Question

Fig. 15.18 shows a sector of a circle of radius r cm containing an angle ${\theta }^{\circ }$. The area of the sector is $Ac{m}^2$ and the perimeter of the sector is 50 cm. Prove that $A=25r-$ $r^2$

Answer 1

$\Rightarrow$ Perimeter of the sector $=50cm$

$\Rightarrow$ $r$ is radius of a sector and $\theta$ is angle of the sector.

$\Rightarrow$ Perimeter of the sector $=\frac{\theta }{{360}^o}\times 2\pi r+2r$

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

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

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

$\therefore$ $\frac{(25-r)\times {360}^o}{\pi r}=\theta$

$\therefore$ $\theta =\frac{{360}^o}{\pi }(\frac{25}{r}-\frac{r}{r})$

$\therefore$ $\theta =\frac{{360}^o}{\pi }(\frac{25}{r}-1)$

$\Rightarrow$ Area of a sector $A=\frac{\theta }{{360}^o}\times \pi r^2$

Now, substituting value of $\theta$ we get,

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

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

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

$\therefore$ $A=25r-r^2$

Hence Proved