Browse Questions

# Prove that the area of the right-angled triangle of given hypotenuse is maximum when the triangle is isosceles.

Toolbox:
• Area of the triangle $A=\large\frac{1}{2}$$\times\; x\sqrt{h^2-x^2} Step 1: Let h be the hypotenuse of the right angled triangle,and let x be its altitude. \therefore Base of the triangle =\sqrt{h^2-x^2} The area of the triangle is A=\large\frac{1}{2}$$\times\;\sqrt{h^2-x^2}$
Step 2:
Now differentiating w.r.t $x$ we get,
$\large\frac{dA}{dx}=\frac{1}{2}$$\big[1.\sqrt{h^2-x^2}+\large\frac{x}{2}$$(h^2-x^2)^{-\Large\frac{1}{2}}(-2x)\big]$
$\qquad=\large\frac{1}{2}$$\big[\sqrt{h^2-x^2}-\large\frac{x^2}{h^2-x^2}\big] \qquad=\large\frac{1}{2}$$\big[\large\frac{h^2-2x^2}{\sqrt{h^2-x^2}}\big]$
For maximum or minimum,we have
$\large\frac{dA}{dx}$$=0 (i.e) \large\frac{1}{2}\big[\large\frac{h^2-2x^2}{\sqrt{h^2-x^2}}\big]$$=0$
$\Rightarrow h^2=2x^2$
$\therefore x=\large\frac{h}{\sqrt{2}}$
Step 3:
Again differentiating w.r.t $x$
$\large\frac{d^2A}{dx^2}=\frac{1}{2}$$\big[-4x\large\frac{1}{\sqrt{h^2-x^2}}$$+(h^2-2x^2)(-\large\frac{-1}{2})$$(h^2-x^2)^{-\Large\frac{3}{2}}(-2x)\big] \qquad=\large\frac{1}{2}\big[\large\frac{-4x}{\sqrt{h^2-x^2}}+\frac{x(h^2-2x^2)}{(h^2-x^2)^{\Large\frac{3}{2}}}\big] When x=\large\frac{h}{\sqrt 2} \large\frac{d^2A}{dx^2}=\large\frac{1}{2}\big[\large\frac{-4(h/\sqrt 2)}{\sqrt{h^2-h^2/2}}+\frac{x(h^2-2(h/\sqrt 2)^2)}{(h^2-(h/\sqrt 2)^2)^{\Large\frac{3}{2}}}\big] Step 4: On simplifying we get, \large\frac{d^2A}{dx^2}$$=-2<0$
Thus $A$ is maximum when $x=\large\frac{h}{\sqrt 2}$
$\therefore$ Base =$\sqrt{h^2-\large\frac{h^2}{2}}$
$\Rightarrow \large\frac{h}{\sqrt{2}}$
$\therefore A$ is maximum when the triangle is isosceles.