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(D) Let the side length of the square $ABCD$ be $3x$. Thus,$AB=BC=CD=AD=3x$.Given $DP:PC=1:2$,we have $DP=x$ and $PC=2x$.In $	riangle DAP$,$AP = \sqr

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$\frac{41}{60}$

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(D) Let the side length of the square $ABCD$ be $3x$. Thus,$AB=BC=CD=AD=3x$.Given $DP:PC=1:2$,we have $DP=x$ and $PC=2x$.In $	riangle DAP$,$AP = \sqrt{AD^2 + DP^2} = \sqrt{(3x)^2 + x^2} = \sqrt{10}x$.Let $\angle DAP = \alpha$. Then $\angle APD = 90^{\circ} - \alpha$. Since $\angle BQP = 90^{\circ}$,in $	riangle ABQ$,$\angle ABQ = 90^{\circ} - \angle BAQ = 90^{\circ} - \alpha$.Thus,$	riangle DAP \sim 	riangle QBA$ by $AA$ similarity.From similarity,$\frac{AD}{QB} = \frac{AP}{AB} = \frac{DP}{AQ}$.$\frac{3x}{QB} = \frac{\sqrt{10}x}{3x} = \frac{x}{AQ}$.$QB = \frac{9x}{\sqrt{10}}$ and $AQ = \frac{3x}{\sqrt{10}}$.Area of $	riangle DAP = \frac{1}{2} 	imes 3x 	imes x = 1.5x^2$.Area of $	riangle ABQ = \frac{1}{2} 	imes AB 	imes QB 	imes \sin(\angle ABQ) = \frac{1}{2} 	imes 3x 	imes \frac{9x}{\sqrt{10}} 	imes \frac{3}{\sqrt{10}} = \frac{81x^2}{20} = 4.05x^2$.Area of quadrilateral $PQBC = 	ext{Area}(ABCD) - 	ext{Area}(	riangle DAP) - 	ext{Area}(	riangle ABQ) = 9x^2 - 1.5x^2 - 4.05x^2 = 3.45x^2 = \frac{69}{20}x^2$.Ratio $= \frac{69x^2/20}{9x^2} = \frac{69}{180} = \frac{23}{60}$.Wait,re-evaluating: Area of $	riangle ABQ = \frac{1}{2} 	imes AQ 	imes QB = \frac{1}{2} 	imes \frac{3x}{\sqrt{10}} 	imes \frac{9x}{\sqrt{10}} = \frac{27x^2}{20} = 1.35x^2$.Area of $PQBC = 9x^2 - 1.5x^2 - 1.35x^2 = 6.15x^2 = \frac{123}{20}x^2$.Ratio $= \frac{123/20}{9} = \frac{123}{180} = \frac{41}{60}$.

Let $ABCD$ be a square and let $P$ be a point on segment $CD$ such that $DP:PC=1:2$. Let $Q$ be a point on segment $AP$ such that $\angle BQP=90^{\circ}$. Then,the ratio of the area of quadrilateral $PQBC$ to the area of the square $ABCD$ is

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