KVPY SA Exam 2013 Question Paper

In ΔAPQ
AP = AQ = √2 , ∠APQ = 30°
In ΔSRP
SR = SP = 1, ∠RSP = 30°
∠FAB = 120°
∠BAS = ∠FAB - ∠FAS = 120° - 90° = 30°
∠SAR = ∠BAR - ∠BAS = 60°
In ΔARS
cos‌60°=

1+1−SR2

2.1.1

[∵ AR = AS = 1]
⇒ SR = 1
Now, ∠RSP = ∠ASP - ∠ASR
= 90° - 60° = 30° [∵ ASR is equilateral]
Now, from ΔSRP ⇒ RP =

√3−1

2√2

In ΔAPR
cos‌∠RAP=

(√2)2+12−PR2

2√2

⇒ cos ∠RAP = 15°
∠PAQ = ∠RAQ - ∠RAP = 45° - 15° = 30°
Now,

ar(Δ‌APQ)

ar(Δ‌SRP)

=

1 2 ×√2×√2‌sin‌30°

1 2 ×1×1×sin‌30°

=2