These NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Exercise 12.2

Question 1.

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

Solution:

Radius of the sector (r) = 6 cm

Central angle of the sector = 60°

We know that

Therefore, the area of required circle is \(\frac { 132 }{ 7 }\) cm²

Question 2.

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

Solution:

Angle of quadrant 0 = 90°

Circumference of the circle = 2πr

22 = 2πr

r = \(\frac { 11 }{ π }\)cm

Question 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 made by minute hand in 60 minute is

∴ Angle made by minute hand in 1 minute is = \(\frac{360^{\circ}}{60^{\circ}}\) = 6°

Length of minute of hand (r) = 14 cm

We know that

Area swept by minute hand in 5 minutes is \(\frac { 153 }{ 3 }\) cm²

Question 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 segment (Use π = 3.14)

Solution:

Given: radius of the circle = 10 cm

Angle subtended by chord at centre = 90°

(i)

(ii) Area of the major segment = Area of the circle – Area of the minor segment

= πr^{2} – 28.5 = 3.14 x 10 x 10-28.5

= 314-28.5 = 285.5 cm^{2}

Question 5.

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

(i) length of the arc.

(ii) area of the sector formed by the arc.

(iii) area of the segment formed by the corresponding chord.

Solution:

We have,

Radius of the circle = 21 cm

Angle at the centre = 60°

(i) We know that

Therefore, length of arc AB = 22 cm.

(iii) Area of segment = Area of sector – Area of equilateral ∆OAB

Therefore, area of sector made by arc AB = is 231 cm².

Question 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. (Use π = 3.14 and \(\sqrt{3}\) = 1.73)

Solution:

Radius of the circle = 15 cm

Angle subtended by chord at centre = 60°

Area of the sector = \(\frac{\pi r^{2} \theta}{360^{\circ}}\) = 3.14 x \(\frac{15 \times 15 \times 60^{\circ}}{360^{\circ}}\) = 117.75 cm^{2}

Area of the triangle formed by radii and chord = \(\frac { 1 }{ 2 }\)r^{2}θ

= \(\frac { 1 }{ 2 }\)(15)^{2} sin 60° = \(\frac { 1 }{ 2 }\) x 15 x 15 x \(\frac{\sqrt{3}}{2}\) = 97.31 cm^{2}

Area of the minor segment = Area of the sector – Area of the triangle formed by radii and chord

= 117.75 – 97.31 = 20.44 cm^{2}

Area of the circle = πr^{2} = 3.14 x 15 x 15 = 706.5 cm^{2}

Area of the circle – Area of the minor segment

= 706.5 – 20.44 = 686.06 cm^{2}

Therefore, area of major segment is 686.0625 cm²

Question 7.

A chord of a circle of the 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 \(\sqrt{3}\) = 1.73).

Solution:

Construct a circle C with radius 12 cm and centre O

Area of the corresponding segment = Area of sector – Area of ∆AOB = 150.72 – 62.28 cm²

= 88.44 cm²

Question 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 (see figure). Find

(i) the area of that 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. (Use π = 3.14)

Solution:

(i) Length of the rope = Radius of the sector grazed by horse = 5 m

Here, angle of the sector = 90°

Hence, the area of the part of the field which is grazed by horse is 19625 m².

(ii) Now, radius = 10m

Angle = 90°

Question 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 as shown in figure.

Find:

(i) the total length of the silver wire required.

(ii) the area of each sector of the brooch.

Solution:

One diameter of the circle in 35 mm long The length of 5 diameters = 5 x 35 = 175 mm

The required length of the silver wire = circumference of circle + length of 5 diameters

= 110 + 175 = 285 mm

(ii) Let O be the centre of circle and radius is

The angle made by the diameters of one part

Hence the area of each sector of the brooch is \(\frac { 385 }{ 4 }\) mm²

Question 10.

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 the umbrella.

Solution:

Let O be the centre of the umbrella and the radius is 45 cm

The angle between two consecutive ribs \(\frac { 360° }{ 8 }\)

Area between two consecutive ribs

= \(\frac{360^{\circ}}{360^{\circ} \times 8} \times \frac{22}{7} \times(45)^{2}\)

= \(\frac{22275}{28} \mathrm{~cm}^{2}\)

Question 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:

Given: length of blade of wiper = radius of sector sweep by blade = 25 cm

Area cleaned by each sweep of the blade = area of sector sweep by blade

Angle of the sector formed by blade of wiper = 115°

Hence, the total area cleaned at each sweep is \(\frac { 158125 }{ 126 }\) cm².

Question 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. (Use π = 3.14)

Solution:

Angle of the sector = 80°

Distance covered = 16.5 km

Radius of the sector formed = 16.5 km

Hence the area of sea over which the ships are warned is 189.97 km².

Question 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. (Use \(\sqrt{3}\) = 1.7)

Solution:

Area of the design = Area of sector AOB – Area of ∆AOB

= 410.67 – 333.20

= 77.47

There are six designs in the circle

Area of six designs = 77.47 x 6

= 464.82 cm²

The cost of making the design at the rate of 0.35 per cm²

₹ (464.82 x 0.35) = ₹ 162.68

Hence the required cost of making the design is ₹ 162.68

Question 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) \(\frac{p}{180^{\circ}}\) × 2πR

(b) \(\frac{p}{180^{\circ}}\) × πR^{2}

(c) \(\frac{p}{360^{\circ}}\) × 2πR

(d) \(\frac{p}{720^{\circ}}\) × 2πR^{2}

Solution:

Sector angle is p in degrees

Radius of the circle = R

Area of the sector = \(\frac{\pi \mathrm{R}^{2} p}{361^{6}}\) = \(\frac{\left(\pi R^{2} p\right) 2}{720^{\circ}}\)

= \(\frac{p}{720^{\circ}}\) × 2πR^{2}

Hence (D) is the correct answer.