Question

Prove that the area of a right-angle of given hypotenuse is maximum when the triangle is isosceles.

Answer 1

Let $\Delta ABC$ be a right-angled triangle:

Let $AB=h$

and $AC=x$

By Pythagoras theorem,

$AB=\sqrt{h^2-x^{2}1}$

Area $\left(\Delta ABC\right)=\frac{1}{2}\times Base\times Height$

$A=\frac{1}{2}x\sqrt{h^2-x^2}$

Differentiating wrt $X$, we get -

$\frac{dA}{dx}=\frac{1}{2}\sqrt{h^2-x^2}+\frac{x}{2}\left(\frac{(-2x)}{2\sqrt{h^2-x^2}}\right)$ [By chain rule]

$=\frac{1}{2}\left(\frac{h^2-x^2-x^2}{\sqrt{h^2-x^2}}\right)=\frac{1}{2}\left(\frac{h^2-2x^2}{\sqrt{h^2-x^2}}\right)$

To maximum the area, we put $\frac{dA}{dx}=0$

$\Rightarrow \frac{1}{2}\left(\frac{h^2-2x^2}{\sqrt{h^2-2x^2}}\right)=0$
$\Rightarrow x=\frac{h}{\sqrt{2}}$

Area is maximum if & only if $\frac{d^{2}A}{d{x}^2}$ at $x=\frac{h}{\sqrt{2}}$ is negative

So, $\frac{d^{2}A}{d{x}^2}=\frac{1}{2}\left[\frac{\sqrt{h^2-x^2}(-4x)-(h^2-2x^2)\frac{1}{2}(h^2-x^2{)}^{-1/2}(-2x)}{(h^2-x^2)}\right]$ [product rule]

$=\frac{1}{2}[\frac{-4x}{\sqrt{h^2-x^2}}+\frac{x(h^2-h^2)}{(h^2-x^2{)}^{3/2}}]$

$\frac{d^{2}A}{d{x}^2}=\frac{1}{2}[\frac{-4/\frac{h}{\sqrt{2}}}{\sqrt{h^2-\frac{h^2}{2}}}+\frac{\frac{h}{\sqrt{h}}(h^2-h^2)}{{(h^2-\frac{h^2}{2})}^{3/2}}]$

$=\frac{1}{2}[-4+0]=-2<0$

$\therefore$ area is maximum when $x=\frac{h}{\sqrt{2}}$

Base $=\sqrt{h^2-x^2}=\sqrt{h^2-\frac{h^2}{2}}=\frac{h}{\sqrt{2}}$

$\therefore \Delta ABC=$ Isosceles triangle