CAT 2019 Shift 2 Paper

Refer to the figure:For this right angle triangle, we have the following relations
${\mathit{a}}^2+{\mathit{b}}^2={20}^2=400....\left(1\right)$ and $\mathit{A}\mathit{P}=\frac{\mathit{a}\mathit{b}}{20}....\left(2\right)$
For maximum value of $\mathit{A}\mathit{P}$, we have to maximize the product ab.
Applying $\mathit{A}\mathit{M}\ge \mathit{G}\mathit{M}$ inequality we get
$\frac{{\mathit{a}}^2+{\mathit{b}}^2}{2}\ge \sqrt{{\mathit{a}}^2\times {\mathit{b}}^2}$
$\Rightarrow \frac{400}{2}\ge \mathit{a}\mathit{b}$
$\Rightarrow \mathit{a}\mathit{b}\le 200$
Hence the maximum value of $\mathit{a}\mathit{b}=200$.
Therefore, the maximum value of $\mathit{A}\mathit{P}=\frac{200}{20}=10$