Question

Find the area of the minor segment of a circle of radius 14 cm, when its central angle is ${60}^{\circ }$.Also find the area of the corresponding major segment. [Use $\pi =\frac{22}{7}$]

Answer 1

Let ACB be the given arc subtending an angle of ${60}^{\circ }$ at the centre.

$\because \theta ={60}^{\circ }$ and OA = OB.
$\therefore △$ OAB is an equilateral triangle.

Here,r= 14 cm and $\theta ={60}^{\circ }$.
Area of the minor segment ACBA=(Area of the sector OACBO)-(Area of $△$ OAB)

=$\frac{\pi r^2\theta }{{360}^{\circ }}-\frac{\sqrt{3}}{4}{r}^2$

=$\frac{22}{7}\times 14\times 14\times \frac{{60}^{\circ }}{{360}^{\circ }}-\frac{\sqrt{3}}{4}\times 14\times 14$


=$\frac{308}{3}-49\sqrt{3}$

$=17.79c{m}^2$

Area of the major segment BDAB=Area of circle-Area of minor segment ACBA

=$\pi r^2-17.89$

=$\frac{22}{7}\times 14\times 14-17.89$

= 616 - 17.89

= $598.11=598c{m}^2$.