Question

A chord AB of a circle, of radius 14 cm makes an angle of ${60}^{\circ }$ at the centre of the circle. Find the area of the minor segment of the circle. (Use $\pi =\frac{22}{7}$)

Answer 1

Area of the minor segment is given by $\pi r^2\times \frac{\theta }{{360}^o}-r^{2}sin\frac{\theta }{2}cos\frac{\theta }{2}$

$=\frac{22}{7}\times 14\times 14\times \frac{60}{360}-14\times 14\times sin{30}^o\times cos{30}^o$

$=\frac{22}{7}\times 14\times 14\times \frac{60}{360}-14\times 14\times \frac{1}{2}\times \frac{\sqrt{3}}{2}$

$=102.67-84.77=17.9c{m}^2$