In the given figure,OP = PQ = 14 , cm and sem | vedclass

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(A) Given: $OP = 14 \, cm$ and $PQ = 14 \, cm$. Therefore,$OQ = OP + PQ = 14 + 14 = 28 \, cm$.The shaded region is bounded by three semicircular arcs:

Vedclass · Accepted Answer

$88$

Vedclass · Answer

(A) Given: $OP = 14 \, cm$ and $PQ = 14 \, cm$. Therefore,$OQ = OP + PQ = 14 + 14 = 28 \, cm$.The shaded region is bounded by three semicircular arcs:$1$. The arc of the semicircle with diameter $OQ$ (length = $\frac{1}{2} \pi 	imes 28 = 14\pi \, cm$).$2$. The arc of the semicircle with diameter $OP$ (length = $\frac{1}{2} \pi 	imes 14 = 7\pi \, cm$).$3$. The arc of the semicircle with diameter $PQ$ (length = $\frac{1}{2} \pi 	imes 14 = 7\pi \, cm$).The perimeter of the shaded area is the sum of the lengths of these three arcs:Perimeter = $14\pi + 7\pi + 7\pi = 28\pi \, cm$.Using $\pi = \frac{22}{7}$,we get:Perimeter = $28 	imes \frac{22}{7} = 4 	imes 22 = 88 \, cm$.

In the given figure,$OP = PQ = 14 \, cm$ and semicircles are drawn on $OP$,$PQ$,and $OQ$ as diameters. Then,the perimeter of the shaded area is . . . . . . $cm$.

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