# Heron's formula

In geometry, Heron's formula (or Hero's formula) gives the area A of a triangle in terms of the three side lengths a, b, c. If ${\textstyle s={\tfrac {1}{2}}(a+b+c)}$ is the semiperimeter of the triangle, the area is,

$A={\sqrt {s(s-a)(s-b)(s-c)}}.$ It is named after first-century engineer Heron of Alexandria (or Hero) who proved it in his work Metrica, though it was probably known centuries earlier.

## Example

Let ABC be the triangle with sides a = 4, b = 13 and c = 15. This triangle’s semiperimeter is

$s={\frac {a+b+c}{2}}={\frac {4+13+15}{2}}=16$ and so the area is

{\begin{aligned}A&={\sqrt {s\left(s-a\right)\left(s-b\right)\left(s-c\right)}}={\sqrt {16\cdot (16-4)\cdot (16-13)\cdot (16-15)}}\\&={\sqrt {16\cdot 12\cdot 3\cdot 1}}={\sqrt {576}}=24.\end{aligned}} In this example, the side lengths and area are integers, making it a Heronian triangle. However, Heron's formula works equally well in cases where one or more of the side lengths are not integers.

## Alternate expressions

Heron's formula can also be written in terms of just the side lengths instead of using the semiperimeter, in several ways,

{\begin{aligned}A&={\tfrac {1}{4}}{\sqrt {(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}\\[6mu]&={\tfrac {1}{4}}{\sqrt {2(a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2})-(a^{4}+b^{4}+c^{4})}}\\[6mu]&={\tfrac {1}{4}}{\sqrt {(a^{2}+b^{2}+c^{2})^{2}-2(a^{4}+b^{4}+c^{4})}}\\[6mu]&={\tfrac {1}{4}}{\sqrt {4(a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2})-(a^{2}+b^{2}+c^{2})^{2}}}\\[6mu]&={\tfrac {1}{4}}{\sqrt {4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}}.\end{aligned}} After expansion, the expression under the square root is a quadratic polynomial of the squared side lengths a2, b2, c2.

The same relation can be expressed using the Cayley–Menger determinant,

$-16A^{2}={\begin{vmatrix}0&a^{2}&b^{2}&1\\a^{2}&0&c^{2}&1\\b^{2}&c^{2}&0&1\\1&1&1&0\end{vmatrix}}.$ ## History

The formula is credited to Heron (or Hero) of Alexandria (fl. 60 AD), and a proof can be found in his book Metrica. Mathematical historian Thomas Heath 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.

A formula equivalent to Heron's, namely,

$A={\frac {1}{2}}{\sqrt {a^{2}c^{2}-\left({\frac {a^{2}+c^{2}-b^{2}}{2}}\right)^{2}}}$ was discovered by the Chinese. It was published in Mathematical Treatise in Nine Sections (Qin Jiushao, 1247).

## Proofs

There are many ways to prove Heron's formula, for example using trigonometry as below, or the incenter and one excircle of the triangle, or as a special case of De Gua's theorem (for the particular case of acute triangles).

### Trigonometric proof using the law of cosines

A modern proof, which uses algebra and is quite different from the one provided by Heron, follows. Let a, b, c be the sides of the triangle and α, β, γ the angles opposite those sides. Applying the law of cosines we get

$\cos \gamma ={\frac {a^{2}+b^{2}-c^{2}}{2ab}}$ From this proof, we get the algebraic statement that

$\sin \gamma ={\sqrt {1-\cos ^{2}\gamma }}={\frac {\sqrt {4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}}{2ab}}.$ The altitude of the triangle on base a has length b sin γ, and it follows

{\begin{aligned}A&={\tfrac {1}{2}}({\mbox{base}})({\mbox{altitude}})\\[6mu]&={\tfrac {1}{2}}ab\sin \gamma \\[6mu]&={\frac {ab}{4ab}}{\sqrt {4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}}\\[6mu]&={\tfrac {1}{4}}{\sqrt {-a^{4}-b^{4}-c^{4}+2a^{2}b^{2}+2a^{2}c^{2}+2b^{2}c^{2}}}\\[6mu]&={\tfrac {1}{4}}{\sqrt {(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}\\[6mu]&={\sqrt {\left({\frac {a+b+c}{2}}\right)\left({\frac {-a+b+c}{2}}\right)\left({\frac {a-b+c}{2}}\right)\left({\frac {a+b-c}{2}}\right)}}\\[6mu]&={\sqrt {s(s-a)(s-b)(s-c)}}.\end{aligned}} ### Algebraic proof using the Pythagorean theorem

The following proof is very similar to one given by Raifaizen. By the Pythagorean theorem we have b2 = h2 + d2 and a2 = h2 + (cd)2 according to the figure at the right. Subtracting these yields a2b2 = c2 − 2cd. This equation allows us to express d in terms of the sides of the triangle:

$d={\frac {-a^{2}+b^{2}+c^{2}}{2c}}.$ For the height of the triangle we have that h2 = b2d2. By replacing d with the formula given above and applying the difference of squares identity we get

{\begin{aligned}h^{2}&=b^{2}-\left({\frac {-a^{2}+b^{2}+c^{2}}{2c}}\right)^{2}\\&={\frac {(2bc-a^{2}+b^{2}+c^{2})(2bc+a^{2}-b^{2}-c^{2})}{4c^{2}}}\\&={\frac {{\big (}(b+c)^{2}-a^{2}{\big )}{\big (}a^{2}-(b-c)^{2}{\big )}}{4c^{2}}}\\&={\frac {(b+c-a)(b+c+a)(a+b-c)(a-b+c)}{4c^{2}}}\\&={\frac {2(s-a)\cdot 2s\cdot 2(s-c)\cdot 2(s-b)}{4c^{2}}}\\&={\frac {4s(s-a)(s-b)(s-c)}{c^{2}}}.\end{aligned}} We now apply this result to the formula that calculates the area of a triangle from its height:

{\begin{aligned}A&={\frac {ch}{2}}\\&={\sqrt {{\frac {c^{2}}{4}}\cdot {\frac {4s(s-a)(s-b)(s-c)}{c^{2}}}}}\\&={\sqrt {s(s-a)(s-b)(s-c)}}.\end{aligned}} ### Trigonometric proof using the law of cotangents

If r is the radius of the incircle of the triangle, then the triangle can be broken into three triangles of equal altitude r and bases a, b, and c. Their combined area is

$A={\tfrac {1}{2}}ar+{\tfrac {1}{2}}br+{\tfrac {1}{2}}cr=rs,$ where ${\textstyle s={\tfrac {1}{2}}(a+b+c)}$ is the semiperimeter.

The triangle can alternately be broken into six triangles (in congruent pairs) of altitude r and bases sa, sb, and sc, of combined area (see law of cotangents)

{\begin{aligned}A&=r(s-a)+r(s-b)+r(s-c)\\[2mu]&=r^{2}\left({\frac {s-a}{r}}+{\frac {s-b}{r}}+{\frac {s-c}{r}}\right)\\[2mu]&=r^{2}\left(\cot {\frac {\alpha }{2}}+\cot {\frac {\beta }{2}}+\cot {\frac {\gamma }{2}}\right)\\[3mu]&=r^{2}\left(\cot {\frac {\alpha }{2}}\cot {\frac {\beta }{2}}\cot {\frac {\gamma }{2}}\right)\\[3mu]&=r^{2}\left({\frac {s-a}{r}}\cdot {\frac {s-b}{r}}\cdot {\frac {s-c}{r}}\right)\\[3mu]&={\frac {(s-a)(s-b)(s-c)}{r}}.\end{aligned}} The middle step above is ${\textstyle \cot {\tfrac {\alpha }{2}}+\cot {\tfrac {\beta }{2}}+\cot {\tfrac {\gamma }{2}}=\cot {\tfrac {\alpha }{2}}\cot {\tfrac {\beta }{2}}\cot {\tfrac {\gamma }{2}},}$ the triple cotangent identity, which applies because the sum of half-angles is ${\textstyle {\tfrac {\alpha }{2}}+{\tfrac {\beta }{2}}+{\tfrac {\gamma }{2}}={\tfrac {\pi }{2}}.}$ Combining the two, we get

$A^{2}=s(s-a)(s-b)(s-c),$ from which the result follows.

## Numerical stability

Heron's formula as given above is numerically unstable for triangles with a very small angle when using floating-point arithmetic. A stable alternative involves arranging the lengths of the sides so that abc and computing

$A={\frac {1}{4}}{\sqrt {{\big (}a+(b+c){\big )}{\big (}c-(a-b){\big )}{\big (}c+(a-b){\big )}{\big (}a+(b-c){\big )}}}.$ The brackets in the above formula are required in order to prevent numerical instability in the evaluation.

## Similar triangle-area formulae

Three other formulae for the area of a general triangle have a similar structure as Heron's formula, expressed in terms of different variables.

First, if ma, mb, and mc are the medians from sides a, b, and c respectively, and their semi-sum is $\sigma ={\tfrac {1}{2}}(m_{a}+m_{b}+m_{c}),$ then

$A={\frac {4}{3}}{\sqrt {\sigma (\sigma -m_{a})(\sigma -m_{b})(\sigma -m_{c})}}.$ Next, if ha, hb, and hc are the altitudes from sides a, b, and c respectively, and semi-sum of their reciprocals is $H={\tfrac {1}{2}}{\bigl (}h_{a}^{-1}+h_{b}^{-1}+h_{c}^{-1}{\bigr )},$ then

$A^{-1}=4{\sqrt {H{\bigl (}H-h_{a}^{-1}{\bigr )}{\bigl (}H-h_{b}^{-1}{\bigr )}{\bigl (}H-h_{c}^{-1}{\bigr )}}}.$ Finally, if α, β, and γ are the three angle measures of the triangle, and the semi-sum of their sines is $S={\tfrac {1}{2}}(\sin \alpha +\sin \beta +\sin \gamma ),$ then

{\begin{aligned}A&=D^{2}{\sqrt {S(S-\sin \alpha )(S-\sin \beta )(S-\sin \gamma )}}\\[5mu]&={\tfrac {1}{2}}D^{2}\sin \alpha \,\sin \beta \,\sin \gamma ,\end{aligned}} where D is the diameter of the circumcircle, ${\textstyle D={\frac {a}{\sin \alpha }}={\frac {b}{\sin \beta }}={\frac {c}{\sin \gamma }}.}$ This last formula coincides with the standard Heron formula when the circumcircle has unit diameter.

## 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.

Brahmagupta's formula gives the area K of a cyclic quadrilateral whose sides have lengths a, b, c, d as

$K={\sqrt {(s-a)(s-b)(s-c)(s-d)}}$ where s, the semiperimeter, is defined to be

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

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

### 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

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

{\begin{aligned}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{aligned}} ### Heron formulae in non-Euclidean geometries

There are also formulae for the area of a triangle in terms of its side lengths for triangles in the sphere or the hyperbolic plane.  For a triangle in the sphere with side lengths $a,b,c$ , half perimeter $s=(a+b+c)/2$ and area $S$ such a formula is

$\tan ^{2}{\frac {S}{4}}=\tan {\frac {s}{2}}\tan {\frac {s-a}{2}}\tan {\frac {s-b}{2}}\tan {\frac {s-c}{2}}$ while for the hyperbolic plane we have
$\tan ^{2}{\frac {S}{4}}=\tanh {\frac {s}{2}}\tanh {\frac {s-a}{2}}\tanh {\frac {s-b}{2}}\tanh {\frac {s-c}{2}}.$ 