The maximum area of a right angled trangle with hypotenuse  $h$  is

Let base $= b\Rightarrow$ Altitude (or perpendicular)

$=\sqrt{h^2-b^2}$

Area, $A=\frac{1}{2}\times \text{base}\times \text{altitude}=\frac{1}{2}\times b\times \sqrt{h^2-b^2}$

Put $\frac{dA}{db}=0,\Rightarrow b=\frac{h}{\sqrt{2}}$ Maximum area

$=\frac{1}{2}\times \frac{h}{\sqrt{2}}\times \sqrt{h^2-\frac{h^2}{2}}=\frac{h^2}{4}$