As shown in the diagram,the side length of sq | vedclass

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(D) The shaded portion is the union of two congruent segments formed by the arcs $\widehat{APC}$ and $\widehat{AQC}$ within the square $ABCD$.Area of

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$252$

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(D) The shaded portion is the union of two congruent segments formed by the arcs $\widehat{APC}$ and $\widehat{AQC}$ within the square $ABCD$.Area of segment $\widehat{APC}$ (with respect to sector $BAPC$):Area $= 	ext{Area of sector } BAPC - 	ext{Area of } \Delta ABC$$= \frac{90^\circ}{360^\circ} 	imes \pi 	imes (21)^2 - \frac{1}{2} 	imes 21 	imes 21$$= \frac{1}{4} 	imes \frac{22}{7} 	imes 441 - 220.5$$= 346.5 - 220.5 = 126 \ cm^2$.Since the two segments are congruent,the total shaded area is:Area $= 2 	imes 126 = 252 \ cm^2$.

As shown in the diagram,the side length of square $ABCD$ is $21 \ cm$. $\widehat{APC}$ is an arc of $\odot(B, BA)$ and $\widehat{AQC}$ is an arc of $\odot(D, DA)$. Find the area of the shaded portion. (in $cm^2$)

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