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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 $90^{\circ}$.
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