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Figure below shows a sector of a circle of radius $r\ cm$ containing an angle $\theta^o$. The area of the sector is $A\ cm^2$ and perimeter of the sector is $50\ cm$. Prove that $ \theta=\frac{360}{\pi}\left(\frac{25}{r}-1\right) $
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Given:
A sector of a circle of radius $r\ cm$ containing an angle $\theta^o$.
The area of the sector is $A\ cm^2$ and the perimeter of the sector is $50\ cm$.
To do:
We have to prove that \( \theta=\frac{360}{\pi}\left(\frac{25}{r}-1\right) \).
Solution:
From the figure,
Radius of the sector of the circle $= r\ cm$
Angle at the centre $= \theta$
Area of the sector $OAB = A\ cm^2$
Perimeter of the sector $OAB = 50\ cm$
Area of the sector $=\pi r^{2} (\frac{\theta}{360^{\circ}})$
$\Rightarrow A=\pi r^{2} (\frac{\theta}{360^{\circ}})$
Therefore,
Perimeter $=\mathrm{OA}+\operatorname{arc} \mathrm{AB}+\mathrm{OB}$
$\Rightarrow 50=2 r+2 \pi r \times(\frac{\theta}{360^{\circ}})$
$\Rightarrow 50-2 r=2 \pi r(\frac{\theta}{360^{\circ}})$
$\Rightarrow \frac{\theta}{360^{\circ}}=\frac{50-2 r}{2 \pi r}$
$\Rightarrow \frac{\theta}{360^{\circ}}=\frac{50}{2 \pi r}-\frac{2 r}{2 \pi r}$
$\Rightarrow \frac{\theta}{360^{\circ}}=\frac{25}{\pi r}-\frac{1}{\pi}$
$\Rightarrow \theta=360(\frac{25}{\pi r}-\frac{1}{\pi})$
$\Rightarrow \theta=\frac{360^{\circ}}{\pi}(\frac{25}{r}-1)$
Hence proved.
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