NTSE Stage 2 SAT 2017 Question Paper

Reason :
Let $\mathrm{OQ}=\mathrm{x}\mathrm{cm}$
Given that the area of square is $2{\mathrm{cm}}^2$
side of square ⇒ $\sqrt{2}\mathrm{cm}$
so $\mathrm{OP}=\frac{\sqrt{2}}{2}\mathrm{cm}$
so in Δ OPQ, by using Pythagoras theorem, ${\mathrm{OQ}}^2={\mathrm{OP}}^2+{\mathrm{PQ}}^2$
⇒ ${\mathrm{x}}^2={\left(\sqrt{2}\right)}^2+{\left(\sqrt{2}\right)}^2$
⇒ $\mathrm{x}=\sqrt{\frac{5}{2}}$ Now, Diameter $\mathrm{AC}=2\sqrt{\frac{5}{2}}\mathrm{cm}$
Let $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}=\mathrm{a}\mathrm{cm}$
⇒ ${\mathrm{a}}^2+{\mathrm{a}}^2={\left(2\sqrt{\frac{5}{2}}\right)}^2$
${\mathrm{a}}^2=5$
⇒ area of square ABCD $=5{\mathrm{cm}}^2$
Hence option (1) is correct