Question

The chord of a circle $84cm$ in diameter subtends an angle of ${60}^o$ at the centre of the circle. Find the area of the minor segment corresponding to the chord. (Take $\sqrt{3}=1.73$)

Answer 1

For the given minor segment $\overline{PR}$,
Diameter $=84cm$ $\Rightarrow$ $r=42cm$
Also $m\angle POR={60}^o=\theta$
Area of minor sector $OPQR$
$=\frac{\pi r^2\theta }{360}$
$=\frac{22}{7}\times \frac{42\times 42\times 60}{360}$
$=924{cm}^2$
In $△OPR,m\angle O={60}^o$
and $OP=OR=42$
$\therefore$ $△OPR$ is an equilateral triangle
Area of equilateral $△OPR=\frac{\sqrt{3}}{4}{a}^2$
$=\frac{1.73}{4}={\left(42\right)}^2=762.93{cm}^2$