Heron's formula explained

In geometry, Heron's formula (or Hero's formula) gives the area of a triangle in terms of the three side lengths Letting be the semiperimeter of the triangle,

s = \tfrac 12 (a + b + c), the area is^[1] 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 be the triangle with sides

a = 4 , b = 13 , and c = 15 . This triangle's semiperimeter is s = \tfrac 12 (a + b + c) = {} \tfrac 12 ( 4+13+15 ) = 16 and so the area is\begin{align} A & = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{ 16 ⋅ ( 16-4 ) ⋅ ( 16-13 ) ⋅ ( 16-15 )} \\ & = \sqrt{ 16 ⋅ 12 ⋅ 3 ⋅ 1 } = \sqrt{ 576 } = 24 . \end{align}

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{align} A & = \tfrac{ 1 }{ 4 }\sqrt{(a + b + c)( - a + b + c)(a - b + c)(a + b - c)} \\[ 6 mu] & = \tfrac{ 1 }{ 4 }\sqrt{ 2 (a ^{2} b ^{2+a} ^{2c} ^{2+b} ^{2c} ^{2)-(a} ^{4+b} ^{4+c} ^{4)}} \\[ 6 mu] & = \tfrac{ 1 }{ 4 }\sqrt{(a ^{2+b} ^{2+c} ^{2)} ^{2-2(a} ^{4+b} ^{4+c} ^{4)}} \\[ 6 mu] & = \tfrac{ 1 }{ 4 }\sqrt{ 4 (a ^{2b} ^{2+a} ^{2c} ^{2+b} ^{2c} ^{2)-(a} ^{2+b} ^{2+c} ^{2)} ^{2}} \\[ 6 mu] & = \tfrac{ 1 }{ 4 }\sqrt{ 4 a ^{2b} ^{2-(a} ^{2+b} ^{2-c} ^{2)} ^{2}. \end{align}}

After expansion, the expression under the square root is a quadratic polynomial of the squared side lengths,, .

The same relation can be expressed using the Cayley–Menger determinant,^[2]

-16 A ^{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 (60 AD),^[3] and a proof can be found in his book Metrica. Mathematical historian Thomas Heath suggested that Archimedes knew the formula over two centuries earlier,^[4] 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 = (1 { 2 }\sqrt{a)/(^{2}) c ^{2} - \left( (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] or as a special case of Brahmagupta's formula (for the case of a degenerate cyclic quadrilateral).

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 be the sides of the triangle and the angles opposite those sides.Applying the law of cosines we get

\cos \gamma = (a ^{2+b} ^{2-c} ^{2})/(2 ab)

From this proof, we get the algebraic statement that

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

The altitude of the triangle on base has length, and it follows

\begin{align} A & = \tfrac 12 ( base ) ( altitude ) \\[ 6 mu] & = \tfrac 12 ab\sin \gamma \\[ 6 mu] & = (ab)/(4 ab) \sqrt{ 4 a ^{2} b ^{2} - (a ^{2} + b ^{2} - c ^{2)} ^{2}} \\[ 6 mu] & = \tfrac 14 \sqrt{ - a ^{4} - b ^{4} - c ^{4} + 2 a ^{2} b ^{2} + 2 a ^{2} c ^{2} + 2 b ^{2} c ^{2}} \\[ 6 mu] & = \tfrac 14 \sqrt{(a + b + c)( - a + b + c)(a - b + c)(a + b - c)} \\[ 6 mu] & = \sqrt{ \left( (a + b + c)/(2) \right) \left( (- a + b + c)/(2) \right) \left( (a - b + c)/(2) \right) \left( (a + b - c)/(2) \right)} \\[ 6 mu] & = \sqrt{s(s - a)(s - b)(s - c)}. \end{align}

Algebraic proof using the Pythagorean theorem

The following proof is very similar to one given by Raifaizen.^[9] By the Pythagorean theorem we have

b ^{2} = h ^{2} + d ^{2} and a ^{2} = h ^{2} + (c - d) ^{2} according to the figure at the right. Subtracting these yields a ^{2} - b ^{2} = c ^{2} - 2 cd. This equation allows us to express in terms of the sides of the triangle:d = (- a ^{2} + b ^{2} + c ^{2})/(2 c) . For the height of the triangle we have that h ^{2} = b ^{2} - d ^{2.} By replacing with the formula given above and applying the difference of squares identity we get\begin{align} h ^{2} & = b ^{2- \left( (- a ^{2} + b ^{2} + c ^{2})/(2 c)} \right) ^{2} \\ & = (( 2 bc - a ^{2} + b ^{2} + c ^{2)(2bc} + a ^{2} - b ^{2} - c ^{2)})/(4 c ^{2}) \\ & = (( (b + c) ^{2} - a ^{2)(a} ^{2} - (b - c) ^{2)})/(4 c ^{2}) \\ & = ((b + c - a)(b + c + a)(a + b - c)(a - b + c))/(4 c ^{2}) \\ & = (2 (s - a) ⋅ 2 s ⋅ 2 (s - c) ⋅ 2 (s - b))/(4 c ^{2}) \\ & = (4 s(s - a)(s - b)(s - c ))/(c ^{2}) . \end{align}

We now apply this result to the formula that calculates the area of a triangle from its height:

\begin{align} A & = (ch)/(2) \\ & = \sqrt{ (c ^{2})/(4) ⋅ (4 s(s - a)(s - b)(s - c))/(c ^{2}) } \\ &= \sqrt.\end

Trigonometric proof using the law of cotangents

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

A = \tfrac 12 ar + \tfrac 12 br + \tfrac 12 cr = rs, where s = \tfrac 12 (a + b + c) is the semiperimeter.

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

\begin{align} A & = r(s - a) + r(s - b) + r(s - c) \\[ 2 mu] & = r ^{2 \left( (s - a)/(r)} + (s - b)/(r) + (s - c)/(r) \right) \\[ 2 mu] & = r ^{2 \left(\cot{ (\alpha)/(2)} } + \cot + \cot\right) \\[3mu] &= r^2\left(\cot \cot \cot\right)\\[3mu] &= r^2\left(\frac \cdot \frac \cdot \frac\right) \\[3mu] &= \frac.\end

The middle step above is $\backslash cot\; +\; \backslash cot\; +\; \backslash cot\; =$

\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 $\backslash tfrac\backslash alpha2\; +\; \backslash tfrac\backslash beta2\; +\; \backslash tfrac\backslash gamma2\; =\; \backslash tfrac\backslash pi2.$

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 \ge b \ge c and computing^{[10] [11]} A = \tfrac 14 \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.

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 and are the medians from sides and respectively, and their semi-sum is

\sigma = \tfrac 12 (m _{a} + m _{b} + m _{c),} then^[12] A = (4)/(3) \sqrt{\sigma (\sigma - m _{a)(\sigma} - m _{b)(\sigma} - m _{c)}.}

Next, if,, and are the altitudes from sides and respectively, and semi-sum of their reciprocals is

H = \tfrac 12 l (h ^{-1} _{a} + h ^{-1} _{b} + h ^{-1} _{c} r ), then^[13] A ^{-1} = 4 \sqrt{H l (H - h ^{-1} _{a} r ) l (H - h ^{-1} _{b} r ) l (H - h ^{-1} _{c} r )}.

Finally, if and are the three angle measures of the triangle, and the semi-sum of their sines is

S = \tfrac 12 (\sin\alpha + \sin\beta + \sin\gamma), then^{[14] [15]} \begin{align} A & = D ^{2} \sqrt{S(S - \sin \alpha)(S - \sin \beta)(S - \sin \gamma)} \\[ 5 mu] & = \tfrac 12 D ^{2} \sin \alpha\sin \beta\sin \gamma, \end{align}

where is the diameter of the circumcircle,

D = a/{\sin \alpha} = b/{\sin \beta} = 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 of a cyclic quadrilateral whose sides have lengths as

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

where

s = \tfrac 12 (a + b + c + d) is the semiperimeter.

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 = (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 are lengths of edges of the tetrahedron (first three form a triangle; opposite to and so on), then^[17]

volume = (\sqrt {( - a + b + c + d)(a - b + c + d)(a + b - c + d)(a + b + c - d))/() } 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}

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 and the semiperimeter

s = \tfrac 12 (a + b + c) and area, such a formula is \tan ^{2} (S)/(4) = \tan (s 2)/(\tan \tan (s - a)/(2)) (s - b)/(\tan 2) (s - c)/(2) while for the hyperbolic plane we have\tan ^{2} (S)/(4) = \tanh (s 2)/(\tanh \tanh (s - a)/(2)) (s - b)/(\tanh 2) (s - c)/(2 .)

See also

• Shoelace formula

References

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2. Havel . Timothy F. . 10.1016/S0747-7171(08)80120-4 . 5-6 . . 579–593 . Some examples of the use of distances as coordinates for Euclidean geometry . 11 . 1991. free .
3. Id . Yusuf . Kennedy . E. S. . . 27958225 . 256819 . 585–587 . A medieval proof of Heron's formula . 62 . 1969. 7 . 10.5951/MT.62.7.0585 .
4. Book: Heath, Thomas L.. A History of Greek Mathematics. II. Oxford University Press. 1921. 321–323.
5. Book: 秦, 九韶 . 數學九章 (四庫全書本) . zh . 1773 . 卷三上, 三斜求积 . https://zh.wikisource.org/zh-hant/%E6%95%B8%E5%AD%B8%E4%B9%9D%E7%AB%A0_(%E5%9B%9B%E5%BA%AB%E5%85%A8%E6%9B%B8%E6%9C%AC)/%E5%85%A8%E8%A6%BD#%E4%B8%89%E6%96%9C%E6%B1%82%E7%A9%8D.
6. Web site: 15 December 1997. Personal email communication between mathematicians John Conway and Peter Doyle. 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. en. 10.1007/s00283-020-09996-8. 0343-6993. free.
8. Book: Niven, Ivan. Maxima and Minima Without Calculus. registration. The Mathematical Association of America. 1981. 7–8.
9. Raifaizen . Claude H. . A Simpler Proof of Heron's Formula . Mathematics Magazine . 44 . 1 . 27–28 . 1971. 10.1080/0025570X.1971.11976093 .
10. Book: Pat H. . Sterbenz . Floating-Point Computation . 1974-05-01 . 1st . Prentice-Hall Series in Automatic Computation . . Englewood Cliffs, New Jersey, USA . 0-13-322495-3.
11. Web site: Miscalculating Area and Angles of a Needle-like Triangle . William M. Kahan . 24 March 2000.
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. . A Heron-type area formula in terms of sines . Mathematical Gazette . 93 . 2009 . 108–109 . 10.1017/S002555720018430X. 132042882 .
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17. W. Kahan, "What has the Volume of a Tetrahedron to do with Computer Programming Languages?", http://www.cs.berkeley.edu/~wkahan/VtetLang.pdf, pp. 16–17.
18. Book: Alekseevskij. D. V. . Vinberg. E. B. . Solodovnikov. A. S. . Geometry of spaces of constant curvature. Gamkrelidze. R. V. . Vinberg. E. B. . Geometry. II: Spaces of constant curvature. Encycl. Math. Sci. . 29. 66. Springer-Verlag. 1993. 1-56085-072-8.

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