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Heron's formula

From Wikipedia, the free encyclopedia
Triangle area in terms of side lengths
This article is about calculating the area of a triangle. For calculating a square root, seeSquare root algorithms § Heron's method.
A triangle with sidesa,b, andc

Ingeometry,Heron's formula (orHero's formula) gives thearea of a triangle in terms of the three side lengths⁠a,{\displaystyle a,}⁠⁠b,{\displaystyle b,}⁠⁠c.{\displaystyle c.}⁠ Letting⁠s{\displaystyle s}⁠ be thesemiperimeter of the triangle,⁠s=12(a+b+c){\displaystyle s={\tfrac {1}{2}}(a+b+c)}⁠, the area⁠A{\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 engineerHeron of Alexandria (or Hero) who proved it in his workMetrica, though it was probably known centuries earlier.

Example

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Area calculator
a3
b4
c5
s6
Area[2]6.000
51

Let⁠△ABC{\displaystyle \triangle ABC}⁠ be the triangle with sides⁠a=4{\displaystyle a=4}⁠,⁠b=13{\displaystyle b=13}⁠, and⁠c=15{\displaystyle c=15}⁠.This triangle's semiperimeter iss=12(a+b+c)={\displaystyle s={\tfrac {1}{2}}(a+b+c)={}}12(4+13+15)=16{\displaystyle {\tfrac {1}{2}}(4+13+15)=16} therefore⁠s−a=12{\displaystyle s-a=12}⁠,⁠s−b=3{\displaystyle s-b=3}⁠,⁠s−c=1{\displaystyle s-c=1}⁠, and the area isA=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 areintegers, making it aHeronian triangle. However, Heron's formula works equally well when the side lengths arereal numbers. As long as they obey the stricttriangle inequality, they define a triangle in theEuclidean 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=14(a+b+c)(−a+b+c)(a−b+c)(a+b−c)=142(a2b2+a2c2+b2c2)−(a4+b4+c4)=14(a2+b2+c2))2−2(a4+b4+c4)=144(a2b2+a2c2+b2c2)−(a2+b2+c2))2=144a2b2−(a2+b2−c2))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 aquadratic polynomial of the squared side lengths⁠a2{\displaystyle \textstyle a^{2}}⁠,⁠b2{\displaystyle \textstyle b^{2}}⁠,⁠c2{\displaystyle \textstyle c^{2}}⁠.

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

−16A2=|0a2b21a20c21b2c2011110|.{\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 toHeron (or Hero) of Alexandria (fl. 60 AD),[4] and a proof can be found in his bookMetrica. Mathematical historianThomas Heath suggested thatArchimedes knew the formula over two centuries earlier,[5] and sinceMetrica 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=12a2c2−(a2+c2−b22)2,{\displaystyle A={\frac {1}{2}}{\sqrt {a^{2}c^{2}-\left({\frac {a^{2}+c^{2}-b^{2}}{2}}\right)^{2}}},}

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

Proofs

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There are many ways to prove Heron's formula, for example usingtrigonometry as below, or theincenter and oneexcircle of the triangle,[8] or as a special case ofDe Gua's theorem (for the particular case of acute triangles),[9] or as a special case ofBrahmagupta'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 usesalgebra and is quite different from the one provided by Heron, follows.[10]Let⁠a,{\displaystyle a,}⁠⁠b,{\displaystyle b,}⁠⁠c{\displaystyle c}⁠ be the sides of the triangle and⁠α,{\displaystyle \alpha ,}⁠⁠β,{\displaystyle \beta ,}⁠⁠γ{\displaystyle \gamma }⁠ theangles opposite those sides.Applying thelaw of cosines we get

cos⁡γ=a2+b2−c22ab{\displaystyle \cos \gamma ={\frac {a^{2}+b^{2}-c^{2}}{2ab}}}

A triangle with sidesa,b andc

From this proof, we get the algebraic statement that

sin⁡γ=1−cos2⁡γ=4a2b2−(a2+b2−c2))22ab.{\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}}.}

Thealtitude of the triangle on base⁠a{\displaystyle a}⁠ has length⁠bsin⁡γ{\displaystyle b\sin \gamma }⁠, and it follows

A=12(base)(altitude)=12absin⁡γ=ab4ab4a2b2−(a2+b2−c2))2=14−a4−b4−c4+2a2b2+2a2c2+2b2c2=14(a+b+c)(−a+b+c)(a−b+c)(a+b−c)=(a+b+c2)(−a+b+c2)(a−b+c2)(a+b−c2)=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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Triangle with altitudeh cutting basec intod + (c −d)

The following proof is very similar to one given by Raifaizen.[11]By thePythagorean theorem we haveb2=h2+d2{\displaystyle b^{2}=h^{2}+d^{2}} anda2=h2+(c−d)2{\displaystyle a^{2}=h^{2}+(c-d)^{2}} according to the figure at the right. Subtracting these yieldsa2−b2=c2−2cd.{\displaystyle a^{2}-b^{2}=c^{2}-2cd.} This equation allows us to express⁠d{\displaystyle d}⁠ in terms of the sides of the triangle:d=−a2+b2+c22c.{\displaystyle d={\frac {-a^{2}+b^{2}+c^{2}}{2c}}.}For the height of the triangle we have thath2=b2−d2.{\displaystyle h^{2}=b^{2}-d^{2}.} By replacing⁠d{\displaystyle d}⁠ with the formula given above and applying thedifference of squares identity we geth2=b2−(−a2+b2+c22c)2=(2bc−a2+b2+c2)(2bc+a2−b2−c2)4c2=((b+c)2−a2)(a2−(b−c)2)4c2=(b+c−a)(b+c+a)(a+b−c)(a−b+c)4c2=2(s−a)⋅2s⋅2(s−c)⋅2(s−b)4c2=4s(s−a)(s−b)(s−c)c2.{\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=ch2=c24⋅4s(s−a)(s−b)(s−c)c2=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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Geometrical significance ofs −a,s −b, ands −c. See thelaw of cotangents for the reasoning behind this.

If⁠r{\displaystyle r}⁠ is the radius of theincircle of the triangle, then the triangle can be broken into three triangles of equal altitude⁠r{\displaystyle r}⁠ and bases⁠a,{\displaystyle a,}⁠⁠b,{\displaystyle b,}⁠ and⁠c.{\displaystyle c.}⁠ Their combined area isA=12ar+12br+12cr=rs,{\displaystyle A={\tfrac {1}{2}}ar+{\tfrac {1}{2}}br+{\tfrac {1}{2}}cr=rs,}wheres=12(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 altitude⁠r{\displaystyle r}⁠ and bases⁠s−a,{\displaystyle s-a,}⁠⁠s−b,{\displaystyle s-b,}⁠ and⁠s−c{\displaystyle s-c}⁠ of combined area (seelaw of cotangents)A=r(s−a)+r(s−b)+r(s−c)=r2(s−ar+s−br+s−cr)=r2(cot⁡α2+cot⁡β2+cot⁡γ2)=r2(cot⁡α2cot⁡β2cot⁡γ2)=r2(s−ar⋅s−br⋅s−cr)=(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 iscot⁡α2+cot⁡β2+cot⁡γ2={\textstyle \cot {\tfrac {\alpha }{2}}+\cot {\tfrac {\beta }{2}}+\cot {\tfrac {\gamma }{2}}={}}cot⁡α2cot⁡β2cot⁡γ2{\displaystyle \cot {\tfrac {\alpha }{2}}\cot {\tfrac {\beta }{2}}\cot {\tfrac {\gamma }{2}}}, thetriple cotangent identity, which applies because the sum of half-angles isα2+β2+γ2=π2.{\textstyle {\tfrac {\alpha }{2}}+{\tfrac {\beta }{2}}+{\tfrac {\gamma }{2}}={\tfrac {\pi }{2}}.}

Combining the two, we getA2=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 isnumerically unstable for triangles with a very small angle, causinground-off error when computing with limited precision such as when usingfloating-point arithmetic. Such triangles have one or two sides whose length is very close to the semiperimeter, leading tocatastrophic cancellation. A stable alternative involves arranging the lengths of the sides so thata≥b≥c{\displaystyle a\geq b\geq c} and computing[12]A=14(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, if⁠ma,{\displaystyle m_{a},}⁠⁠mb,{\displaystyle m_{b},}⁠ and⁠mc{\displaystyle m_{c}}⁠ are themedians from sides⁠a,{\displaystyle a,}⁠⁠b,{\displaystyle b,}⁠ and⁠c{\displaystyle c}⁠ respectively, and their semi-sum is⁠σ=12(ma+mb+mc){\displaystyle \sigma ={\tfrac {1}{2}}(m_{a}+m_{b}+m_{c})}⁠, then[13]A=43σ(σ−ma)(σ−mb)(σ−mc).{\displaystyle A={\tfrac {4}{3}}{\sqrt {\sigma (\sigma -m_{a})(\sigma -m_{b})(\sigma -m_{c})}}.}

Next, if⁠ha{\displaystyle h_{a}}⁠,⁠hb{\displaystyle h_{b}}⁠, and⁠hc{\displaystyle h_{c}}⁠ are thealtitudes from sides⁠a,{\displaystyle a,}⁠⁠b,{\displaystyle b,}⁠ and⁠c{\displaystyle c}⁠ respectively, and semi-sum of their reciprocals is⁠H=12(ha−1+hb−1+hc−1){\displaystyle \textstyle H={\tfrac {1}{2}}{\bigl (}h_{a}^{-1}+h_{b}^{-1}+h_{c}^{-1}{\bigr )}}⁠, then[14]A−1=4H(H−ha−1)(H−hb−1)(H−hc−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, if⁠α,{\displaystyle \alpha ,}⁠⁠β,{\displaystyle \beta ,}⁠ and⁠γ{\displaystyle \gamma }⁠ are the three angle measures of the triangle, and the semi-sum of theirsines is⁠S=12(sin⁡α+sin⁡β+sin⁡γ){\displaystyle S={\tfrac {1}{2}}(\sin \alpha +\sin \beta +\sin \gamma )}⁠, then[15][16]A=D2S(S−sin⁡α)(S−sin⁡β)(S−sin⁡γ)=12D2sin⁡αsin⁡βsin⁡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}}}

where⁠D{\displaystyle D}⁠ is the diameter of thecircumcircle,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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Cyclic Quadrilateral

Heron's formula is a special case ofBrahmagupta's formula for the area of acyclic quadrilateral. Heron's formula and Brahmagupta's formula are both special cases ofBretschneider's formula for the area of aquadrilateral. 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 area⁠K{\displaystyle K}⁠ of acyclic quadrilateral whose sides have lengths⁠a,{\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)}}}

wheres=12(a+b+c+d){\displaystyle s={\tfrac {1}{2}}(a+b+c+d)} is thesemiperimeter.

Heron's formula is also a special case of theformula 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 aCayley–Menger determinant in terms of the squares of thedistances between the three given vertices,A=14−|0a2b21a20c21b2c2011110|{\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 toTartaglia's formula for thevolume of athree-simplex.

Another generalization of Heron's formula to pentagons and hexagons inscribed in a circle was discovered byDavid 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 adegenerate 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 thetriangle inequality and do not describe the sides of a Euclidean triangle. In this case, Heron's formula gives animaginary result. For example if⁠a=3{\displaystyle a=3}⁠ and⁠b=c=1{\displaystyle b=c=1}⁠, then⁠A=354i{\displaystyle \textstyle A={\tfrac {3{\sqrt {5}}}{4}}i}⁠. This can be interpreted using a triangle in thecomplex coordinate plane⁠C2{\displaystyle \mathbb {C} ^{2}}⁠, where "area" can be a complex-valued quantity, or as a triangle lying in apseudo-Euclidean plane with one space-like dimension and one time-like dimension.[18]

Volume of a tetrahedron

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If⁠U,{\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 to⁠U{\displaystyle U}⁠ and so on), then[19]volume=(−a+b+c+d)(a−b+c+d)(a+b−c+d)(a+b+c−d)192uvw{\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}}}

Tetrahedron with base △UVW

wherea=yxYZ,b=XyZ,c=yXYz,d=Xxyz,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 inspherical geometry to its side lengths. For aspherical triangle with side lengths⁠a,{\displaystyle a,}⁠⁠b,{\displaystyle b,}⁠ and⁠c{\displaystyle c}⁠, semiperimeter⁠s=12(a+b+c){\displaystyle s={\tfrac {1}{2}}(a+b+c)}⁠, and area⁠S{\displaystyle S}⁠,[20]tan2⁡S4=tan⁡s2tan⁡s−a2tan⁡s−b2tan⁡s−c2{\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 inhyperbolic geometry the analogous formula istan2⁡S4=tanh⁡s2tanh⁡s−a2tanh⁡s−b2tanh⁡s−c2.{\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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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 fromthe original on 2024-05-29. Retrieved2021-12-27.
  2. ^The formula used here is thenumerically stable formula (relabeled for⁠a≤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, with⁠a=3{\displaystyle a=3}⁠,⁠b=4{\displaystyle b=4}⁠,⁠c=6.999{\displaystyle c=6.999}⁠, the correct area is⁠0.205{\displaystyle 0.205}⁠ but the naive implementation produces⁠0.000{\displaystyle 0.000}⁠ instead.
    The area is reported as "Not a triangle" when the side lengths fail thetriangle inequality. When the area is equal to zero, the three side lengths specify adegenerate 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. Retrieved25 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. Retrieved2025-10-28.
  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.

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