In Fig.,ABCD is a square of side 14 , cm. Wit | vedclass

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(D) The area of each of the $4$ sectors is equal to each other,and each is a sector of $90^{\circ}$ in a circle of $7 \, cm$ radius.Area of each secto

Vedclass · Accepted Answer

$42$

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(D) The area of each of the $4$ sectors is equal to each other,and each is a sector of $90^{\circ}$ in a circle of $7 \, cm$ radius.Area of each sector $= \frac{90^{\circ}}{360^{\circ}} 	imes \pi(7)^{2}$$= \frac{1}{4} 	imes \frac{22}{7} 	imes 7 	imes 7$$= \frac{77}{2} \, cm^{2} = 38.5 \, cm^{2}$Area of square $ABCD = (	ext{Side})^{2} = (14)^{2} = 196 \, cm^{2}$Area of shaded portion = Area of square $ABCD - 4 	imes$ (Area of each sector)$= 196 - 4 	imes \frac{77}{2}$$= 196 - 154$$= 42 \, cm^{2}$Therefore,the area of the shaded portion is $42 \, cm^{2}$.

In $Fig.$,$ABCD$ is a square of side $14 \, cm$. With centres $A, B, C$ and $D$,four circles are drawn such that each circle touches externally two of the remaining three circles. Find the area of the shaded region (in $cm^2$). $\left[\text{Use } \pi = \frac{22}{7}\right]$

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