In the given figure a square OABC is inscribed in a quadrant OPBQ. IF OA = 10 cm, then the perimeter of arc. QPB is

5π

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10π

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Solution

The correct option is B

$R a d i u s o f q u a r t e r c i r c l e w i l l b e \sqrt{10^{2} + 10^{2}} = 10 \sqrt{2}$ hence the length of arc QPB will be $\frac{π r}{2} = \frac{π 10 \sqrt{2}}{2} = 5 \sqrt{2} π$

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