Question

In the figure given below, $ABCDEF$ is a regular hexagon of side length $1,AFPS$ and $ABQR$ are squares. Then, the ratio $ar(APQ)$ lar $(SRP)$ equals

• A. $\frac{\sqrt{2}+1}{2}$
• B. $\sqrt{2}$
• C. $\frac{3\sqrt{3}}{4}$
• D. 2

Answer 1

Answer: (D)

Given,

$ABCDEF$ is a regular hexagon of side length $1.$
$ABQR$ and $AFPS$ is a square of each side length also $1$ .
$ADCDEF$ is a regular hexagon
$\therefore \angle FAB=120^{\circ }$
In square $ABQR$,
$AB=BQ=1$
$AQ$ is a diagonal of square
$\therefore AQ=\sqrt{A{B}^2+B{Q}^2}=\sqrt{2}$
$\Rightarrow \angle BAS=\angle FAB-\angle FAS$
$=120^{\circ }-90^{\circ }=30^{\circ }$
$\Rightarrow \angle SAR=\angle BAR-\angle BAS$
$=90^{\circ }-30^{\circ }=60^{\circ }$
$\Rightarrow \angle ASR=60^{\circ }$
$[\because △ARS$ is an equilateral triangle $]$
$\Rightarrow \angle RSP=\angle ASP-\angle ASR$
$=90^{\circ }-60^{\circ }=30^{\circ }$
$\Rightarrow \angle FAB=\angle FAP+\angle PAQ+\angle QAB$
$\Rightarrow 120^{\circ }=45^{\circ }+\angle PAQ+45^{\circ }$
$[\because \angle FAP=\angle QAB=45^{\circ }$
$FA=FP$ and $AB=BQ]$
$\therefore \angle PAQ=30^{\circ }$
$\therefore \frac{\text{ Area of }△PAQ}{\text{ Area of }△RSP}=\frac{\frac{1}{2}\times AQ\times AP\times \sin 30^{\circ }}{\frac{1}{2}\times RS\times PS\times \sin 30^{\circ }}$
$=\frac{\sqrt{2}\times \sqrt{2}}{1}=2$
$[\because AQ=AP=\sqrt{2},RS=PS=1]$