# The area of a sector of a circle of radius 5 cm

Question:

The area of a sector of a circle of radius $5 \mathrm{~cm}$ is $5 \pi \mathrm{cm}^{2}$. Find the angle contained by the sector.

Solution:

We know that the area A of a sector of an angle θ in the circle of radius r is given by

$A=\frac{\theta}{360^{\circ}} \times \pi r^{2}$

It is given that radius $r=5 \mathrm{~cm}$ and area $A=5 \pi \mathrm{cm}^{2}$.

Now we substitute the value of r and A in above formula to find the value of θ,

$5 \pi=\frac{\theta}{360^{\circ}} \times \pi \times 5 \times 5$

$\theta=\frac{360^{\circ} \times 5 \pi}{\pi \times 5 \times 5}$

$=72^{\circ}$