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 formula (sometimes called Hero's formula), named after Hero of Alexandria, gives the area of a triangle when the length of all three sides are known. Unlike other triangle area formulae, there is no need to calculate angles or other distances in the triangle first.

Formulation

Heron's formula states that the area of a triangle whose sides have lengths a, b, and c is

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

where s is the semi-perimeter of the triangle; that is,

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

Heron's formula can also be written as

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

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

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

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

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

Example

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

s = 1/2(a + b + c) = 1/2(4 + 13 + 15) = 16, and 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.

History

The formula is credited to Heron (or Hero) of Alexandria, and a proof can be found in his book Metrica, written around AD 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,

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

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

Proofs

There are many different 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 (in his book Metrica), 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

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

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

Algebraic proof using the Pythagorean theorem

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

The following proof is very similar to one given by Raifaizen. By the Pythagorean theorem we have b² = h² + d² and a² = h² + (c − d)² according to the figure at the right. Subtracting these yields a² − b² = 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 = b² − d². 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

Geometrical significance of s − a, s − b, and s − c. See the law of cotangents for the reasoning behind this.

From the first part of the law of cotangents proof, we have that the triangle's area is both

${\begin{aligned}A&=r{\big (}(s-a)+(s-b)+(s-c){\big )}=r^{2}\left({\frac {s-a}{r}}+{\frac {s-b}{r}}+{\frac {s-c}{r}}\right)\\&=r^{2}\left(\cot {\frac {\alpha }{2}}+\cot {\frac {\beta }{2}}+\cot {\frac {\gamma }{2}}\right)\end{aligned}}$

and A = rs, but, since the sum of the half-angles is π/2, the triple cotangent identity applies, so the first of these is

${\begin{aligned}A&=r^{2}\left(\cot {\frac {\alpha }{2}}\cot {\frac {\beta }{2}}\cot {\frac {\gamma }{2}}\right)=r^{2}\left({\frac {s-a}{r}}\cdot {\frac {s-b}{r}}\cdot {\frac {s-c}{r}}\right)\\&={\frac {(s-a)(s-b)(s-c)}{r}}.\end{aligned}}$

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 a ≥ b ≥ c 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.

Other area formulae resembling Heron's formula

Three other area formulae have the same structure as Heron's formula but are expressed in terms of different variables. First, denoting the medians from sides a, b, and c respectively as m_a, m_b, and m_c and their semi-sum 1/2(m_a + m_b + m_c) as σ, we have

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

Next, denoting the altitudes from sides a, b, and c respectively as h_a, h_b, and h_c, and denoting the semi-sum of the reciprocals of the altitudes as H = 1/2(h⁻¹
_a + h⁻¹
_b + h⁻¹
_c) we have

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

Finally, denoting the semi-sum of the angles' sines as S = 1/2(sin α + sin β + sin γ), we have

$A=D^{2}{\sqrt {S(S-\sin \alpha )(S-\sin \beta )(S-\sin \gamma )}}$

where D is the diameter of the circumcircle: D = a/sin α = b/sin β = c/sin γ.

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,

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

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

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

See also

• Shoelace formula

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
• An alternative proof of Heron's Formula without words
• Factoring Heron