Question

Figure shows a sector of a circle of radius r cm containing an angle $\theta {}^{\circ }$. The area of the sector is A $c{m}^2$ and perimeter of the sector is 50 cm. Prove that

(i) $\theta =\frac{360}{\pi }(\frac{25}{r}-1)$

(ii) $A=25r-r^2$

Answer 1

(i) Length of arc of the sector $=2\pi r\times \frac{\theta }{{360}^o}$

Now, perimeter of the sector = 50

$\Rightarrow 2\pi r\times \frac{\theta }{{360}^o}+2r=50$

$\Rightarrow 2\pi r\times \frac{\theta }{{360}^o}=50-2r$

$\Rightarrow \theta =(50-2r)\times \frac{{360}^o}{2\pi r}=(\frac{25}{r}-1)\frac{{360}^o}{\pi }$

Hence, proved.

(ii) Area of the sector, $A=\pi r^2\times \frac{\theta }{{360}^o}$

$=\pi r^2\times (\frac{25}{r}-1)\times \frac{1}{\pi }$

$=r^2(\frac{25}{r}-1)$

$=25r-r^2$

Hence, proved.