Question

A chord PQ of a circle with radius 15 cm subtends an angle of 60^° with the centre of the circle. Find the area of the minor as well as the major segment. ($\mathrm{\pi }$ = 3.14,$\sqrt{3}$= 1.73)

Answer 1

Radius of the circle, r = 15 cm

Let O be the centre and PQ be the chord of the circle.



∠POQ = θ = 60º

Area of the minor segment = Area of the shaded region

$$\begin{gathered} =r^2\left(\frac{\mathrm{\pi }\theta }{360°}-\frac{\sin \theta }{2}\right) \\ ={\left(15\right)}^2\times \left(\frac{3.14\times 60°}{360°}-\frac{\sin 60°}{2}\right) \\ =225\times \frac{3.14}{6}-225\times \frac{\sqrt{3}}{4} \\ =117.75-97.31 \\ =20.44{\mathrm{cm}}^2 \end{gathered}$$
Now,

Area of the circle = $\mathrm{\pi }{r}^2=3.14\times {\left(15\right)}^2=3.14\times 225$ = 706.5 cm²

∴ Area of the major segment = Area of the circle − Area of the minor segment = 706.5 − 20.44 = 686.06 cm²

Thus, the areas of the minor segment and major segment are 20.44 cm² and 686.06 cm², respectively.