# Area of the triangle?

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.

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Are you familiar with trigonometry? Use cosine rule to find one angle and then use $P=\frac12ab\sin\alpha$ to find area. – Lazar Ljubenović Apr 1 '12 at 8:03
What is a,b and c? – Quixotic Apr 1 '12 at 8:05
@Lazar i wish there is accepted button for comments also.Thanks.Got the answer.:) – vikiiii Apr 1 '12 at 8:12

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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 How did you get term $\frac{ab}{2}$ in final formula ? – pedja Apr 1 '12 at 8:40 I've fixed that now. – Sidharth Iyer Apr 1 '12 at 8:53

A good exercise would be proving that Heron's formula could be written as:

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

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