Question

What is the area of the minor segment of a circle of radius 14 cm , when the angle of the corresponding sector is ${60}^{\circ }$

• A. $(\frac{308}{3}-49\sqrt{3})c{m}^2$
• B. $(\frac{320}{3}-50\sqrt{3})c{m}^2$
• C. $49c{m}^2$
• D. $\frac{308}{3}c{m}^2$

Answer 1

The correct option is A $(\frac{308}{3}-49\sqrt{3})c{m}^2$

Given that , radius of circle (r ) = 14 cm

Angle of the corresponding sector. i.e central angle
$\left(\theta \right)={60}^{\circ }$

$\because$ In $\Delta$ AOB
$OA=OB=Radiusofcircle$
$\therefore \Delta$ AOB is isosceles.

$\Rightarrow \angle OAB=\angle OBA=\theta$

Now, in $\Delta$ OAB
$\angle AOB+\angle OAB+\angle OBA={180}^{\circ }$

[Since , sum of interior angles of any triangle is ${180}^{\circ }$]

$\Rightarrow {60}^{\circ }+\theta +\theta ={180}^{\circ }[given,\angle AOB={60}^{\circ }]$

$\Rightarrow 2\theta ={120}^{\circ }$

$\Rightarrow \theta ={60}^{\circ }$

i.e $\angle AOB=\angle OBA={60}^{\circ }=\angle AOB$

Since, all angles of $\Delta AOB$ are equal to ${60}^{\circ }$. So, $\Delta AOB$ is an equilateral triangle.

Also, area of $\Delta AOB$
$=\frac{\sqrt{3}}{4}\times (14{)}^2$
$[\because Areaofanequilateraltriangle=\frac{\sqrt{3}}{4}(side{)}^2]$

$=\frac{\sqrt{3}}{4}\times 196$
$=49\sqrt{3}c{m}^2$

Area of sector OBAO
$=\frac{\pi r^2}{{360}^{\circ }}\times \theta$
$=\frac{22}{7}\times \frac{14\times 14}{360}\times {60}^{\circ }$
$=\frac{22\times 2\times 14}{6}$
$=\frac{22\times 14}{3}$
$=\frac{308}{3}c{m}^2$

$\therefore$ The area of the minor segment
= Area of sector OBAO - Area of the equilateral triangle
$=(\frac{308}{3}-49\sqrt{3})c{m}^2$