# NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 – Area related to Circles

Download NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 – Area related to Circles. This Exercise contains 14 questions, for which detailed answers have been provided in this note. In case you are looking at studying the remaining Exercise for Class 10 for Maths NCERT solutions for other Chapters, you can click the link at the end of this Note.

### NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 – Area related to Circles

NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 – Area related to Circles

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

Solution:

Area of sector = $\cfrac { \theta }{ { 360 }^{ \circ } }$ × πr2
= $\cfrac { { 60 }^{ \circ } }{ { 360 }^{ \circ } }$ x $\cfrac { 22 }{ 7 }$ x 6²
= $\cfrac { 1 }{ 6 }$ x $\cfrac { 22 }{ 7 }$ x 6²
= $\cfrac { 22 }{ 7 }$ x 6
= $\cfrac { 132 }{ 7 }$ cm2
= 18.8 cm2

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

Solution:

Let the radius of the circle be r
Circumference = 2πr = 22 cm
⇒ 2 × $\cfrac { 22 }{ 7 }$ × r = 22
r = 7/2 cm

Angle of quadrant (θ) = 90°
Area of quadrant = $\cfrac { \theta }{ { 360 }^{ \circ } }$  × πr2
= $\cfrac { { 90 }^{ \circ } }{ { 360 }^{ \circ } }$ x $\cfrac { 22 }{ 7 }$ x r²
= $\cfrac { 1 }{ 4 }$ x $\cfrac { 22 }{ 7 }$ x ${ \left( \cfrac { 7 }{ 2 } \right) }^{ 2 }$
= $\cfrac { 22\times 7\times 7 }{ 4\times 7\times { 2 }^{ 2 } }$
= $\cfrac { 22\times 7 }{ 4\times 4 }$
= $\cfrac { 154 }{ 16 }$ cm2
= 9.625 cm2

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

Solution:

Angle swept by the minute hand in 60 minutes = 360°
Angle swept by the minute hand in 1 minute = $\cfrac { { 360 }^{ \circ } }{ { 60 }^{ \circ } }$ = 6°
Angle swept by the minute hand in 5 minutes (θ) = 5 × 6° = 30°
Area swept by the minute hand = $\cfrac { \theta }{ { 360 }^{ \circ } }$ × πr2
= $\cfrac { { 360 }^{ \circ } }{ { 60 }^{ \circ } }$ x $\cfrac { 22 }{ 7 }$ x 14²
= $\cfrac { 1 }{ 12 }$ x $\cfrac { 22 }{ 7 }$ x 14 x 14
= $\cfrac { 22\times 7 }{ 3 }$
= $\cfrac { 154 }{ 3 }$ cm2
= 51.34 cm2

4. 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

Solution:

Let the chord be AB and the centre of the circle be O

Area of minor segment = Area of minor sector – Area of △OAB
= $\cfrac { { 90 }^{ \circ } }{ { 360 }^{ \circ } }$ × π × (10)2$\cfrac { 1 }{ 2 }$ × 10 × 10
= $\cfrac { 1 }{ 4 }$ × 3.14 × 100 – $\cfrac { 1 }{ 2 }$ × 100
= 78.5 – 50
= 28.5 cm2

Area of major sector = Area of circle – Area of minor sector
= π × (10)2$\cfrac { { 90 }^{ \circ } }{ { 360 }^{ \circ } }$ × π × (10)2
= 3.14 × 100 – $\cfrac { 1 }{ 4 }$ × 3.14 × 100
= 314 – 78.5
= 235.5 cm2

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

i. Length of arc =$\cfrac { \theta }{ { 360 }^{ \circ } } \times 2\pi r$
= $\cfrac { { 60 }^{ \circ } }{ { 360 }^{ \circ } }$ x 2 x $\cfrac { 22 }{ 7 }$ x 21
= $\cfrac { 1 }{ 6 }$ x 2 x 22 x 3
= 22 cm

ii. area of the sector formed by the arc

ii. Area of sector = $\cfrac { \theta }{ { 360 }^{ \circ } } \times \pi { r }^{ 2 }$
= $\cfrac { { 60 }^{ \circ } }{ { 360 }^{ \circ } }$ x $\cfrac { 22 }{ 7 }$ x 21²
= $\cfrac { 1 }{ 6 }$ x $\cfrac { 22 }{ 7 }$ x 21 x 21
= 231 cm2

iii. area of the segment formed by the corresponding chord

iii. Let the chord be AB and the centre of the circle be O
∠AOB = 60°
∵ OA = OB = Radius
∴ ∠OAB = ∠OBA = 60°
⇒ OAB is an equilateral triangle
Area of segment formed by the chord = Area of sector – Area of △OAB
= 231 – $\cfrac { \sqrt { 3 } }{ 4 }$ (21)²
= (231 – $\cfrac { 441\sqrt { 3 } }{ 4 }$) cm2

6. 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.

Solution:

Let the chord be AB and the centre of the circle be O

∠AOB = 60°
∵ OA = OB = Radius
∴ ∠OAB = ∠OBA = 60°
⇒ OAB is an equilateral triangle

Area of minor segment = Area of minor sector – Area of △OAB
= $\cfrac { { 60 }^{ \circ } }{ { 360 }^{ \circ } }$ x π × 15²$\cfrac { \sqrt { 3 } }{ 4 }$ × 152
=$\cfrac { 1 }{ 6 }$ x 3.14 x 225 – $\cfrac { 225\sqrt { 3 } }{ 4 }$
= 117.75 – 97.3125
= 20.4375 cm2

Area of major segment = Area of circle – Area of minor segment
= π × 152 – 20.4375
= 3.14 × 225 – 20.4375
= 706.5 – 20.4375
= 686.0625 cm2

7. A chord of a circle of radius 12 cm subtends an angle 120° at the centre. Find the area of the corresponding segment of the circle.

Solution:

Let AB be the chord and O be the centre of the circle. Draw a perpendicular OC from O to AB

∴ ∠AOC = ∠BOC = 60°
In △OAC
cos 60° = $\cfrac { OC }{ OA }$
$\cfrac { 1 }{ 2 }$ = $\cfrac { OC }{ 12 }$
OC = 6 cm
sin 60° = $\cfrac { AC }{ OA }$
$\cfrac { \sqrt { 3 } }{ 2 }$ = $\cfrac { AC }{ 12 }$
AC = 6√3 cm
AB = 2 × AC (Chord is bisected by the perpendicular from the centre)
AB = 12√3 cm
Area of the segment = Area of the sector – Area of △OAB
= $\cfrac { { 120 }^{ \circ } }{ { 360 }^{ \circ } }$ x π × 122  – $\cfrac { 1 }{ 2 }$ x 12√3 x 6
= $\cfrac { 1 }{ 3 }$ x 3.14 x 144 – 36√3
= 150.72 – 62.28
= 88.44 cm2

8. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope. Find

i. the area of the part of the field in which the horse can graze
ii. the increase in the grazing area if the rope were 10 m long instead of 5 m

Solution:

The horse will be able to graze only a quadrant of the circle with radius equal to the length of the rope

i. Radius of quadrant of circle = Length of rope = 5 m
Grazing area = Area of quadrant
= $\cfrac { { 90 }^{ \circ } }{ { 360 }^{ \circ } }$ x π × 52
= $\cfrac { 1 }{ 4 }$ x 3.14 x 25
= 19.625 m2

ii. Radius of quadrant of circle = Length of rope = 10 m
New grazing area = Area of quadrant
= $\cfrac { { 90 }^{ \circ } }{ { 360 }^{ \circ } }$ x π × 102
= $\cfrac { 1 }{ 4 }$ x 3.14 x 100
= 78.5 m2

Increase in the grazing area = 78.5 – 19.625
= 58.875 m2

9. 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. Find:

i. the total length of the silver wire required
ii. the area of each sector of the brooch

Solution:

Diameter of circle (d) = 35 mm
Radius of circle (r) = ½ Diameter = $\cfrac { 35 }{ 2 }$ mm

i. Length of wire required = Circumference + 5 × Diameter
= 2πr + 5d
= 2 x $\cfrac { 22 }{ 7 }$ x $\cfrac { 35 }{ 2 }$ + 5 x 35
= 110 + 175
= 285 mm

ii. Area of circle = πr2
= $\cfrac { 22 }{ 7 }$ x $\cfrac { 35 }{ 2 }$ x $\cfrac { 35 }{ 2 }$
= $\cfrac { 3850 }{ 4 }$ mm2

∵ The brooch is divided into 10 equal sectors
∴ Area of 10 sectors = Area of circle
Area of 1 sector = $\cfrac { 1 }{ 10 }$ Area of circle
= $\cfrac { 385 }{ 4 }$ mm2

10. An umbrella has 8 ribs which are equally spaced. Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Solution:

∵ The umbrella is a flat circle of radius 45 cm
∴ The portion between the two consecutive ribs will be a sector of the circle
The umbrella has 8 sectors
∴ Angle of each sector = $\cfrac { { 360 }^{ \circ } }{ 8 }$ = 45°
Area between two consecutive ribs = Area of a sector
= $\cfrac { { 45 }^{ \circ } }{ { 360 }^{ \circ } }$ x π × 452
= $\cfrac { 1 }{ 8 }$ x $\cfrac { 22 }{ 7 }$ x 45 x 45
= $\cfrac { 22275 }{ 28 }$ cm2

11. 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.

Solution:

The wipers sweep two sectors of angle 115° each
Radius of the circle = Length of wipers = 25 cm

Area cleaned at each sweep of each blade = Area of sector formed by wiper
=$\cfrac { { 115 }^{ \circ } }{ { 360 }^{ \circ } }$ x π × 252
= $\cfrac { 23 }{ 72 }$ x $\cfrac { 22 }{ 7 }$ x 625
= $\cfrac { 158125 }{ 252 }$ cm2

Area cleaned at each swipe of both blades = 2 × Area cleaned by each blades
= 2 × $\cfrac { 158125 }{ 252 }$
= $\cfrac { 158125 }{ 126 }$ cm2

12. 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.

Solution:

The light is spread over an area of a sector of angle 80° and radius 16.5 km

Area of the sea over which ships are warned = Area of the sector
= $\cfrac { { 80 }^{ \circ } }{ { 360 }^{ \circ } }$ x π × 16.52
= $\cfrac { 2 }{ 9 }$ x 3.14 x 16.5 x 16.5
= 189.97 km2

13. A round table cover has six equal designs as shown in the figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per cm2.

Solution:

Consider only one of the six sectors of the cover as shown in the figure

Angle of each sector = $\cfrac { { 360 }^{ \circ } }{ 6 }$ = 60°
∠AOB = 60°
∵ OA = OB = Radius
∴ ∠OAB = ∠OBA = 60°
⇒ OAB is an equilateral triangle
Area of the segment = Area of sector – Area of △OAB
= $\cfrac { { 60 }^{ \circ } }{ { 360 }^{ \circ } }$ x π × 28² – $\cfrac { \sqrt { 3 } }{ 4 }$ x 28²
= $\cfrac { 1 }{ 6 }$ x $\cfrac { 22 }{ 7 }$ x 28 x 28  – $\cfrac { 1.7 }{ 4 }$ x 28 x 28
= ($\cfrac { 1232 }{ 3 }$ – 333.2) cm2

Area of complete design = 6 × Area of one segment
= 6 x ($\cfrac { 1232 }{ 3 }$ – 333.2)
= 2 x 1232 – 6 x 333.2
= 2464 – 1999.2
= 464.8 cm2

Cost of making 1 cm2 of design = ₹ 0.35
∴ Cost of making 464.8 cm2 of design = ₹ 0.35 × 464.8 = ₹ 162.68

14. Tick the correct answer in the following:

Area of a sector of angle p (in degrees) of a circle with radius R is

A. $\cfrac { p }{ 180 }$×2πr
B. $\cfrac { p }{ 180 }$×πr2
C. $\cfrac { p }{ 360 }$×2πr
D. $\cfrac { p }{ 720 }$×2πr2

Area of sector with central angle p (in degrees)
= $\cfrac { p }{ 360 }$ x πR²
= $\cfrac { p }{ 720 }$ x 2πR²(Multiplying numerator and denominator by 2)

Therefore, the correct option is (D)

NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.2 – Area related to Circles, has been designed by the NCERT to test the knowledge of the student on the topic – Areas of Sector and Segment of a Circle

Download NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 – Area related to Circles