FormulaProved

Heron's formula

Statement

The area $S$ of a triangle with side lengths $a$ , $b$ , $c$ and semiperimeter $s = \dfrac{a + b + c}{2}$ is given by $S = \sqrt{s(s - a)(s - b)(s - c)}$ .

Why is it true?

Three side lengths $a$ , $b$ , $c$ determine a rigid triangle up to congruence, so its area $S$ must be completely determined by $a$ , $b$ , and $c$ without needing to measure an altitude. Each factor $s - a = \dfrac{b + c - a}{2}$ measures how much slack the triangle inequality $b + c > a$ has: as one side approaches the sum of the other two, the corresponding factor $s - a$ shrinks to $0$ and the triangle flattens into a degenerate segment of area $0$ .

Proof sketch

Combine the area formula $S = \tfrac{1}{2}ab\sin\gamma$ with the law of cosines $\cos\gamma = \dfrac{a^2 + b^2 - c^2}{2ab}$ : squaring and using $\sin^2\gamma = (1 - \cos\gamma)(1 + \cos\gamma)$ gives $16S^2 = 4a^2b^2 - (a^2 + b^2 - c^2)^2 = (c^2 - (a - b)^2)((a + b)^2 - c^2)$ . Factoring both differences of squares yields $16S^2 = (a + b + c)(b + c - a)(a + c - b)(a + b - c) = 16s(s - a)(s - b)(s - c)$ , and taking square roots gives $S = \sqrt{s(s - a)(s - b)(s - c)}$ .

Topics that use this theorem

Triangles and their special lines

Related theorems

Law of cosines Pythagorean theorem

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

Thomas L. Heath (1921). A History of Greek Mathematics, Vol. 2: From Aristarchus to Diophantus
H. S. M. Coxeter (1969). Introduction to Geometry · DOI:10.1088/0031-9112/13/7/027