Question

A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use $\pi = 3.14$ and $\sqrt3 = 1.73$)

Answer 1

Radius (r) of circle = $15$ cm
Area of sector $OPRQ$ = $\frac{60°}{360°} \times \pi r^2$
= $\frac{1}{6} \times 3.14 \times (15)^2$
= $117.75\,cm^2$
In $∆OPQ$,
$∠OPQ = ∠OQP$ (As $OP = OQ$)
$∠OPQ + ∠OQP + ∠POQ = 180°$
$2∠OPQ = 120°$
$∠OPQ = 60°$
$∆OPQ$ is an equilateral triangle.
Area of $∆OPQ$ = $\frac{\sqrt3 } 4 \times (side)^2$
= $\frac{\sqrt3 } 4 \times (15)^2$

= $\frac{225\sqrt3 } 4 cm^2$

= $56.25 \sqrt3$
= $97. 3125 \, cm^2$
Area of segment $PRQ$ = Area of sector $OPRQ$ − Area of $∆OPQ$
= $117.75 − 97.3125$ = $20.4375$ $cm^2$
Area of major segment $PSQ$ = Area of circle − Area of segment $PRQ$
= $\pi (15)^2 \- 20.4375$
= $3.14 \times 225 - 20.4375$
= $706.5 - 20.4375$
= $686. 0625$ $cm^2$

So, the answer is $686. 0625\ cm^2$.