Question

In the given figure, O is the centre of the circle. m(arc PQR)= ${60}^0$ OP= 10 cm. Find the area of the shaded region.
($\pi =3.14,\sqrt{3}=1.73$)

Answer 1

Here, $r=10cmand\theta ={60}^o$

Area of the shaded region $=$ area of the segment

$=[\frac{\pi \theta }{360}-sin\frac{\theta }{2}\times cos\frac{\theta }{2}]r^2$

$=[\frac{3.14\times 60}{360}-sin{30}^o\times cos{30}^o](10{)}^2$

$=[\frac{3.14}{6}-\frac{\sqrt{3}}{4}]100$

$=0.902\times 100=90.2c{m}^2$