Question

A chord PQ of length 12 cm subtends an angle of 120° at the centre of a circle. Find the area of the minor segment cut off by the chord PQ.

Answer 1

Length of chord PQ = 12 cm 

Angle subtend at the center = 120° 

Let radius of circle = r cm 

Area of sector = \(\frac{120}{360}πr^2\) = \(\frac{πr^2}3\) cm²

Length of triangle POQ = r cos 60

= r x \(\frac{1}2\) = \(\frac{r}2\) cm

Length of base PQ = 2 × RQ

= 2 x r sin 60 = 2 x r x \(\frac{\sqrt3}2\) = \(\sqrt{3r}\)

Put value of r in respective place, 

Area of minor segment = area of sector – area of ∆POQ