# Heron's formula

A triangle with sides a, b, and c

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,[1]

${\displaystyle A={\sqrt {s(s-a)(s-b)(s-c)}}.}$

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

${\displaystyle s={\frac {a+b+c}{2}}={\frac {4+13+15}{2}}=16}$

and so the area is

{\displaystyle {\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,

{\displaystyle {\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,

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

## History

The formula is credited to Heron (or Hero) of Alexandria (fl. 60 AD),[2] and a proof can be found in his book Metrica. Mathematical historian Thomas Heath suggested that Archimedes knew the formula over two centuries earlier,[3] 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.[4]

A formula equivalent to Heron's, namely,

${\displaystyle A={\frac {1}{2}}{\sqrt {a^{2}c^{2}-\left({\frac {a^{2}+c^{2}-b^{2}}{2}}\right)^{2}}}}$

was discovered by the Chinese. It was published in Mathematical Treatise in Nine Sections (Qin Jiushao, 1247).[5]

## 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,[6] or as a special case of De Gua's theorem (for the particular case of acute triangles).[7]

### Trigonometric proof using the law of cosines

A modern proof, which uses algebra and is quite different from the one provided by Heron, follows.[8] 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

${\displaystyle \cos \gamma ={\frac {a^{2}+b^{2}-c^{2}}{2ab}}}$

From this proof, we get the algebraic statement that

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

The altitude of the triangle on base a has length b sin γ, and it follows

{\displaystyle {\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

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

The following proof is very similar to one given by Raifaizen.[9] 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:

${\displaystyle d={\frac {-a^{2}+b^{2}+c^{2}}{2c}}.}$

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

{\displaystyle {\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:

{\displaystyle {\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

Geometrical significance of sa, sb, and sc. See the law of cotangents for the reasoning behind this.

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

${\displaystyle A={\tfrac {1}{2}}ar+{\tfrac {1}{2}}br+{\tfrac {1}{2}}cr=rs,}$

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)

{\displaystyle {\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

${\displaystyle A^{2}=s(s-a)(s-b)(s-c),}$

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[10][11] involves arranging the lengths of the sides so that abc and computing

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

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 ${\displaystyle \sigma ={\tfrac {1}{2}}(m_{a}+m_{b}+m_{c}),}$ then[12]

${\displaystyle A={\frac {4}{3}}{\sqrt {\sigma (\sigma -m_{a})(\sigma -m_{b})(\sigma -m_{c})}}.}$

Next, if ha, hb, and hc are the altitudes from sides a, b, and c respectively, and semi-sum of their reciprocals is ${\displaystyle H={\tfrac {1}{2}}{\bigl (}h_{a}^{-1}+h_{b}^{-1}+h_{c}^{-1}{\bigr )},}$ then[13]

${\displaystyle A^{-1}=4{\sqrt {H{\bigl (}H-h_{a}^{-1}{\bigr )}{\bigl (}H-h_{b}^{-1}{\bigr )}{\bigl (}H-h_{c}^{-1}{\bigr )}}}.}$

Finally, if α, β, and γ are the three angle measures of the triangle, and the semi-sum of their sines is ${\displaystyle S={\tfrac {1}{2}}(\sin \alpha +\sin \beta +\sin \gamma ),}$ then[14][15]

{\displaystyle {\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

Cyclic Quadrilateral

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

${\displaystyle K={\sqrt {(s-a)(s-b)(s-c)(s-d)}}}$

where s, the semiperimeter, is defined to be

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

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,

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

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.[16]

### 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[17]

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

where

{\displaystyle {\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. [18] For a triangle in the sphere with side lengths ${\displaystyle a,b,c}$, half perimeter ${\displaystyle s=(a+b+c)/2}$ and area ${\displaystyle S}$ such a formula is

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

## References

1. ^ Kendig, Keith (2000). "Is a 2000-year-old formula still keeping some secrets?". The American Mathematical Monthly. 107 (5): 402–415. doi:10.1080/00029890.2000.12005213. JSTOR 2695295. MR 1763392. S2CID 1214184.
2. ^ Id, Yusuf; Kennedy, E. S. (1969). "A medieval proof of Heron's formula". The Mathematics Teacher. 62 (7): 585–587. doi:10.5951/MT.62.7.0585. JSTOR 27958225. MR 0256819.
3. ^ Heath, Thomas L. (1921). A History of Greek Mathematics. Vol. II. Oxford University Press. pp. 321–323.
4. ^
5. ^ 秦, 九韶 (1773). "卷三上, 三斜求积". 數學九章 (四庫全書本) (in Chinese).
6. ^ "Personal email communication between mathematicians John Conway and Peter Doyle". 15 December 1997. Retrieved 25 September 2020.
7. ^ Lévy-Leblond, Jean-Marc (2020-09-14). "A Symmetric 3D Proof of Heron's Formula". The Mathematical Intelligencer. 43 (2): 37–39. doi:10.1007/s00283-020-09996-8. ISSN 0343-6993.
8. ^ Niven, Ivan (1981). Maxima and Minima Without Calculus. The Mathematical Association of America. pp. 7–8.
9. ^ Raifaizen, Claude H. (1971). "A Simpler Proof of Heron's Formula". Mathematics Magazine. 44 (1): 27–28. doi:10.1080/0025570X.1971.11976093.
10. ^ Sterbenz, Pat H. (1974-05-01). Floating-Point Computation. Prentice-Hall Series in Automatic Computation (1st ed.). Englewood Cliffs, New Jersey, USA: Prentice Hall. ISBN 0-13-322495-3.
11. ^ William M. Kahan (24 March 2000). "Miscalculating Area and Angles of a Needle-like Triangle" (PDF).
12. ^ Benyi, Arpad, "A Heron-type formula for the triangle," Mathematical Gazette 87, July 2003, 324–326.
13. ^ Mitchell, Douglas W., "A Heron-type formula for the reciprocal area of a triangle," Mathematical Gazette 89, November 2005, 494.
14. ^ Mitchell, Douglas W. (2009). "A Heron-type area formula in terms of sines". Mathematical Gazette. 93: 108–109. doi:10.1017/S002555720018430X. S2CID 132042882.
15. ^ Kocik, Jerzy; Solecki, Andrzej (2009). "Disentangling a triangle" (PDF). American Mathematical Monthly. 116 (3): 228–237. doi:10.1080/00029890.2009.11920932. S2CID 28155804.
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17. ^ W. Kahan, "What has the Volume of a Tetrahedron to do with Computer Programming Languages?", [1], pp. 16–17.
18. ^ Page 66 in Alekseevskij, D. V.; Vinberg, E. B.; Solodovnikov, A. S. (1993), "Geometry of spaces of constant curvature", in Gamkrelidze, R. V.; Vinberg, E. B. (eds.), Geometry. II: Spaces of constant curvature, Encycl. Math. Sci., vol. 29, Springer-Verlag, pp. 1–138, ISBN 1-56085-072-8