Question

Find the area of the minor segment of a circle of radius 14 cm, when the angle of the corresponding sector is ${60}^{\circ }$.

Answer 1

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

$=\pi r^2\times \frac{{60}^o}{{360}^o}-r^{2}sin{30}^{o}cos{30}^o$

$=\frac{22}{7}\times 14\times 14\times \frac{1}{6}-14\times 14\times \frac{1}{2}\times \frac{\sqrt{3}}{2}$

$=(\frac{308}{3}-49\sqrt{3})c{m}^2$