  Question

Find the area of the segment of a circle of radius 12 cm whose corresponding sector has a central angle of $${ 60 }^{ \circ }$$.

Solution

Area of minor segment = Area of sector- Area of $$\Delta OAB$$In $$\Delta OAB$$Refer image,$$\theta=60^0$$$$OA=OB=r=12cm$$$$\angle B=\angle A=x[\angle s$$ opp. to equal sides are equal$$\Rightarrow \angle A+\angle B+\angle O=180^0$$$$\Rightarrow x+x+60^0=180^0$$$$\Rightarrow 2x=180^0-60^0$$$$\Rightarrow x=\dfrac{120^0}{2}=60^0$$$$\therefore \Delta OAB$$ is equilateral $$\Delta$$ with each side (a) $$=12cm$$Area of the equilateral $$\Delta=\dfrac{\sqrt{3}}{4}a^2$$Area of minor segment = Area of the sector - Area of $$\Delta OAB$$$$=\dfrac{\pi r^2\theta}{360^0}-\dfrac{\sqrt{3}}{4}a^2$$$$=\dfrac{3.14\times 12\times 12\times 60^0}{360^0}-\dfrac{\sqrt{3}}{4}\times 12\times 12$$$$=6.28\times 12-36\sqrt{3}$$$$\therefore$$ Area of minor segment $$=(75.36-36\sqrt{3})cm^2$$ Mathematics

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