Question

A sector of angle $40°$ is cut out from a circle. If the area of the sector is $\frac{77}{18}{\mathrm{m}}^2$, find the radius of the circle.

Answer 1

Determine the radius of the circle.

The area of a sector $\left(A\right)$ is calculated by $A=\frac{\theta }{360°}\times {\mathrm{\pi r}}^2$ where $r$ represents the radius of the circle and $\theta$ represents the central angle of the circle.

$$\begin{gathered} \Rightarrow \frac{77}{18}=\frac{40°}{360°}\times \frac{22}{7}\times r^2 \\ \Rightarrow r^2=\frac{77\times 360\times 7}{18\times 22\times 40} \\ \Rightarrow r=\sqrt{\frac{77\times 360\times 7}{18\times 22\times 40}}[\mathrm{Radius}\mathrm{cannot}\mathrm{be}\mathrm{negative}] \\ r=3.5\mathrm{m} \end{gathered}$$

Hence, the radius of the circle is $3.5\mathrm{m}$.