# CAT 2019 Quant Question with Solution 61

Question:
Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm. If AP is perpendicular on BC, then the maximum possible length of AP, in cm, is

1. 10
2. $6 \sqrt{2}$
3. $8 \sqrt{2}$
4. 5

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 =$\frac{200}{20}=10$

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