Question

A chord of circle of radius 10 cm subtends a right angle at the centre. Find area of the minor sector

Answer 1

Here, $r=10cm$ and $\theta$ $={90}^{\circ }$

We have,
Area of the minor sector $OAPB=$ $\frac{\theta }{360}\times \pi r^2$

$=\frac{90}{360}\times 3.14\times 10\times 10c{m}^2$

$=78.5c{m}^2$