Question

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

• A. $\frac{h^2}{2\sqrt{2}}$
• B. $\frac{h^2}{2}$
• C. $\frac{h^2}{\sqrt{2}}$
• D. $\frac{h^2}{4}$

Answer 1

The correct option is D

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

Explanation for the correct option:

Maximize the area of triangle:

Given that the hypotenuse of the right-angled triangle is $h$.

Let base $=b$

Then altitude $=\sqrt{(h^2-b^2)}$

Area of triangle $=\frac{1}{2}$ base $\times$ altitude $=\frac{1}{2}\times b\times \sqrt{(h^2-b^2)}$

We know that for maximum area, $\frac{dA}{db}=0$

$$\begin{gathered} \Rightarrow \frac{dA}{db}=0 \\ \Rightarrow \frac{1}{2}\left[\sqrt{(h^2-b^2)}+b\times -\frac{2b}{2\sqrt{(h^2-b^2)}}\right]=0 \\ \Rightarrow \frac{1}{2}\left[\sqrt{(h^2-b^2)}-\frac{b^2}{\sqrt{(h^2-b^2)}}\right]=0 \\ \Rightarrow \frac{1}{2}\left[\frac{h^2-2b^2}{\sqrt{(h^2-b^2)}}\right]=0 \\ \Rightarrow \frac{1}{2}\left[\frac{h^2-2b^2}{\sqrt{(h^2-b^2)}}\right]=0 \\ \Rightarrow [h^2-2b^2]=0 \\ \Rightarrow h^2=2b^2 \\ \Rightarrow b=\frac{h}{\sqrt{2}} \end{gathered}$$

So, the area of the triangle when $b=\frac{h}{\sqrt{2}}$ is

$$\begin{gathered} A=\frac{1}{2}\times b\times \sqrt{h^2-b^2} \\ =\frac{1}{2}\times \frac{h}{\sqrt{2}}\times \sqrt{h^2-\frac{h^2}{2}} \\ =\frac{h^2}{4} \end{gathered}$$

Hence, option(C) is the correct answer.