Question:
A sector of $56^{\circ}$ cut out from a circle contains area $4.4 \mathrm{~cm}^{2}$. Find the radius of the circle.
Solution:
We know that the area A of a sector of circle at an angle θ of radius r is given by
$A=\frac{\theta}{360^{\circ}} \pi r^{2}$
It is given that, Area of a sector $A=4.4 \mathrm{~cm}^{2}$ and angle $\theta=56^{\circ}$.
We can find the value of r by substituting these values in above formula,
$A=\frac{56^{\circ}}{360^{\circ}} \times \frac{22}{7} r^{2}$
$4.4=\frac{56^{\circ}}{360^{\circ}} \times \frac{22}{7} r^{2}$
$r^{2}=\frac{360^{\circ}}{56^{\circ}} \times \frac{7}{22} \times 4.4$
$r^{2}=9$
$r=\sqrt{9}$
$r=3 \mathrm{~cm}$