Question

Find the area of the sector of a circle of whose radius is $14cm$ and angle of sector is ${45}^{\circ }$.

Answer 1

Given:

$r=14cm$

$\theta ={45}^{\circ }$

We know that,

$A=\frac{\theta }{360}\times \pi r^2$

Therefore,

$A=\frac{45}{360}\times \frac{22}{7}\times (14{)}^{2}c{m}^2$

$=\frac{1}{8}\times \frac{22\times 2\times 14}{1}$

$=77c{m}^2$

Hence, this is the required area.