Heron's formula
Template:Short description Template:About
In geometry, Heron's formula (or Hero's formula) gives the area of a triangle in terms of the three side lengths Template:Tmath Template:Tmath Template:Tmath Letting Template:Tmath be the semiperimeter of the triangle, <math>s = \tfrac12(a + b + c),</math> the area Template:Tmath is<ref>Template:Cite journal</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.
ExampleEdit
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| Area<ref>The formula used here is the numerically stable formula (relabeled for Template:Tmath), not simply Template:Tmath. For example, with Template:Tmath, Template:Tmath, Template:Tmath, the correct area is Template:Tmath but the naive implementation produces Template:Tmath instead. Template: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> | Template:Calculator |
Let Template:Tmath be the triangle with sides Template:Tmath, Template:Tmath, and Template:Tmath. This triangle's semiperimeter is <math>s = \tfrac12(a+b+c)= {}</math><math>\tfrac12(4+13+15) = 16</math> therefore Template:Tmath, Template:Tmath, Template:Tmath, 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 integers, making it a Heronian triangle. However, Heron's formula works equally well when the side lengths are real numbers. 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 expressionsEdit
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 Template:Tmath, Template:Tmath, Template:Tmath.
The same relation can be expressed using the Cayley–Menger determinant,<ref>Template:Cite journal</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>
HistoryEdit
The formula is credited to Heron (or Hero) of Alexandria (Template:Fl. 60 AD),<ref>Template:Cite journal</ref> and a proof can be found in his book Metrica. Mathematical historian Thomas Heath suggested that Archimedes knew the formula over two centuries earlier,<ref>Template:Cite book</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>{{#invoke:Template wrapper|{{#if:|list|wrap}}|_template=cite web |_exclude=urlname, _debug, id |url = https://mathworld.wolfram.com/{{#if:HeronsFormula%7CHeronsFormula.html}} |title = Heron's Formula |author = Weisstein, Eric W. |website = MathWorld |access-date = |ref = Template:SfnRef }}</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>Template:Cite book</ref>
ProofsEdit
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>{{#invoke:citation/CS1|citation |CitationClass=web }}</ref> or as a special case of De Gua's theorem (for the particular case of acute triangles),<ref>Template:Cite journal</ref> or as a special case of Brahmagupta's formula (for the case of a degenerate cyclic quadrilateral).
Trigonometric proof using the law of cosinesEdit
A modern proof, which uses algebra and is quite different from the one provided by Heron, follows.<ref>Template:Cite book</ref> Let Template:Tmath Template:Tmath Template:Tmath be the sides of the triangle and Template:Tmath Template:Tmath Template:Tmath the angles opposite those sides. Applying the law of cosines we get
<math display=block>\cos \gamma = \frac{a^2+b^2-c^2}{2ab}</math>
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 of the triangle on base Template:Tmath has length Template:Tmath, 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 theoremEdit
The following proof is very similar to one given by Raifaizen.<ref>Template:Cite journal</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 Template:Tmath 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 Template:Tmath 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 cotangentsEdit
If Template:Tmath is the radius of the incircle of the triangle, then the triangle can be broken into three triangles of equal altitude Template:Tmath and bases Template:Tmath Template:Tmath and Template:Tmath 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 Template:Tmath and bases Template:Tmath Template:Tmath and Template:Tmath 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 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 stabilityEdit
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 <math>a \ge b \ge c</math> and computing<ref>Template:Cite book</ref><ref>{{#invoke:citation/CS1|citation |CitationClass=web }}</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 formulaeEdit
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 Template:Tmath Template:Tmath and Template:Tmath are the medians from sides Template:Tmath Template:Tmath and Template:Tmath respectively, and their semi-sum is <math>\sigma = \tfrac12(m_a + m_b + m_c),</math> then<ref>Template:Cite journal</ref> <math display=block>A = \frac{4}{3} \sqrt{\sigma (\sigma - m_a)(\sigma - m_b)(\sigma - m_c)}.</math>
Next, if Template:Tmath, Template:Tmath, and Template:Tmath are the altitudes from sides Template:Tmath Template:Tmath and Template:Tmath 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>Template:Cite journal</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 Template:Tmath Template:Tmath and Template:Tmath are the three angle measures of the triangle, and the semi-sum of their sines is <math>S = \tfrac12(\sin\alpha + \sin\beta + \sin\gamma),</math> then<ref>Template:Cite journal</ref><ref>Template:Cite journal</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 Template:Tmath 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.
GeneralizationsEdit
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 Template:Tmath of a cyclic quadrilateral whose sides have lengths Template:Tmath Template:Tmath Template:Tmath Template:Tmath 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 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, <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 three-simplex.
Another generalization of Heron's formula to pentagons and hexagons inscribed in a circle was discovered by David P. Robbins.<ref>Template:Cite journal</ref>
Degenerate and imaginary trianglesEdit
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 result. For example if Template:Tmath and Template:Tmath, then Template:Tmath. This can be interpreted using a triangle in the complex coordinate plane Template:Tmath, where "area" can be a complex-valued quantity, or as a triangle lying in a pseudo-Euclidean plane with one space-like dimension and one time-like dimension.<ref>Template:Cite journal</ref>
Volume of a tetrahedronEdit
If Template:Tmath Template:Tmath Template:Tmath Template:Tmath Template:Tmath Template:Tmath are lengths of edges of the tetrahedron (first three form a triangle; Template:Tmath opposite to Template:Tmath and so on), then<ref>{{#invoke:citation/CS1|citation |CitationClass=web }}</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 geometryEdit
L'Huilier's formula relates the area of a triangle in spherical geometry to its side lengths. For a spherical triangle with side lengths Template:Tmath Template:Tmath and Template:Tmath, semiperimeter Template:Tmath, and area Template:Tmath,<ref>Template:Cite book </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 alsoEdit
Notes and referencesEdit
<references/>
External linksEdit
- 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
- Template:MathPages
- A Geometric Proof of Heron's Formula
- An alternative proof of Heron's Formula without words
- Factoring Heron