CAT 2019 Quant Question with Solution 61

Correct Answer: Option: 1

Refer to the figure:

For this right angle triangle, we have the following relations

${{a}^{2}}+{{b}^{2}}={{20}^{2}}=400....(1)$ and

$AP=\frac{ab}{20}....(2)$

For maximum value of AP, we have to maximize the product ab.

Applying $AM\ge GM$ inequality we get

$\frac{{{a}^{2}}+{{b}^{2}}}{2}\ge \sqrt{{{a}^{2}}\times {{b}^{2}}}$

$\Rightarrow \frac{400}{2}\ge ab$

$\Rightarrow ab\le 200$

Hence the maximum value of ab =200.

Therefore, the maximum value of AP =$\frac{200}{20}=10$