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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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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,
$\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.
answered Oct 1, 2013 by sreemathi.v
 

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