Editing Heron's formula

{{Short description|Triangle area in terms of side lengths}}
{{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''' (or '''Hero's formula''') gives the [[area of a triangle]] in terms of the three side lengths {{tmath|a,}} {{tmath|b,}} {{tmath|c.}} Letting {{tmath|s}} be the [[semiperimeter]] of the triangle, <math>s = \tfrac12(a + b + c),</math> the area {{tmath|A}} is<ref>{{cite journal
 | last = Kendig
 | first = Keith
 | doi = 10.1080/00029890.2000.12005213
 | issue = 5
 | journal = [[The American Mathematical Monthly]]
 | jstor = 2695295
 | mr = 1763392
 | pages = 402–415
 | title = Is a 2000-year-old formula still keeping some secrets?
 | url = https://www.maa.org/programs/maa-awards/writing-awards/is-a-2000-year-old-formula-still-keeping-some-secrets
 | volume = 107
 | year = 2000
 | s2cid = 1214184
 | access-date = 2021-12-27
 | archive-date = 2024-05-29
 | archive-url = https://web.archive.org/web/20240529215533/https://maa.org/programs/maa-awards/writing-awards/is-a-2000-year-old-formula-still-keeping-some-secrets
 | url-status = dead
 }}</ref>

<math display=block>A = \sqrt{s(s-a)(s-b)(s-c)}.</math>

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==
<div class="calculator-container" data-calculator-refresh-on-load="true" style="
    border: 1px solid grey;
    float: right;
    margin: 3px;">
{|- class="wikitable floatleft" style="margin: 0;"
! colspan=2 style="padding: 0 1em" | Area calculator
|-
| {{calculator label|label={{font color|orange|{{mvar|a}}}}|for=a}} || {{calculator|size=3|default=3|min=1|step=1|id=a}}
|-
| {{calculator label|label={{font color|green|{{mvar|b}}}}|for=b}} || {{calculator|size=3|default=4|min=1|step=1|id=b}}
|-
| {{calculator label|label={{font color|purple|{{mvar|c}}}}|for=c}} || {{calculator|size=3|default=5|min=1|step=1|id=c}}
|-
| {{mvar|s}} || {{calculator|type=plain|default=6|formula=(a+b+c)/2|NaN-text=&#32;|id=s}}
|-
| Area<ref>The formula used here is the [[#Numerical stability|numerically stable formula]] (relabeled for {{tmath|a \leq b \leq c}}), not simply {{tmath|\textstyle ~\!\!\sqrt{s(s-a)(s-b)(s-c)}\! }}. For example, with {{tmath|1= a=3}}, {{tmath|1= b=4}}, {{tmath|1= c=6.999}}, the correct area is {{tmath|0.205}} but the naive implementation produces {{tmath|0.000}} instead. {{pb}} The area is reported as "Not a triangle" when the side lengths fail the [[triangle inequality]]. When the area is equal to zero, the three side lengths specify a [[degenerate triangle]] with three colinear points.</ref>
|style="max-width: 4em" | {{calculator|type=plain|decimals=3|default=6.000|formula=sqrt((c+(b+a))*(a-(c-b))*(a+(c-b))*(c+(b-a)))/4|NaN-text=Not a triangle}}
|}{{calculator|type=hidden|default=5|formula=max(a,b,c)|id=maxabc}}{{calculator|type=hidden|default=1|formula=1-ifpositive(min(a,b,c),0,1)|id=allnonneg}}<!--
  Figure of the triangle using CSS. If allnonneg is 0, then the figure will be hidden.
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      width: 240px;
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      position: relative;
      display: inline-block;">
    <div style="
        width: calc(var(--calculator-c,5) * var(--scale));
        border-top: var(--thickness) solid purple;
        left: 40px;
        top: 150px;
        position: absolute;"></div>
    <div style="
        width: calc(var(--calculator-b,4) * var(--scale));
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                (var(--calculator-c,5) * var(--calculator-c,5)
                    + var(--calculator-b,4) * var(--calculator-b,4)
                    - var(--calculator-a,3) * var(--calculator-a,3)
                ) / 2 / var(--calculator-b,4) / var(--calculator-c,5) 
            )));
        transform-origin: 0px calc(var(--thickness) / 2); /* rotate from the center of the line rather than the top */
        left: 40px;
        top: 150px;
        position: absolute;"></div>
    <div style="
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                    + var(--calculator-a,3) * var(--calculator-a,3)
                    - var(--calculator-b,4) * var(--calculator-b,4)
                ) / 2 / var(--calculator-a,3) / var(--calculator-c,5) 
            )));
        transform-origin: calc(var(--calculator-a,3) * var(--scale)) calc(var(--thickness) / 2);
        left: calc(40px + var(--calculator-c,5) * var(--scale) - var(--calculator-a,3) * var(--scale));
        top: 150px;
        position: absolute;"></div>
  </div>
</div>

Let {{tmath|\triangle ABC}} be the triangle with sides {{tmath|1= a = 4}}, {{tmath|1= b = 13}}, and {{tmath|1= c = 15}}.
This triangle's semiperimeter is <math>s = \tfrac12(a+b+c)= {}</math><math>\tfrac12(4+13+15) = 16</math> therefore {{tmath|1= s-a = 12}}, {{tmath|1= s-b =3}}, {{tmath|1= s-c =1}}, and the area is
<math display=block>\begin{align}
A &= {\textstyle \sqrt{s(s-a)(s-b)(s-c)}} \\[3mu]
&= {\textstyle \sqrt{16 \cdot 12 \cdot 3 \cdot 1 \vphantom)} } \\[3mu]
&= 24.
\end{align}</math>

In this example, the triangle's side lengths and area are [[integer]]s, making it a [[Heronian triangle]]. However, Heron's formula works equally well when the side lengths are [[real number]]s. As long as they obey the strict [[triangle inequality]], they define a triangle in the [[Euclidean plane]] whose area is a positive real number.

== Alternate expressions ==

Heron's formula can also be written in terms of just the side lengths instead of using the semiperimeter, in several ways,

<math display=block>\begin{align}
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^2c^2+b^2c^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^2b^2+a^2c^2+b^2c^2)-(a^2+b^2+c^2)^2} \\[6mu]
&=\tfrac{1}{4}\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}.
\end{align}</math>

After expansion, the expression under the square root is a [[quadratic polynomial]] of the squared side lengths {{tmath|a^2}}, {{tmath|b^2}}, {{tmath|c^2}}.

The same relation can be expressed using the [[Cayley–Menger determinant]],<ref>{{cite journal
 | last = Havel | first = Timothy F.
 | doi = 10.1016/S0747-7171(08)80120-4
 | issue = 5–6
 | journal = [[Journal of Symbolic Computation]]
 | pages = 579–593
 | title = Some examples of the use of distances as coordinates for Euclidean geometry
 | volume = 11
 | year = 1991| doi-access = free
 }}</ref>

<math display=block> -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}. </math>

== History ==

The formula is credited to [[Hero of Alexandria|Heron (or Hero) of Alexandria]] ({{fl.}} 60 AD),<ref>{{cite journal
 | last1 = Id | first1 = Yusuf
 | last2 = Kennedy | first2 = E. S.
 | journal = [[The Mathematics Teacher]]
 | jstor = 27958225
 | mr = 256819
 | pages = 585–587
 | title = A medieval proof of Heron's formula
 | volume = 62
 | year = 1969| issue = 7
 | doi = 10.5951/MT.62.7.0585
 }}</ref> and a proof can be found in his book ''Metrica''. Mathematical historian [[Thomas Heath (classicist)|Thomas Heath]] suggested that [[Archimedes]] knew the formula over two centuries earlier,<ref>{{cite book
| author=Heath, Thomas L.
| title=A History of Greek Mathematics
| volume=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 was discovered by Chinese mathematician Qin Jiushao:

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

published in ''[[Mathematical Treatise in Nine Sections]]'' ([[Qin Jiushao]], 1247).<ref>{{Cite book |title=數學九章 (四庫全書本) |lang=zh |last=秦 |first=九韶 |year=1773 |chapter=卷三上, 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 ==
There are many ways to prove Heron's formula, for example using [[trigonometry]] as below, or the [[incenter]] and one [[excircle]] of the triangle,<ref>{{Cite web|date=15 December 1997|title=Personal email communication between mathematicians John Conway and Peter Doyle|url=https://math.dartmouth.edu/~doyle/docs/heron/heron.txt|access-date=25 September 2020}}</ref> or as a special case of [[De Gua's theorem]] (for the particular case of acute triangles),<ref>{{Cite journal|last=Lévy-Leblond|first=Jean-Marc|date=2020-09-14|title=A Symmetric 3D Proof of Heron's Formula|journal=The Mathematical Intelligencer|volume=43|issue=2|pages=37–39|language=en|doi=10.1007/s00283-020-09996-8|issn=0343-6993|doi-access=free}}</ref> 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.<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 {{tmath|a,}} {{tmath|b,}} {{tmath|c}} be the sides of the triangle and {{tmath|\alpha,}} {{tmath|\beta,}} {{tmath|\gamma}} the [[angle]]s opposite those sides.
Applying the [[law of cosines]] we get

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

[[File:Triangle with notations 2 without points.svg|thumb|A triangle with sides {{mvar|a}}, {{mvar|b}} and {{mvar|c}}]]
From this proof, we get the algebraic statement that

<math display=block>\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 {{tmath|a}} has length {{tmath|b\sin\gamma}}, and it follows

<math display=block>\begin{align}
A
&= \tfrac12 (\mbox{base}) (\mbox{altitude}) \\[6mu]
&= \tfrac12 ab\sin \gamma \\[6mu]
&= \frac{ab}{4ab}\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2} \\[6mu]
&= \tfrac14\sqrt{-a^4 - b^4 - c^4 + 2 a^2 b^2 + 2 a^2 c^2 + 2 b^2 c^2} \\[6mu]
&= \tfrac14\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{align}</math>

===Algebraic proof using the Pythagorean theorem===
[[Image:Triangle with notations 3.svg|thumb|270px|Triangle with altitude {{mvar|h}} cutting base {{mvar|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
| doi = 10.1080/0025570X.1971.11976093
 }}</ref>
By the [[Pythagorean theorem]] we have <math>b^2 = h^2 + d^2</math> and <math>a^2 = h^2 + (c - d)^2</math> according to the figure at the right. Subtracting these yields <math>a^2 - b^2 = c^2 - 2cd.</math> This equation allows us to express {{tmath|d}} in terms of the sides of the triangle:
<math display=block>d = \frac{-a^2 + b^2 + c^2}{2c}.</math>
For the height of the triangle we have that <math>h^2 = b^2 - d^2.</math> By replacing {{tmath|d}} with the formula given above and applying the [[difference of squares]] identity we get
<math display=block>
\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{\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{align}
</math>

We now apply this result to the formula that calculates the area of a triangle from its height:
<math display=block>\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.]]
If {{tmath|r}} is the radius of the [[incircle]] of the triangle, then the triangle can be broken into three triangles of equal altitude {{tmath|r}} and bases {{tmath|a,}} {{tmath|b,}} and {{tmath|c.}} Their combined area is
<math display=block>A = \tfrac12ar + \tfrac12br + \tfrac12cr = rs,</math>
where <math>s = \tfrac12(a + b + c)</math> is the semiperimeter.

The triangle can alternately be broken into six triangles (in congruent pairs) of altitude {{tmath|r}} and bases {{tmath|s - a,}} {{tmath|s - b,}} and {{tmath|s - c}} of combined area (see [[law of cotangents]])
<math display=block>\begin{align}
 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{align}</math>

The middle step above is <math display=inline>\cot{\tfrac{\alpha}{2}} + \cot{\tfrac{\beta}{2}} + \cot{\tfrac{\gamma}{2}} = {}</math><math>\cot{\tfrac{\alpha}{2}}\cot{\tfrac{\beta}{2}}\cot{\tfrac{\gamma}{2}},</math> the [[Proofs of trigonometric identities#Miscellaneous – the triple cotangent identity|triple cotangent identity]], which applies because the sum of half-angles is <math display=inline>\tfrac\alpha2 + \tfrac\beta2 + \tfrac\gamma2 = \tfrac\pi2.</math>

Combining the two, we get
<math display=block>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 involves arranging the lengths of the sides so that <math>a \ge b \ge c</math> and computing<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>
<math display=block>A = \tfrac14 \sqrt{\big(a + (b + c)\big) \big(c - (a - b)\big) \big(c + (a - b)\big) \big(a + (b - c)\big)}.</math>
The extra brackets indicate the order of operations required to achieve numerical stability 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 {{tmath|m_a,}} {{tmath|m_b,}} and {{tmath|m_c}} are the [[median (geometry)|median]]s from sides {{tmath|a,}} {{tmath|b,}} and {{tmath|c}} respectively, and their semi-sum is <math>\sigma = \tfrac12(m_a + m_b + m_c),</math> then<ref>{{cite journal|last=Bényi |first=Árpád |title=A Heron-type formula for the triangle |journal=Mathematical Gazette |volume=87 |date=July 2003 |pages=324–326 |doi=10.1017/S0025557200172882}}</ref>
<math display=block>A = \frac{4}{3} \sqrt{\sigma (\sigma - m_a)(\sigma - m_b)(\sigma - m_c)}.</math>

Next, if {{tmath|h_a}}, {{tmath|h_b}}, and {{tmath|h_c}} are the [[altitude (triangle)|altitude]]s from sides {{tmath|a,}} {{tmath|b,}} and {{tmath|c}} respectively, and  semi-sum of their reciprocals is <math>H = \tfrac12\bigl(h_a^{-1} + h_b^{-1} + h_c^{-1}\bigr),</math> then<ref>{{cite journal|last=Mitchell |first=Douglas W. |title=A Heron-type formula for the reciprocal area of a triangle |journal=Mathematical Gazette |volume=89 |date=November 2005 |page=494 |doi=10.1017/S0025557200178532}}</ref>
<math display=block>
A^{-1} = 4 \sqrt{H\bigl(H-h_a^{-1}\bigr)\bigl(H-h_b^{-1}\bigr)\bigl(H-h_c^{-1}\bigr)}.
</math>

Finally, if {{tmath|\alpha,}} {{tmath|\beta,}} and {{tmath|\gamma}} are the three angle measures of the triangle, and the semi-sum of their [[sine]]s is <math>S = \tfrac12(\sin\alpha + \sin\beta + \sin\gamma),</math> then<ref>{{cite journal |last=Mitchell |first=Douglas W. |title=A Heron-type area formula in terms of sines |journal=Mathematical Gazette |volume=93 |year=2009 |pages=108–109 |doi=10.1017/S002555720018430X|s2cid=132042882 }}</ref><ref>{{cite journal |last1=Kocik |first1=Jerzy |last2=Solecki |first2=Andrzej |date=2009 |title=Disentangling a triangle |journal=American Mathematical Monthly |volume=116 |number=3 |pages=228–237 |doi=10.1080/00029890.2009.11920932 |s2cid=28155804 |url=http://lagrange.math.siu.edu/Kocik/triangle/monthlyTriangle.pdf}}</ref>
<math display=block>\begin{align}
A &= D^{2} \sqrt{S(S-\sin \alpha)(S-\sin \beta)(S-\sin \gamma)} \\[5mu]
  &= \tfrac12 D^{2} \sin \alpha\,\sin \beta\,\sin \gamma,
\end{align}</math>

where {{tmath|D}} is the diameter of the [[circumcircle]], <math>D =  a/{\sin \alpha} = b/{\sin \beta} = c/{\sin \gamma}.</math> This last formula coincides with the standard Heron formula when the circumcircle has unit diameter.

== Generalizations ==
[[File:Cyclicquadrilateral.png|right|Cyclic Quadrilateral|thumb|192x192px]]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 {{tmath|K}} of a [[cyclic quadrilateral]] whose sides have lengths {{tmath|a,}} {{tmath|b,}} {{tmath|c,}} {{tmath|d}} as

<math display=block>K=\sqrt{(s-a)(s-b)(s-c)(s-d)}</math>

where <math>s = \tfrac12(a + b + c + d)</math> is the [[semiperimeter]].

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 display=block> 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>{{cite journal|first=D. P. |last=Robbins |author-link=David P. Robbins |title=Areas of Polygons Inscribed in a Circle |journal=Discrete & Computational Geometry |volume=12 |pages=223–236 |date=1994 |issue=2 |doi=10.1007/BF02574377}}</ref>

===Degenerate and imaginary triangles===
If one of three given lengths is equal to the sum of the other two, the three sides determine a [[degenerate triangle]], a line segment with zero area. In this case, the semiperimeter will equal the longest side, causing Heron's formula to equal zero.

If one of three given lengths is greater than the sum of the other two, then they violate the [[triangle inequality]] and do not describe the sides of a Euclidean triangle. In this case, Heron's formula gives an [[imaginary number|imaginary]] result. For example if {{tmath|1= a=3}} and {{tmath|1= b=c=1}}, then {{tmath|1=\textstyle A = \tfrac {3\sqrt5}4i }}. This can be interpreted using a triangle in the [[complex coordinate space|complex coordinate plane]] {{tmath|\C^2}}, where "area" can be a complex-valued quantity, or as a triangle lying in a [[pseudo-Euclidean space|pseudo-Euclidean plane]] with one space-like dimension and one time-like dimension.<ref>{{Cite journal |last=Schwartz |first=Mark |date=2007 |title=Review of Conics |url=http://www.jstor.org/stable/27642242 |journal=The American Mathematical Monthly |volume=114 |issue=5 |pages=461–464 |jstor=27642242 |issn=0002-9890}}</ref>

=== Volume of a tetrahedron ===
If {{tmath|U,}} {{tmath|V,}} {{tmath|W,}} {{tmath|u,}} {{tmath|v,}} {{tmath|w}} are lengths of edges of the tetrahedron (first three form a triangle; {{tmath|u}} opposite to {{tmath|U}} and so on), then<ref>{{cite web |first=William |last=Kahan |author-link=William Kahan |title=What has the Volume of a Tetrahedron to do with Computer Programming Languages? |url=http://www.cs.berkeley.edu/~wkahan/VtetLang.pdf |pages=16–17}}</ref>
<math display=block>
\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 display=block>\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>

===Spherical and hyperbolic geometry===
[[L'Huilier's formula]] relates the area of a triangle in [[spherical geometry]] to its side lengths. For a [[spherical triangle]] with side lengths {{tmath|a,}} {{tmath|b,}} and {{tmath|c}}, semiperimeter {{tmath|1= s=\tfrac12(a+b+c)}}, and area {{tmath|S}},<ref>{{cite book
| last1  = Alekseevskij
| first1 = D. V. 
| last2  = Vinberg
| first2 = E. B. 
| last3  = Solodovnikov
| first3 = A. S. 
| chapter = Geometry of spaces of constant curvature
| editor-last1  = Gamkrelidze
| editor-first1 = R. V. 
| editor-last2  = Vinberg
| editor-first2 = E. B. 
| title  = Geometry. II: Spaces of constant curvature
|series= Encyclopaedia of Mathematical Sciences
|volume=29
|page=66
| publisher = Springer-Verlag
| publication-date = 1993
| isbn = 1-56085-072-8
}} </ref>
<math display=block>
\tan^2 \frac S 4 = \tan \frac s2 \tan\frac{s-a}2 \tan\frac{s-b}2 \tan\frac{s-c}2
</math>

For a triangle in [[hyperbolic geometry]] the analogous formula is
<math display=block>
\tan^2 \frac S 4 = \tanh \frac s2 \tanh\frac{s-a}2 \tanh\frac{s-b}2 \tanh\frac{s-c}2. 
</math>

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

== Notes and 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]

{{Ancient Greek mathematics}}

{{DEFAULTSORT:Heron's Formula}}
[[Category:Theorems about triangles]]
[[Category:Articles containing proofs]]
[[Category:Area]]
[[Category:Greek mathematics]]
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