$A$ chord of a circle of radius $15 \, cm$ subtends an angle of $60^{\circ}$ at the centre. Find the areas of the corresponding minor and major segments of the circle.
(Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)

(N/A) Radius $(r)$ of the circle $= 15 \, cm$.
Area of sector $OPRQ = \frac{60^{\circ}}{360^{\circ}} \times \pi r^{2}$
$= \frac{1}{6} \times 3.14 \times (15)^{2}$
$= \frac{1}{6} \times 3.14 \times 225 = 117.75 \, cm^{2}$.
In $\triangle OPQ$,since $OP = OQ = 15 \, cm$,the angles opposite to these sides are equal,i.e.,$\angle OPQ = \angle OQP$.
Since $\angle POQ = 60^{\circ}$,the sum of the other two angles is $180^{\circ} - 60^{\circ} = 120^{\circ}$.
Thus,$\angle OPQ = \angle OQP = 60^{\circ}$.
Since all angles are $60^{\circ}$,$\triangle OPQ$ is an equilateral triangle.
Area of $\triangle OPQ = \frac{\sqrt{3}}{4} \times (\text{side})^{2}$
$= \frac{1.73}{4} \times (15)^{2} = \frac{1.73}{4} \times 225 = 97.3125 \, cm^{2}$.
Area of minor segment $PRQ = \text{Area of sector } OPRQ - \text{Area of } \triangle OPQ$
$= 117.75 - 97.3125 = 20.4375 \, cm^{2}$.
Area of major segment $PSQ = \text{Area of circle} - \text{Area of minor segment } PRQ$
$= \pi r^{2} - 20.4375$
$= 3.14 \times 225 - 20.4375 = 706.5 - 20.4375 = 686.0625 \, cm^{2}$.