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