CAT 2020 Exam Shift 1 Solved Paper

Given the circle is inscribed in the rhombus of diagonals 12 and 16 . Let $\mathit{O}$ be the point of intersection of the diagonals of the rhombus. Then, OE (radius) $\mathit{⟂}$ DC.
Also $\mathit{D}\mathit{C}=\sqrt{6^2+8^2}=10$
As area of $∆\mathit{O}\mathit{D}\mathit{C}$ should be the same, we have, $\frac{1}{2}\times 6\times 8=\frac{1}{2}\times \mathit{O}\mathit{E}\times 10$
$\Rightarrow \mathit{O}\mathit{E}=4.8$
$\therefore$ Required ratio of areas $=\frac{\mathit{\pi }{\left(4.8\right)}^2}{\frac{1}{2}\times 12\times 16}=\frac{6\mathit{\pi }}{25}$