Question

Question 10
Find the area of the minor segment of a circle of radius 14 cm , when the angle of the corresponding sector is ${60}^{\circ }$

Answer 1

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

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

Since, in $\Delta AOB,OA=OB=Radiusofcircle$. i.e., $\Delta AOB$ 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 ${60}^{\circ }$ i.e $\Delta AOB$ is an equilateral triangle.

Also, $OA=\frac{\sqrt{3}}{4}(side{)}^2$

$=\frac{\sqrt{3}}{4}\times \left(14{)}^2\right[\because Areaofanequilateraltriangle=\frac{\sqrt{3}}{4}\left(side{)}^2\right]$

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

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
$(\frac{308}{3}-49\sqrt{3})c{m}^2$