Question

The area of a right-angled triangle of given hypotenuse is maximum when the triangle is

• A. scalene
• B. equilateral
• C. isosceles
• D. None of these

Answer 1

Answer: (C)

Let $h$ be the hypotenuse of the right-angled triangle, and
let $x$ be its altitude. Then,
Base of the triangle $=\sqrt{h^2-x^2}$
Let $A$ be the area of the triangle.
then,

$A=\frac{1}{2}x\sqrt{h^2-x^2}$
$\Rightarrow \frac{dA}{dx}=\frac{1}{2}\left\{1\cdot \sqrt{h^2-x^2}+x\frac{1}{2}{\left(h^2-x^2\right)}^{-\frac{1}{2}}\frac{d}{dx}\left(h^2-x^2\right)\right\}$
$\Rightarrow \frac{dA}{dx}=\frac{1}{2}\left\{\sqrt{h^2-x^2}-\frac{x^2}{\sqrt{h^2-x^2}}\right\}$
$=\frac{1}{2}\left\{\frac{h^2-2x^2}{\sqrt{h^2-x^2}}\right\}$
For maximum or minimum, we have
$\frac{dA}{dx}=0\Rightarrow \frac{1}{2}\left\{\frac{h^2-2x}{\sqrt{h^2-x^2}}\right\}=0$
$\Rightarrow h^2=2x^2$
$\Rightarrow x=\frac{h}{\sqrt{2}}$
Now,
$\Rightarrow \frac{d^{2}A}{d{x}^2}=\frac{1}{2}\left\{\left(-4x\right)\frac{1}{\sqrt{h^2-x^2}}+\left(h^2-2x^2\right)\left(-\frac{1}{2}\right)\times {\left(h^2-x^2\right)}^{-3/2}\frac{d}{dx}\left(h^2-x^2\right)\right\}$
$\Rightarrow \frac{d^{2}A}{d{x}^2}=\frac{1}{2}\left\{\frac{-4x}{\sqrt{h^2-x^2}}+\frac{x\left(h^2-2x\right)}{{\left(h^2-x^2\right)}^{3/2}}\right\}$
$\Rightarrow {\left(\frac{d^{2}A}{d{x}^2}\right)}_{x=\frac{h}{\sqrt{2}}}=-2<0$.
$\left[\text{neglecting}x=\frac{-h}{\sqrt{2}}\because \frac{d^{2}A}{d{x}^2}{|}_{x=\frac{-h}{\sqrt{2}}}>0\right]$
Thus, $A$ is maximum when $x=\frac{h}{\sqrt{2}}$.
$\therefore$ Base $=\sqrt{h^2-\left(\frac{h^2}{2}\right)}=\frac{h}{\sqrt{2}}$
Hence, $A$ is maximum when the triangle is isosceles.