Heron's Formula/Proof 4

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 $\triangle ABC$.

$\ds \AA$

$=$

$\ds \dfrac {a b \sin C} 2$

$\ds $

$=$

$\ds \dfrac {a b} 2 \cdot 2 \sin \dfrac C 2 \cos \dfrac C 2$

$\ds $

$=$

$\ds a b \sqrt {\dfrac {\paren {s - a} \paren {s - b} } {a b} } \sqrt {\dfrac {s \paren {s - c} } {a b} }$

$\ds $

$=$

$\ds \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$

simplifying

This entry was named for Heron of Alexandria.