Question

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

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

Answer 1

Answer: (D)

$2$

$ABQR,ARPS$ are squares then $AB=BQ=QR=AR=1=AF=FP=AS$ and $AQ,AP$ are diagonals of length $\sqrt{2}$ each.

In regular hexagon, each interioi angle $=\frac{(6-2)\times 180}{6}={120}^o$

$\angle FAB={120}^o$

$\angle RAB={90}^o=\angle FAS$ as $ABQR$ and $AFPS$ are squares.

$\angle FAB=\angle RAB+\angle FAS-\angle RAS$

$\therefore {120}^o={90}^o+{90}^o-\angle RAS$

$\angle RAS={60}^o\to$ $△RAS$ is equvilateral triangle with side $=1$

$SR=1$

Similarly $PS=1$

Also $\angle RAS=\angle RAQ+\angle PAS-\angle PAQ$

$\therefore {60}^o={45}^o+{45}^o-\angle PAQ$

$\therefore \angle PAQ={30}^o$

$\angle ASR={60}^o$ and $\angle ASP={90}^o$, then $\angle RSP=30$

$\frac{area△APQ}{area△SRP}=\frac{\frac{1}{2}AP\times AQ\sin {30}^o}{\frac{1}{2}RS\times PS\sin {30}^o}=\frac{\sqrt{2}\times \sqrt{2}}{1\times 1}=2$