A wire when bent in the form of a circle encl | vedclass

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Question

(A) Area of the circle $= 1386 \, cm^2$.Using the formula for the area of a circle,$\pi r^2 = 1386$.$\frac{22}{7} 	imes r^2 = 1386 \implies r^2 = \fr

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$484 \sqrt{3} \, cm^2$

Vedclass · Answer

(A) Area of the circle $= 1386 \, cm^2$.Using the formula for the area of a circle,$\pi r^2 = 1386$.$\frac{22}{7} 	imes r^2 = 1386 \implies r^2 = \frac{1386 	imes 7}{22} = 63 	imes 7 = 441$.Thus,$r = \sqrt{441} = 21 \, cm$.The length of the wire is equal to the circumference of the circle: $C = 2 \pi r = 2 	imes \frac{22}{7} 	imes 21 = 132 \, cm$.When this wire is bent into an equilateral triangle,its perimeter remains $132 \, cm$.Let the side of the equilateral triangle be $a$. Then $3a = 132$,which gives $a = 44 \, cm$.The area of an equilateral triangle is given by $\frac{\sqrt{3}}{4} a^2$.Area $= \frac{\sqrt{3}}{4} 	imes 44 	imes 44 = \sqrt{3} 	imes 11 	imes 44 = 484 \sqrt{3} \, cm^2$.

$A$ wire when bent in the form of a circle encloses an area of $1386 \, cm^2$. What will be the enclosed area when the same wire is bent into the form of an equilateral triangle?

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