Question

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 √3 =1.73]

Answer 1

Radius (r) of circle = 15 cm

Area of sector OPRQ =

 

=1/6 x 3.14 x (15)²

=117.75 cm²

= 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