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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 $$=(75.36-36\sqrt{3})cm^2$$

1791269_1542541_ans_e7a0d002c67e433987b92d6bfe8a5ea2.png

Mathematics

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