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

Let base $=b$=b

Altitude (or perpendicular) $=\sqrt{h^2-b^2}$=h2−b2​ Area, $\mathrm{A}=\frac{1}{2} \times$A=21​× base $\times$× altitude $=\frac{1}{2} \times b \times \sqrt{h^2-b^2}$=21​×b×h2−b2​ $\Rightarrow \frac{d \mathrm{~A}}{d b}=\frac{1}{2}\left[\sqrt{h^2-b^2}+b \cdot \frac{-2 b}{2 \sqrt{h^2-b^2}}\right]$⇒dbd A​=21​[h2−b2​+b⋅2h2−b2​−2b​] $=\frac{1}{2}\left[\frac{h^2-2 b^2}{\sqrt{h^2-b^2}}\right]$=21​[h2−b2​h2−2b2​] Put $\frac{d \mathrm{~A}}{d b}=0, \Rightarrow b=\frac{h}{\sqrt{2}}$dbd A​=0,⇒b=2​h​ Maximumarea $=\frac{1}{2} \times \frac{h}{\sqrt{2}} \times \sqrt{h^2-\frac{h^2}{2}}=\frac{h^2}{4}$=21​×2​h​×h2−2h2​​=4h2​