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Question

(B) Given,radius of the circle $(r) = 14 \, cm$ and central angle $(	heta) = 60^{\circ}$.The area of the minor segment is given by the formula: Area

Vedclass · Accepted Answer

$49 \sqrt{3}$

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(B) Given,radius of the circle $(r) = 14 \, cm$ and central angle $(	heta) = 60^{\circ}$.The area of the minor segment is given by the formula: Area of sector - Area of triangle.Area of sector $= \frac{	heta}{360^{\circ}} 	imes \pi r^{2} = \frac{60^{\circ}}{360^{\circ}} 	imes \frac{22}{7} 	imes 14 	imes 14 = \frac{1}{6} 	imes 22 	imes 2 	imes 14 = \frac{308}{3} \approx 102.67 \, cm^{2}$.Since the central angle is $60^{\circ}$ and the two sides are radii,the triangle formed is an equilateral triangle.Area of equilateral triangle $= \frac{\sqrt{3}}{4} 	imes (	ext{side})^{2} = \frac{\sqrt{3}}{4} 	imes 14^{2} = \frac{\sqrt{3}}{4} 	imes 196 = 49 \sqrt{3} \, cm^{2} \approx 84.87 \, cm^{2}$.Area of minor segment $= \frac{308}{3} - 49 \sqrt{3} \, cm^{2} \approx 102.67 - 84.87 = 17.8 \, cm^{2}$.Note: The provided options seem to represent the area of the triangle rather than the segment. Based on the standard calculation for the area of the minor segment,the result is $\frac{308}{3} - 49 \sqrt{3}$. However,if the question implies the area of the triangle formed by the chord,the answer is $49 \sqrt{3}$.

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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