Areas related to Circles Class 10

Ex 11.1, 1 teackoo
Find the area of a sector of a circle with radius 6 cm if angle of the
sector is 60°.
Given that,
Radius = r=6cm
& Angle of the sector = @ = 60°
We know that,
Area of sector of circle = —— x nr?
360°
= 9 yy eye
“360 %7 * (6)
atx x 36
6° 7
_ 6x22
“7

Ex 11.1, 2 teachoo
Find the area of a quadrant of a circle whose circumference is 22 cm.
Area of quadrant =2 x Area of circle
= 1 x ner?
4
Now, we need to find r.
It is given that
Circumference = 2mr
22 = 2nr
22
—=nr
2
11=nr

Ex11.1,3 teachoo
The length of the minute hand of a clock is 14 cm. Find the area
swept by the minute hand in 5 minutes. 5 1 2 4 >
r=14cm
2) -
i 6 8
Minute hand completes full circle degree in one hour
Now,
Angle swept by minute hand in 1 hour (i.e 60 minutes ) = 360°
. : . 360
Angle swept by minutes hand in 1 minutes = 7 6°
Angle swept by minute hand in 5 minutes = 6° x 5 = 30°
Hence, 6 = 30°, r=14cm
Area swept by minutes hand = Area of sector

Ex 11.1, 4 teackhoo
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 (Use m= 3.14)

Given that (we
OA = OB = radius = 10 cm A gos /
6=90° Pp

Now,

Area of segment APB = Area of sector OAPB — Area of AAOB

Area of sector OAPB

Area of sector OAPB = = x Tr?

=~ x 3.14 x (10)?
360

teackhoo
Ex 11.1,5
In a circle of radius 21 cm, an arc subtends an angle of 60° at the
centre. Find:
(i) the length of the arc
Length of Arc APB = 2 x (2nr) <> 8
360 Pp
= x2x2x 21
360° 7
1
=x 2x 22x 3
= 22cm

Ex 11.1, 6 teackoo
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 n= 3.14 and V3 = 1.73)
In a given circle,
Radius (r) = 15 cm L»>

And, @ = 60° A B

P

Now,

Area of segment APB = Area of sector OAPB — Area of AOAB
Finding Area of sector OAPB
Area of sector OAPB = = x ar?

Ex 11.1, 7 teackoo
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 m = 3.14 and v3 = 1.73)
In a given circle, (cae
Radius (r) = 12 cm A B
P

And, @ = 120°
Now,

Area of segment APB = Area of sector OAPB-— Area of AOAB
Finding Area of sector OAPB
Area of sector OAPB= x mr?

360

Ex 11.1, 8 teackoo
A horse is tied to a peg at one corner of a square shaped grass field of
side 15 m by means of a5 m long rope (see figure). Find
(i) the area of that part of the field in which the horse can graze.(Use
A D
m= 3.14) or: we 7? Pees
Stet Saeteyts vy
ese sel eset ite
tats ftette tat,
Qa 332 +s +s tot Ye *
Let ABCD be square field eterraen,
hecae eel
‘ _
And, length of rope =5 m r=5m rae YS
“debe” +”
° = wwe
~r=5m B . ¢
We need to find area of field which horse can graze,
i.e. Area of sector QBP a
r=5m
Since In square all angles are 90°.
B P
Hence, Z QBP = 90°

Ex 11.1, 9 teackoo
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 figure. Find :

(i) the total length of the silver wire required
fo LV%e\
ee,

@) ®)

Brooch is made with silver wire in form of a circle. GAO

Diameter of the brooch = 35 mm

‘ 35
Radius of brooch = > mm

Since wire is used in making 5 diameter & circle.

So,

Total wire is used = Length of wire in circle + Wire used in S diameters

Ex 11.1, 10 teackoo
An umbrella has 8 ribs which are equally spaced (see figure).
Assuming umbrella to be a flat circle of radius 45 cm, find the area
between the two consecutive ribs of t .
Area of rib = F x Area of umbrella SB cl
Radius =r =45 cm
Area of umbrella = Area of circle

= nr?

= 7 (45)

== x 45 x 45

Ex 11.1, 11 feachoo
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.
Radius =r =25 cm
Sweeping angle = 115°
Total area cleaned by the two wipers
= 2 x Area cleaned by one wiper
=2x Area of sector with angle 115°
=2x x ar
360
=2xtBy 2 x (25)?
360° 7

Ex 11.1, 12 teackoo
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 m= 3.14)
Radius (r)= 16.5 km r=16.5km
And, 8 = 80°
Area over which ships are warned = Area of sector with angle 80°
= x ar?
360°
= ry B y (16.5)?
360° 7
=~ 2x2 x 16.5 x 16.5
360 °° 7
2 22 165 — 165
== x= x x
9° 7 10 10

Ex 11.1, 13 teackoo
A round table cover has six equal designs as shown in figure. If the
radius of the cover is 28 cm, find the cost of making the designs at
the rate of Rs 0.35 per cm. (Use V3 = 1.7) SE Sp.
LE aN
f A
In the question a round table cover KT { |) 60° s)
ra RRS 5
4 3 ey
is given which has six equal designs. SP,
Bb aA
“ele fe
Let us join ends of the design pA
Since the designs are equal,
the angle made by the designs at point O will be equal.
We know that
Sum of angles at a point = 360°
6 x Angle made by one design = 360°

teackhoo
Ex 11.1, 14
Tick the correct answer in the following :
Area of a sector of angle p (in degrees) of a circle with radius R is
= Pe 2 Pe Pe 2
(A) ian * 2nR (B) iao * mR? (C) 300 * 2nR (D) 70 * 2nR
Area of tor=—-x ar?
ea of asector=>7 x mr
Where 8 = angle, r = radius of circle
Here, we have
0 =pand radius =R
Putting these values in formula
Area of sector = 2 x R?
ea of sector = =~ x 1

Example 1 (Method 1) teachoo
Find the area of the sector of a circle with radius 4 cm and of
angle 30°. Also, find the area of the corresponding major sector
(Use n = 3.14).
Given
: A\
Radius =r = 4cm, @ = 30° Lip
Ap B
Now,
0
Area of sector = — x mr?
360
= 30 2
=3e0 * 3.14 x (4)
1
=—x3.14x4x4
12
1
=3% 3.14 x 4

Example 2 teackoo
Find the area of the segment AYB shown in figure, if radius of the
circle is 21 cm and ZAOB = 120°. (Usen= = ) So’
‘ B
—

In a given circle,
Radius (r} = 21cm

And, @ = 120°
Now,

Area of segment AYB = Area of sector OAYB — Area of AOAB
Finding Area of sector OAYB
Area of sector OAYB = = x ar?

teachoo.
Example 1 eames
The cost of fencing a circular field at the rate of Rs 24 per metre is Rs
5280. The field is to be ploughed at the rate of Rs 0.50 per m2. Find
the cost of ploughing the field (Take nm = a ).
Total cost of fencing a circular field = Rs 5280
Cost of fencing per meter = Rs 24
So, total length of field = Total cost of fencing
cost of fencing per meter

_ 5280

“24

= 220m
Here,
Total length of the field would be circumference of the circular field .

Total length = circumference of circular field

teachoo.
Example 4(Method 1) eames
In figure, two circular flower beds have been shown on two sides of a
square lawn ABCD of side 56 m. If the centre of each circular flower
bed is the point of intersection O of the diagonals of the square lawn,
find the sum of the areas of the lawn and the flower beds.
AB
Area of lawn + Flower bed NK 7
= Area of lawn(square of side 56 m) 56m
+ Area of flower bed AB & CD D aN Ic
_”
Area of lawn = Area of square of side 56 m
= (Side)?
= (56)?
= 3136 m2

Example 5 teachoo.com
Find the area of the shaded region in figure, where ABCD is a square
of side 14 cm.
Area of shaded region = Area of square ABCD — Area of 4 circles
Area of square ABCD 14cm
. ‘7A S———pB
Side of square = 14 cm
bramerrYoramey
Area of square = {side}? NS
= (14)?
=14«14 KZA LA
= 196 cm?
Area of 4 circles
Since ABCD is a square.
Hence, AB = BC=CD=AD=14cm

teachoo.
Example 6 (Method 1) eaenee nom
Find the area of the shaded design in figure, where ABCD isa
square of side 10 cm and semicircles are drawn with each side of
the square as diameter. (Use m= 3.14)
A B
Given y
Side of square ABCD = 10 cm
Area of square ABCD = {side}?
= (10)
= 100 cm? D Cc
Given semicircle is drawn with side of square as diameter,
So, Diameter of semicircle = Side of square = 10 cm
Radius of semicircle = = = > =5cm

This question is
inspired from
Ex 12.3, 6 - Chapter 12 Class 10
- Areas Related to Circle
Ex 12.3, 11 - Chapter 12 Class 10
- Areas Related to Circles
Pookalam is the flower bed or flower pattern designed during Onam in Kerala. It is similar as Rangoli in North India and Kolam in Tamil Nadu. During the festival of Onam , your school is planning to conduct a Pookalam competition. Your friend who is a partner in competition , suggests two designs given below.
Observe these carefully
Design I:
This design is made with a circle of radius 32cm leaving equilateral triangle ABC in the middle as shown in the given figure.
Design II:
This Pookalam is made with 9 circular design each of radius 7cm.
Refer Design I:
Question 1
The side of equilateral triangle is
(a) 12√3 cm (b) 32√3 cm (c) 48 cm (d) 64 cm
Question 2
The altitude of the equilateral triangle is
(a) 8 cm (b) 12 cm (c) 48 cm (d) 52 cm
Refer Design II:
Question 3
The area of square is
(a) 1264 cm
2
(b) 1764 cm
2
(c) 1830 cm
2
(d) 1944 cm
2
Question 4
Area of each circular design is
(a) 124 cm
2
(b) 132 cm
2
(c) 144 cm
2
(d) 154 cm
2
Question 5
Area of the remaining portion of the square ABCD is
(a) 378 cm
2
(b) 260 cm
2
(c) 340 cm
2
(d) 278 cm
2

This question is
inspired from
Ex 12.2, 9 - Chapter 12 Class 10
- Areas Related to Circle
Ex 12.1, 4 - Chapter 12 Class 10
- Areas Related to Circles
A brooch is a small piece of jewellery which has a pin at the back so it can be fastened on a dress, blouse or coat. Designs of some brooch are shown below. Observe them carefully.
Design A:
Brooch A is made with silver wire in the form of a circle with diameter 28mm. The wire used for making 4 diameters which divide the circle into 8 equal parts.
Design B:
Brooch b is made two colours_Gold and silver. Outer part is made with Gold. The circumference of silver part is 44mm and the gold part is 3mm wide everywhere.
Refer to Design A
Question 1
The total length of silver wire required is
(a) 180 mm (b) 200 mm
(c) 250 mm (d) 280 mm
Question 2
The area of each sector of the brooch is
(a) 44 mm
2
(b) 52 mm
2
(c) 77 mm
2
(d) 68 mm
2
Refer Design B:
Question 3
The circumference of outer part (golden) is
(a) 48.49 mm (b) 82.2 mm
(c) 72.50 mm (d) 62.86 mm
Question 4
The difference of areas of golden and silver parts is
(a) 18 π (b) 44 π
(c) 51 π (d) 64 π
Question 5
A boy is playing with brooch B. He makes revolution with it along its edge. How many complete revolutions must it take to cover 80 π mm?
(a) 2 (b) 3 (c) 4 (d) 5

Question A horse is tied to a peg at one corner of a square shaped grass field of sides 15 m by means of a 5 m long rope (see the given figure). Labeling our figure
Question 1 What is the area of the grass field? (a) 225 m2 (b) 225 m (c) 255 m2 (d) 15 m
Area of grass field
= Area of square of side 15 m
= (Side)2
= 152
= 225 m2

Question In a workshop, 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 the given figure. Question 1 What is the radius of the circle? (A) 35/2 mm (B) 5/2 mm (C) 35 mm (C) 10 mm Given
Diameter = 35 mm

Question In a class activity, Sheena does a block painting on a square handkerchief as shown in figure. She made nine designer circles each of radius 7 cm Question 1 What is the area of the nine circles? (A) 154 cm2 (B) 145 cm2 (C) 1386 cm2 (C) 1836 cm2 Given
Radius of circle = 7 cm

If the area of a circle is 154 cm2, then its perimeter is
(A) 11 cm (B) 22 cm (C) 44 cm (D) 55 cm
Now,
Area of circle = 𝜋r2
154 = 𝟐𝟐/𝟕 × r2
154 × 7/22 = r2
7 × 7 = r2
49 = r2
r2 = 49
r2 = 72
r = 7 cm

If θ is the angle (in degrees) of a sector of a circle of radius r,
then area of the sector is
(A) (𝜋𝑟^2 𝜃)/360 (B) (𝜋𝑟^2 𝜃)/180 (C) 2𝜋𝑟𝜃/360 (D) 2𝜋𝑟𝜃/180

If the sum of the areas of two circles with radii R1 and R2 is equal to the area of a circle of radius R, then
(A) R1 + R2 = R (B) R_1^2 + R_2^2 = R2
(C) R1 + R2 < R (D) R_1^2 + R_2^2 < R2
Given that
sum of the areas of two circles with radii R1 and R2 is equal to the area of a circle of radius R

If the sum of the circumferences of two circles with radii R1
and R2 is equal to the circumference of a circle of radius R, then
R1 + R2 = R
(B) R1 + R2 > R
(C) R1 + R2 < R
(D) Nothing definite can be said about the relation among R1 , R2
and R.
Given that
sum of the circumferences of two circles with radii R1 and R2 is equal to the circumference of a circle of radius R

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 the circle and square.
Let Radius of circle = r
Side of square = a

Area of the largest triangle that can be inscribed in a semi-circle of radius r units is
(A) 𝑟^2 sq. units (B) 1/2 𝑟^2 sq. units
(C) 2𝑟^2 sq. units (D) √2 𝑟^2 sq. units
Let’s consider semi-circle with radius r
With Diameter AB

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
Let Radius of circle = r
Side of square = a

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
Now,
Area of 8m circle + Area of 6m circle = Area of circle with radius r
𝜋82 + 𝜋62 = 𝜋r2
82 + 62 = r2
64 + 36 = r2
100 = r2
r2 = 100
r2 = 102
r = 10 m

The area of the circle that can be inscribed in a square of side 6 cm is
(A) 36 π cm2 (B) 18 π cm2 (C) 12 π cm2 (D) 9 π cm2
Since circle is inscribed in the square
Diameter of circle = Side of square
= 6 cm

The area of the square that can be inscribed in a circle of radius 8 cm is
(A) 256 cm2 (B) 128 cm2 (C) 64 2 cm2 (D) 64 cm2
Now,
Diagonal of square = Diameter of circle
= 16 cm

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
Now,
Circumference of 18 cm circle + Circumference of 10 cm circle = Circumference of circle with radius r
2𝜋(18) + 2𝜋(10) = 2𝜋r
18 + 10 = r
28 = r
r = 28 cm

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
Now,
Area of 24 cm circle + Area of 7cm circle = Area of circle with radius r
𝜋(24)2 + 𝜋(7)2 = 𝜋r2
(24)2 + 72 = r2
Now,
Area of 24 cm circle + Area of 7cm circle = Area of circle with radius r
𝜋(24)2 + 𝜋(7)2 = 𝜋r2
(24)2 + 72 = r2
576 + 49 = r2
625 = r2
r2 = 625
r2 = 252
r = 25 cm

If the area of a circle is equal to the sum of the areas of two circles of diameters 10 cm and 24 cm, then diameter of the larger circle (in cm) is (A) 34 (B) 26 (C) 17 (D) 14
Now,
Area of 5 cm circle + Area of 12 cm circle = Area of circle with radius r
𝜋(5)2 + 𝜋(12)2 = 𝜋r2
52 + (12)2 = r2
25 + 144 = r2
169 = r2
r2 = 169
r = √169
r = √("132" )
r = 13 cm

If 𝜋 is taken as 22/7, the distance (in metres) covered by a wheel of diameter 35 cm, in one revolution, is (A) 2.2 (B) 1.1 (C) 9.625 (D) 96.25
Diameter = 35 cm
Radius = 𝟑𝟓/𝟐 cm

Ex 12.1,1 teachoo.com
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.
Given that,
Radius of 1% circle =r, = 9cm
Radius of 2"4 circle =r, = 19 cm
Let the radius of required circle = rcm
According to question,
Circumference of required circle = Sum of circumference of two circles
Circumference of small circle Circumference of larger circle
= 2nr, = 201,
=2m X9 =2xXmX19
=187 =387

Ex 12.1, 2 teachoo.com
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.
Given that,
Radius of 1% circle =r, = 8cm
Radius of 2" circle =r, =6cm
Let the radius of required circle = r cm
According to question,
Area of required circle = Sum of the area of both the circles
Area of first circle = r,? Area of second circle = zr,?
= 1 (8) = (6)
= m(8 X 8) = 1(6 x 6)
=641 = 367

teachoo.com
Ex 12.1, 3
Given figure depicts an archery target marked with its five scoring
areas 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 five scoring regions.
WHITE

Diameter of gold circle = 21 cm
So, Radius of gold = — = ae cm = 10.5 cm
Now,
Area of gold = 1 g91p7

== «10.5 x 10.5

= 346.5 cm?

teachoo.co
Ex 12.1, 4 ”
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?
. Total distance
Number of revolutions = 2 WL.
Distance covered in 1 revolution
Diameter of circle = 80 cm @
: 80
radius = r=— =40cm Ue
2 Circumference = 2mr

Distance covered in one revolution = Circumference of wheel

=27r

=2xmx40

= 80R cm

Ex 12.1, 5 teachoo.com
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) 7 units (C) 4 units (D} 7 units
It is given that
Perimeter = area of a circle
2nr = nr?
an rt
Tt ~ Yr
2=r
r= 2 units
Hence A is correct.

Ex 12.3, 1 teachoo.com
Find the area of the shaded region in figure, if PQ = 24 cm, PR= 7 cm
and O is the centre of the circle.
Q
Area of shaded region
= Area of semicircle — Area of APQR
Since , QR is diameter, R
Pa {7

It forms a semicircle. 77)
We know that angle in a semicircle is a right angle.
Hence , Z RPQ = 90°
Hence, ARPQ is right triangle
Now , as per Pythagoras theorem
(Hypotenuse)? = (Height)? + (Base)?

(QR)? = (PQ)? + {PR}?

Ex 12.3, 2 teachoo.com
Find the area of the shaded region in figure, if radii of the two
concentric circles with centre O are 7 cm and 14 cm respectively and
ZAOC = 40°.
Area of shaded region
= Area of sector AOC — Area of sector BOD Bi
2 "
Area of sector BOD Cc
In smaller circle, @ = 40° and r= 7 cm
Area of sector BOD = — x mr?
360
= 40 y 22 2
“360% 7 * 7)
=tx2x7x7
9° 7
1
==X22X7
9
_ 154 yy
9

Ex 12.3, 3 teachoo.com
Find the area of the shaded region in figure, if ABCD is a square of
side 14 cm and APD and BPC are semicircles.
A B
Area of shaded region \
= Area of square ABCD
— Area of semicircle APD
— Area of semicircle BPC /\
D Cc
Area of square ABCD
Side of square = 14cm
Area of square = Side x Side
=14x 14
= 196 cm?

Ex 12.3, 4 teachoo.com
Find the area of the shaded region in figure, 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 shaded region ,
= Area of circle with radius 6 cm N
+ Area of equilateral triangle with side 12 cm \ “4
— Area of sector ODE J \
B
A 12cm
Area of circle
Radius of circle = r=6 cm
Area of circle = mr?
2 2
=> x (6)
== x36
7
= 792 omy?
7

Ex 12.3, 5 feachoo.com
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 figure. Find the area of the remaining portion of the square.
F y, 4cm ur
Area of remaining portion (e]
= Area of square r=icm
— Area of middle circle — Area of 4 quadrants

For square ABCD
Area of square = (side)?

= (4?

=4x4

=16

teachoo.com
Ex 12.3, 6
In a circular table cover of radius 32 cm, a design is formed leaving
an equilateral triangle ABC in the middle as shown in figure. Find
the area of the design (shaded region).
A
Area of design [WwW ZN
= Area of circle — area of triangle ABC SS |
LS ROANTIPARRAD ASIA S
For circle AU) Nor
Radius = r = 32 cm
Area of circle = mr?
=? x 322
7
== x 32x 32
= 22528
7

Ex 12.3, 7 teachoo.com
In figure, 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.
Since square has all angles 90° OrG)
Area of shaded region |
= Area of square ABCD a
— Area of 4 quadrants of circle

Area of square ABCD
Given side of square = 14 cm
Area of the square = (Side)*

= (14)

=(14 x 14)

= 196 cm?

Ex 12.3, 8 teachoo.com
Given figure depicts a racing track whose left and right ends are semi-
circular. 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, find :
(i) the distance around the track along its inner edge
k 106m
D Cc
|
106m
It is given that two inner line segments are parallel,
So, EF II HG
Here, EF = AB = 106 m
HG = DC=106m
EH=FG=60m
AE =BF=HD=GC=10m

Ex 12.3, 9 feachoo.com
In figure, 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, find the area of the shaded region. B
Area of shaded region
= Area of circle with diameter OD D c

+ Area of semicircle ACB

— Area of triangle ABC *<
Area of circle with diameter OD
Since OD & OA are radius of circle with centre O
OD = OA=7cm
Diameter = OD = 7 cm
radius = r= 2S? = 7 om

2 2

Ex 12.3, 10 teachoo.com
The area of an equilateral triangle ABC is 17320.5 cm*. With each
vertex of the triangle as centre, a circle is drawn with radius equal
to half the length of the side of the triangle (see figure). Find the
area of the shaded region. (Use m= 3.14 and V3= 1.73205)
Area of shaded region
= Area of equilateral triangle 5 /\,

— Area of sector ADE /Y\

— Area of sector BDF

— Area of sector CFE
It is given that,
A ABC is an equilateral triangle,
So, sides of an triangle = AB = BC = AC

Ex 12.3, 11 teachoo.com
On a square handkerchief, nine circular designs each of radius 7 cm
are made {see figure). Find the area of the remaining portion of the
handkerchief. A B
BOQ
ORR
Area of remaining portion SS
; BI NDSYNISSY
= Area of square — Area of 9 circles is A <A |
pREALYM
Here, handkerchief is made up of contributing 9 circles.
So, Side of square = 3 x Diameter of circle
Here,
Radius of circle = 7 cm
Diameter of circle = 7x 2=14cm
So, Side of square = 3 x 14
=42cm

Ex 12.3, 12 teachoo.com
In figure, OACB is a quadrant of a circle with centre O and radius
3.5 cm. If OD = 2 cm, find the area of the
(i) quadrant OACB, A
Cc
It is given that,
OACB is a quadrant 2em
. B'
Given radius = 3.5 cm —<———
r=3.5cm
We know that
Area of Quadrant OACB = : x Area of a circle
Area of quadrant OACB = omer?
aly 22 2
=Ux>™ (G.5)
=ty2 35x35
4° 7
= 9.625 cm?

teachoo.com
Ex 12.3, 13
In figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20
cm, find the area of the shaded region. (Use m= 3.14}
Q
Area of shaded region
c B
= Area of quadrant OPBQ
— Area of square OABC
Or P
20 cm
Area of square
Side of square = OA = 20 cm
Area of square = (side)?
= (20)
= 20x 20
= 400 cm?

teachoo.com
Ex 12.3, 14
AB and CD are respectively arcs of two concentric circles of radii 21
cm and 7 cm and centre O (see figure). If ZAOB = 30°, find the area
of the shaded region.
A, B
Area of shaded region
= Area of sector AOB — Area of sector COD
21cm
Area of sector AOB Ns
radius =r= 21cm _
& 8 = 30° oO
Area of sector AOB = -— x mr?
360
ay 2% 21x21
360 °° 7
=-1+ x22x3x21
12
= 231 on
2

Ex 12.3, 15 teachoo.com
In figure, 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.
Area of shaded region B
= Area of semicircle BEC
— {Area of quadrant ABDC — Area of A ABC)
Area quadrant ABDC AG ec
14cm
Radius = 14cm
Area of quadrant ABDC = : (area of circle)
=i 2
=a% (mtr?)
aly 22 2
=7*%>%* (14)
=1yxy2x 14x14
4° 7
= 154 cm?

Ex 12.3, 16 teachoo.com
Calculate the area of the designed region in figure common
between the two quadrants of circles of radius 8 cm each.
8cem
Area of designed region ToS
"4 a
= Area of 1% quadrant A SY
8cm WY w\ Scm
+ Area of 2"¢ quadrant SSS
SD TA
rata
— Area of square Hy
8cm
8
Area of 1** quadrant = a x mr? —
= y 2% y ge
360° 7 Som 8cm
=tx2x8x8
4° 7
-2x2x8
7 8cm
352
= cm?
7