# Heron's formula

A triangle with sides a, b, and c.

In geometry, Heron's (or Hero's) formula, named after Heron of Alexandria,[1] states that the area T of a triangle whose sides have lengths a, b, and c is

$T = \sqrt{s(s-a)(s-b)(s-c)}$

where s is the semiperimeter of the triangle:

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

Heron's formula can also be written as:

$T={\ \sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)\ \over 16}\,}$
$T={\ \sqrt{2(a^2 b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4)\ \over 16}\,}$
$T=\frac{1}{4}\sqrt{(a^2 + b^2 + c^2)^2 - 2(a^4 + b^4 + c^4)}.$

Heron's formula is distinguished from other formulas for the area of a triangle, such as half the base times the height or half the modulus of a cross product of two sides, by requiring no arbitrary choice of side as base or vertex as origin.

## History

The formula is credited to Heron (or Hero) of Alexandria, and a proof can be found in his book, Metrica, written c. A.D. 60. It has been suggested that Archimedes knew the formula over two centuries earlier, and since Metrica is a collection of the mathematical knowledge available in the ancient world, it is possible that the formula predates the reference given in that work.[2]

A formula equivalent to Heron's namely:

$T=\frac1{2}\sqrt{a^2c^2-\left(\frac{a^2+c^2-b^2}{2}\right)^2}$, where $a \ge b \ge c$

was discovered by the Chinese independently of the Greeks. It was published in Shushu Jiuzhang (“Mathematical Treatise in Nine Sections”), written by Qin Jiushao and published in A.D. 1247.

## Proof

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

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

by the law of cosines. From this proof get the algebraic statement:

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

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

\begin{align} T & = \frac{1}{2} (\mbox{base}) (\mbox{altitude}) \\ & = \frac{1}{2} ab\sin \widehat C \\ & = \frac{1}{4}\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2} \\ & = \frac{1}{4}\sqrt{(2a b -(a^2 +b^2 -c^2))(2a b +(a^2 +b^2 -c^2))} \\ & = \frac{1}{4}\sqrt{(c^2 -(a -b)^2)((a +b)^2 -c^2)} \\ & = \sqrt{\frac{(c -(a -b))(c +(a -b))((a +b) -c)((a +b) +c)}{16}} \\ & = \sqrt{\frac{(b + c - a)}{2}\frac{(a + c - b)}{2}\frac{(a + b - c)}{2}\frac{(a + b + c)}{2}} \\ & = \sqrt{\frac{(a + b + c)}{2}\frac{(b + c - a)}{2}\frac{(a + c - b)}{2}\frac{(a + b - c)}{2}} \\ & = \sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}. \end{align}

The difference of two squares factorization was used in two different steps.

## Proof using the Pythagorean theorem

Triangle with altitude h cutting base c into d + (c − d).

Heron's original proof made use of cyclic quadrilaterals, while other arguments appeal to trigonometry as above, or to the incenter and one excircle of the triangle [2]. The following argument reduces Heron's formula directly to the Pythagorean theorem using only elementary means.

We wish to prove 4T2 = 4s(s - a)(s - b)(s - c). The left-hand side equals

$4 T^2 = (c h)^2 = c^2(b^2-d^2) = (c b)^2 - (c d)^2$

while the right-hand side equals

$4s(s-a)(s-b)(s-c) = [s(s-a)+(s-b)(s-c)]^2 - [s(s-a)-(s-b)(s-c)]^2$

via the identity (p + q)2 - (p - q)2 = 4pq. It therefore suffices to show

$cb=s(s-a)+(s-b)(s-c)$

and

$cd=s(s-a)-(s-b)(s-c).$

Substituting 2s = (a + b + c) into the former,

$s(s-a)+(s-b)(s-c)=\frac{1}{4}(a+b+c)(-a+b+c) + \frac{1}{4}(a-b+c)(a+b-c) = \frac{1}{4}[(b+c)^2-a^2] + \frac{1}{4}[a^2-(b-c)^2] = \frac{1}{4}[(b+c)^2 - (b-c)^2] = cb$

as desired. Similarly, the latter expression becomes

$s(s-a)-(s-b)(s-c)=\frac{1}{4}[(b+c)^2-a^2] - \frac{1}{4}[a^2-(b-c)^2] = \frac{1}{2}(b^2+c^2-a^2).$

Using the Pythagorean theorem twice, b2 = d2 + h2 and a2 = (c - d)2 + h2, which allows us to simplify the expression to

$\frac{1}{2}(b^2+c^2-a^2) = \frac{1}{2}[d^2+c^2-(c-d)^2] = cd.$

The result follows.

## Numerical stability

Heron's formula as given above is numerically unstable for triangles with a very small angle. A stable alternative [3] [4] involves arranging the lengths of the sides so that $a \ge b \ge c$ and computing

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

The brackets in the above formula are required in order to prevent numerical instability in the evaluation.

## Generalizations

Heron's formula is a special case of Brahmagupta's formula for the area of a cyclic quadrilateral. Heron's formula and Brahmagupta's formula are both special cases of Bretschneider's formula for the area of a quadrilateral. Heron's formula can be obtained from Brahmagupta's formula or Bretschneider's formula by setting one of the sides of the quadrilateral to zero.

Heron's formula is also a special case of the formula for the area of a trapezoid or trapezium based only on its sides. Heron's formula is obtained by setting the smaller parallel side to zero.

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

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

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

Another generalization of Heron's formula to pentagons and hexagons inscribed in a circle was discovered by David P. Robbins.[5]

### Heron-type formula for the volume of a tetrahedron

If U, V, W, u, v, w are lengths of edges of the tetrahedron (first three form a triangle; u opposite to U and so on), then[6]

$\text{volume} = \frac{\sqrt {\,( - a + b + c + d)\,(a - b + c + d)\,(a + b - c + d)\,(a + b + c - d)}}{192\,u\,v\,w}$

where

\begin{align} a & = \sqrt {xYZ} \\ b & = \sqrt {yZX} \\ c & = \sqrt {zXY} \\ d & = \sqrt {xyz} \\ X & = (w - U + v)\,(U + v + w) \\ x & = (U - v + w)\,(v - w + U) \\ Y & = (u - V + w)\,(V + w + u) \\ y & = (V - w + u)\,(w - u + V) \\ Z & = (v - W + u)\,(W + u + v) \\ z & = (W - u + v)\,(u - v + W). \end{align}