A wire bent in the form of a circle of radius | vedclass

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(D) The circumference of the circle is given by $C = 2 \pi r$.Substituting $r = 42 \, cm$,we get $C = 2 	imes \frac{22}{7} 	imes 42 = 264 \, cm$.Sin

Vedclass · Accepted Answer

$60$

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(D) The circumference of the circle is given by $C = 2 \pi r$.Substituting $r = 42 \, cm$,we get $C = 2 	imes \frac{22}{7} 	imes 42 = 264 \, cm$.Since the wire is reshaped into a rectangle,the perimeter of the rectangle equals the circumference of the circle.Let the sides of the rectangle be $6x$ and $5x$ (based on the ratio $6:5$).The perimeter of the rectangle is $2(6x + 5x) = 2(11x) = 22x$.Equating the perimeters: $22x = 264$.Solving for $x$: $x = \frac{264}{22} = 12$.The smaller side is $5x = 5 	imes 12 = 60 \, cm$.

$A$ wire bent in the form of a circle of radius $42 \, cm$ is bent in the form of a rectangle whose sides are in the ratio of $6:5$. The smaller side of the rectangle is (in $cm$)

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