Question

In the figure given below, $A B C D E F$ is a regular hexagon of side length $1, A F P S$ and $A B Q R$ are squares. Then, the ratio $ar(A P Q)$ lar $(S R P)$ equals

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

Answer 1

Answer: (D)

Given,

$A B C D E F$ is a regular hexagon of side length $1 .$
$A B Q R$ and $A F P S$ is a square of each side length also $1$ .
$A D C D E F$ is a regular hexagon
$\therefore \angle F A B=120^{\circ}$
In square $A B Q R$,
$A B=B Q=1$
$A Q$ is a diagonal of square
$\therefore A Q=\sqrt{A B^{2}+B Q^{2}}=\sqrt{2}$
$\Rightarrow \angle B A S=\angle F A B-\angle F A S$
$=120^{\circ}-90^{\circ}=30^{\circ}$
$\Rightarrow \angle S A R=\angle B A R-\angle B A S$
$=90^{\circ}-30^{\circ}=60^{\circ}$
$\Rightarrow \angle A S R=60^{\circ}$
$[\because \triangle A R S$ is an equilateral triangle $]$
$\Rightarrow \angle R S P=\angle A S P-\angle A S R$
$=90^{\circ}-60^{\circ}=30^{\circ}$
$\Rightarrow \angle F A B=\angle F A P+\angle P A Q+\angle Q A B$
$\Rightarrow 120^{\circ}=45^{\circ}+\angle P A Q+45^{\circ}$
$\left[\because \angle F A P=\angle Q A B=45^{\circ}\right.$
$ F A=F P$ and $A B=B Q]$
$\therefore \angle PAQ = 30^{\circ}$
$ \therefore \frac{\text { Area of } \triangle P A Q}{\text { Area of } \triangle R S P} =\frac{\frac{1}{2} \times A Q \times A P \times \sin 30^{\circ}}{\frac{1}{2} \times R S \times P S \times \sin 30^{\circ}} $
$=\frac{\sqrt{2} \times \sqrt{2}}{1}=2$
$ {[\because A Q} =A P=\sqrt{2}, R S=P S=1] $