NTSE Stage 2 SAT 2018 Exam Question Paper

Let the side of square be a.
Then, radius of inner circle = half of side square $=\frac{\mathit{a}}{2}$
Therefore, the area of inner circle $=\mathit{\pi }{\left(\mathrm{radius}\right)}^2$
$=\mathit{\pi }{\left(\frac{\mathit{a}}{2}\right)}^2=\mathit{\pi }\frac{{\mathit{a}}^2}{4}$
Since, the diagonal of the square$=\sqrt{2}\mathit{a}$
Then, radius of outer circle = half of diagonal of square $=\frac{\sqrt{2}\mathit{a}}{2}$
Hence, the area of outer circle $=\mathit{\pi }{\left(\frac{\sqrt{2}\mathit{a}}{2}\right)}^2=\mathit{\pi }\frac{2{\mathit{a}}^2}{4}$
Hence, the required ratio of the areas of inner circle to outer circle is $\frac{{\mathit{a}}^2}{4}:\frac{2{\mathit{a}}^2}{4}=1:2$