Question

Find the areas of both the segments of a circle of radius 42 cm with central angle 120°.

Answer 1

$$\begin{gathered} \mathrm{Area}oftheminorsector=\frac{120}{360}\times \mathrm{\pi }\times 42\times 42 \\ =\frac{1}{3}\times \mathrm{\pi }\times 42\times 42 \\ =\mathrm{\pi }\times 14\times 42 \\ =1848{\mathrm{cm}}^2 \end{gathered}$$

Area of the triangle = $\frac{1}{2}{R}^2\sin \theta$
Here, R is the measure of the equal sides of the isosceles triangle and θ is the angle enclosed by the equal sides.
Thus, we have:
$$\begin{gathered} \frac{1}{2}\times 42\times 42\times \sin \left(120°\right) \\ =762.93{\mathrm{cm}}^2 \end{gathered}$$

Area of the minor segment = Area of the sector $-$ Area of the triangle
=$1848-762.93=1085.07{\mathrm{cm}}^2$

Area of the major segment = Area of the circle $-$ Area of the minor segment
$$\begin{gathered} =\left(\mathrm{\pi }\times 42\times 42\right)-1085.07 \\ =5544-1085.07 \\ =4458.93{\mathrm{cm}}^2 \end{gathered}$$