A sector of 56° cut out from a circle contains

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,

$A=\frac{56^{\circ}}{360^{\circ}} \times \frac{22}{7} r^{2}$

$4.4=\frac{56^{\circ}}{360^{\circ}} \times \frac{22}{7} r^{2}$

$r^{2}=\frac{360^{\circ}}{56^{\circ}} \times \frac{7}{22} \times 4.4$

$r^{2}=9$

$r=\sqrt{9}$

$r=3 \mathrm{~cm}$