Question

In the figure 7.44, $O$ is the centre of the circle. $m\left(\text{arc}PQR\right)={60}^{\circ }$, $OP=10$ cm. Find the area of the shaded region .

Answer 1

REF.Image

$\because$ the angle is ${60}^{\circ }$, hence, the triangle is equilateral triangle

Area of $\Delta POR=\frac{\sqrt{3}}{4}\times {10}^2=25\sqrt{3}=43.3$

Area of sector = $\frac{{60}^{\circ }}{{360}^{\circ }}\times \pi r^2=\frac{\pi r^2}{6}=52.36$

$\therefore$ shaded area = 52.36-43.3

$=10.06c{m}^2$