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

• A. 10
• B. \(6\sqrt{2}\)
• C. \(8\sqrt{2}\)
• D. 5

Answer 1

The Correct Option is A

From the figure above
For this right-angled triangle, the following relations are established:
\(a^2+b^2=20^2=400.......(1) \)
\(AP=\frac{ab}{20}.........​(2)\)
To achieve the maximum value of \(AP\), the product ab must be maximized.

Applying the \(AM ≥ GM\) inequality, we get:
\(\frac{a^2+b^2}{2}≥\sqrt{a^2+b^2}\)
\(⇒\frac{400}{2}≥ab\)
\(⇒ab≤200\)

Hence, the maximum value of ab is 200.
Therefore, the maximum value of AP is: \(\frac{200}{20}=10.\)