Question

A chord PQ of the circle of radius 10 cm subtends an angle of 60 degrees at the centre of the circle. Find the area of minor and major segment of the circle.

Answer 1

The radius of the circle = 10 cm.

Chord PQ subtends an angle = 60 degrees at the centre.

PQO is an equilateral triangle, O being the centre of the circle.

Area of minor segment = θ/360 x π r² – 1/2 r² sinθ

= 60/360 x 22/7 x 10 x 10 – 1/2 x 10 x 10 x √3/2

= 9.03

Area of circle = πr² = 22/7 x 10 x 10 = 314 (approx.)

Area of minor segment = Area of circle – Area of minor segment

= 314 – 9.03

= 304.97 cm² (approx.)