Heron's Formula/Proof 2

From ProofWiki
Jump to navigation Jump to search

Theorem

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


Then the area $A$ of $\triangle ABC$ is given by:

$A = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$

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


Proof

A triangle can be considered as a cyclic quadrilateral one of whose sides has degenerated to zero.

From Brahmagupta's Formula, the perimeter of a cyclic quadrilateral is given by:

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

where $s$ is the semiperimeter:

$s = \dfrac {a + b + c + d} 2$

The result follows by letting $d$ tend to zero.

$\blacksquare$


Source of Name

This entry was named for Heron of Alexandria.


Sources