A chord AB of a circle, of radius 14 cm makes an angle of 60° at the

1.	A chord AB of a circle, of radius 14 cm makes an angle of 60° at the centre of the circle. Find the area of the minor segment of the circle.(Take π = \(\frac{22}7\))
Answer» Radius of circle = 14 cm Angle = 60° Area of sector = \(\frac{θ}{360}πr^2\) = \(\frac{60}{360}π\times14\times14\) = \(\frac{98}3π\) = 102.57 cm² Area of triangle OAB = \(\frac{1}2r^2\) sin θ = \(\frac{1}{2}\times14\times14\times{sin}θ\) = \(\frac{1}{2}\times14\times14\times\frac{\sqrt3}2\) = \(49\sqrt3\) = 84.77 cm² So, Area of minor segment = 102.57 – 84.77 = 17.80 cm²

A chord AB of a circle, of radius 14 cm makes an angle of 60° at the centre of the circle. Find the area of the minor segment of the circle.(Take π = \(\frac{22}7\))

Answer»

Radius of circle = 14 cm

Angle = 60°

Area of sector = \(\frac{θ}{360}πr^2\)

= \(\frac{60}{360}π\times14\times14\) = \(\frac{98}3π\) = 102.57 cm²

Area of triangle OAB = \(\frac{1}2r^2\) sin θ

= \(\frac{1}{2}\times14\times14\times{sin}θ\)

= \(\frac{1}{2}\times14\times14\times\frac{\sqrt3}2\)

= \(49\sqrt3\) = 84.77 cm²

So,

Area of minor segment = 102.57 – 84.77 = 17.80 cm²