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 $44.8 \mathrm{~m}^{2}$. 

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Updated on: 10-Oct-2022

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