Question

A chord of a circle of radius 12 cm subtends an angle of ${120}^{\circ }$ at the center. Find the area of the corresponding segment of the circle. (Use $\pi =3.14and\sqrt{3}=1.73)$

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Radius of the circle, r = 12 cm
Draw a perpendicular OD to chord AB. It will bisect AB.
$\angle A={180}^{\circ }-({90}^{\circ }+{60}^{\circ })={30}^{\circ }$
$cos{30}^{\circ }=\frac{AD}{OA}$ $\Rightarrow \frac{\sqrt{3}}{2}=\frac{AD}{12}$
$\Rightarrow AD=6\sqrt{3}cm$
$\Rightarrow AB=2\times AD=12\sqrt{3}cm$
$sin{30}^{\circ }=\frac{OD}{OA}$
$\Rightarrow \frac{1}{2}=\frac{OD}{12}$
$\Rightarrow OD=6cm$
$Areaof\Delta AOB:$
$=\frac{1}{2}\times base\times height$
$=\frac{1}{2}\times 12\sqrt{3}\times 6=36\sqrt{3}cm$
$=36\times 1.73=62.28c{m}^2$

Angle made by Minor sector $={120}^{\circ }$
Area of the sector making angle ${120}^{\circ }$
$=\left(\frac{{120}^{\circ }}{{360}^{\circ }}\right)\times \pi r^{2}c{m}^2$
$=\left(\frac{1}{3}\right)\times {12}^2\pi c{m}^2=\frac{144}{3}\pi c{m}^2$
$=48\times 3.14c{m}^2=150.72c{m}^2$
$\therefore$ Area of the corresponding Minor segment:
= Area of the Minor sector - Area of $\Delta AOB$
$=150.72c{m}^2-62.28c{m}^2$
$=88.44c{m}^2$