Question

Question 6

In the figure given, the radius of the circle is 15 cm. The angle subtended by the chord AB at the centre O is ${60}^{\circ }$. Find the area of the major and minor segments.

Answer 1

Radius of the circle = 15 cm
$\Delta AOB$ is an isosceles triangle as two sides are equal.
$\therefore \angle A=\angle B$
Now, sum of all angles of triangle $={180}^{\circ }$
$\Rightarrow$$\angle A+\angle B+\angle O={180}^{\circ }$
$\Rightarrow 2\angle A={180}^{\circ }-{60}^{\circ }$
$\Rightarrow \angle A=\frac{{120}^{\circ }}{2}$
$\Rightarrow \angle A={60}^{\circ }$

Hence, the 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 subtended at the centre by minor segment $={60}^{\circ }$
Area of the sector making angle $\theta$
$=\left(\frac{\theta }{{360}^{\circ }}\right)\times \pi r^2$
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 $\Delta AOB$
$=588.75c{m}^2+97.3c{m}^2=686.05c{m}^2$