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

Answer: (A)

10

Answer 2

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