The perimeter of a sector of a circle of radius $ 5.7 \mathrm{~m} $ is $ 27.2 \mathrm{~m} $. Find the area of the sector.

Given:

Radius of the circle $r=5.7 \mathrm{~m}$.

Perimeter of a sector $=27.2 \mathrm{~m}$.

To do:

We have to find the area of the sector.

Solution:

Let $\theta$ be the angle subtended at the centre.

Length of the arc $=$ Perimeter $- 2r$

$= 27.2 - 2 (5.7)\ m$
$= 27.2 - 11.4\ m$

$= 15.8\ m$
Therefore,

$2 \pi r \times \frac{\theta}{360^{\circ}}=15.8$

$\Rightarrow 2 \times \frac{22}{7} \times 5.7 \times \frac{\theta}{360^{\circ}}=15.8$

$\Rightarrow \frac{\theta}{360^{\circ}}=\frac{15.8 \times 7}{2 \times 22 \times 5.7}$

$\Rightarrow \frac{\theta}{360^{\circ}}=\frac{110.6}{250.8}$............(i)

Area of the sector $=\pi r^{2} \times \frac{\theta}{360^{\circ}}$

$=\frac{22}{7} \times(5.7)^{2} \times \frac{110.6}{250.8}$                  [From (i)]

$=\frac{22}{7} \times \frac{32.49 \times 110.6}{250.8}$

$=\frac{32.49 \times 15.8}{11.4}$

$=45.03 \mathrm{~m}^{2}$

The area of the sector is $45.03 \mathrm{~m}^{2}$.