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
- 10
- $6 \sqrt{2}$
- $8 \sqrt{2}$
- 5
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 =$\frac{200}{20}=10$
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