A chord of a circle of radius 15 cm subtends an angle of 60° at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle.
(Use $\pi = 3.14$ and $\sqrt{3} = 1.73$.)                                                                                                                                  [Page No. 148]

Chord subtends $60^\circ$, $r = 15$ cm

$$\begin{aligned} \text{Area of sector} &= \frac{60}{360^\circ} \times 3.14 \times 15^2 \\ &= \frac{1}{6} \times 706.5 \\ &= 117.75 \text{ cm}^2 \end{aligned}$$

Triangle formed is equilateral.

$$\begin{aligned} \text{Area of triangle} &= \frac{\sqrt{3}}{4} a^2 \\ &= 1.734 \times 15^2 \\ &= 97.31 \text{ cm}^2 \end{aligned}$$

So, minor segment area = $97.31 \text{ cm}^2$

$$117.75 - 97.31 = 20.44 \text{ cm}^2$$

Area of circle: $3.14 \times 15^2 = 706.5 \text{ cm}^2$

$$706.5 - 20.44 = 686.06 \text{ cm}^2$$

So, major segment area = $686.06 \text{ cm}^2$

Caution
Students should clearly distinguish a sector from a segment. A segment’s area equals the sector’s area minus the triangle’s area; forgetting to subtract the triangle is a common error.