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The perimeter of a certain sector of a circle of radius \( 5.6 \mathrm{~m} \) is \( 27.2 \mathrm{~m} \). Find the area of the sector.
Given:
Radius of the circle $r=5.6 \mathrm{~m}$.
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 $- 2r$
$= 27.2 - 2 (5.6)\ m$
$= 27.2 - 11.2\ m$
$= 16\ m$
Therefore,
$2 \pi r \times \frac{\theta}{360^{\circ}}=16$
$\Rightarrow 2 \times \pi \times 5.6 \times \frac{\theta}{360^{\circ}}=16$
$\Rightarrow \frac{\theta}{360^{\circ}}=\frac{16}{2 \times \pi \times 5.6 }$
$\Rightarrow \frac{\theta}{360^{\circ}}=\frac{1}{0.7\pi}$............(i)
Area of the sector $=\pi r^{2} \times \frac{\theta}{360^{\circ}}$
$=\pi \times(5.6)^{2} \times \frac{1}{0.7\pi}$ [From (i)]
$=44.8 \mathrm{~m}^{2}$
The area of the sector is $44.8 \mathrm{~m}^{2}$.