Find the area of the minor segment of a circl | vedclass

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Question

Given that, radius of circle $(r)=14 \,cm$

and angle of the corresponding sector i.e., central angle $(	heta)=60^{\circ}$

since, in $	riangle

Vedclass · Accepted Answer

$49 \sqrt{3}$

Vedclass · Answer

Given that, radius of circle $(r)=14 \,cm$

and angle of the corresponding sector i.e., central angle $(	heta)=60^{\circ}$

since, in $	riangle AOB , OA = OB =$ Radius of circle i.e., $	riangle AOB$ is isosceles.

$\Rightarrow$ $\angle O A B=\angle O B A=	heta$

Now, in $	riangle O A B \quad \angle A O B+\angle O A B+\angle O B A=180^{\circ}$ [since,sum of interior angles of any triangle is $\left.180^{\circ}ight]$

$\Rightarrow$ $60^{\circ}+	heta+	heta=180^{\circ}$ [given, $\left.\angle A O B=60^{\circ}ight]$

$\Rightarrow$ $2 	heta=120^{\circ}$

$\Rightarrow \quad 	heta=60^{\circ}$

i.e. $\angle O A B=\angle O B A=60^{\circ}=\angle A O B$

since, all angles of $	riangle A O B$ are equal to $60^{\circ}$ i.e., $	riangle A O B$ is an equilateral triangle.

Also, $O A_{5}=O B=A B=14 cm$

So, Area of $	riangle O A B=\frac{\sqrt{3}}{4}(	ext { side })^{2}$

$=\frac{\sqrt{3}}{4} 	imes(14)^{2}$ [ $\because$ area of an equilateral triangle $=\frac{\sqrt{3}}{4}$ (side) $^{2}$ ]

$=\frac{\sqrt{3}}{4} 	imes 196=49 \sqrt{3} \,cm ^{2}$

Find the area of the minor segment of a circle of radius $14\,cm$, when the angle of the corresponding sector is $60^{\circ} .$ (in $cm ^{2}$)

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