# proof of Heron’s formula

Let $\alpha$ be the angle between the sides $b$ and $c$, then we get from the cosines law:

 $\cos\alpha=\frac{b^{2}+c^{2}-a^{2}}{2bc}.$

Using the equation

 $\sin\alpha=\sqrt{1-\cos^{2}\alpha}$

we get:

 $\sin\alpha=\frac{\sqrt{-a^{4}-b^{4}-c^{4}+2b^{2}c^{2}+2a^{2}b^{2}+2a^{2}c^{2}}% }{2bc}.$

Now we know:

 $\Delta=\frac{1}{2}bc\sin\alpha.$

So we get:

 $\displaystyle\Delta$ $\displaystyle=$ $\displaystyle\frac{1}{4}\sqrt{-a^{4}-b^{4}-c^{4}+2b^{2}c^{2}+2a^{2}b^{2}+2a^{2% }c^{2}}$ $\displaystyle=$ $\displaystyle\frac{1}{4}\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}$ $\displaystyle=$ $\displaystyle\sqrt{s(s-a)(s-b)(s-c)}.$

This is Heron’s formula. $\Box$

Title proof of Heron’s formula ProofOfHeronsFormula 2013-03-22 12:41:38 2013-03-22 12:41:38 mathwizard (128) mathwizard (128) 5 mathwizard (128) Proof msc 51-00