Heron s formula

In geometry, Heron's formula (sometimes given as Hero's formula) states that the area of a triangle whose sides have lengths a, b, c is given by

$\mathrm{area} = \sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\,$

where

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

(see also square root). Heron's formula can also be written in the form

$\mathrm{area}={\ \sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)\,}\ \over 4}.\,$

Contents

History

The formula is credited to Heron of Alexandria in the 1st century AD, and a proof can be found in his book Metrica. It is now believed that Archimedes already knew the formula, and it is of course possible that it has been known long before.

Proof

A modern proof, which uses algebra and trigonometry and is quite unlike the one provided by Heron, follows. Let a, b, c be the sides of the triangle and A, B, C the angles opposite those sides. We have

$\cos(C) = \frac{a^2+b^2-c^2}{2ab}$

by the law of cosines. From this we get with some algebra

$\sin(C) = \sqrt{1-\cos^2(C)} = \frac{\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2 }}{2ab}$ .

The altitude of the triangle on base a has length bsin(C), and it follows

$S = \frac{1}{2} (\mbox{base}) (\mbox{altitude})$

$\qquad = \frac{1}{2} ab\sin(C)$

$\qquad = \frac{1}{4}\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2}$

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

Here the simple algebra in the last step was omitted.

Generalizations

The formula is in fact a special case of Brahmagupta's formula for the area of a cyclic quadrilateral; both of which are special cases of Bretschneider's formula for the area of a quadrilateral.

Expressing Heron's formula with a determinant in terms of the squares of the distances between the three given vertices,

$S = \frac{1}{4} \sqrt{ \begin{vmatrix} 0 & a^2 & b^2 & 1 \\ a^2 & 0 & c^2 & 1 \\ b^2 & c^2 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{vmatrix} }$

illustrates its similarity to Tartaglia's formula for the volume of a four-simplex.