Question

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.

Answer 1

Step 1. Calculate the angle $\left(\theta \right)$ of the sector.

Let the radius of circle be $r$ and the arc length is $l$.

The perimeter of a sector is given as,

$$\begin{gathered} \therefore P=2r+r\theta \left[\because l=r\theta \right] \\ \Rightarrow 27.2=(2\times 5.6)+5.6\theta \left[\because r=5.6\mathrm{m}\mathrm{and}P=27.2\mathrm{m}\right] \\ \Rightarrow 27.2=11.2+5.6\theta \\ \Rightarrow \theta =\frac{16}{5.6} \\ \Rightarrow \theta =\frac{160}{56} \\ \Rightarrow \theta =\frac{20}{7}\mathrm{radian} \\ \Rightarrow \theta =\frac{20}{7}\times \frac{180}{\mathrm{\pi }}\mathrm{degrees}\left[\because 1\mathrm{radian}=\frac{180}{\mathrm{\pi }}\mathrm{degrees}\mathrm{and}\mathrm{\pi }=\frac{22}{7}\right] \\ \Rightarrow \theta =163.7\mathrm{degrees} \end{gathered}$$

Step 2. Calculate the area of the sector.

The area of the sector is given as,

$\begin{array}{rcl}\therefore \mathrm{Area}& =& \frac{\theta (indegrees)}{360°}\times \mathrm{\pi }{r}^2\\ & =& \frac{163.7}{360}\times \frac{22}{7}\times {(5.6)}^2\\ & =& 44.82{\mathrm{m}}^2\end{array}$

Hence, the area of the sector is $44.82{\mathrm{m}}^2$.