In the figure given below, $\mathrm{ABCDEF}$ is a regular hexagon of side length $1$, $\mathrm{AFPS}$ and $\mathrm{ABQR}$ are squares. Then the ratio Area $\mathrm{(APQ)}/$Area $\mathrm{(SRP)}$ equals

• A

  $\frac{\sqrt{2}+1}{2}$
• B

  $\sqrt{2}$
• C

  $\frac{3 \sqrt{3}}{4}$
• Correct answer
  D

  $2$

Solution

In $\triangle \mathrm{APQ}$ $\mathrm{AP}=\mathrm{AQ}=\sqrt{2}, \angle \mathrm{APQ}=30^{\circ}$ In $\Delta$ $\mathrm{SRP}$ $\mathrm{S R=S P=1}, \angle \mathrm{R S P=30}^{\circ}$ $\angle \mathrm{FAB}=120^{\circ}$ $\angle \mathrm{B A S}=\angle \mathrm{F A B}-\angle \mathrm{F A S}=120^{\circ}-90^{\circ}=30^{\circ}$ $\angle \mathrm{S A R}=\angle \mathrm{B A R}-\angle \mathrm{B A S}=60^{\circ}$ In $\triangle \mathrm{A R S}$ $\begin{array}{l} \cos 60^{\circ}=\frac{1+1-\mathrm{SR}^{2}}{2.1 .1}[\because \mathrm{AR}=\mathrm{AS}=1] \\ \Rightarrow \mathrm{SR}=1 \end{array}$ Now, $\angle \mathrm{RSP}=\angle \mathrm{ASP}-\angle \mathrm{ASR}$ $=90^{\circ}-60^{\circ}=30^{\circ}[\because$ $\mathrm{ASR}$ is equilateral] Now, from $\Delta \mathrm{SRP} \Rightarrow \mathrm{RP}=\frac{\sqrt{3}-1}{2 \sqrt{2}}$ In $\triangle \mathrm{APR}$ $\cos \angle \mathrm{RAP}=\frac{(\sqrt{2})^{2}+1^{2}-P R^{2}}{2 \sqrt{2}}$ $\Rightarrow \cos \angle \mathrm{RAP}=15^{\circ}$ $\angle \mathrm{PAQ}=\angle \mathrm{RAQ}-\angle \mathrm{RAP}=45^{\circ}-15^{\circ}=30^{\circ}$ Now, $\frac{\operatorname{ar}(\Delta \mathrm{APQ})}{\operatorname{ar}(\Delta \mathrm{SRP})}=\frac{\frac{1}{2} \times \sqrt{2} \times \sqrt{2} \sin 30^{\circ}}{\frac{1}{2} \times 1 \times 1 \times \sin 30^{\circ}}=2$

Hence, the correct answer is option: 'D'.