 ), 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
_ ... t2_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 coneshaped 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 coneshaped 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 "' h48 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 coneshaped 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 icecream cone is 12 cmand its surface area is 113.04 sq cm. What is the radius of the base icecream 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 coneshaped paper hat is 24 cm and the radi the base is 7 cm. How m""n pap"r will be n""n"* ,0.' ""nrr".*" rrihats?
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 .'. l25 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
9v'.:
+l3m 4M2 (104)
5?nt3
n2:4n2
(3) 20c2 + L5c:!c (5) 40a2 :(10c) 
2yx5y10y,
...
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 35rs : 7x2
35:7 x5and x5:x2xx3 .". 35.ts :7x2 x Sx3 '.
3515
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+a3S
(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) x5 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:4a6
2m
Bm)24m4
=7b3 +
:
(l)
3m3
+ r3b3
0

Quotient :2y
2b2)t8b4 tgb4
2yu
0 *.r
(6) su'z+u
4a6 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 :: \