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

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  • 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,
For maximum or minimum,we have
(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 $x$
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,
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.
answered Oct 1, 2013 by sreemathi.v

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