In the figure,a square OABC is inscribed in a | vedclass

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(B) In $	riangle OAB$,since $OABC$ is a square,$\angle OAB = 90^{\circ}$.By Pythagoras theorem,$OB^2 = OA^2 + AB^2$.Since $OA = AB = 20 \, cm$,we hav

Vedclass · Accepted Answer

$228$

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(B) In $	riangle OAB$,since $OABC$ is a square,$\angle OAB = 90^{\circ}$.By Pythagoras theorem,$OB^2 = OA^2 + AB^2$.Since $OA = AB = 20 \, cm$,we have $OB^2 = 20^2 + 20^2 = 400 + 400 = 800$.Thus,the radius $(r)$ of the quadrant is $OB = \sqrt{800} = 20\sqrt{2} \, cm$.The area of the quadrant $OPBQ = \frac{1}{4} 	imes \pi r^2 = \frac{1}{4} 	imes 3.14 	imes (20\sqrt{2})^2$.$= \frac{1}{4} 	imes 3.14 	imes 800 = 3.14 	imes 200 = 628 \, cm^2$.The area of the square $OABC = (	ext{side})^2 = 20^2 = 400 \, cm^2$.The area of the shaded region = Area of quadrant $OPBQ$ - Area of square $OABC$.$= 628 \, cm^2 - 400 \, cm^2 = 228 \, cm^2$.

In the figure,a square $OABC$ is inscribed in a quadrant $OPBQ$. If $OA = 20 \, cm,$ find the area of the shaded region. (in $cm^2$) (Use $\pi = 3.14$)

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