Question

In the figure given, the radius of the circle is 15cm. The angle subtended by the chord AB at the centre O is ${60}^{\circ }$. Find the area of the Major segment.
1. $588.75c{m}^2$
2. $117.75c{m}^2$
3. $686.05c{m}^2$
4. $97.3c{m}^2$

Answer 1

Radius of the circle = 15 cm
$\Delta AOB$ is isosceles as two sides are equal.
$\therefore \angle A=\angle B$
Sum of all angles of triangle
$={180}^{\circ }$
$\angle A+\angle B+\angle C={180}^{\circ }$
$\Rightarrow 2\angle A={180}^{\circ }-{60}^{\circ }$
$\Rightarrow \angle A=\frac{{120}^{\circ }}{2}$
$\Rightarrow \angle A={60}^{\circ }$
Triangle is equilateral as $\angle A=\angle B=\angle C={60}^{\circ }$ $\therefore OA=OB=AB=15cm$
Area of equilateral $\Delta AOB=\frac{\sqrt{3}}{4}\times (OA{)}^2=\frac{\sqrt{3}}{4}\times {15}^2$ $=\left(225\frac{\sqrt{3}}{4}\right)c{m}^2=97.3c{m}^2$

Angle subtend at the centre by minor segment $={60}^{\circ }$
Area of Minor sector making angle ${60}^{\circ }=\left(\frac{{60}^{\circ }}{{360}^{\circ }}\right)\times \pi r^{2}c{m}^2$ = $\left(\frac{1}{6}\right)\times {15}^2\pi c{m}^2=\frac{225}{6}\pi c{m}^2$
$=\left(\frac{225}{6}\right)\times 3.14c{m}^2=117.75c{m}^2$
Area of the minor segment = Area of Minor sector - Area of equilateral $\Delta AOB$ $=117.75c{m}^2-97.3c{m}^2=20.4c{m}^2$
Angle made by Major sector $={360}^{\circ }-{60}^{\circ }={300}^{\circ }$
Area of the sector making angle ${300}^{\circ }=\left(\frac{{300}^{\circ }}{{360}^{\circ }}\right)\times \pi r^{2}c{m}^2$ $=\left(\frac{5}{6}\right)\times {15}^2\pi c{m}^2=\frac{1125}{6}\pi c{m}^2$
$=\left(\frac{1125}{6}\right)\times 3.14c{m}^2=588.75c{m}^2$
Area of major segment = Area of Major sector + Area of equilateral $\Delta AOB$ $=588.75c{m}^2+97.3c{m}^2=686.05c{m}^2$