KVPY SA Exam 2013 Question Paper

In ΔAPQ
AP = AQ = $\sqrt{2}$ , ∠APQ = 30°
In ΔSRP
SR = SP = 1, ∠RSP = 30°
∠FAB = 120°
∠BAS = ∠FAB - ∠FAS = 120° - 90° = 30°
∠SAR = ∠BAR - ∠BAS = 60°
In ΔARS
$\cos 60°=\frac{1+1-{\mathrm{SR}}^2}{2.1.1}$ [∵ AR = AS = 1]
⇒ SR = 1
Now, ∠RSP = ∠ASP - ∠ASR
= 90° - 60° = 30° [∵ ASR is equilateral]
Now, from ΔSRP ⇒ RP $=\frac{\sqrt{3}-1}{2\sqrt{2}}$
In ΔAPR
$\cos \angle \mathrm{RAP}=\frac{{\left(\sqrt{2}\right)}^2+1^2-{\mathrm{PR}}^2}{2\sqrt{2}}$
⇒ cos ∠RAP = 15°
∠PAQ = ∠RAQ - ∠RAP = 45° - 15° = 30°
Now, $\frac{\mathrm{ar}\left(\mathrm{\Delta }\mathrm{APQ}\right)}{\mathrm{ar}\left(\mathrm{\Delta }\mathrm{SRP}\right)}$ $=\frac{\frac{1}{2}\times \sqrt{2}\times \sqrt{2}\sin 30°}{\frac{1}{2}\times 1\times 1\times \sin 30°}=2$