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

Area of the minor sector $=\frac{120}{360} \times \pi \times 42 \times 42$

$=\frac{1}{3} \times \pi \times 42 \times 42$

$=\pi \times 14 \times 42$

$=1848 \mathrm{~cm}^{2}$

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:

$\frac{1}{2} \times 42 \times 42 \times \sin \left(120^{\circ}\right)$

$=762.93 \mathrm{~cm}^{2}$

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

$=(\pi \times 42 \times 42)-1085.07$

$=5544-1085.07$

$=4458.93 \mathrm{~cm}^{2}$