Question

A circular disc of radius 6 cm is divided into three sectors with central angles ${90}^{\circ },{120}^{\circ }and{150}^{\circ }$. What part of the whole circle is the sector with central angle ${150}^{\circ }$? Also, calculate the ratio of the areas of the three sectors.

Answer 1

Radius of disc (r) = 6 cm

When θ = 90°

Area of a sector I = $\frac{\theta }{{360}^{\circ }}\times \pi \times r^2$
= $\frac{{90}^{\circ }}{{360}^{\circ }}\times \pi \times r^2$
= $\frac{1}{4}\times \pi \times r^2$

When θ = 120°

Area of a sector II =$\frac{{120}^{\circ }}{{360}^{\circ }}\times \pi \times r^2$
=$\frac{1}{3}\times \pi \times r^2$

When θ = 150°

Area of a sector III = $\frac{150\circ }{{360}^{\circ }}\times \pi \times r^2=\frac{5}{12}\times \pi \times r^2$

Required ratio = Area of sector I : area of sector II : area of sector III

=$\frac{1}{4}\times \pi \times r^2:\frac{1}{3}\times \pi \times r^2:\frac{5}{12}\times \pi \times r^2$

=$\frac{1}{4}:\frac{1}{3}:\frac{5}{12}$

= 3:4:5

Hence, required ratio = 3 : 4 : 5