Question

A chord AB of circle of radius 14 cm makes an angle of 60° at the centre of a circle. Find the area of the minor segment of the circle.

Answer 1

Given,

Radius of the circle (r) = 14 cm = OA = OB

Angle subtended by the chord with the centre of the circle, θ = 60°

In triangle AOB, angle A = angle B [angle opposite to equal sides OA and OB = x]

By angle sum property,

∠A + ∠B + ∠O = 180

x + x + 60° = 180°

2x = 120°, x = 60°

All angles are 60° so the triangle OAB is an equilateral with OA = OB = AB

Area of the minor segment = area of sector – area of triangle OAB

= θ/360 × πr² – √3/4 (side)²

= 60/360 x (22/7) 14² – √3/4 (14)²

= [1/6 x 22 x 2 x 14] – √3 x (7)²

= 102.7 – 84.9 = 17.8

Therefore, the area of the minor segment = 17.80 cm²