# 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.

Thanks in advance.

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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

## 3 Answers

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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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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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