Question

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$={90}^o$. Then the ratio of the area of quadrilateral PQBC to the area of the square ABCD is$?$

• A. $\frac{31}{60}$
• B. $\frac{37}{60}$
• C. $\frac{39}{60}$
• D. $\frac{41}{60}$

Answer 1

The correct option is D $\frac{41}{60}$

$ar\square PBCQ=ar\square ABCD-ar△AQB-ar△PDAar△PDA=\frac{1}{2}\times 3x\times 3x=\frac{{9x}^2}{2}In△PDA,\sin \alpha =\frac{PD}{PA}=\frac{x}{\sqrt{10}x}=\frac{1}{\sqrt{10}}In△AQB,\sin \alpha =\frac{AQ}{AB}=\frac{AQ}{3x}=\frac{1}{\sqrt{10}}\Rightarrow AQ=\frac{3x}{\sqrt{10}}\cos \alpha =\frac{QB}{AB}=\frac{QB}{3x}=\frac{3}{\sqrt{10}}ar△AQB=\frac{1}{2}\times QB\times AQ=\frac{1}{2}\times \frac{3x}{\sqrt{10}}\times \frac{9x}{\sqrt{10}}ar△AQB=\frac{27x^2}{20}ar\square PQBC=ar\square ABCD-ar△AQB-ar△PDAar\square PQBC=9x^2-\frac{27x^2}{20}-\frac{{9x}^2}{2}ar\square PQBC=9x^2-\frac{57x^2}{20}=\frac{123x^2}{20}\therefore \frac{ar\square PBCQ}{ar\square ABCD}=\frac{123x^2}{20}\times \frac{1}{{9x}^2}\Rightarrow \frac{ar\square PBCQ}{ar\square ABCD}=\frac{41}{60}$