SSC CHSL 6 Dec 2015 Shift 2 Paper

Let side of square ABCD = 1 unit
⇒ Diagonal AC = $\sqrt{1^2+1^2}=\sqrt{2}$ units
It is given that △QBC ∼ △PAC
Ratio of areas of two similar triangles is equal to the ratio of squares of corresponding sides.
⇒ $\frac{\mathrm{Area}\mathrm{of}△\mathrm{QBC}}{\mathrm{Area}\mathrm{of}△\mathit{P}\mathit{A}\mathit{C}}$ $=\frac{{\left(\mathit{B}\mathit{C}\right)}^2}{{\left(\mathit{A}\mathit{C}\right)}^2}$
= ${\left(\frac{1^2}{\sqrt{2}}\right)}^2=\frac{1}{2}$