JAMB 1985 · UME · Q47

In the figure, a sector of a circle of radius 3 has an angle of $120^\circ$ at the centre. The area of the shaded segment is

Worked solution (try it first)

The sector is $\frac{120}{360} = \frac13$ of the circle: $\frac13 \times \pi \times 3^2 = 3\pi$ .
The triangle $OAB$ has two sides of 3 with $120^\circ$ between them: area $= \frac12 \times 3 \times 3 \times \sin120^\circ$
$= \frac{9\sqrt3}{4}$ .
Segment = sector − triangle: $3\pi - \frac{9\sqrt3}{4} = 3\left(\pi - \frac{3\sqrt3}{4}\right)$ , option C.