Unit 4- Area and Circumference of a circleExercise - 4 .
Given diameter of circle =
1:Smsd
.Circumerenre of irle - nmd
3-14 x IN-5
36-N n
b) Giuen diameter of crcle
d5 r
t
.Cirrumferene e of drcde Td
3-14 5.
= I6 3R cn
)Gven vadius of civcle r 44 cm Y
Ciunfevence of civcle : AY
A314x 49
30 77 Cm
b) Given radius o1 irele
.
67mm
Grcufecence ot cirele c
Y
3 H 67
= 490 76 mn
7a) The given shape include semicire)e
and recargle
.
perimeter of given shape Circular 2
Part ot smicvcle +
3 sides of rectunele
TTY* 6+98
n+6t3+3
3.14x G +G+3*?
9-43+6+t8
31:4 CM
b perimeter ot g iven shap er jo+ 1o+
10C
4pevimeter of
eirele)
0 + (*)
20+ Ax3-14 10)
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b pevimeter of given shap e 10t10
4Pevimeter of
circle)
90+ (207)
0t
x 3-14x 1o)
20+ 1s.7
35: 7 C
c pevimeter of glven shape
R (M
F1Rtl+a (perimeter of
circle
+(pevimeter of cirele*
4 t (Ttd)
4 *(3-14 XG)
4t L1g. 84)
= 4 t 3763
= G68cM
peimeten of given shape - 36m
perimeter Ot cireular part
of semcircle + 35t0 + 35
TY + 35 R0+ 35
3-14x 20 + 90
31.4t90
1 1 4 mm
)perimeter of given shape :5tm
45+4.5+ Llperimeter o
cirele
4
x 3-14 x 4 s)
+(.26)
peimeter of given
shape 45tm
45+4.5tLlperimeter of
órele)
4+Lx 3-14 X4 5) = shole
+(s.6)
4 4. 71
13.7) cm
pevimetrr of giver shape
St15
(pevimeter of cirele)
30+ Canr)
30+ 3 (4% 3.14 X15)
30+
14.2)
30+ 70 65
100 65CM
4Diameter of bicycle toheel 75 Cm
a) Civumfevente of bicyele wheel = *nY
3 - 4 X 75
35 S (m
bkm 10o000 (
1 f a gir tovevs 1 km wih bicy
e then
No. of \mes he wheel
wilrotate -l00000
35.5
= 100000
355
=434.6
4 5 Eimes
5Inney radivs of hole =44 cme ,
Outer tadius of hole 50 cm =Va
. uvtr circumference of hole T
*3-14 x50
l00 x 3 4
311 cm
5) Tnner radius of hole = 44 cm ,
Outr Adjus of hole 50 cm Ya
Oute cireumference of hole
x3-14 x50
l00 x 3 14
314 cm
4 Innc cirrumterenee of hole
= 307 7 Cm
Diffeceoce betoeen botn circurferenees =
34-307 7
= 6:98 C
6 3 Cn
. Ouber tireumfevehte is 63 chlongev than
nner circumterence
Radius of Inner tratk = 4Om =*
Radius of Outev track = 46m = Ta
Circumterence ot Tnner track =
20Y,
3 - 4 x yo
5| m
Cirrumterenc e of Outer brack = RTIT2
x 3:14X 6
8 83 m
. DHfevce bettoen the distance covercd
too riends who suns on the tracks
88 38-251
37-69m
ere se-42
Given radius of circle = cm =Y
A e a of he cirele e
T
3 J 4 x (a)
R54. 34 Cm
Erere) se-y2
aivnadivs of circle = 9cm =Y
. Area of he circle e Tr
3 4 x (a)*
54 34 Cm
2543 cm
venradius ot ctrele T 04 cn
. Arca of the cirele Ir
3.14 x (6-4)
F 5434 Cm
5 Cm
C Given radius of ôrcle = Y: 8. mn
Area of the cirele = r 2
43.16 Mm
d)Given Yadius of cirele43 mm
= v 99cm
Area of the civcle Tr2 KEECM
3:1 x 89)
24 316 16 cm
c 4 316 R cm
eGven Tadius of circle =Y* 4.4 cCm
Area otthe circle Tr
3 1 4 X (4:4) 4-4c
G0-79 cm
- 60.8 Cm
f Given radius of Circle : T0.44 Cm
Area ohe cirele = TY2
.
3:14 x(0-44)
0:6079 CM
0.6cn*
Gien vadivs of semitirele 8on
Area of the
shape = TY
3:4x(3)
2
= 0:6Cn
iven adius o4 semici rcle =8
.Area of the shape = TY*
3-)4x (3)
00-9
2
100-48 cm
b)Gven radius T:]0C
Axea of the shape (*
314 xo0
3
78 5 cM
CAnqle of the shape = 30
5 radius of the shape I5 Cm
.. Area ot theshape
2 3 0 x IR =3Go
L3kich is whole
3-l4 15 XI5 Cirele
706 5
2
58-38 ch
d Area of the shape Area of
c
tArea of semicirele
rectangle
LB + (
3x6+ (3-1HX 3X3 CM
3Cm
h3 4-13
613 Cm
e)Tf we extend the sides of shape
we
get Square and One quad rant
Of circle
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S .'d=6
Y 3Cm
484 13
6 1 3 Cm
e) Tf we extend the sides of shape
we
get squave and ohe qugd rant
of circle
. Area O+
shape Area ot
=
$au arec 3
Arca ot one quadrant
of circle
(side)'- p
(s- (3-143a)
-(38
= 4-7-07
: 9 3 cm
fPrea of the shape 2{Prea of semiardes
wth redi us dem)+
(Avea of semcircle wth
yadius 3 Cm)+ Area
frectangle
m
=
3:14(1+ 3-1H X(3+1
- 43.4 cm
37 Accevding to Kiaosi area ot tircle = 7=A
Ohere r= radius
And aconding to Ameha area of cirele A =1p
ere D= diameter
Acco rding of Kwasi Area of circle A= TIr
Lhere
Deaiameto
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here Dr diameter
Acco rdiha4 0f Kwasi Aea of cirele A = TTr
neto
Arca of cirele
aCcording to
Amelia
Henee formulac ven byboth are Corect
6xereise h
Da) Arca of rertg le not
covered by disc Area ot
Rctangle- Arca ot
crcle
LxB- Tr
6R4
-
3.14 C2)*
z24~ 3-4X4
a 4 -1R 56
l44 Cm cirele
ot
Area covered by disc =Area
3-4x ( )
3.14X4
1 56 Cm
Let be the lenyth o he rectan ge so hat
covered by
qrea covered by the dis c area not
the disc
Avea covered by he dise = 1:56
' 5 6 43-12-56
4I S»
length of the vectangle shoul d b
So thart area covered by disc is
as the area not rovercd by the dis
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area covered by he dis c
rea nt coverea u
Hhe disc
he disc = 1R:56
Area covered by LX
1.5G **4-3-14 x
56 4x-12-
. 4I R5-14
ofhe vectangle should
be . 8 cm
length he same
So that area covered by disc is
as he area not coveYCdby he disc
a Acea COvere y YY Y_Y3
the semicireular pieces O
chocolate 4 (Aveo o
Semi àrde )
Chrea of cirele)
S Radius = cm
(34x )
[ 8 - l 4 x4)
(1R.56)
5.1 Cm
of he box hot covered
b)Area of he base
Area ot base of bOx Area
by hocolade =
of base covered by choco late
Lx B-25.1 Slength of bor
= 16xA - 45-1R
c4diameter of
3 -5 Sei circle
fec cetoge of he base of he box nof
coveeby the cho Co lat 68 I00 3Atbox
of base
57
) a) To ird: Area of squive not coveved
by the dsc
Area ot square =
(Side)
-(5)
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A ot bo X
3
e : 5 7.
3 a) To ind Avea of squáre not covered
by the disc
Area ot square =(Side)
-(5)
Pkea ot d i s c T r
D 6Cm
F
3-14 x (2.5)* Y = -5 Cr
3:14x G5
963 C
. Area of he sauare not covered by the
disc Area of sauare - Area of disc
519 63
5 37
Axea of squave (side)
5 Cm
Aca ot
Square covered by disC
Hx Avea of circle
Mx (ar*)
SD:*:. 5(
Ax(s-14x95
4x (3-14x | 56)
5 Cm
4x (4.)
19:6 C
Ared of Hhe squave hot covered by disC
Aca of sauave - Aea covered hy disc
5 19-6
5.4 Cm
Prea ot squave covered by disc
=a5( T))
Dof di S =\cn
Radius of disc » :o-5c
. Avea o4 souaYe (oveed by dist :
*5
5
-95 (
5 (0.
14 63 cm
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- Araa of he 6quare hot cove«ed b isc
Arca of sayuare - Avea covered by disc
5 - 196
5.4 Cm
Y
Area ot sauave coveved by disc
25(tr)
Dof di 20 -\c
Radius o disc :0-5c =
.Avea of sauare (oued by dist = 5(TY)
5(3-4 x (o5)*)
5 (3H Xo 26)
5 (o-785)
14 63 cm
Prea of
sqare not toveved by the disC =
*5-19-3
537 cm
Area suare not covered by one disC oTadius
.5 Cfrea ot sauare not covered
o adius
0.5 cm
by 5 disesS
. wnSwev in
part (a) = answer in part (c)
Also 5.47537
- FAvea of
squave ttovevtd by H dis(s t
radj us
5 Avea of square no covered by 5
discs 0f radius 0-5 m
NS OC in part (6) ansuaer in part (c)
The given Spira s ade kom
Hve quadrants ot
drcles
difterent
The Raabus of smallest
Hheius of each
9uaerant s
icCm.
in Size
odjacend quadrants diuble
A)Aea of spiral Area of all quad vants
Area o nakann orth
+Ara of radius lcm
*Avea of ayackant wth
+Avea ot quadvant wih
rea of qnad rar w
qyadrant wth
