Formule van Heron

Definieer ${\displaystyle s={\frac {a+b+c}{2}}}$

De oppervlakte van ${\displaystyle \triangle ABC}$ is gegeven door:

${\displaystyle \mathrm {oppervlakte} ={\frac {1}{2}}ab\sin \gamma }$

Volgens de cosinusregel is:

${\displaystyle c^{2}=a^{2}+b^{2}-2ab\cos \gamma }$

zodat:

${\displaystyle \cos \gamma ={\frac {a^{2}+b^{2}-c^{2}}{2ab}}}$

Substitutie geeft:

${\displaystyle \mathrm {oppervlakte} =}$

${\displaystyle ={\frac {1}{2}}ab\sin \gamma }$

${\displaystyle ={\frac {1}{2}}ab{\sqrt {1-\cos ^{2}\gamma }}}$

${\displaystyle ={\frac {1}{2}}ab{\sqrt {(1-\cos \gamma )(1+\cos \gamma )}}}$

${\displaystyle ={\frac {1}{2}}ab{\sqrt {\left(1-{\frac {c^{2}-a^{2}-b^{2}}{2ab}}\right)\left(1+{\frac {c^{2}-a^{2}-b^{2}}{2ab}}\right)}}}$

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

${\displaystyle ={\sqrt {s\left(s-a\right)\left(s-b\right)\left(s-c\right)}}}$