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Find the area of the triangle whose sides are given by:

$ {(b^2+c^2)}^{0.5} , {(c^2+a^2)}^{0.5}, {(a^2+b^2)}^{0.5} $

I tried it by using hero's Formula but the equation becomes too complicated and cant get solved.

Is there is any easier method by which we can get the simplified answer.

Thanks in advance.

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    @Lazar i wish there is accepted button for comments also.Thanks.Got the answer.:)2012-04-01

3 Answers 3

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Let's denote :

$x=\sqrt{b^2+c^2}$

$y=\sqrt{a^2+c^2}$

$z=\sqrt{a^2+b^2}$

and let's denote as $\alpha$ angle opposite to the side $z$ of the triangle ,then :

$A=\frac{1}{2} xy\sin \alpha$

According to Cosine rule :

$\cos\alpha =\frac{x^2+y^2-z^2}{2xy}$

Now , use Pythagorean trigonometry identity :

$\sin^2 \alpha + \cos^2 \alpha =1$

to express $\sin \alpha$ in terms of $x,y,z$ and then substitute it into formulae for area .

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As Lazar suggests we use the cosine rule which gives: $a^2+b^2=(a^2+c^2)+(b^2+c^2)-2\sqrt{(a^2+c^2)(b^2+c^2)}\cos{\alpha}$
$\therefore \cos({\alpha})=\frac{c^2}{\sqrt{(a^2+c^2)(b^2+c^2)}}$, which gives $\sin({\alpha})=\sqrt\frac{a^2b^2+(a^2+b^2)c^2}{(a^2+c^2)(b^2+c^2)}$
So, the area is $\frac{\sqrt{a^2b^2+(a^2+b^2)c^2}}{2}$

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    I've fixed that now.2012-04-01
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A good exercise would be proving that Heron's formula could be written as:

$S=\frac{1}{4}\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}$