Question

Find the area of both the segments of a circle of radius 42 cm with central angle ${120}^{\circ }$. [ Given, sin ${120}^{\circ }=\frac{\sqrt{3}}{2}and\sqrt{3}$ = 1.73. ]

Answer 1

Area of the minor sector $=\frac{120}{360}\times \pi \times 42\times 42=\frac{1}{3}\times \pi \times 42\times 42=1848c{m}^2$

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

Area of the triangle $=\frac{1}{2}\times 42\times 42\times sin\left({120}^o\right)=762.93c{m}^2$

Area of the minor segment = Area of the sector – Area of the triangle

$=1848–762.93=1085.07c{m}^2$

Area of the major segment = Area of the circle – Area of the minor segment

$=(\pi \times 42\times 42)–1085.07=5544–108.07=4458.93c{m}^2$