 Lesson: Area related to circles
Exercise 11.1 (10)
Choose the correct answer from the given four
options:
Question: 1
If the sum of the areas of two circles with radii
1
R
and is equal to the area of a circle of radius ,
2
R
R
then
a.
1 2
R R R
b.
2 2 2
1 2
R R R
c.
1 2
R +R R<
d.
2 2 2
1 2
R R R
Solution:
(b)
Sum of areas of two circles of radii and = area
1
R
2
R
R
Hence,
2 2 2
1 2
πR πR πR
Take common from the first expression.
π
R
2 2 2
1 2
π(R R ) π
2 2 2
1 2
R R R Thus, the correct option is (B).
Question: 2
If the sum of the circumferences of two circles with
radii and is equal to the circumference of a
1
R
2
R
R
a.
1 2
R R R
b.
1 2
R R R
c.
1 2
R R R
d. Nothing definite can be said about the
relation among , , and .
R
1
R
2
R
Solution:
(a)
Given: Circumference of circle with radius +
1
R
Circumference of circle of radius =
2
R
R
1 2
2π(R R ) 2πR
1 2
R R R
Thus, the correct option is (A).
Question: 3 If the circumference of a circle and the perimeter of
a square are equal, then
a. Area of the circle = Area of the square
b. Area of the circle > Area of the square
c. Area of the circle < Area of the square
d. Nothing definite can be said about the relation
between the areas of circle and square.
Solution:
(b)
Circumference of the circle with radius, =
R
2πR
Perimeter of square of side, =
a
4a
Circumference Perimeter
of circle of square
2πR 4a
=
2πR
a
4
πR
a
2
Area of circle =
2 2
22
πR R
7
area of square,
a
2
2
πR
2 2
2
22
R
7 2
2
121
R
49
2
2
Area of circle
Area of squ
22
R
7
121
R
4
are
9
22 49
121 7
14
11
1.27
Area of circle = 1.27 Area of square.
So, Area of circle > Area of square.
Thus, the correct option is (B).
Question: 4
Area of the largest triangle that can be inscribed in
a semi-circle of radius units is
r
a. sq. unit
2
r
b. sq. unit
2
1
r
2
c. sq. unit
2
2r
d. sq. unit
2
2r Solution:
(a)
Let PQR be a triangle inscribed in a semicircle of
area .
r
The base of the triangle QR = Diameter of
semicircle =
2r
Altitude of the triangle OP = radius of the
semicircle =
r
1
Area of trianglePQR ×base×height
2
1
2r r
2
2
r sq.unit
Thus, the area of triangle is .
2
r sq u. nits
Question: 5 If the perimeter of a circle is equal to that of a
square, then the ratio of their areas is
a.
22:7
b.
14:11
c.
7: 22
d.
11:14
Solution:
(b)
Perimeter of circle of radius, =
r
2πr
Perimeter of square of side, =
a
4a
Perimeter Perimeter
of square of circle
4a 2πr
2πr
a
4
πr
2
Area of circle =
2
πr
2
22
r
7
Area of square,
2
2
πr
a
2
2
2
22
r
2 7 2
121
r
49
2
2
Ratio of area of circle
to that
2
of
2
r
7
12
s
1
r
4
u
9
q are
22 49
121 7
14
11
Thus, the ratio of area of circle to that of square is
14:11.
Question: 6
It is proposed to build a single circular park equal in
area to the sum of areas of two circular parks of
diameters 16 m and 12 m in a locality. The radius of
the new park would be
a. 10 m
b. 15 m
c. 20 m
d. 24 m
Solution:
(a) Let be the radius of new park.
R
Area of the Area of Area of
new park park 1 park 2
2 2 2
1 2
πR πr πr
2 2 2
1 2
R r r
1
1
D 16
r 8m
2 2
2
2
D 12
r 6m
2 2
2 2 2
1 2
R r r
2 2
8 6
64 36
100
So,
2
R 100
R 100
10m
Thus, the radius of the new park is 10 m.
Question: 7
The area of the circle that can be inscribed in a
square of side 6 cm is
a. 36 cm
2
π
b. 18 cm
2
π c. 12 cm
2
π
d. 9 cm
2
π
Solution:
(d)
Side of given a square,
a 6cm
Radius of circle that can be inscribed inside the
square,
a 6
r 3cm
2 2
Area of the circle inscribed
2 2 2
πr π 3 9π cm
Thus, the area of the square is .
2
9π cm
Question: 8 The area of the square that can be inscribed in a
circle of radius 8 cm is
a. 256 cm
2
b. 128 cm
2
c. 64 cm
2
2
d. 64 cm
2
Solution:
(b)
Diagonal of a square, SQ =
2r 2 8 16cm
Applying Pythagoras theorem,
2 2 2
SR RQ SQ
2 2 2
a a 16
2
2a 256
2
a 128 Thus, area of square that can be inscribed in the
circle is .
2
128cm
Question: 9
The radius of a circle whose circumference is equal
to the sum of the circumferences of the two circles
of diameters 36 cm and 20 cm is
a. 56 cm
b. 42 cm
c. 28 cm
d. 16 cm
Solution:
(c)
Diameters of two circles are 36 cm and 20 cm.
Let and be the radii of two circles.
1
r
2
r
1
Circumference of first circle 2πr
1
D
2π
2
36
2π
2
36π
2
Circumference of second circle 2πr
2
D
2π
2 20
2π
2
20π
Circumference of Circumference Circumference
the new circle of circle 1 of circle 2
2πr 36π 20π
2πr 56π
56
r
2
28cm
Thus, radius of the circle is 28 cm.
Question: 10
The diameter of a circle whose area is equal to the
sum of the areas of the two circles of radii 24 cm
and 7 cm is
a. 31 cm
b. 25 cm
c. 62 cm
d. 50 cm
Solution:
(d)
Radius of two circles = 24 cm and 7 cm Area of Area of Area of
the circle circle 1 circle 2
2 2 2
1 2
πr πr πr
2 2 2
r 24 7
2
r 625
r 625
r 25cm
Diameter of the circle
2r 2 25 50cm
Thus, diameter of the circle is 50 cm.
Exercise 11.2 (14)
Question: 1
Is the area of the circle inscribed in a square of side
a
2 2
πa cm
Solution: a
r cm
2
Area of the circle
2
πr
2
a
π
2
2
2
πa
cm
4
So,
2
2 2 2
πa
cm πa cm
4
Hence, the above statement is false.
Question: 2
Will it be true to say that the perimeter of a square
circumscribing a circle of radius cm is 8a cm?
a
Solution:
True. Side of the square =
2a cm
Perimeter of the square
4 sideof square
4 2a
8a cm
Hence, the above statement is true.
Question: 3
In the adjoining figure, a square is inscribed in a
circle of diameter and another square is
d
circumscribing the circle. Is the area of the outer
square four times the area of the inner square? Give Solution:
No, the outer square area is not four times the area
of inner square.
Diagonal of inner square = diameter of the circle =
d.
Apply Pythagoras theorem,
2 2 2
EF EH FH
2 2 2
a a d
2
2
d
a
2 Area of inner square
2
2
d
a
2
Area of outer square =
2
2
side d
So, area of larger square is two times the area of
smaller square.
Thus, the given statement is false.
Question: 4
Is it true to say that area of a segment of a circle is
less than the area of its corresponding sector? Why?
Solution:
Area of major segment PQR > Area of
corresponding sector OPQR
Whereas Area of minor segment ACB < Area of
corresponding sector OACB.
Thus, the above statement is wrong. Question: 5
Is it true that the distance travelled by a circular
wheel of diameter d cm in one revolution is
2πd
cm? Why?
Solution:
False
Distance travelled by wheel in one revolution =
Circumference of the circular wheel
Circumference of the circle
2πr
d
2π
2
πd=
Hence, the distance travelled by wheel in one
revolution =
πd
Thus, the above statement is false.
Question: 6
In covering a distance m, a circular wheel of
s
radius m makes revolutions. Is this statement
r
s
2πr
true? Why?
Solution:
True.
Let n be the number of revolutions. Total distance covered by wheel in n revolutions,
s 2πrn
Number of revolutions,
s
n
2πr
Thus, the above statement is true.
Question: 7
The numerical value of the area of a circle is greater
than the numerical value of its circumference. Is
this statement true? Why?
Solution:
False.
Let radius of circle be 1 cm.
Area of circle
2 2 2
πr π 1 π cm
Circumference of the circle =
2πr 2π 1 2π cm.
Thus, area of circle is not always greater than its
circumference.
Question: 8
If the length of an arc of a circle of radius is equal
r
to that of an arc of a circle of radius , then the
2r
angle of the corresponding sector of the first circle is double the angle of the corresponding sector of
the other circle. Is this statement false? Why?
Solution:
Length of arc of a circle
θ
2πr
360
Length of arc of first circle,
1
1
θ
l 2πr
360
Length of arc of second circle,
2
2
θ
l 2π 2r
360
2
θ
4πr
360
Both the lengths are equal.
1 2
l l
1 2
θ θ
2πr 4πr
360 360
1
2
θ
2
θ
Thus, the given statement is true.
Question: 9
The areas of two sectors of two different circles
with equal corresponding arc lengths are equal. Is
this statement true? Why? Solution:
False.
Area of sector
2
θ
πr
360
Area of sector of circle 1,
2
1 1
1
πr θ
A
360
Area of sector of circle 2,
2
2 2
2
πr θ
A
360
Length of arc of the two circles are equal.
1 1 2 2
2πr θ 2πr θ
360 360
1 1 2 2
r θ r θ
Let
1 1 2 2
r θ r θ a
Now, find the ratio of area of the two sectors.
2
1 1
1
2
2 2
2
πr θ
A
360
πr θ
A
360
2
1 1
2
2 2
r θ
r θ
1
2
r a
r a
1
2
r
r So, the areas will only be equal, when the two
Thus, the above statement is False.
Question: 10
The areas of two sectors of two different circles are
equal. Is it necessary that their corresponding arc
lengths are equal? Why?
Solution:
False.
Ratio of areas of the two circles
2
1 1 1
2
2 2 2
A r θ
A r θ
1 2
A A
2 2
1 1 2 2
r θ r θ
2
1 2
2
2 1
θ r
θ r
1
2
Ratio of length of arcs
of the two circles
l
l
1 1
2 2
2πr θ
2πr θ
2
2 1
2
1 2
r r
r r 2
1
r
r
Length of arc will be equal when the two circle will
Thus, the above statement is false.
Question: 11
Is the area of the largest circle that can be drawn
inside a rectangle of length a cm and breadth b cm
(a > b) is ? Why?
2 2
πb cm
Solution:
False.
Diameter of largest circle = breadth of the
rectangle.
Length of rectangle = cm
a
b
Therefore, diameter of circle = cm
b
Area of the circle
2
2
πd
πr
4
2
2
πb
cm
4
Thus, the above statement is false. Question: 12
Circumferences of two circles are equal. Is it
necessary that their areas be equal? Why?
Solution:
True.
Circumference of two circles are equal.
1 2
2πr 2πr
1 2
r r
Circle having same radius will have similar areas.
Thus, the above statement is true.
Question: 13
Areas of two circles are equal. Is it necessary that
their circumferences are equal? Why?
Solution:
True.
Area of the two circles are equal.
2 2
1 2
πr πr
2 2
1 2
r r
1 2
r r
Circles with same radius will have same
circumference.
Thus, the above statement is true. Question: 14
Is it true to say that area of a square inscribed in a
circle of diameter p cm is ? Why?
2 2
p cm
Solution:
Diagonal of the square inscribed = Diameter of the
circle = p cm
2 2 2
PS PQ QS
2 2 2
a a p
2
2
p
a
2
Area of the square
2
2 2
p
a cm
2
Thus, the given statement is false.
Exercise 11.3 (16) Question: 1
Find the radius of a circle whose circumference is
equal to the sum of the circumferences of two
circles of radii 15 cm and 18 cm.
Solution:
Radius of two circles = 15 cm and 18 cm
Circumference Circumference Circumference
of the circle of circle1 of circle2
1 1
2πr 2πr 2πr
1 2
r r r
15 18
33cm
Thus, the radius of the circle is 33 cm.
Question: 2
In the adjoining figure, a square of diagonal 8 cm is
inscribed in a circle. Find the area of the shaded
region. Solution:
Let a be the side of the square.
Diagonal of the inscribed square = Diameter of the
circle = 8 cm
2 2 2
PS PQ QS
2 2 2
a a 8
2 2
64
a 32cm
2 Therefore, area of the square
2 2
a 32cm
Area of the circle
2
2
πd
πr
4
2
22 8
7 4
2
22 8
7 4
2
352
cm
7
Area of the Area of Area of
2 2
πr a
352
32
7
352 224
7
2
128
cm
7
Thus, area of shaded region is .
2
128
cm
7
Question: 3
Find the area of a sector of a circle of radius 28 cm
and central angle .
45 Solution:
r 28cm
Central angle,
θ 45
Area of the sector
2
πr θ
360
2
π 28 45
360
2
308cm
Thus, area of the sector is 308 cm
2
.
Question: 4
The wheel of a motor cycle is of radius 35 cm. How
many revolutions per minute must the wheel make
so as to keep a speed of 66 km/h? Solution:
Distance travelled by the wheel in one revolution
2πr 2π 0.35
0.7π cm
Speed of the wheel
5
66km / h m / s
18
55
m/s
3
Let, total number of revolution in 1 min be .
n
Distance
Speed
Time
Total distance travelled in revolutions = 0.7
n
π n
Total time taken = 1 min = 60 sec.
55 0.7π n
3 60
55 60 7
n
3 22 0.7
500rev/min
The, the total number of revolutions in 1 min are
500.
Question: 5
A cow is tied with a rope of length 14 m at the
corner of a rectangular field of dimensions . Find the area of the field in which the
20m 16m
cow can graze.
Solution:
Cow will graze in the area corresponding to sector
STU of the rectangle PQRS.
Radius of the area grazed = 14 m
Central angle =
90
2
Area grazed Area of
by cow the sector
πr θ
360
2
22 14 90
7 360
22 14 14
7 4
2
154m
Thus, the grazed area is 154 m
2
. Question: 6
Find the area of the flower bed (with semi-circular
ends) shown in figure.
Solution:
r
10
cm 5cm
2
Length of the rectangle, = 38 cm
l
Breadth of the rectangle, = 10 cm
b
Now, find the area of the flower bed.
Area of the Area of the Area of the
flower bed two semicircles rectangle PQRS
2
πr
l b
2
2
2
πr 38 10
2
25π 380 cm
Thus, area of the flower bed is .
2
25π 380 cm Question: 7
In the adjoining figure, AB is a diameter of the
circle, AC = 6 cm and BC = 8 cm. Find the area of
the shaded region (Use π = 3.14).
Solution: In ,
ABC
Base, AC =
6cm
Height, BC =
8cm
Now, find the area of triangle.
2
25π 380 cm
1
6 8
2
2
24 cm
Apply Pythagoras theorem,
2 2 2
AC BC AB
2 2 2
AB 6 8
AB 36 64
100
10cm
So, diameter of the circle is .
10cm
Therefore, radius of the circle = 5 cm
2
Area of
the cir le
πr
c
2
3.14 5
314
5 5
100
7850
100 2
78.5cm
Area of the Area of Area of
78.5 24
54.5cm
Thus, area of the shaded region is 54.5 cm.
Question: 8
Find the area of the shaded field shown in the given
figure.
Solution: Length of rectangle = 8 m
Breadth of rectangle = 6 m
Radius of the semicircle = 6 – 4 = 2 m
Area of the Area of the Area of the
2
πr
l b
2
2
π (2)
4 8
2
2
2π 32 m
Thus, area of the shaded region is .
2
32 2π m
Question: 9
Find the area of the shaded region in the given
figure. Solution:
Length of triangle PQRS,
1
l 12m
2
b 26cm
Diameter of semicircle
12 4 4
4m
4
r 2m
2
Length of the rectangle TUVW,
2
l 26 (3 2 2 2)  
16m
2
b 12 8
4m Area of
rectangle PQRS
Area of the Area of two
Area of
rectangle TUVW
2
1 1 2 2
πr
l b 2 l b
2
π 2 2
12 26 2 4 16
2
2
312 2 2π 64 m
2
248 4π m
Thus, area of the required region is .
2
248 4π m
Question: 10
Find the area of the minor segment of a circle of
radius 14 cm, when the angle of the corresponding
sector is .
60
Solution: r 14cm
Angle of corresponding sector,
θ 60
In triangle PQR,
PR = PQ = (radii of the circle)
r
Let, (angles opposite to equal sides
R Q x
are equal)
By the Angle P Q R 18 sum property0
x x 60 180
2x 180 60
x 60
So, the triangle PQR is an equilateral triangle.
Side of the triangle PQR,
a r 14cm
Area of minor Area of Area of
segment sector PQR 2
2
πr θ 3
r
360 4
2
2
π 14 60 3
14
360 4
22 14 14 60 3
14 14
7 360 4
2
102.67 84.87 cm
2
17.8cm
Thus, area of the minor segment is .
2
17.8cm
Question: 11
figure, where arcs drawn with centres P, Q, R and S
intersect in pairs at mid-points W, T, U and V of the
sides PQ, PS, RS and QR, respectively of a square
PQRS (Use π = 3.14). Solution:
Side of the square,
a 12cm
a 12
6cm
2 2
Central angle of the sector =
90
Area of the Area of Area of the
shaded part the square 4 sectors
2
2
πr θ
a 4
360
2
2
3.14 6 90
12 4
360
2
3.14 6 6 90
12 4
360
2
144 4 28.26 cm
2
144 113.04cm
2
30.96cm
Thus, area of shaded region is .
2
30.96cm Question: 12
In the adjoining figure, arcs are drawn by taking
vertices A, B and C of an equilateral triangle of side
10 cm. to intersect the sides BC, CA and AB at their
respective mid-points D, E and F. Find the area of
the shaded region (Use π = 3.14).
Solution:
The sides of an equilateral triangle are equal.
And, all the angles of an equilateral triangle,
θ 60 Now, arcs are drawn by taking the vertices P, Q and
R of the equilateral triangle.
Each side of equilateral triangle = 10 cm
According to the question,
1
2
1
10
2
5cm
3 Area of sector madeby arc
region
2
πr θ
3
360
2
3.14 5 60
3
360
2
3 13.08 cm
2
39.25cm
Thus, area of the shaded region is 39.25 .
2
cm
Question: 13
In the given figure, arcs have been drawn with radii
14 cm each and with centres P, Q and R. Find the Solution:
It is given that,
Radius of each arc, = 14 cm
r
Now, from the above figure,
(Interior angles of a
1 2 2
θ θ θ 180
triangle) Area of Sum of areasof
2 2 2
1 2 3
πr θ πr θ πr θ
360 360 360
2
1 2 3
22
14
7
θ θ θ
360
22 14 2
180
360
2
308cm
Thus, area of shaded region is .
2
308cm
Question: 14
A circular park is surrounded by a road 21 m wide.
If the radius of the park is 105 m, find the area of
Solution: 1
r 105m
21m
2
r 105 21
126m
Now, find the area of the road.
Area of the Area of road Area of
2 2
2 1
πr -πr
2 2
π 126 - π 105
2 2
π 126 105
22
4851
7
2
15246m
Thus, area of the road is 15246 .
2
m
Question: 15
In the given figure, arcs have been drawn of radius
21 cm each with vertices P, Q, R and S of quadrilateral PQRS as centres. Find the area of the
Solution:
r
21cm
1 2 3 4
θ θ θ θ
360
Now, find the area of the shaded region. Area of shaded Sum of areas of
region all the 4 arcs.
2 2 2 2
1 2 3 4
πr θ πr θ πr θ πr θ
360 360 360 360
2
1 2 3 4
22
21
7
θ θ θ θ
360
22 21 3
360
360
2
3186cm
Thus, area of the shaded region is .
2
1386cm
Question: 16
A piece of wire 20 cm long is bent into the form of
an arc of a circle subtending an angle of at its
60
centre. Find the radius of the circle.
Solution: Length of the arc = length of the wire =
20cm
Angle made by the arc, =
θ
60
Now, find the radius of the arc.
Length of
the arc
2πrθ
360
2πrθ
20
360
20 360
r
2π 60
60
r cm
π
Thus, radius of the circle = .
60
cm
π
Exercise 11.4 (20) Question: 1
The area of a circular playground is 22176 . Find
2
m
the cost of fencing this ground at the rate of Rs 50
per meter.
Solution:
Area of the circular playground =
2
22176m
Now, find the radius of the playground,
Area 22176
2
πr 22176
2
22176 7
r
22
2
155232
r
22
r 7056
84m
Now, find the length of fencing.
Length of Circumference of
fencing the playground
2πr
22
2 84
7
2 22 12
528m Cost of fencing per meter
Rs.50
Cost of fencing 528 meter =
528 50 26400Rs.
Thus, the cost of fencing is Rs. 26400.
Question: 2
The diameters of front and rear wheels of a tractor
are 80 cm and 2 m respectively. Find the number of
revolutions that rear wheel will make in covering a
distance in which the front wheel makes 1400
revolutions.
Solution:
Diameter of front wheel
80cm 0.8m
So, radius of the front wheel,
1
0.8
r m
2
Number of revolutions made by front wheel,
1
n 1400
Diameter of rear wheel
2m
So, radius of the rear wheel,
2
2
r m
2
Let, number of revolutions made by rear wheel be
.
2
n Total distance Total distance
covered by front wheel covered by rear wheel
1 1 2 2
2πr n 2πr n
2
22 0.8 22 2
2 1400 2 n
7 2 7 2
2
7040 7
n
2 22 2
560rev.
Thus, the number of revolutions made by rear
wheel is 560.
Question: 3
Sides of a triangular field are 15 m, 16 m and 17 m.
With the three corners of the field a cow, a buffalo
and a horse are tied separately with ropes of length
7 m each to graze in the field. Find the area of the
field which cannot be grazed by the three animals.
Solution: Radius of the sector = length of ropes
1 2 3
r r r 7m
Let, be the semi-perimeter and , and be the
s
a
b
c
sides of the triangle PQR.
Here,
a 15m
b 16m
c 17m
a b c
Semiperimeter, s
2
17 16 15
2
48
2
24m
1 2 3
by the trian
θ θ
gle
θ 180 Now, find the area not grazed by the animals.
Area not grazed by Area of triangle Area of the
the three animals PQR three sectors.
2 2 2
1 1 2 2 3 3
πr θ πr θ πr θ
s s a s b s c
360 360 360
2
1 2 3
22
7
7
24 24 17 24 16 24 15 θ θ θ
360
22 7
24 7 8 9 180
360
2 2
24 21m 77m
2
24 21 77 m
Thus, area of region not grazed by animals is
.
2
24 21 77 m
Question: 4
Find the area of the segment of a circle of radius 12
cm whose corresponding sector has a central angle
of (Use π = 3.14).
60
Solution: r 12cm
Central angle,
θ 60
Now, from the figure,
In triangle PQR,
PR = PD = (radius of the circle)
r
Let, (angle opposite to equal sides
R Q x
are equal)
(Angle sum property)
P Q R 180
x x 60 180
2x 180 60
x 60
So, triangle PQR is an equilateral triangle.
Now, find the area of the minor segment. Area of minor Area of Area of
segment sector triangle PQR
2
2
πr θ 3
r
360 4
2
2
3.14 12 60 3
12
360 4
2
75.36 36 3 cm
Thus, area of the minor segment is
.
2
75.36 36 3 cm
Question: 5
A circular pond is 17.5 m in diameter. It is
surrounded by a 2 m wide path. Find the cost of
constructing the path at the rate of Rs 25 per
2
m
Solution: 1
17.5
r m 8.75m
2
Width of the path
2m
Radius of pond and path combined,
2
r 8.75 2
10.75m
Area of the Area of pond Area of the
path and path pond
2 2
2 1
πr πr
2 2
π 10.75 π 8.75
2 2
π 10.75 8.75
22
39
7
2
122.57m
Cost of constructing 1 of path
2
m
25Rs. Cost of constructing path
2
122.57m
25 122.57
3064.25Rs.
Thus, cost of constructing the path is
3064.29Rs.
Question: 6
In the adjoining figure, ABCD is a trapezium with
AB || DC, AB = 18 cm, DC = 32 cm and distance
between AB and DC = 14 cm. If arcs of equal radii 7
cm with centres A, B, C and D have been drawn,
then find the area of the shaded region of the
figure.
Solution: 7cm
Length of the two parallel sides of the trapezium
are,
AB 18 cm
DC 32 cm
Distance between the parallel sides of trapezium
14cm
1 2 3 4
Sum of angles of
θ θ θ θ
r l
3 0
a
6
Now, find the area of the shaded region. Area of shaded Area of Sum of areas of
region trapezium all the 4 arcs
2
1
2
2
2
3
2
4
πr θ
360
πr θ
1
360
sum of parallel sides height
2
πr θ
360
πr θ
360
2
1 2 3 4
22
7
1
7
18 32 14 θ θ θ θ
2 360
22 7
350 360
360
2
350 154 cm
2
196cm
Thus, the area of shaded region is .
2
196cm
Question: 7
Three circles each of radius 3.5 cm are drawn in
such a way that each of them touches the other
two. Find the area enclosed between these circles.
Solution: From the figure,
(Equilateral triangle)
θ 60
3.5cm
So, side of equilateral triangle,
a 7cm
Area of shaded Area of equilateral Area of
region triangle STU 3 sectors
2
2
3 πr θ
a 3
4 360
2
2
3 3.14 3.5 60
7 3
4 360
2
21.22 19.25 cm
2
1.97cm
Thus, the required area is .
2
1.97cm Question: 8
Find the area of the sector of a circle of radius 5 cm,
if the corresponding arc length is 3.5 cm.
Solution:
Length of the arc =
3.5cm
r
5cm
Now, find the angle subtended by the arc.
θ
Length of
the ar
2πrθ
3.
6
c
5
3 0
3.5 360
θ
2π 5
3.5 36
θ
π
Now, find the area of the sector. 2
Area of the
secto
πr θ
36
r
0
2
π 5 3.5 36
360 π
2
8.75cm
Thus, the required area of the circle is .
2
8.75cm
Question: 9
Four circular cardboard pieces of radii 7 cm are
placed on a paper in such a way that each piece
touches other two pieces. Find the area of the
portion enclosed between these pieces.
Solution:
According to the above figure,
At the contact point P, UP and PV are radii. So, if a tangent pass through point P, UV will be
perpendicular to that tangent.
So, the interior angles of quadrilateral UVWT are
each.
90
7cm
Side of the square,
a 14 cm
Now, find the area of the shaded region.
Area of shaded Area of square Area of the4
region UTVW sectors
2
2
πr θ
4
360
a
2
2
3.14 7 90
14 4
360
2
196 154cm
2
42cm
Thus, the required area is .
2
42cm
Question: 10
On a square cardboard sheet of area 784 four
2
cm
congruent circular plates of maximum size are
placed such that each circular plate touches the
other two plates and each side of the square sheet is tangent to two circular plates. Find the area of
the square sheet not covered by the circular plates.
Solution:
Let us consider each side of the square be .
a
2
Area of thesquarePQRS 784 cm
2
a 784
a 784
a 28cm
Now, according to the above figure.
4r RQ 28
28
r
4
r 7cm
Therefore, radius of the circle is 7 cm.
Now, find the area of the region not covered by
circular plates. Area of region not covered Area of Area of the
by circular plates the square 4 circles
2
784 4πr
2
784 4π 7
784 616
2
168cm
Thus, the required area is .
2
168cm
Question: 11
Floor of a room is of dimensions 5 m × 4 m and it is
covered with circular tiles of diameters 50 cm each
as shown in the given figure. Find the area of floor
that remains uncovered with tiles. (Use π = 3.14). Solution:
Diameter of each tile = 50 cm = 0.5 m
0.5
r 0.25m
2
Number of tiles along the length =
5
10
0.5
Number of tiles along the width =
4
8
0.5
Total number of tiles =
10 8 80tiles
Total area covered by tiles
2
πr 80
2
3.14 0.25 80
2
15.7m
Now, area of the rectangular floor
5 4
2
20m
So, area not covered by circular tiles
20 15.7
2
4.3m
Thus, the area not covered by circular tiles is .
2
4.3m
Question: 12
All the vertices of a rhombus lie on a circle. Find the
area of the rhombus, if area of the circle is 1256
2
cm
. (Use π = 3.14). Solution:
Area of circle =
2
1256cm
Now, find the radius of the circle.
2
πr 1256
1256
r
3.14
20cm
Now, length of both the diagonals =
2r
Therefore, length of each diagonal of the rhombus =
2 20 40cm
Now, find the area of the rhombus.
Area of the rhombus
1 2
1
d d
2
1
40 40
2
2
800cm
Thus, the required area is .
2
800cm Question: 13
An archery target has three regions formed by three
concentric circles as shown in the given figure. If
the diameters of the concentric circles are in the
ratio , then find the ratio of the areas of three
1:2:3
regions.
Solution: Given: Ratio of diameters of three circles =
1:2:3
Hence, ratio of their radii will be .
1:2:3
Let,
1
r r,
2
r 2r,
3
r 3r
Now, find the areas of the three circles.
2 2
1 1
A πr πr
2 2
2 2
A πr 4πr
2 2
3 3
A πr 9πr
Now, according to the question,
2 2
Area of region between
first and second circle
4πr πr
2
3πr
2 2
Area of region between
second and third circle
9πr 4πr
2
5πr
Required ratio of areas is .
2 2 2
πr :3πr :5πr 1:3:5
Thus, the ratio is .
1:3:5
Question: 14
The length of the minute hand of a clock is 5 cm.
Find the area swept by the minute hand during the
time-period 6:05 am and 6:40 am.
Solution: Length of minute hand = radius =
5cm
Area swept by minute hand in 60 minutes =
2
πr
Area swept in 1 minute
2
πr
60
Time difference =
35min
2
Area swept by minute
hand in 35 minutes
πr
35
60
22 5 5
35
7 60
2
5
45 cm
6
Thus, the required area is .
2
5
45 cm
6
Question: 15
Area of a sector of central angle of a circle is
200
770 Find the length of the corresponding arc of
2
cm
this sector.
Solution:
Central angle,
θ 200
Area of sector
2
2
πr θ
770cm
360
Now, find the radius of the circle. 770
l
360
r
π 200
e,
r 441
r 21cm
Length of the corresponding arc
2πrθ
360
2 3.14 21 200
360
1
73 cm.
3
Thus, length of the required arc is .
1
73 cm
3
Question: 16
The central angles of two sectors of circles of radii 7
cm and 21 cm are respectively
and
. Find the
120
40
areas of the two sectors as well as the lengths of the
corresponding arcs. What do you observe?
Solution:
For the first circle, 7cm
Central angle =
120
Length of arc
2πrθ
360
2 3.14 7 120
360
44
cm
3
Area of the sector
2
πr θ
360
22 7 7 120
7 360
2
154
cm
3
For the second circle,
21cm
Central angle
40
Length of arc
2πrθ
360 2 3.14 21 40
360
44
cm
3
Area of the sector
2
πr θ
360
22 21 21 40
7 360
2
154 cm
Thus, it can be observed that, the lengths of arcs of
the two circles are equal, whereas areas of the
corresponding sectors are not equal.
Question: 17
Find the area of the shaded region given in the
given figure. Solution:
Area of square PQRS
2
2
14
196cm
Inner figure consists of 4 semicircles of radius and
r
square JKLM of side .
2r
So, find the radius of semicircle.
According to the question,
2r 2r 6 14
4r 8
r 2cm
Area covered by Area of square Area of the4
inner figure JKLM semicircles
2
2
πr
2r 4
2
2
2
2 2 2π 2 2
16 8π cm
Area of Area of Area covered
shaded region squarePQRS by inner figure
196 16 8π
2
180 8π cm
Thus, the area of shaded region is .
2
180 8π cm
Question: 18
Find the number of revolutions made by a circular
wheel of area 1.54 in rolling a distance of 176 m.
2
m
Solution:
Distance covered by the wheel
2πrn 176m
Here, is the number of revolutions made.
n
Area of wheel =
2 2
πr 1.54m
Now, find the radius of the wheel.
1.54
rw ,
π
heel
1.54
r
3.14
0.7m
Now, find the number of revolutions made by the
wheel. 176 7
n
2 22 0.7
40revolutions.
Thus, the circular wheel made 40 revolutions in
covering a distance of 176 m.
Question: 19
Find the difference of the areas of two segments of
a circle formed by a chord of length 5 cm
subtending an angle of at the centre.
90
Solution:
Let us consider, chord SR = 5 cm
It divides the circle in minor segment SPR and
major segment SQR.
Also, let the radius of the circle be .
r Now, in triangle SOR,
Apply Pythagoras theorem.
2 2 2
OS OR SR
2 2 2
r r 5
2
25
r
2
Now, find the area of the minor segment.
Area of minor Area of sector Area of triangle
segment ORSP SOR
2
πr θ 1
r r
360 2
2 2
πr r
4 2
2
π 25 25
cm
8 4
Now, find the area of the major segment.
Area of major Area of Area of minor
segment circle segment
2 2
π 25 25
πr cm
8 4
2
π 25 π 25 25
cm
2 8 4 Area of major
segment
Difference in areas of
major and minor segments
Area of minor
segment
2
π 25 25 π 25 25
πr
8 4 8 4
25π 25π 25 25π 25
2 8 4 8 4
2
25π 25
cm
4 2
Thus, the difference in area of major segment and
minor segment is .
2
25π 25
cm
4 2
Question: 20
Find the difference of the areas of a sector of angle
and its corresponding major sector of a circle
120
Solution: Let, be the angle made by major segment and
2
θ
1
θ
be the angle made by minor segment.
(Given)
1
θ 120
2
θ 360 120
240
21cm
2
2 1
Difference in area of major
and mi
πr
θ
nor segment
θ
360
2
π 21
240 120
360
2
462cm
Thus, the required area is .
2
462cm