CBSE Class X: 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°.

Let OACB be a sector of the circle making 60° angle at centre O of the circle.
Area of sector of angle θ =

Area of sector OACB =

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

Question 2:

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

Quadrant of circle will subtend 90° angle at the centre of the circle.
Area of such quadrant of the circle

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.

We know that in 1 hour (i.e., 60 minutes), the minute hand rotates 360°.
In 5 minutes, minute hand will rotate =

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 =

Area of sector of 30°

Therefore, the area swept by the minute hand in 5 minutes is

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 sector

[Use π = 3.14]

Let AB be the chord of the circle subtending 90° angle at centre O of the circle.
Area of major sector OADB =

Area of minor sector OACB =

Area of ΔOAB =

= 50 cm²
Area of minor segment ACB = Area of minor sector OACB −
Area of ΔOAB = 78.5 − 50 = 28.5 cm²

Question 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

(ii) Area of the sector formed by the arc

(iii) Area of the segment forced by the corresponding chord

Radius (r) of circle = 21 cm
Angle subtended by the given arc = 60°
Length of an arc of a sector of angle θ =

Length of arc ACB =

= 22 cm
Area of sector OACB =

In ΔOAB,
∠OAB = ∠OBA (As OA = OB)
∠OAB + ∠AOB + ∠OBA = 180°
2∠OAB + 60° = 180°
∠OAB = 60°
Therefore, ΔOAB is an equilateral triangle.
Area of ΔOAB =

Area of segment ACB = Area of sector OACB − Area of ΔOAB

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

]

Radius (r) of circle = 15 cm
Area of sector OPRQ =

In ΔOPQ,
∠OPQ = ∠OQP (As OP = OQ)
∠OPQ + ∠OQP + ∠POQ = 180°
2 ∠OPQ = 120°
∠OPQ = 60°
ΔOPQ is an equilateral triangle.
Area of ΔOPQ =

Area of segment PRQ = Area of sector OPRQ − Area of ΔOPQ
= 117.75 − 97.3125
= 20.4375 cm²
Area of major segment PSQ = Area of circle − Area of segment PRQ

Question 7:

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]

Let us draw a perpendicular OV on chord ST. It will bisect the chord ST.
SV = VT
In ΔOVS,

Area of ΔOST =

Area of sector OSUT =

Area of segment SUT = Area of sector OSUT − Area of ΔOST
= 150.72 − 62.28
= 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 the given figure). Find

(i) The area of that part of the field in which the horse can graze.

(ii) The increase in the grazing area of the rope were 10 m long instead of 5 m.

[Use π = 3.14]

From the figure, it can be observed that the horse can graze a sector of 90° in a circle of 5 m radius.
Area that can be grazed by horse = Area of sector OACB

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

Increase in grazing area = (78.5 − 19.625) m²
= 58.875 m²

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

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

Circumference of brooch = 2πr

= 110 mm
Length of wire required = 110 + 5 × 35
= 110 + 175 = 285 mm
It can be observed from the figure that each of 10 sectors of the circle is subtending 36° at the centre of the circle.

Therefore, area of each sector =

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.

There are 8 ribs in an umbrella. The area between two consecutive ribs is subtending

at the centre of the assumed flat circle.

Area between two consecutive ribs of circle =

Question 11:

A car has two wipers which do not overlap. Each wiper has 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 figure that each blade of wiper will sweep an area of a sector of 115° in a circle of 25 cm radius.
Area of such sector =

Area swept by 2 blades =

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 warned. [Use π = 3.14]

It can be observed from the figure that the lighthouse spreads light across a
sector of 80° in a circle of 16.5 km radius.
Area of sector OACB =

Question 13:

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 cm². [Use]

It can be observed that these designs are segments of the circle.
Consider segment APB. Chord AB is a side of the hexagon. Each chord will substitute

at the centre of the circle.
In ΔOAB,
∠OAB = ∠OBA (As OA = OB)
∠AOB = 60°
∠OAB + ∠OBA + ∠AOB = 180°
2∠OAB = 180° − 60° = 120°
∠OAB = 60°
Therefore, ΔOAB is an equilateral triangle.
Area of ΔOAB =