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Figure below shows a sector of a circle, centre $O$, containing an angle $\theta$. Prove that:Area of the shaded region is $\frac{r^{2}}{2}(\tan \theta-\frac{\pi \theta}{180})$
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Given:
A sector of a circle with centre $O$, containing an angle $\theta$.
To do:
We have to prove that the area of the shaded region is $\frac{r^{2}}{2}(\tan \theta-\frac{\pi \theta}{180})$.
Solution:
From the figure,
Radius of the circle $= r$
Arc $AC$ subtends $\theta$ at the centre of the circle.
$\angle OAB$ is a right triangle.
In right angled triangle $OAB$,
$\tan \theta=\frac{\mathrm{AB}}{\mathrm{OA}}$
$\Rightarrow \mathrm{AB}=\mathrm{OA} \times \tan \theta$
$=r \tan \theta$
Area of $\Delta \mathrm{OAB}=\frac{1}{2} \mathrm{OA} \times \mathrm{AB}$
$=\frac{1}{2} \times r \times (r \tan \theta)$
$=\frac{1}{2} r^{2} \tan \theta$
Area of sector $\mathrm{OAC}=\pi r^{2} (\frac{\theta}{360^{\circ}})$
$=\frac{\pi r^{2} \theta}{360}$
Therefore,
Area of the shaded portion $=$ Area of $\Delta \mathrm{OAB}-$ Area of sector $OAC$
$=\frac{1}{2} r^{2} \tan \theta-\frac{\pi r^{2} \theta}{360}$
$=\frac{1}{2} r^{2}(\tan \theta-\frac{2 \pi \theta}{360})$
$=\frac{1}{2} r^{2}(\tan \theta-\frac{\pi \theta}{180})$
$=\frac{r^{2}}{2}(\tan \theta-\frac{\pi \theta}{180})$
Hence proved.
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