Heron's Formula

Theorem
Let $\triangle ABC$ be a triangle with sides $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.

Then the area $\AA$ of $\triangle ABC$ is given by:
 * $\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$

where $s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$.