Heron's Formula - Wikipedia

Triangle area in terms of side lengths This article is about calculating the area of a triangle. For calculating a square root, see Square root algorithms § Heron's method.

In geometry, Heron's formula (or Hero's formula) gives the area of a triangle in terms of the three side lengths ⁠ a , {\displaystyle a,} ⁠ ⁠ b , {\displaystyle b,} ⁠ ⁠ c . {\displaystyle c.} ⁠ Letting ⁠ s {\displaystyle s} ⁠ be the semiperimeter of the triangle, ⁠ s = 1 2 ( a + b + c ) {\displaystyle s={\tfrac {1}{2}}(a+b+c)} ⁠, the area ⁠ A {\displaystyle A} ⁠ is[1]

A = s ( s − a ) ( s − b ) ( s − c ) . {\displaystyle 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

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Area calculator
a	3
b	4
c	5
s	6
Area[2]	6.000

51

Let ⁠ △ A B C {\displaystyle \triangle ABC} ⁠ be the triangle with sides ⁠ a = 4 {\displaystyle a=4} ⁠, ⁠ b = 13 {\displaystyle b=13} ⁠, and ⁠ c = 15 {\displaystyle c=15} ⁠. This triangle's semiperimeter is s = 1 2 ( a + b + c ) = {\displaystyle s={\tfrac {1}{2}}(a+b+c)={}} 1 2 ( 4 + 13 + 15 ) = 16 {\displaystyle {\tfrac {1}{2}}(4+13+15)=16} therefore ⁠ s − a = 12 {\displaystyle s-a=12} ⁠, ⁠ s − b = 3 {\displaystyle s-b=3} ⁠, ⁠ s − c = 1 {\displaystyle s-c=1} ⁠, and the area is A = s ( s − a ) ( s − b ) ( s − c ) = 16 ⋅ 12 ⋅ 3 ⋅ 1 ) = 24. {\displaystyle {\begin{aligned}A&={\textstyle {\sqrt {s(s-a)(s-b)(s-c)}}}\\[3mu]&={\textstyle {\sqrt {16\cdot 12\cdot 3\cdot 1{\vphantom {)}}}}}\\[3mu]&=24.\end{aligned}}}

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 expressions

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

A = 1 4 ( a + b + c ) ( − a + b + c ) ( a − b + c ) ( a + b − c ) = 1 4 2 ( a 2 b 2 + a 2 c 2 + b 2 c 2 ) − ( a 4 + b 4 + c 4 ) = 1 4 ( a 2 + b 2 + c 2 ) ) 2 − 2 ( a 4 + b 4 + c 4 ) = 1 4 4 ( a 2 b 2 + a 2 c 2 + b 2 c 2 ) − ( a 2 + b 2 + c 2 ) ) 2 = 1 4 4 a 2 b 2 − ( a 2 + b 2 − c 2 ) ) 2 . {\displaystyle {\begin{aligned}A&={\tfrac {1}{4}}{\sqrt {(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}\\[6mu]&={\tfrac {1}{4}}{\sqrt {2{\bigl (}a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2}{\bigr )}-{\bigl (}a^{4}+b^{4}+c^{4}{\bigr )}}}\\[6mu]&={\tfrac {1}{4}}{\sqrt {{\bigl (}a^{2}+b^{2}+c^{2}{\bigr )}{\vphantom {)}}^{2}-2{\bigl (}a^{4}+b^{4}+c^{4}{\bigr )}}}\\[6mu]&={\tfrac {1}{4}}{\sqrt {4{\bigl (}a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2}{\bigr )}-{\bigl (}a^{2}+b^{2}+c^{2}{\bigr )}{\vphantom {)}}^{2}}}\\[6mu]&={\tfrac {1}{4}}{\sqrt {4a^{2}b^{2}-{\bigl (}a^{2}+b^{2}-c^{2}{\bigr )}{\vphantom {)}}^{2}}}.\end{aligned}}}

After expansion, the expression under the square root is a quadratic polynomial of the squared side lengths ⁠ a 2 {\displaystyle \textstyle a^{2}} ⁠, ⁠ b 2 {\displaystyle \textstyle b^{2}} ⁠, ⁠ c 2 {\displaystyle \textstyle c^{2}} ⁠.

The same relation can be expressed using the Cayley–Menger determinant,[3]

− 16 A 2 = | 0 a 2 b 2 1 a 2 0 c 2 1 b 2 c 2 0 1 1 1 1 0 | . {\displaystyle -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}}.}

History

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The formula is credited to Heron (or Hero) of Alexandria (fl. 60 AD),[4] and a proof can be found in his book Metrica. Mathematical historian Thomas Heath suggested that Archimedes knew the formula over two centuries earlier,[5] 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.[6]

A formula equivalent to Heron's was discovered by Chinese mathematician Qin Jiushao:

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

published in Mathematical Treatise in Nine Sections (Qin Jiushao, 1247).[7]

Proofs

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There are many ways to prove Heron's formula, for example using trigonometry as below, or the incenter and one excircle of the triangle,[8] or as a special case of De Gua's theorem (for the particular case of acute triangles),[9] or as a special case of Brahmagupta's formula (for the case of a degenerate cyclic quadrilateral).

Trigonometric proof using the law of cosines

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A modern proof, which uses algebra and is quite different from the one provided by Heron, follows.[10] Let ⁠ a , {\displaystyle a,} ⁠ ⁠ b , {\displaystyle b,} ⁠ ⁠ c {\displaystyle c} ⁠ be the sides of the triangle and ⁠ α , {\displaystyle \alpha ,} ⁠ ⁠ β , {\displaystyle \beta ,} ⁠ ⁠ γ {\displaystyle \gamma } ⁠ the angles opposite those sides. Applying the law of cosines we get

cos ⁡ γ = a 2 + b 2 − c 2 2 a b {\displaystyle \cos \gamma ={\frac {a^{2}+b^{2}-c^{2}}{2ab}}}

From this proof, we get the algebraic statement that

sin ⁡ γ = 1 − cos 2 ⁡ γ = 4 a 2 b 2 − ( a 2 + b 2 − c 2 ) ) 2 2 a b . {\displaystyle \sin \gamma ={\sqrt {1-\cos ^{2}\gamma }}={\frac {\sqrt {4a^{2}b^{2}-{\bigl (}a^{2}+b^{2}-c^{2}{\bigr )}{\vphantom {)}}^{2}}}{2ab}}.}

The altitude of the triangle on base ⁠ a {\displaystyle a} ⁠ has length ⁠ b sin ⁡ γ {\displaystyle b\sin \gamma } ⁠, and it follows

A = 1 2 ( base ) ( altitude ) = 1 2 a b sin ⁡ γ = a b 4 a b 4 a 2 b 2 − ( a 2 + b 2 − c 2 ) ) 2 = 1 4 − a 4 − b 4 − c 4 + 2 a 2 b 2 + 2 a 2 c 2 + 2 b 2 c 2 = 1 4 ( a + b + c ) ( − a + b + c ) ( a − b + c ) ( a + b − c ) = ( a + b + c 2 ) ( − a + b + c 2 ) ( a − b + c 2 ) ( a + b − c 2 ) = s ( s − a ) ( s − b ) ( s − c ) . {\displaystyle {\begin{aligned}A&={\tfrac {1}{2}}({\mbox{base}})({\mbox{altitude}})\\[6mu]&={\tfrac {1}{2}}ab\sin \gamma \\[6mu]&={\frac {ab}{4ab}}{\sqrt {4a^{2}b^{2}-{\bigl (}a^{2}+b^{2}-c^{2}{\bigr )}{\vphantom {)}}^{2}}}\\[6mu]&={\tfrac {1}{4}}{\sqrt {-a^{4}-b^{4}-c^{4}+2a^{2}b^{2}+2a^{2}c^{2}+2b^{2}c^{2}}}\\[6mu]&={\tfrac {1}{4}}{\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{aligned}}}

Algebraic proof using the Pythagorean theorem

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The following proof is very similar to one given by Raifaizen.[11] By the Pythagorean theorem we have b 2 = h 2 + d 2 {\displaystyle b^{2}=h^{2}+d^{2}} and a 2 = h 2 + ( c − d ) 2 {\displaystyle a^{2}=h^{2}+(c-d)^{2}} according to the figure at the right. Subtracting these yields a 2 − b 2 = c 2 − 2 c d . {\displaystyle a^{2}-b^{2}=c^{2}-2cd.} This equation allows us to express ⁠ d {\displaystyle d} ⁠ in terms of the sides of the triangle: d = − a 2 + b 2 + c 2 2 c . {\displaystyle d={\frac {-a^{2}+b^{2}+c^{2}}{2c}}.} For the height of the triangle we have that h 2 = b 2 − d 2 . {\displaystyle h^{2}=b^{2}-d^{2}.} By replacing ⁠ d {\displaystyle d} ⁠ with the formula given above and applying the difference of squares identity we get h 2 = b 2 − ( − a 2 + b 2 + c 2 2 c ) 2 = ( 2 b c − a 2 + b 2 + c 2 ) ( 2 b c + 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 . {\displaystyle {\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: A = c h 2 = c 2 4 ⋅ 4 s ( s − a ) ( s − b ) ( s − c ) c 2 = s ( s − a ) ( s − b ) ( s − c ) . {\displaystyle {\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

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If ⁠ r {\displaystyle r} ⁠ is the radius of the incircle of the triangle, then the triangle can be broken into three triangles of equal altitude ⁠ r {\displaystyle r} ⁠ and bases ⁠ a , {\displaystyle a,} ⁠ ⁠ b , {\displaystyle b,} ⁠ and ⁠ c . {\displaystyle c.} ⁠ Their combined area is A = 1 2 a r + 1 2 b r + 1 2 c r = r s , {\displaystyle A={\tfrac {1}{2}}ar+{\tfrac {1}{2}}br+{\tfrac {1}{2}}cr=rs,} where s = 1 2 ( a + b + c ) {\displaystyle s={\tfrac {1}{2}}(a+b+c)} is the semiperimeter.

The triangle can alternately be broken into six triangles (in congruent pairs) of altitude ⁠ r {\displaystyle r} ⁠ and bases ⁠ s − a , {\displaystyle s-a,} ⁠ ⁠ s − b , {\displaystyle s-b,} ⁠ and ⁠ s − c {\displaystyle s-c} ⁠ of combined area (see law of cotangents) A = r ( s − a ) + r ( s − b ) + r ( s − c ) = r 2 ( s − a r + s − b r + s − c r ) = r 2 ( cot ⁡ α 2 + cot ⁡ β 2 + cot ⁡ γ 2 ) = r 2 ( cot ⁡ α 2 cot ⁡ β 2 cot ⁡ γ 2 ) = r 2 ( s − a r ⋅ s − b r ⋅ s − c r ) = ( s − a ) ( s − b ) ( s − c ) r . {\displaystyle {\begin{aligned}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{aligned}}}

The middle step above is cot ⁡ α 2 + cot ⁡ β 2 + cot ⁡ γ 2 = {\textstyle \cot {\tfrac {\alpha }{2}}+\cot {\tfrac {\beta }{2}}+\cot {\tfrac {\gamma }{2}}={}} cot ⁡ α 2 cot ⁡ β 2 cot ⁡ γ 2 {\displaystyle \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 α 2 + β 2 + γ 2 = π 2 . {\textstyle {\tfrac {\alpha }{2}}+{\tfrac {\beta }{2}}+{\tfrac {\gamma }{2}}={\tfrac {\pi }{2}}.}

Combining the two, we get A 2 = s ( s − a ) ( s − b ) ( s − c ) , {\displaystyle A^{2}=s(s-a)(s-b)(s-c),} from which the result follows.

Numerical stability

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Heron's formula as given above is numerically unstable for triangles with a very small angle, causing round-off error when computing with limited precision such as when using floating-point arithmetic. Such triangles have one or two sides whose length is very close to the semiperimeter, leading to catastrophic cancellation. A stable alternative involves arranging the lengths of the sides so that a ≥ b ≥ c {\displaystyle a\geq b\geq c} and computing[12] A = 1 4 ( a + ( b + c ) ) ( c − ( a − b ) ) ( c + ( a − b ) ) ( a + ( b − c ) ) . {\displaystyle A={\tfrac {1}{4}}{\sqrt {{\big (}a+(b+c){\big )}{\big (}c-(a-b){\big )}{\big (}c+(a-b){\big )}{\big (}a+(b-c){\big )}}}.} The extra parentheses indicate the order of operations required to achieve numerical stability in the evaluation.

Similar triangle-area formulae

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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 ⁠ m a , {\displaystyle m_{a},} ⁠ ⁠ m b , {\displaystyle m_{b},} ⁠ and ⁠ m c {\displaystyle m_{c}} ⁠ are the medians from sides ⁠ a , {\displaystyle a,} ⁠ ⁠ b , {\displaystyle b,} ⁠ and ⁠ c {\displaystyle c} ⁠ respectively, and their semi-sum is ⁠ σ = 1 2 ( m a + m b + m c ) {\displaystyle \sigma ={\tfrac {1}{2}}(m_{a}+m_{b}+m_{c})} ⁠, then[13] A = 4 3 σ ( σ − m a ) ( σ − m b ) ( σ − m c ) . {\displaystyle A={\tfrac {4}{3}}{\sqrt {\sigma (\sigma -m_{a})(\sigma -m_{b})(\sigma -m_{c})}}.}

Next, if ⁠ h a {\displaystyle h_{a}} ⁠, ⁠ h b {\displaystyle h_{b}} ⁠, and ⁠ h c {\displaystyle h_{c}} ⁠ are the altitudes from sides ⁠ a , {\displaystyle a,} ⁠ ⁠ b , {\displaystyle b,} ⁠ and ⁠ c {\displaystyle c} ⁠ respectively, and semi-sum of their reciprocals is ⁠ H = 1 2 ( h a − 1 + h b − 1 + h c − 1 ) {\displaystyle \textstyle H={\tfrac {1}{2}}{\bigl (}h_{a}^{-1}+h_{b}^{-1}+h_{c}^{-1}{\bigr )}} ⁠, then[14] A − 1 = 4 H ( H − h a − 1 ) ( H − h b − 1 ) ( H − h c − 1 ) . {\displaystyle A^{-1}=4{\sqrt {H{\bigl (}H-h_{a}^{-1}{\bigr )}{\bigl (}H-h_{b}^{-1}{\bigr )}{\bigl (}H-h_{c}^{-1}{\bigr )}}}.}

Finally, if ⁠ α , {\displaystyle \alpha ,} ⁠ ⁠ β , {\displaystyle \beta ,} ⁠ and ⁠ γ {\displaystyle \gamma } ⁠ are the three angle measures of the triangle, and the semi-sum of their sines is ⁠ S = 1 2 ( sin ⁡ α + sin ⁡ β + sin ⁡ γ ) {\displaystyle S={\tfrac {1}{2}}(\sin \alpha +\sin \beta +\sin \gamma )} ⁠, then[15][16] A = D 2 S ( S − sin ⁡ α ) ( S − sin ⁡ β ) ( S − sin ⁡ γ ) = 1 2 D 2 sin ⁡ α sin ⁡ β sin ⁡ γ , {\displaystyle {\begin{aligned}A&=D^{2}{\sqrt {S(S-\sin \alpha )(S-\sin \beta )(S-\sin \gamma )}}\\[5mu]&={\tfrac {1}{2}}D^{2}\sin \alpha \,\sin \beta \,\sin \gamma ,\end{aligned}}}

where ⁠ D {\displaystyle D} ⁠ is the diameter of the circumcircle, D = a / sin ⁡ α = b / sin ⁡ β = c / sin ⁡ γ . {\displaystyle 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

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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 ⁠ K {\displaystyle K} ⁠ of a cyclic quadrilateral whose sides have lengths ⁠ a , {\displaystyle a,} ⁠ ⁠ b , {\displaystyle b,} ⁠ ⁠ c , {\displaystyle c,} ⁠ ⁠ d {\displaystyle d} ⁠ as

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

where s = 1 2 ( a + b + c + d ) {\displaystyle s={\tfrac {1}{2}}(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 − | 0 a 2 b 2 1 a 2 0 c 2 1 b 2 c 2 0 1 1 1 1 0 | {\displaystyle 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.[17]

Degenerate and imaginary triangles

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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 ⁠ a = 3 {\displaystyle a=3} ⁠ and ⁠ b = c = 1 {\displaystyle b=c=1} ⁠, then ⁠ A = 3 5 4 i {\displaystyle \textstyle A={\tfrac {3{\sqrt {5}}}{4}}i} ⁠. This can be interpreted using a triangle in the complex coordinate plane ⁠ C 2 {\displaystyle \mathbb {C} ^{2}} ⁠, 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.[18]

Volume of a tetrahedron

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If ⁠ U , {\displaystyle U,} ⁠ ⁠ V , {\displaystyle V,} ⁠ ⁠ W , {\displaystyle W,} ⁠ ⁠ u , {\displaystyle u,} ⁠ ⁠ v , {\displaystyle v,} ⁠ ⁠ w {\displaystyle w} ⁠ are lengths of edges of the tetrahedron (first three form a triangle; ⁠ u {\displaystyle u} ⁠ opposite to ⁠ U {\displaystyle U} ⁠ and so on), then[19] volume = ( − a + b + c + d ) ( a − b + c + d ) ( a + b − c + d ) ( a + b + c − d ) 192 u v w {\displaystyle {\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 a = y x Y Z , b = X y Z , c = y X Y z , d = X x y z , X = ( − U + v + w ) ( U + v + w ) , x = ( U − v + w ) ( U + v − w ) , Y = ( − V + w + u ) ( V + w + u ) , y = ( V − w + u ) ( V + w − u ) , Z = ( − W + u + v ) ( W + u + v ) , z = ( W − u + v ) ( W + u − v ) . {\displaystyle {\begin{aligned}a&={\sqrt {{\vphantom {y}}xYZ}},\qquad b={\sqrt {XyZ}},&c&={\sqrt {{\vphantom {y}}XYz}},\qquad d={\sqrt {{\vphantom {X}}xyz}},\\[4mu]X&=(-U+v+w)\,(U+v+w),&x&=(U-v+w)\,(U+v-w),\\Y&=(-V+w+u)\,(V+w+u),&y&=(V-w+u)\,(V+w-u),\\Z&=(-W+u+v)\,(W+u+v),&z&=(W-u+v)\,(W+u-v).\end{aligned}}}

Spherical and hyperbolic geometry

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L'Huilier's formula relates the area of a triangle in spherical geometry to its side lengths. For a spherical triangle with side lengths ⁠ a , {\displaystyle a,} ⁠ ⁠ b , {\displaystyle b,} ⁠ and ⁠ c {\displaystyle c} ⁠, semiperimeter ⁠ s = 1 2 ( a + b + c ) {\displaystyle s={\tfrac {1}{2}}(a+b+c)} ⁠, and area ⁠ S {\displaystyle S} ⁠,[20] tan 2 ⁡ S 4 = tan ⁡ s 2 tan ⁡ s − a 2 tan ⁡ s − b 2 tan ⁡ s − c 2 {\displaystyle \tan ^{2}{\frac {S}{4}}=\tan {\frac {s}{2}}\tan {\frac {s-a}{2}}\tan {\frac {s-b}{2}}\tan {\frac {s-c}{2}}}

For a triangle in hyperbolic geometry the analogous formula is tan 2 ⁡ S 4 = tanh ⁡ s 2 tanh ⁡ s − a 2 tanh ⁡ s − b 2 tanh ⁡ s − c 2 . {\displaystyle \tan ^{2}{\frac {S}{4}}=\tanh {\frac {s}{2}}\tanh {\frac {s-a}{2}}\tanh {\frac {s-b}{2}}\tanh {\frac {s-c}{2}}.}

See also

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• Shoelace formula

Notes and references

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1. ^ Kendig, Keith (2000). "Is a 2000-year-old formula still keeping some secrets?". The American Mathematical Monthly. 107 (5): 402–415. doi:10.1080/00029890.2000.12005213. JSTOR 2695295. MR 1763392. S2CID 1214184. Archived from the original on 2024-05-29. Retrieved 2021-12-27.
2. ^ The formula used here is the numerically stable formula (relabeled for ⁠ a ≤ b ≤ c {\displaystyle a\leq b\leq c} ⁠), not simply ⁠   s ( s − a ) ( s − b ) ( s − c ) {\displaystyle \textstyle ~\!\!{\sqrt {s(s-a)(s-b)(s-c)}}\!} ⁠. For example, with ⁠ a = 3 {\displaystyle a=3} ⁠, ⁠ b = 4 {\displaystyle b=4} ⁠, ⁠ c = 6.999 {\displaystyle c=6.999} ⁠, the correct area is ⁠ 0.205 {\displaystyle 0.205} ⁠ but the naive implementation produces ⁠ 0.000 {\displaystyle 0.000} ⁠ instead. 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.
3. ^ Havel, Timothy F. (1991). "Some examples of the use of distances as coordinates for Euclidean geometry". Journal of Symbolic Computation. 11 (5–6): 579–593. doi:10.1016/S0747-7171(08)80120-4.
4. ^ Id, Yusuf; Kennedy, E. S. (1969). "A medieval proof of Heron's formula". The Mathematics Teacher. 62 (7): 585–587. doi:10.5951/MT.62.7.0585. JSTOR 27958225. MR 0256819.
5. ^ Heath, Thomas L. (1921). A History of Greek Mathematics. Vol. II. Oxford University Press. pp. 321–323.
6. ^ Weisstein, Eric W. "Heron's Formula". MathWorld.
7. ^ 秦, 九韶 (1773). "卷三上, 三斜求积". 數學九章 (四庫全書本) (in Chinese).
8. ^ "Personal email communication between mathematicians John Conway and Peter Doyle". 15 December 1997. Retrieved 25 September 2020.
9. ^ Lévy-Leblond, Jean-Marc (2020-09-14). "A Symmetric 3D Proof of Heron's Formula". The Mathematical Intelligencer. 43 (2): 37–39. doi:10.1007/s00283-020-09996-8. ISSN 0343-6993.
10. ^ Niven, Ivan (1981). Maxima and Minima Without Calculus. The Mathematical Association of America. pp. 7–8.
11. ^ Raifaizen, Claude H. (1971). "A Simpler Proof of Heron's Formula". Mathematics Magazine. 44 (1): 27–28. doi:10.1080/0025570X.1971.11976093.
12. ^ Kahan, William M. (1983). "Mathematics Written in Sand – the hp-15C, Intel 8087, etc." (PDF). Proceedings of the American Statistical Association, Statistical Computing Section. pp. 12–26. See "the area of a triangle", pp. 10–11 of reformatted postprint version.
13. ^ Bényi, Árpád (July 2003). "A Heron-type formula for the triangle". Mathematical Gazette. 87: 324–326. doi:10.1017/S0025557200172882.
14. ^ Mitchell, Douglas W. (November 2005). "A Heron-type formula for the reciprocal area of a triangle". Mathematical Gazette. 89: 494. doi:10.1017/S0025557200178532.
15. ^ Mitchell, Douglas W. (2009). "A Heron-type area formula in terms of sines". Mathematical Gazette. 93: 108–109. doi:10.1017/S002555720018430X. S2CID 132042882.
16. ^ Kocik, Jerzy; Solecki, Andrzej (2009). "Disentangling a triangle" (PDF). American Mathematical Monthly. 116 (3): 228–237. doi:10.1080/00029890.2009.11920932. S2CID 28155804.
17. ^ Robbins, D. P. (1994). "Areas of Polygons Inscribed in a Circle". Discrete & Computational Geometry. 12 (2): 223–236. doi:10.1007/BF02574377.
18. ^ Schwartz, Mark (2007). "Review of Conics". The American Mathematical Monthly. 114 (5): 461–464. ISSN 0002-9890. JSTOR 27642242.
19. ^ Kahan, William (3 April 2012). "What has the Volume of a Tetrahedron to do with Computer Programming Languages?" (PDF). pp. 16–17. Retrieved 2025-10-28.
20. ^ Alekseevskij, D. V.; Vinberg, E. B.; Solodovnikov, A. S. (1993). "Geometry of spaces of constant curvature". In Gamkrelidze, R. V.; Vinberg, E. B. (eds.). Geometry. II: Spaces of constant curvature. Encyclopaedia of Mathematical Sciences. Vol. 29. Springer-Verlag. p. 66. ISBN 1-56085-072-8.

External links

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• 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 Archived 2018-09-08 at the Wayback Machine
• An alternative proof of Heron's Formula without words
• Factoring Heron

v t e Ancient Greek mathematics
Mathematicians(timeline)	Anaxagoras Anthemius Apollonius Archimedes Archytas Aristaeus the Elder Aristarchus Autolycus Bion Bryson Callippus Carpus Chrysippus Cleomedes Conon Ctesibius Democritus Dicaearchus Dinostratus Diocles Dionysodorus of Caunus Dionysodorus of Amisene Diophantus Domninus Eratosthenes Euclid Eudemus Eudoxus Eutocius Geminus Heliodorus Heron Hipparchus Hippasus Hippias Hippocrates Hypatia Hypsicles Isidore of Miletus Leon Marinus Menaechmus Menelaus Metrodorus Nicomachus Nicomedes Nicoteles Oenopides Pandrosion Pappus Perseus Philolaus Philon Philonides Porphyry of Tyre Posidonius Proclus Ptolemy Pythagoras Serenus Sosigenes Sporus Thales Theaetetus Theodorus Theodosius Theon of Alexandria Theon of Smyrna Thymaridas Xenocrates Zeno of Elea Zeno of Sidon Zenodorus
Treatises	Almagest Arithmetica Conics (Apollonius) Catoptrics Data (Euclid) Elements (Euclid) Little Astronomy Measurement of a Circle On Conoids and Spheroids On the Sizes and Distances (Aristarchus) On Sizes and Distances (Hipparchus) On the Moving Sphere (Autolycus) Optics (Euclid) On Spirals On the Sphere and Cylinder Ostomachion Phaenomena (Euclid) Planisphaerium Spherics (Theodosius) Spherics (Menelaus) The Quadrature of the Parabola The Sand Reckoner
Conceptsand definitions	Chord Circles of Apollonius Apollonian circles Apollonian gasket Problem of Apollonius Commensurability Diophantine equation Euclidean geometry Golden ratio Lune of Hippocrates Method of exhaustion Parallel postulate Platonic solid Regular polygon Straightedge and compass construction Angle trisection Doubling the cube Squaring the circle Quadratrix of Hippias Neusis construction
Results	In Elements Angle bisector theorem Exterior angle theorem Euclidean algorithm Euclid's theorem Geometric mean theorem Hinge theorem Inscribed angle theorem Intercept theorem Intersecting chords theorem Intersecting secants theorem Law of cosines Pons asinorum Pythagorean theorem Tangent-secant theorem Thales's theorem Theorem of the gnomon Apollonius's theorem Aristarchus's inequality Heron's formula Law of sines Menelaus's theorem Pappus's area theorem Problem II.8 of Arithmetica Ptolemy's inequality Ptolemy's table of chords Ptolemy's theorem Spiral of Theodorus
Centers/Schools	Cyrene Platonic Academy Pythagoreanism School of Chios
Related	Ancient Greek astronomy Attic numerals Greek numerals History of A History of Greek Mathematics by Thomas Heath Archimedes Palimpsest algebra timeline arithmetic timeline calculus timeline geometry timeline logic timeline mathematics timeline numbers prehistoric counting numeral systems list Other cultures Arabian/Islamic Babylonian Chinese Egyptian Inca Indian Japanese
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