Question
A chord PQ of a circle of radius 10 cm subtends an angle of 60° at the centre of the circle. Find the area of major and minor segments of the circle.
Solution
Radius of the circle, r = 10 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.
So, area of ΔOPQ
Area of minor segment PRQ
= Area of sector OPRQ − Area of ΔOPQ
= 52.33 − 43.30
= 9.03 cm2
Area of major segment PSQ
= Area of circle − Area of minor segment PRQ
=π(10)2−9.03=314−9.03=304.97 cm2
