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Let $\mathrm{ABCD}$ be a square and let $\mathrm{P}$ be point on segment $\mathrm{CD}$ such that $\mathrm{D P: P C=1: 2}$. Let $\mathrm{Q}$ be a point on segment $\mathrm{AP}$ such that $\angle \mathrm{BQP}=90^{\circ}$. Then the ratio of the area of quadrilateral $\mathrm{PQBC}$ to the area of the square $\mathrm{ABCD}$ is
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$\frac{41}{60} 3$
$\mathrm{CP}=\frac{2}{3} \mathrm{a} \quad \quad \mathrm{PD}=\frac{\mathrm{a}}{3}$
Let $\angle \mathrm{PAD}=\phi \quad \tan \phi=\frac{1}{3}$ (In $\left.\triangle \mathrm{APD}\right)$
Now, $\angle \mathrm{DAP}=\angle \mathrm{QBA}=\phi$
Required ratio $=\frac{\text { area of } P Q B C}{a^{2}}=\frac{a^{2}-(\text { area of } \Delta A D P+\text { area of } \Delta A Q B)}{a^{2}}$
$\Rightarrow 1-\left(\frac{1}{6}+\frac{3}{20}\right)=\frac{41}{60}$
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