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

Question

Aicharya · Accepted Answer

Let base $=b$

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

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

Solution