The area of a sector of a circle of radius $ 5 \mathrm{~cm} $ is $ 5 \pi \mathrm{cm}^{2} $. Find the angle contained by the sector.

Given:

Radius of the circle $=5 \mathrm{~cm}$.

Area of the sector $=5\pi \mathrm{cm}^{2}$

To do:

We have to find the angle contained by the sector.

Solution:

Let $\theta$ be the angle subtended by the sector.

We know that,

Area of the sector subtending $\theta$ at the centre $=\pi r^{2} \times \frac{\theta}{360^{\circ}}$

Therefore,

$5\pi \mathrm{cm}^{2}=\pi(5)^{2} \times \frac{\theta}{360^{\circ}} \mathrm{cm}^{2}$

$\Rightarrow 25 \pi \times \frac{\theta}{360^{\circ}}=5\pi$

$\Rightarrow \frac{\theta}{360^{\circ}}=\frac{5\pi}{25 \pi}$

$\Rightarrow \frac{\theta}{360^{\circ}}=\frac{1}{5}$

$\Rightarrow \theta=\frac{360^{\circ}}{5}=72^{\circ}$ 

The angle contained by the sector is $72^{\circ}$.