NCERT Solutions for Class 10 Maths Chapter 11 Area Related to Circles

NCERT
SOLUTIONS
CLASS - 10th

aglase .co

Class : 10th
Subject : Maths
Chapter : 12
Chapter Name :Area related to circles

Exercise 12.1

Q1 The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has
circumference equal to the sum of the circumferences of the two circles.

Answer. Radius (r ) of the 1st circle = 19cm
1

Radius (r ) of the 1st circle = 9 cm
2

Let the radius of 3rd circle be r.
Circumference of 1st circle = 2πr 1 = 2π(19) = 38π
Circumference of 2nd= 2πr 1 = 2π(9) = 18π
Circumference of 3rd circle =2rπ
That given ,
Circumference of 3rdcircle = Circumference of 2nd circle = Circumference of 1st circle
2πr = 38π + 18π = 56π
56π
r = = 28
2π

Therefore, the radius of the circle which has circumference equal to the sum of the circumference
of the given two circles is 28 cm.

Page : 225 , Block Name : Exercise 12.1

Q2 The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area
equal to the sum of the areas of the two circles.

Answer. Radius (r ) of the 1st circle = 8 cm
1

Radius (r ) of the 1st circle = 6 cm
2

Let the radius of 3rd circle be r.
Area of 1ts circle = πr = π(8) = 64π
2
1
2

Area of 1ts circle = πr = π(6) = 36π
2
1
2

That given,
Area of 3rd circle =Area of 2nd circle =Area of 1st circle
2 2 2
πr = πr + πr
1 2
2
πr = 64π + 36π
2
πr = 100π

r = ±10

However, the radius cannot be negative. Therefore, the radius of the circle having area equal to the
sum of the areas of the two circles is 10 cm.

Page : 225 , Block Name : Exercise 12.1

Q3 Fig. depicts an archery target marked with its ve scoring regions from the centre outwards as
Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and
each of the other bands is 10.5 cm wide. Find the area of each of the ve scoring regions.

Radius (r ) of gold region (i.e. 1st circle) =
1
21

2
= 10.5cm

Given that each circle is 10.5 cm wider than the previous circle.
Therefore . radius r of 2nd circle =10.5 + 10.5 =21 cm
2

radius r of 2nd circle =21 + 10.5 =31.5 cm
3

radius r of 2nd circle =31.5 + 10.5 =42 cm
4

radius r of 2nd circle =42 + 10.5 =52.5 cm
5

Area of gold region = Area of 1st circle =πr = π(10.5) = 346.5cm
2
1
2 2

Area of red region = Area of 2nd circle - Area of 1st circle
=πr − πr
2
2
2
1

=π(21) − π(10.5)
2 2

=441π − 110.25π = 330.75π
= 1039.5cm 2

Area of blue region = Area of 3rd circle - Area of 2nd circle
=πr − πr
2
3
2
2

=π(31.5) − π(21)
2 2

=992.25π − 441π = 551.25π
= 1732.5cm 2

Area of black region = Area of 4th circle - Area of 3rd circle
=πr − πr
2

3
2

2

=π(42) 2
− π(31.5)
2

=1764π − 992.25π = 771.75π
= 2425.5cm 2

Area of white region = Area of 5th circle - Area of 4th circle
=πr − πr
2
3
2
2

=π(52.5) − π(42) 2 2

=2756.25π − 1764π = 992.25π
= 3118.5cm 2

Therefore, areas of gold, red, blue, black, and white regions are 346.5cm , 1039.5cm ,
2 2

1732.5cm , 2425.5cm , 3118.5cm respectively .
2 2 2

Page : 225 , Block Name : Exercise 12.1

Q4 The wheels of a car are of diameter 80 cm each. How many complete revolutions does each
wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?

Answer. Diameter of the wheel of the car = 80 cm
Radius (r) of the wheel of the car = 40 cm
Circumference of wheel =2πr
= 2π(40) = 80πcm

Speed of car = 66 km/hr
66×100000
= cm/min
60
35000
= = 4375
8

Therefore , each wheel of the car will make 4375 revolutions.

Page : 226 , Block Name : Exercise 12.1

Q5 Tick the correct answer in the following and justify your choice : If the perimeter and the area
of a circle are numerically equal, then the radius of the circle is
(A) 2 units
(B) π units
(C) 4 units
(D) 7 units

Answer. Let the radius of the circle be r.
Circumference of circle = 2πr
Area of circle =πr 2

Given that, the circumference of the circle and the area of the circle equal.
This implies 2πr = πr 2

r=2
Therefore, the radius of the circle is 2 units.
Hence, the correct answer is A.

Page : 226 , Block Name : Exercise 12.1

Exercise 12.2

Q1 Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60 . ∘

Area of sector of angle θ = 360
θ
∘ × πr
2

∘

Area of sector OACB = 60

360
∘
×
22

7
× (6)
2

1 22 132 2
= × × 6 × 6 = cm
6 7 7

Therefore , the area of sector of the circle making 60 at the centre of the circle is ∘ 132

7
cm
2

Page : 230 , Block Name : Exercise 12.2

Q2 Find the area of a quadrant of a circle whose circumference is 22 cm.

Let the radius of the circle be r.
Circumference = 22 cm
2πr = 22
22 11
r = =
2π π

Quadrant of circle will subtend 60 at the centre of the circle .∘

∘

Area of such quadrant of the circle = × π × r
360
90
∘
2

2
1 11
= × π × ( )
4π

121 121×7
= =
4π 4×22
77 2
= cm
8

Page : 230 , Block Name : Exercise 12.2

Q3 The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5
minutes.

We know that in 1 hour (i, e. , 60 minutes), the minute hand rotates 3600.
∘

In 5 minutes, minute hand will rotate = × 5 = 30
360

60
∘

Therefore, the area swept by the minute hand in 5 minutes will be the area of a sector of 30 in a ∘

circle of 14 cm radius.
Area of sector of angle θ = × πr
θ

360
∘
2

∘

Area of sector of 30 = ∘ 30

360
∘
×
22

7
× 14 × 14

22
= × 2 × 14
12
11×14
=
3
154 2
= cm
3

Therefore , the area swept by the minute hand in 5 minutes =
154 2
cm
3

Page : 230 , Block Name : Exercise 12.2

Q4 A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the
corresponding :
(i) minor segment
(ii) major sector. (Use π = 3.14)

Let AB be the chord of the subtending 90 angle at centre O of the circle.
∘

∘ ∘

Area of major sector OADB = (
360 −90 2
∘ ) × πr
360

∘

=( 270

360
∘ ) πr
2

3
= × 3.14 × 10 × 10
4
2
= 78.5cm

Area of triangle OAB = 1

2
× OA × OB =
1

2
× 10 × 10

=50cm 2

Area of minor segment Acb = ARea of minor sector OACB
Area of triangle OAB = 78.5 - 50 = 28.5 cm 2

Page : 230 , Block Name : Exercise 12.2

Q5 In a circle of radius 21 cm, an arc subtends an angle of 60 at the centre. Find: ∘

(i) the length of the arc
(ii) area of the sector formed by the arc
(iii) area of the segment formed by the corresponding chord .

∘

Length of arc ACB = 60

360
∘
× 2 ×
22

7
× 21

= 1

6
× 2 × 22 × 3

= 22 cm
∘

Area of sector OACB = 60

360
∘
× πr
2

= 1

6
×
22

7
× 21 × 21

=231cm 2

In ∠OAB
∠OAB = ∠OBA(AsOA = OB)
∘
∠OAB + ∠AOB + ∠OBA = 180
∘ ∘
2∠OAB + 60 = 180
∘
∠OAB = 60

Therefore , triangle OAB is an equilateral triangle.
√3
Area of triangle OAB== 4
× ( Side )
2

√3 441√3
2 2
= × (21) = cm
4 4

Area of segment ACB = Area of sector OACB = Area of triangle OAB
44[√3
2
= (231 − ) cm
4

Page : 230 , Block Name : Exercise 12.2

Q6 A chord of a circle of radius 15 cm subtends an angle of 60 at the centre. Find the areas of the
∘

corresponding minor and major segments of the circle. (Use π = 3.14 and 3 = 1.73)

Radius (r) of circle =15 cm
∘

Area of sector OPRQ = 60

360
∘ × πr
2

1 2
= × 3.14 × (15)
6

2
= 117.75cm

In triangle OPQ ,
∠OP Q = ∠OQP (AsOP = OQ)
∘
∠OP Q + ∠OQP + ∠P OQ = 180
∘
2∠OPQ = 120
∘
∠OP Q = 60

△OP Q is an equilateral triangle .
√3
Area of ΔOP Q = 4
x( side )
2

√3 225√3
2 2
= × (15) = cm
4 4

56.25√3
2
97.3125cm

Area of major segment PSQ = Area of circle — Area of segment PRQ
=π(15) − 20.4375
2

=3.14 × 225 − 20.4375
=706.5 − 20.4375
=686.0625cm 2

Page : 230 , Block Name : Exercise 12.2

Q7 A chord of a circle of radius 12 cm subtends an angle of 120 at the centre. Find the area of the
∘

corresponding segment of the circle. (Use π = 3.14 and 3 = 1.73)

Let us draw a perpendicular OV on chord ST. Lt will bisect the chord ST.
SV = VT
In triangle OVS ,

OV ∘
= cos 60
OS

OV 1
=
12 2

OV = 6 cm
SV ∘ √3
= sin 60 =
SO 2

SV √3
=
12 2

SV = 6√3cm

ST = 2SV = 2 × 6√3 = 12√3cm

Area of angle OST = 1

2
× ST × OV

1
= × 12√3 × 6
2
2
= 36√3 = 36 × 1, 73 = 62.28cm
∘

Area of sector OSUT = 120

360
∘
× π(12)
2

1 2
= × 3, 14 × 144 = 150.72cm
3

Area of segment SUT - Area of sector OSUT - Area of triangle OST
= 150.72 - 62.28
=88.44 cm 2

Page : 230 , Block Name : Exercise 12.2

Q8 A horse is tied to a peg at one corner of a square shaped grass eld of side 15 m by means of a 5
m long rope (see Fig. 12.11). Find (i) the area of that part of the eld in which the horse can graze.
(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14)

Frorn the gure, it be Observed that the horse graze a sector Of 90 in a circle of 5 m radius.
∘

Area that can be grazed by horse = Area of sector OACB
∘

= 90
πr
360
∘
2

= 1

4
× 3.14 × (5)
2

=19.625m 2

Area that can be grazed by the horse when length of rope is 10 m long
∘

= 90
× π × (10)
360
∘
2

= 1

4
× 3.14 × 100

=78.5m 2

Increase in grazing area = (78.5 - 19.625)m 2

Page : 230 , Block Name : Exercise 12.2

Q9 A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also
used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. 12.12.
Find :
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.

Answer. Total length of wire required will be the length of 5 diameters and the circumference of
the brooch.
Radius of circle = mm 35

2

Circumference of brooch = 2πr
=2 × 22

7
× (
35

2
)

=110 mm
Length of wire required = 110 +5 x 35 = 285 mm
It can be observed from the gure that each Of 10 sectors Of the circle is subtending 36 at the
∘

centre of the circle.

∘

Therefore , area of each sector = 36
∘ × πv
2

360

= 1

10
×
22

7
× (
35

2
) = ×(
35

2
)

= 385

4
mm
2

Page : 230 , Block Name : Exercise 12.2

Q10 An umbrella has 8 ribs which are equally spaced (see Fig. 12.13). Assuming umbrella to be a
at circle of radius 45 cm, nd the area between the two consecutive ribs of the umbrella.

∘

Answer. There 8 ribs in umbrella. The between two consecutive ribs is subtending 360

8
∘
= 45 at
the centre of the assumed at circle.

∘

Area between two consecutive ribes of circle = 45

360
∘ × πr
2

1 22 2
× × (45)
8 7
11 22275 2
= × 2025 = cm
28 28

Page : 231 , Block Name : Exercise 12.2

Q11 A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping
through an angle of 115 . Find the total area cleaned at each sweep of the blades.
∘

It can be observed from the gure that each blade of wiper will sweep an area of a sector of115 in ∘

a circle of 25 cm radius.
∘

Area of such sector = × π × (25)
115

360
∘
2

= 23

72
×
22

7
× 25 × 25

= 158125

252
cm
2

Area swept by 2 blades = 2 × 158125

252
158125 2
cm
126

Page : 231 , Block Name : Exercise 12.2

Q12 To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of
angle 80 to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π
∘

= 3.14)

It can be observed from the gure that the lighthouse spreads light across a sector of 800 in a circle
of 16.5 km radius.
∘

Area of sector OACB = × πν
80

360
∘
2

= 2

9
× 3.14 × 16.5 × 16.5

= 189.97km 2

Page : 231 , Block Name : Exercise 12.2

Q13 A round table cover has six equal designs as shown in Fig. 12.14. If the radius of the cover is 28
cm, nd the cost of making the designs at the rate of ` 0.35 percm . (Use 3 = 1.7)
2

It can be observed that these designs segments of the circle.
∘

Consider segment APE, Chord AE is a side Of the hexagon. Each chord will substitute 360

6
∘
= 60

at the centre of circle.
In triangle OAB ,
∠OAB = ∠OBA(AsOA = OB)
∘
∠AOB = 60
∘
∠OAB + ∠OBA + ∠AOB = 180
∘ ∘ ∘
2∠OAB = 180 − 60 = 120
∘
∠OAB = 60

Therefore, triangle OAB is an equilateral triangle.
√3
Area of triangle OAB = 4
× ( side )
2

√3
=
4
× (28)
2
= 196√3 = 196 × 1.7 333.2cm = 2

0

Area of sector OAPB = 60

0
× πr
2

360
1 22
× × 28 × 28
6 7
1232 2
cm
3

Area of segment APB = Area of sector OAPB - Area of triangle OAB
=( 1232

3
− 333.2) cm
2

Therefore , area of designs = 6 × ( 1232

3
− 333.2) cm
2

= (2464 − 1999.2)cm 2

= 464.8cm 2

Cost of making 1 cm designs = Rs 0.35 2

Cost of making 464.76 cm designs = Rs162.68 2

Therefore, the cost of making such designs is Rs 162.68,

Page : 231 , Block Name : Exercise 12.2

Q14 Tick the correct answer in the following : Area of a sector of angle p (in degrees) of a circle
with radius R is
p
× 2πR
180
p 2
× πR
180
p
× 2πR
360
p 2
× 2πR
720

We know that area of sector of angle = θ = θ

360
∘ × πR
2

Area of sector of angle P =
p 2
∘
(πR )
360

=(
p 2
∘
) (2πR )
720

Hence, (D) is the correct answer.

Page : 231 , Block Name : Exercise 12.2

Exercise 12.3

Q1 Find the area of the shaded region in Fig., if PQ = 24 cm, PR = 7 cm and O is the centre of the
circle.

Answer. It can be observed that RQ is the diameter of the circle. Therefore, ∠RPQ will be 90 ∘

By applying Pythagoras theorem in triangle PQR,
2 2 2
RP + PQ = RQ
2 2 2
(7) + (24) = RQ

RQ = √625 = 25

Radius of circle, OR =
RQ 25
=
2 2

Since RQ is the diameter of the circle, it divides the circle in two equal parts.
Area of the semi-circle RPQOR = πr 1

2
2

2

= 1

2
π(
25

2
)

= 1

2
×
22

7
×
625

4

= 6875

28
cm
2

Area of triangle PQR = 1

2
× PQ × PR

= 1

2
× 24 × 7

=84cm 2

Area of shaded region = Area of semi-circle RPQOR - Area of ΔPQR
= 6875
− 84
28

=
6875−2352

28

= 4523

28
cm
2

Page : 234 , Block Name : Exercise 12.3

Q2 Find the area of the shaded region in Fig., if radii of the two concentric circles with centre O are
7 cm and 14 cm respectively and ∠ AOC = 40 ∘

Radius of inner circle = 7 cm
Radius of outer circle = 14 cm
Area of shaded region = Area of sector OAFC -Area of sector OBED
∘ ∘

= 40
× π(14) =
360
∘ × π(7)
2

360
40
∘
2

= 1

9
×
22

7
× 14 × 14 −
1

9
×
22

7
× 7 × 7

== 616 154 462
− =
9 9 9

=
154 2
cm
3

Page : 235 , Block Name : Exercise 12.3

Q3 Find the area of the shaded region in Fig. , if ABCD is a square of side 14 cm and APD and BPC
are semicircles.

Answer. It can be observed from the gure that the radius of each semi-circle is 7 cm.

Area of each semi-circle = 1

2
πr
2

= 1

2
×
22

7
× (7)
2

=77cm 2

Area of square ABCD +( Side ) = (14) 2 2

=196cm 2

Area of the shaded region
=Area of square ABCD - Area of semi-circle APD - Area of semi-circle BPC
=196 − 77 − 77 = 196 − 154 = 42cm 2

Page : 235 , Block Name : Exercise 12.3

Q4 Find the area of the shaded region in Fig. , where a circular arc of radius 6 cm has been drawn
with vertex O of an equilateral triangle OAB of side 12 cm as centre.

∘

Area of sector OCDE = 60

360
∘ πr
2

= 1

6
×
22

7
× 6 × 6

= 132

7
cm
2

√3 √3×12×12
Area of triangle OAB = 4
(12)
2
=
4
= 36√3cm
2

Area of circle =πr 2
=
22

7
× 6 × 6 =
792

7
cm
2

Area of shaded region = Area of triangle OAB + Area of circle - Area of sector OCDE
= 36√3 + −
792

7
132

7

660 2
(36√3 + ) cm
7

Page : 235 , Block Name : Exercise 12.3

Q5 From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a

circle of diameter 2 cm is cut as shown in Fig. . Find the area of the remaining portion of the
square.

Each quadrant is a sector of 90 in a circle of 1 cm radius. ∘

∘

Area of each quadrant = πr
90

360
∘
2

= 1

4
×
22

7
× (1)
2
=
22

28
cm
2

Area of square = ( Side ) 2
= (4)
2
= 16cm
2

= 22

7
cm
2

Area of the shaded region = Area of square - Area of circle - 4 x Area of quadrant
= 16 − − 4 ×
22

7
22

28

= = 16 − 22

7
−
22

7
= 16 −
44

7

=
112−44 68 2
= cm
7 7

Page : 235 , Block Name : Exercise 12.3

Q6 In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC
in the middle as shown in Fig. Find the area of the design.

Answer. Radius (r) of circle 32 cm
AD is the median of triangle ABC.
2
AO = AD = 32
3

AD = 48 cm
In triangle ABD,
=AB = AD + BD 2 2 2

2

=AB 2
= (48)
2
+ (
AB

2
)

2

= 3AB

4
= (48)
2

=AB = 48×2
=
96

√3 √3

=32√3cm
Area of equilateral a triangle ,
√3
× 32 × 32 × 3 = 96 × 8 × √3
4
2
768√3cm

Area of circle = πr 2

= × (32)
22

7
2

= 22

7
× 1024

= 22528

7
cm
2

Area of design = area of circle - Area of triangle ABC
=( 22528

7
− 768√3) cm
2

Page : 235 , Block Name : Exercise 12.3

Q7 In Fig., ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such
that each circle touch externally two of the remaining three circles. Find the area of the shaded
region.

Area of each of the 4 sectors is equal to each and is a sector of 90 in a circle of 7 crn radius.
∘

∘

Area of each sector = 90
× π(7)
360
∘
2

= 1

4
×
22

7
× 7 × 7

= 77

2
cm
2

Area of Square of ABCD =( Side ) = (14) = 196cm
2 2 2

Area Of shaded portion = Area Of square ABCD - 4 x Area Of each sector
= 196 − 4 × 77
= 196 − 154
2

=42cm 2

Therefore , the area of shaded portion is 42cm . 2

Page : 236 , Block Name : Exercise 12.3

Q8 Fig. depicts a racing track whose left and right ends are semicircular. The distance between the
two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide,
nd :
(i) the distance around the track along its inner edge
(ii) the area of the track.

Distance around the track along its inner edge =AB + arc BEC + CD + arc DFA
=106 + × 2πr + 106 + × 2πr
1

2
1

2

=212 + 1

2
× 2 ×
22

7
× 30 +
1

2
× 2 ×
22

7
× 30

=212 + 2 × 22

7
× 30

=212 + 1320

7

=
1484+1320 2804
= m
7 7

Area of the track = (Area of GHIJ - Area of ABCD) + (Area of semi-circle HKI - Area
of semi-circle BEC) + (Area of semi-circle GLJ - Area of semi-circle AFD)
=
1 22 2 1 22 2 1 22 2 1 22 2
106 × 80 − 106 × 60 + × × (40) − × × (30) + × × (40) − × × (30)
2 7 2 7 2 7 2 7

=106(80 − 60) + 22

7
× (40)
2
−
22

7
× (30)
2

=106(20) + 22

7
[(40)
2
− (30) ]
2

=2120 + 22

7
(40 − 30)(40 + 30)

=2120 + ( 22

7
) (10)(70)

=2120 + 2200
= 4320m 2

Therefore , the area of shaded region is 4320m 2

Page : 236 , Block Name : Exercise 12.3

Q9 In Fig., AB and CD are two diameters of a circle (with centre O) perpendicular to each other and
OD is the diameter of the smaller circle. If OA = 7 cm, nd the area of the shaded region.

Radius r of larger circle = 7cm
1

Radius r of smaller circle = cm
2
7

2

Area of smaller circle = πr 2
1

= 22

7
×
7

2
×
7

2

= 77

2
cm
2

Area of semi-circle AECFB of larger circle = 1

2
πr
2
2

= 1

2
×
22

7
× (7)
2

= 77cm 2

Area of triangle ABC = 1

2
× AB × OC

= 1

2
× 14 × 7 = 49cm
2

Area of the shaded region =Area of smaller circle + Area of semi-circle AECFB
- Area of triangle ABC
= + 77 − 49
77

2

= 28 +
77 2
= 28 + 38.5 = 66.5cm
2

Page : 236 , Block Name : Exercise 12.3

Q10 The area of an equilateral triangle ABC is 17320.5 cm . With each vertex of the triangle as
2

centre, a circle is drawn with radius equal to half the length of the side of the triangle (see Fig.).
Find the area of the shaded region.

Answer. Let the side of the equilateral triangle be a,
Area of equilateral triangle = 17320.5 cm 2

√3
2
(a) = 17320.5
4
1.73205 2
a = 17320.5
4
2
a = 4 × 10000

A = 200 cm

Each sector of measure 60 ∘

∘

Area of sector ADEF =
60 2
× π × r ∘
360

= 1

6
× π × (100)
2

= 3.14×10000

6

= 15700

3
cm
2

Area Of shaded region Area Of equilateral triangle - 3 x Area Of each sector
15700
17320.5 − 3 ×
3
2
17320.5 − 15700 = 1620.5cm

Page : 236 , Block Name : Exercise 12.3

Q11 On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig). Find
the area of the remaining portion of the handskter chief.

From the gure, it can be observed that the side of the squ«e is 42 cm.
Area of square = ( Side ) = (42) = 1764cm
2 2 2

Area of each circle =9 × 154 = 1386cm 2

Area of the remaining portion of the = 1764 -1386 = 378 cm 2

Page : 237 , Block Name : Exercise 12.3

Q12 In Fig. OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, nd the
area of the
(i) quadrant OACB,
(ii) shaded region.

(i) Since OACB is a quadrant, it will subtend 90 angle at O.
∘

∘

Area of quadrant OACB = × πr
90

360
∘
2

2
1 22 2 1 22 7
= × × (3.5) = × × ( )
4 7 4 7 2

11×7×7 77 2
= cm
2×7×2×2 8

(ii) area of triangle = 1

2
× OB × OD
1
× 3.5 × 2
2
1 7
= × × 2
2 2
7 2
cm
2

Area of the shaded region = Area of OACE - Area of triangle OBO
77 7
= −
8 2
77−28

8
49 2
cm
8

Page : 237 , Block Name : Exercise 12.3

Q13 In Fig. a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, nd the area of the
shaded region.

Answer. In △OAB,
2 2 2
OB = OA + AB
2 2
= (20) + (20)

OB = 20√2

Radius of circle = 20√2cm
∘

Area of quadrant OPBQ =
90 2
∘ × 3.14 × (20√2)
360

1
× 3.14 × 800
4
2
628cm

Area of OABC = ( Side ) = (20) = 400cm2 2 2

Area of shaded region = Area of quadrant OPBQ - Area of OABC
=(628 − 400)cm 2

= 228cm 2

Page : 237 , Block Name : Exercise 12.3

Q14 AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O
(see Fig.). If angle AOB = 30
∘

nd the area of the shaded region.

Area of the shaded region = Area of sector OAEB = Area of sector OCFD
∘ ∘

= 30
× π × (21) −
∘
2 30
× π × (7) ∘
2

360 360

=
1 2 2
× π [(21) − (7) ]
12

=
1 22
× × [(21 − 7)(21 + 7)]
12 7

=
22×14×28

12×7

=
308 2
cm
3

Page : 237 , Block Name : Exercise 12.3

Q15 In Fig. ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as
diameter. Find the area of the shaded region.

As ABC is a quadrant of the circle,angle BAC will be of measure 90 .
0

In triangle ABC,

2 2 2
BC = AC + AB

=(14) 2
+ (14)
2

=BC = 14√2
Radius r Of semi-circle drawn on BC =
14√2
1 = 7√2cm
2

Area of triangle ABC = 1

2
× AB × AC

= 1

2
× 14 × 14

=98cm 2

∘

Area of Sector ABCD = =
90 2
∘ × π
360

=
1 22
× × 14 × 14
4 7

=154cm 2

Area of semi-circle drawn on BC = = 1

2
× π × r
2
1
=
1

2
×
22

7
× (7√2)
2

1 22 2
× × 98 = 154cm
2 7

Area of shaded region = Area of semi-circle - (Area of sector ABCD - Area of triangle ABC) = 154 -
(154-98)
=98cm 2

Page : 237 , Block Name : Exercise 12.3

Q16 Calculate the area of the designed region in Fig. 12.34 common between the two quadrants of
circles of radius 8 cm each.

The designed is the common region between two sectors BAEC and DAFC.
∘

Area of sector BAEC = × × (8)
90

360
∘
22

7
2

= 1
×
22

7
× 64
4

= 22×16

7

=
352 2
cm
7

Area of triangle BAC =
1
× BA × BC
2

= 1

2
× 8 × 8 = 32cm
2

Area of the designed portion = 2 x (Area of segrnent AEC)
= 2 x (Area of sector BAEC - Area of triangle BAC)
=2 × (
352
− 32)
7

=2 (
352−224
)
7

= 2×128

7

=
256 2
cm
7

Page : 238 , Block Name : Exercise 12.3