The maximum area of a right angled triangle with hypotenuse h is

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$\frac{h^{2}}{\sqrt{2}}$

$\frac{h^{2}}{2}$

$\frac{h^{2}}{2 \sqrt{2}}$

$\frac{h^{2}}{4}$

answer is D.

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Detailed Solution

$x^{2} + y^{2} = h^{2} A . M \geq G . M \Rightarrow \frac{x^{2} + y^{2}}{2} \geq \sqrt{x^{2} y^{2}} \Rightarrow x y \leq \frac{h^{2}}{2} \Rightarrow \frac{1}{2} x y \leq \frac{h^{2}}{4} m a x a r e a = \frac{h^{2}}{4}$

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