Heron's Formula/Proof 2

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 $\triangle ABC$.

Proof

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

From Brahmagupta's Formula, the area 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

• 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.1$: The Pythagorean Theorem: Appendix: The Formulas of Heron and Brahmagupta
• 1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): The Area Enclosed Against The Seashore: $31$