A chord of a circle of radius subtends an angle of at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use and )

Areas Related to Circles | Exercise 11.1

Question 6

A chord of a circle of radius $15\text{ cm}$ subtends an angle of $60^\circ$ at the centre. Find the areas of the corresponding minor and major segments of the circle.

(Use $\pi = 3.14$ and $\sqrt{3} = 1.73$ )

Solution

We will find the areas of the minor and major segments.

Step 1 — Calculate Area of Sector

We are given the radius $\mathbf{r = 15\text{ cm}}$ . The angle subtended at the centre is $\mathbf{\theta = 60^\circ}$ . Let's find the area of the sector. The formula is $(\theta/360^\circ) \times \pi r^2$ .

$\text{Area of sector} = \frac{60^\circ}{360^\circ} \times \pi \times (\mathbf{15\text{ cm}})^2$

$= \frac{1}{6} \times \mathbf{3.14} \times \mathbf{225}$

$= \frac{\mathbf{706.5}}{6}$

$\boxed{117.75 \text{ cm}^2}$

Diagram 1

Step 2 — Calculate Area of Minor Segment

Now, we find the area of the triangle $\triangle \text{OPQ}$ . Since $\text{OP} = \text{OQ} = \mathbf{15\text{ cm}}$ (radii) and $\angle \text{POQ} = \mathbf{60^\circ}$ . Therefore, $\triangle \text{OPQ}$ is an equilateral triangle. Its area is $(\sqrt{3}/4) \times \text{side}^2$ . Here, the side is $\mathbf{15\text{ cm}}$ .

$\text{Area of } \triangle \text{OPQ} = \frac{\sqrt{3}}{4} \times (\mathbf{15\text{ cm}})^2$

$= \frac{\mathbf{1.73}}{4} \times \mathbf{225}$

$= \frac{\mathbf{389.25}}{4}$

$= 97.3125 \text{ cm}^2$ Now, we can find the area of the minor segment. It is the area of the sector minus the area of the triangle.

$\text{Area of minor segment} = \text{Area of sector} - \text{Area of } \triangle \text{OPQ}$

$= \mathbf{117.75\text{ cm}^2} - \mathbf{97.3125\text{ cm}^2}$

$\boxed{20.4375 \text{ cm}^2}$

Step 3 — Calculate Area of Major Segment

First, let's find the total area of the circle. The formula for the area of a circle is $\pi r^2$ . Here, $r$ is $\mathbf{15\text{ cm}}$ .

$\text{Area of circle} = \pi \times (\mathbf{15\text{ cm}})^2$

$= \mathbf{3.14} \times \mathbf{225}$

$\boxed{706.5 \text{ cm}^2}$

Now, we can find the area of the major segment. It is the area of the circle minus the area of the minor segment.

$\text{Area of major segment} = \text{Area of circle} - \text{Area of minor segment}$

$= \mathbf{706.5\text{ cm}^2} - \mathbf{20.4375\text{ cm}^2}$

$\boxed{686.0625 \text{ cm}^2}$

Answer

(i) The area of the minor segment is $\mathbf{20.4375\text{ cm}^2}$ . (ii) The area of the major segment is $\mathbf{686.0625\text{ cm}^2}$ .

More questions in Exercise 11.1

Unless stated otherwise, use $\pi = \frac{22}{7}$ .

Find the area of a sector of a circle with radius $6\text{ cm}$ if angle of the sector is $60^\circ$ .

Find the area of a quadrant of a circle whose circumference is $22\text{ cm}$ .

The length of the minute hand of a clock is $14\text{ cm}$ . Find the area swept by the minute hand in $5$ minutes.

A chord of a circle of radius $10\text{ cm}$ subtends a right angle at the centre. Find the area of the corresponding :

(i) minor segment

(ii) major sector. (Use $\pi = 3.14$ )

In a circle of radius $21\text{ cm}$ , an arc subtends an angle of $60^\circ$ 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

A chord of a circle of radius $15\text{ cm}$ subtends an angle of $60^\circ$ at the centre. Find the areas of the corresponding minor and major segments of the circle.

(Use $\pi = 3.14$ and $\sqrt{3} = 1.73$ )

A chord of a circle of radius $12\text{ cm}$ subtends an angle of $120^\circ$ at the centre. Find the area of the corresponding segment of the circle.

(Use $\pi = 3.14$ and $\sqrt{3} = 1.73$ )

A horse is tied to a peg at one corner of a square shaped grass field of side $15\text{ m}$ by means of a $5\text{ m}$ long rope (see Fig. 11.8). 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 $10\text{ m}$ long instead of $5\text{ m}$ . (Use $\pi = 3.14$ )

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. 11.9. Find :

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

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

Q10

An umbrella has 8 ribs which are equally spaced (see Fig. 11.10). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Q11

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

Q12

To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle $80^\circ$ to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use $\pi = 3.14$ )

Q13

A round table cover has six equal designs as shown in Fig. 11.11. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per $\text{cm}^2$ . (Use $\sqrt{3} = 1.7$ )

Q14

Tick the correct answer in the following :

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

(A) $\frac{p}{180} \times 2\pi\text{R}$ (B) $\frac{p}{180} \times \pi\text{R}^2$ (C) $\frac{p}{360} \times 2\pi\text{R}$ (D) $\frac{p}{720} \times 2\pi\text{R}^2$