Question

A chord of a circle of radius 30 cm makes an angle of 60° at the centre of the circle. Find the area of the minor and major segments.

Answer 1

Let AB be the chord of a circle with centre O and radius 30 cm such that $\angle AOB=60°$.
Area of the sector OACBO $=\frac{\mathrm{\pi }{r}^2\theta }{360}$
$$\begin{gathered} =\left(3.14\times 30\times 30\times \frac{60}{360}\right){\mathrm{cm}}^2 \\ =471{\mathrm{cm}}^2 \end{gathered}$$
Area of $∆OAB$$=\frac{1}{2}{r}^2\sin \theta$
$$\begin{gathered} =\left(\frac{1}{2}\times 30\times 30\times \sin 60°\right){\mathrm{cm}}^2 \\ =\left(225\times 1.732\right){\mathrm{cm}}^2 \\ =389.7{\mathrm{cm}}^2 \end{gathered}$$

Area of the minor segment = (Area of the sector OACBO) $-$ (Area of $∆OAB$)
$$\begin{gathered} =\left(471-389.7\right){\mathrm{cm}}^2 \\ =81.3{\mathrm{cm}}^2 \end{gathered}$$

Area of the major segment = (Area of the circle) $-$ (Area of the minor segment)
$$\begin{gathered} =\left|\left(3.14\times 30\times 30\right)-81.3\right|{\mathrm{cm}}^2 \\ =2744.7{\mathrm{cm}}^2 \end{gathered}$$