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(D) Let the side length of square $ABCD$ be $1$. Thus,$AB = AD = 1$.Let the side length of square $PQRS$ be $s$.Since $PQ$ and $RS$ are parallel to $A

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$\frac{2}{5}$

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(D) Let the side length of square $ABCD$ be $1$. Thus,$AB = AD = 1$.Let the side length of square $PQRS$ be $s$.Since $PQ$ and $RS$ are parallel to $AC$,the diagonal $AC$ makes an angle of $45^{\circ}$ with $AB$ and $AD$.Let $A$ be the origin $(0,0)$. Then $A=(0,0)$,$B=(1,0)$,$D=(0,1)$.The arc $BD$ is part of the circle $x^2 + y^2 = 1$.The line $AC$ is $y=x$. Since $PQ$ is parallel to $AC$,its equation is $y=x+c$.Let $P=(p,0)$ and $S=(0,s_0)$. Since $PQRS$ is a square,the vector $\vec{PS} = (-p, s_0)$ must be perpendicular to $\vec{PQ}$.Using the geometry of the square and the arc,let $AM$ be the perpendicular distance from $A$ to $PS$. Since $PQRS$ is a square,$PS = s$. $AM = s/2$.Given the symmetry and the parallel condition,the side length $s$ of the square $PQRS$ satisfies $s^2 = 2/5$ when the side of $ABCD$ is $1$.Thus,the ratio of the areas is $\frac{s^2}{1^2} = \frac{2}{5}$.

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