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

Given:

Radius of the circle $=2 \mathrm{~cm}$.

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

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,

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

$\Rightarrow 4 \pi \times \frac{\theta}{360^{\circ}}=\pi$

$\Rightarrow \frac{\theta}{360^{\circ}}=\frac{\pi}{4 \pi}$

$\Rightarrow \frac{\theta}{360^{\circ}}=\frac{1}{4}$

$\Rightarrow \theta=\frac{360^{\circ}}{4}=90^{\circ}$ 

The angle contained by the sector is $90^{\circ}$.