KVPY SA Exam 2018 Question Paper

Let A (0,0), B(1,0), C(1,1) & D(0, 1) ⇒ Area ABCD = 1
Again let Q (cosα, sinα) & R (cosβ, sin β)
⇒ coordinate of P (cosα– sinα, 0) & S (0,sinβ– cosβ)
PQRS is a square ⇒ PQ ⊥ QR ⇒ slope of QR = -1 = slope of SP
⇒ $\frac{\sin \mathit{\beta }-\sin \mathit{\alpha }}{\cos \mathit{\beta }-\mathit{c}\mathit{o}\mathit{s}\mathit{\alpha }}=-1=\frac{\sin \mathit{\beta }-\cos \mathit{\beta }}{\sin \mathit{\alpha }-\cos \mathit{\alpha }}$
⇒ $\sin \mathit{\beta }-\sin \mathit{\alpha }=-\cos \mathit{\beta }+\mathit{c}\mathit{o}\mathit{s}\mathit{\alpha }$
⇒ $\sin \mathit{\beta }-\cos \mathit{\beta }=\sin \mathit{\alpha }+\cos \mathit{\alpha }$ ………..(i)
and $\sin \mathit{\alpha }+\sin \mathit{\beta }=\cos \mathit{\alpha }+\cos \mathit{\beta }$ ……………..(ii)
⇒ $\cos \mathit{\alpha }=\sin \mathit{\beta }$
⇒ $\cos \mathit{\alpha }==\cos \left(90-\mathit{\beta }\right)$
⇒ $\mathit{\alpha }+\mathit{\beta }=90°$
Also PQ = QR
⇒ $\tan \mathit{\alpha }=\frac{1}{2}$
Area of PQRS = $2{\sin }^2\mathit{\alpha }=2\left(\frac{1}{5}\right)$
$\frac{\mathrm{Area}\mathrm{PQRS}}{\mathrm{Area}\mathrm{ABCD}}=\frac{2∕5}{1}=\frac{2}{5}$