View source for Heron's formula

{{about|calculating the area of a triangle|calculating a square root|Heron's method}}

[[File:Triangle with notations 2 without points.svg|thumb|198px|A triangle with sides ''a'', ''b'', and ''c''.]]

In [[geometry]], '''Heron's formula''' (sometimes called Hero's formula), named after [[Hero of Alexandria]],<ref>{{cite web|title=Fórmula de Herón para calcular el área de cualquier triángulo|url=http://recursostic.educacion.es/descartes/web/materiales_didacticos/formula_heron/formula_de_Heron.htm|language=Spanish|accessdate=30 June 2012}}</ref> 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 {{math|''a''}}, {{math|''b''}}, and {{math|''c''}} is

:<math>A = \sqrt{s(s-a)(s-b)(s-c)},</math>

where {{math|''s''}} is the [[semi-perimeter]] of the triangle; that is,

:<math>s=\frac{a+b+c}{2}.</math><ref>{{cite journal|author=Kendig, Keith|title=Is a 2000-Year-Old Formula Still Keeping Some Secrets?|journal=Amer. Math. Monthly|volume=107|year=2000|pages=402–415|url=http://www.maa.org/programs/maa-awards/writing-awards/is-a-2000-year-old-formula-still-keeping-some-secrets|doi=10.2307/2695295}}</ref>

Heron's formula can also be written as

:<math>A=\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}</math>
:<math>A=\frac{1}{4}\sqrt{2(a^2 b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4)}</math>
:<math>A=\frac{1}{4}\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}</math>
:<math>A=\frac{1}{4}\sqrt{4(a^2b^2+a^2c^2+b^2c^2)-(a^2+b^2+c^2)^2}.</math>

==Example==

Let {{math|△''ABC''}} be the triangle with sides {{math|''a'' {{=}} 4}}, {{math|''b'' {{=}} 13}} and {{math|''c'' {{=}} 15}}. 
The semiperimeter is {{math|''s'' {{=}} {{sfrac|1|2}}(''a'' + ''b'' + ''c'') {{=}} {{sfrac|1|2}}(4 + 13 + 15) {{=}} 16}}, and the area is

:<math>
\begin{align}
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{align}
</math>

In this example, the side lengths and area are all [[integer]]s, making it a [[Heronian triangle]]. However, Heron's formula works equally well in cases where one or all of these numbers is not an integer.

== History ==

The formula is credited to [[Hero of Alexandria|Heron (or Hero) of Alexandria]], and a proof can be found in his book, ''Metrica'', written {{circa}} CE 60. It has been suggested that [[Archimedes]] knew the formula over two centuries earlier,<ref>{{cite book
| author=Heath, Thomas L.
| title=A History of Greek Mathematics (Vol II)
| publisher=Oxford University Press
| year=1921
| pages=321–323}}</ref> 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.<ref>{{MathWorld |urlname=HeronsFormula |title=Heron's Formula}}</ref>

A formula equivalent to Heron's, namely

:<math>A=\frac1{2}\sqrt{a^2c^2-\left(\frac{a^2+c^2-b^2}{2}\right)^2}</math>

was discovered by the Chinese independently{{citation needed|date=October 2017}} of the Greeks. It was published in ''[[Mathematical Treatise in Nine Sections]]'' ([[Qin Jiushao]], 1247).<ref>{{Cite book|title=數學九章 (四庫全書本)|last=秦|first=九韶|publisher=|year=1773|isbn=|location=|pages=|chapter=卷五第二题, 三斜求积|chapter-url=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}}</ref>

== Proofs ==
Heron's original proof made use of [[cyclic quadrilateral]]s, while other arguments appeal to [[trigonometry]] as below, or to the [[incenter]] and one [[excircle]] of the triangle [http://www.math.dartmouth.edu/~doyle/docs/heron/heron.txt].

===Trigonometric proof using the law of cosines===
A modern proof, which uses [[algebra]] and is quite unlike the one provided by Heron (in his book Metrica), follows.<ref>{{cite book
| author=Niven, Ivan
| title=Maxima and Minima Without Calculus
| url=https://archive.org/details/maximaminimawith0000nive
| url-access=registration
| publisher=The Mathematical Association of America
| year=1981
| pages=[https://archive.org/details/maximaminimawith0000nive/page/7 7–8]}}</ref>
Let {{math|''a''}}, {{math|''b''}}, {{math|''c''}} be the sides of the triangle and {{math|''α''}}, {{math|''β''}}, {{math|''γ''}} the [[angle]]s opposite those sides.
Applying the [[law of cosines]] we get

:<math>\cos \gamma = \frac{a^2+b^2-c^2}{2ab}</math>

From this proof we get the algebraic statement that

:<math>\sin \gamma = \sqrt{1-\cos^2 \gamma} = \frac{\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2 }}{2ab}.</math>

The [[altitude (triangle)|altitude]] of the triangle on base {{math|''a''}} has length {{math|''b'' sin ''γ''}}, and it follows

: <math>
\begin{align}
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{(2a b -(a^2 +b^2 -c^2))(2a b +(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{align}
</math>

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

===Algebraic proof using the Pythagorean theorem===
[[Image:Triangle with notations 3.svg|thumb|270px|Triangle with altitude {{math|''h''}} cutting base {{math|''c''}} into {{math|''d'' + (''c'' − ''d'')}}.]]
The following proof is very similar to one given by Raifaizen.<ref>{{Cite journal
  | last = Raifaizen
  | first = Claude H.
  | title = A Simpler Proof of Heron's Formula
  | journal = Mathematics Magazine
  | volume = 44
  | number = 1
  | pages = 27–28
  | year = 1971
}}</ref>
By the [[Pythagorean theorem]] we have {{math|''b''{{sup|2}} {{=}} ''h''{{sup|2}} + ''d''{{sup|2}}}} and {{math|''a''{{sup|2}} {{=}} ''h''{{sup|2}} + (''c'' − ''d''){{sup|2}}}} according to the figure at the right. Subtracting these yields {{math|''a''{{sup|2}} − ''b''{{sup|2}} {{=}} ''c''{{sup|2}} − 2''cd''}}. This equation allows us to express {{math|''d''}} in terms of the sides of the triangle:
:<math>d=\frac{-a^2+b^2+c^2}{2c}</math>
For the height of the triangle we have that {{math|''h''{{sup|2}} {{=}} ''b''{{sup|2}} − ''d''{{sup|2}}}}. By replacing {{math|''d''}} with the formula given above and applying the [[difference of squares]] identity we get
:<math>
\begin{align}
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{((b+c)^2-a^2)(a^2-(b-c)^2)}{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{align}
</math>

We now apply this result to the formula that calculates the area of a triangle from its height:
:<math>
\begin{align}
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{align}
</math>

===Trigonometric proof using the law of cotangents===
[[File:Herontriangle2greek.svg|thumb|270px|right|Geometrical significance of {{math|''s'' − ''a''}}, {{math|''s'' − ''b''}}, and {{math|''s'' − ''c''}}.  See the [[Law of cotangents]] for the reasoning behind this.]]
From the first part of the [[Law of cotangents]] proof,<ref>The second part of the Law of cotangents proof depends on Heron's formula itself, but this article depends only on the first part.</ref> we have that the triangle's area is both
:<math>
\begin{align}
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{align}
</math>
and {{math|''A'' {{=}} ''rs''}}, but, since the sum of the half-angles is {{sfrac|{{pi}}|2}}, the [[Proofs of trigonometric identities#Miscellaneous -- the triple cotangent identity|triple cotangent identity]] applies, so the first of these is
:<math>
\begin{align}
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{align}
</math>

Combining the two, we get
:<math>A^2 = s(s-a)(s-b)(s-c)</math>
from which the result follows.

== Numerical stability ==
Heron's formula as given above is [[Numerical stability|numerically unstable]] for triangles with a very small angle when using floating point arithmetic. A stable alternative<ref>{{cite book |author-first=Pat H. |author-last=Sterbenz |title=Floating-Point Computation |date=1974-05-01 |edition=1st |series=Prentice-Hall Series in Automatic Computation |publisher=[[Prentice Hall]] |location=Englewood Cliffs, New Jersey, USA |isbn=0-13-322495-3<!-- 978-0-13-322495-5 -->}}</ref><ref>{{cite web |url=http://www.cs.berkeley.edu/~wkahan/Triangle.pdf |title=Miscalculating Area and Angles of a Needle-like Triangle |author=William M. Kahan |date=24 March 2000}}</ref> involves arranging the lengths of the sides so that {{math|''a'' ≥ ''b'' ≥ ''c''}} and computing
:<math>A = \frac{1}{4}\sqrt{(a+(b+c)) (c-(a-b)) (c+(a-b)) (a+(b-c))}.</math>
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 {{math|''a''}}, {{math|''b''}}, and {{math|''c''}} respectively as {{math|''m<sub>a</sub>''}}, {{math|''m<sub>b</sub>''}}, and {{math|''m<sub>c</sub>''}} and their semi-sum {{math|{{sfrac|1|2}}(''m<sub>a</sub>'' + ''m<sub>b</sub>'' + ''m<sub>c</sub>'')}} as {{math|''σ''}}, we have<ref>Benyi, Arpad, "A Heron-type formula for the triangle," ''Mathematical Gazette" 87, July 2003, 324–326.</ref>
:<math>A = \frac{4}{3} \sqrt{\sigma (\sigma - m_a)(\sigma - m_b)(\sigma - m_c)}.</math>

Next, denoting the altitudes from sides {{math|''a''}}, {{math|''b''}}, and {{math|''c''}} respectively as {{math|''h<sub>a</sub>''}}, {{math|''h<sub>b</sub>''}}, and {{math|''h<sub>c</sub>''}}, and denoting the semi-sum of the reciprocals of the altitudes as {{math|''H'' {{=}} {{sfrac|1|2}}(''h''{{su|b=''a''|p=−1}} + ''h''{{su|b=''b''|p=−1}} + ''h''{{su|b=''c''|p=−1}})}} we have<ref>Mitchell, Douglas W., "A Heron-type formula for the reciprocal area of a triangle," ''Mathematical Gazette'' 89, November 2005, 494.</ref>
:<math>A^{-1} = 4 \sqrt{H(H-h_a^{-1})(H-h_b^{-1})(H-h_c^{-1})}.</math>

Finally, denoting the semi-sum of the angles' sines as {{math|''S'' {{=}} {{sfrac|1|2}}(sin ''α'' + sin ''β'' + sin ''γ'')}}, we have<ref>Mitchell, Douglas W., "A Heron-type area formula in terms of sines," ''Mathematical Gazette'' 93, March 2009, 108–109.</ref>
:<math>A = D^{2} \sqrt{S(S-\sin \alpha)(S-\sin \beta)(S-\sin \gamma)}</math>

where {{math|''D''}} is the diameter of the circumcircle: {{math|''D'' {{=}} {{sfrac|''a''|sin ''α''}} {{=}} {{sfrac|''b''|sin ''β''}} {{=}} {{sfrac|''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 [[trapezoid#Area|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 [[distance]]s between the three given vertices,
:<math> 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} } </math>
illustrates its similarity to [[Tartaglia's formula]] for the [[volume]] of a [[Simplex|three-simplex]].

Another generalization of Heron's formula to pentagons and hexagons inscribed in a circle was discovered by [[David P. Robbins]].<ref>D. P. Robbins, "Areas of Polygons Inscribed in a Circle", Discr. Comput. Geom. 12, 223-236, 1994.</ref>

=== Heron-type formula for the volume of a tetrahedron ===
If {{math|''U''}}, {{math|''V''}}, {{math|''W''}}, {{math|''u''}}, {{math|''v''}}, {{math|''w''}} are lengths of edges of the tetrahedron (first three form a triangle; {{math|''u''}} opposite to {{math|''U''}} and so on), then<ref>W. Kahan, "What has the Volume of a Tetrahedron to do with Computer Programming Languages?", [http://www.cs.berkeley.edu/~wkahan/VtetLang.pdf], pp.&nbsp;16–17.</ref>
:<math>
\text{volume} = \frac{\sqrt {\,( - a + b + c + d)\,(a - b + c + d)\,(a + b - c + d)\,(a + b + c - d)}}{192\,u\,v\,w}</math>
where
: <math>
    \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} 
</math>

==See also==
*[[Shoelace formula]]

== References ==
<references/>

== External links ==
*[http://www.cut-the-knot.org/pythagoras/herons.shtml A Proof of the Pythagorean Theorem From Heron's Formula] at [[cut-the-knot]]
*[http://www.mathopenref.com/heronsformula.html Interactive applet and area calculator using Heron's Formula]
*[http://www.math.dartmouth.edu/~doyle/docs/heron/heron.txt J.H. Conway discussion on Heron's Formula]
*{{MathPages|id=home/kmath196/kmath196|title=Heron's Formula and Brahmagupta's Generalization}}
*[http://jwilson.coe.uga.edu/EMT668/EMAT6680.2000/Umberger/MATH7200/HeronFormulaProject/GeometricProof/geoproof.html A Geometric Proof of Heron's Formula]
*[http://www.maa.org/sites/default/files/0746834212944.di020798.02p0691h.pdf An alternative proof of Heron's Formula without words]
*[http://www.maa.org/sites/default/files/Pratt-CMJ0902994.pdf Factoring Heron]

{{DEFAULTSORT:Heron's Formula}}
[[Category:Triangle geometry]]
[[Category:Articles containing proofs]]
[[Category:Area]]
[[Category:Theorems in plane geometry]]
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