Heron's formula

This article is about calculating the area of a triangle. For calculating a square root, see Heron's method.

A triangle with sides a, b, and c.

In geometry, Heron's (or Hero's) formula, named after Heron of Alexandria, 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:

$T={\ {\sqrt {(a+b+c)(-a+b+c)(a-b+c)(a+b-c)\ \over 16}}\,}$

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

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

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.

A formula equivalent to Heron's namely:

$T={\frac {1}{2}}{\sqrt {a^{2}c^{2}-\left({\frac {a^{2}+c^{2}-b^{2}}{2}}\right)^{2}}}$, where $a\geq b\geq c$

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

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

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

$4T^{2}=(ch)^{2}=c^{2}(b^{2}-d^{2})=(cb)^{2}-(cd)^{2}$

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 involves arranging the lengths of the sides so that $a\geq b\geq c$ 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 of the area of a trapezoid 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.

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

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

See also

• Heronian triangle

References

• Heath, Thomas L. (1921). A History of Greek Mathematics (Vol II). Oxford University Press. pp. 321–323.

External links

• A Proof of the Pythagorean Theorem From Heron's Formula at cut-the-knot
• Interactive applet and area calculator using Heron's Formula
• J.H. Conway discussion on Heron's Formula
• "Heron's Formula and Brahmagupta's Generalization". MathPages.com.
• A Geometric Proof of Heron's Formula