Question

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

Answer 1

Let the chord be AB. The ends of the chord are connected to the centre of the circle O to give the triangle OAB.

OAB is an isosceles triangle. The angle at the centre is 60$°$

Area of the triangle = $\frac{1}{2}{\left(30\right)}^2\sin 60°=450\times \frac{\sqrt{3}}{2}=389.25{\mathrm{cm}}^2$

Area of the sector OACBO = $\frac{60}{360}\times \mathrm{\pi }\times 30\times 30=150\mathrm{\pi }=471{\mathrm{cm}}^2$

Area of the minor segment = Area of the sector $-$ Area of the triangle
=$471-389.25=81.75{\mathrm{cm}}^2$

Area of the major segment = Area of the circle $-$ Area of the minor segment
$$\begin{gathered} =\left(\mathrm{\pi }\times 30\times 30\right)-81.29 \\ =2744.71{\mathrm{cm}}^2 \end{gathered}$$