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| {{about|calculating the area of a triangle|calculating a square root|Heron's method}}
| | My name: Neville Bernacchi<br>Age: 20<br>Country: Germany<br>Home town: Solingen Wald <br>Post code: 42653<br>Address: Park Str. 3<br><br>my blog; [http://hemorrhoidtreatmentfix.com/external-hemorrhoids external hemorrhoids] |
| [[Image:Triangle with notations 2.svg|thumb|198px|A triangle with sides ''a'', ''b'', and ''c''.]]
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| In [[geometry]], '''Heron's (or 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> states that the [[area]] ''T'' of a [[triangle]] whose sides have lengths ''a'', ''b'', and ''c'' is
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| :<math>T = \sqrt{s(s-a)(s-b)(s-c)}</math> | |
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| where ''s'' is the [[semiperimeter]] of the triangle:
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| :<math>s=\frac{a+b+c}{2}.</math> | |
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| Heron's formula can also be written as:
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| :<math>T=\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}</math> | |
| :<math>T=\frac{1}{4}\sqrt{2(a^2 b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4)}</math>
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| :<math>T=\frac{1}{4}\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}</math> | |
| :<math>T=\frac{1}{4}\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}</math>
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| Heron's formula is distinguished from other formulas for the area of a triangle, such as half the base times the height or half the modulus of a cross product of two sides, by requiring no arbitrary choice of side as base or vertex as origin.
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| ==Example==
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| Let ΔABC be the triangle with sides ''a''=7, ''b''=4 and ''c''=5.
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| The semiperimeter is <math>s=\tfrac{1}{2}(a+b+c)=\tfrac{1}{2}(7+4+5)=8</math> , and the area is
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| : <math>T = \sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}= \sqrt{8 \cdot (8-7) \cdot (8-4) \cdot (8-5)}</math>
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| ::<math>=\sqrt{8 \cdot 1 \cdot 4 \cdot 3}=\sqrt{96}=4\sqrt{6} \approx 9.8</math></div>
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| == History ==
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| The formula is credited to [[Hero of Alexandria|Heron (or Hero) of Alexandria]], and a proof can be found in his book, ''Metrica'', written ''c.'' <small>A.D.</small> 60. It has been suggested that [[Archimedes]] knew the formula over two centuries earlier, 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>
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| A formula equivalent to Heron's namely:
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| :<math>T=\frac1{2}\sqrt{a^2c^2-\left(\frac{a^2+c^2-b^2}{2}\right)^2}</math>, where <math>a \ge b \ge c</math>
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| was discovered by the Chinese independently of the Greeks. It was published in ''Shushu Jiuzhang'' (“[[Mathematical Treatise in Nine Sections]]”), written by [[Qin Jiushao]] and published in <small>A.D.</small> 1247.
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| == Proof ==
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| A modern proof, which uses [[algebra]] and is quite unlike the one provided by Heron (in his book Metrica), follows. Let ''a'', ''b'', ''c'' be the sides of the triangle and ''A'', ''B'', ''C'' the [[angle]]s opposite those sides. We have
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| :<math>\cos \widehat C = \frac{a^2+b^2-c^2}{2ab}</math>
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| by the [[law of cosines]]. From this proof get the algebraic statement:
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| :<math>\sin \widehat C = \sqrt{1-\cos^2 \widehat C} = \frac{\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2 }}{2ab}.</math>
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| The [[altitude (triangle)|altitude]] of the triangle on base ''a'' has length ''b''·sin(''C''), and it follows
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| : <math>
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| \begin{align}
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| T & = \frac{1}{2} (\mbox{base}) (\mbox{altitude}) \\
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| & = \frac{1}{2} ab\sin \widehat C \\
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| & = \frac{1}{4}\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2} \\
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| & = \frac{1}{4}\sqrt{(2a b -(a^2 +b^2 -c^2))(2a b +(a^2 +b^2 -c^2))} \\
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| & = \frac{1}{4}\sqrt{(c^2 -(a -b)^2)((a +b)^2 -c^2)} \\
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| & = \sqrt{\frac{(c -(a -b))(c +(a -b))((a +b) -c)((a +b) +c)}{16}} \\
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| & = \sqrt{\frac{(b + c - a)}{2}\frac{(a + c - b)}{2}\frac{(a + b - c)}{2}\frac{(a + b + c)}{2}} \\
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| & = \sqrt{\frac{(a + b + c)}{2}\frac{(b + c - a)}{2}\frac{(a + c - b)}{2}\frac{(a + b - c)}{2}} \\
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| & = \sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}.
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| \end{align}
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| </math>
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| The [[difference of two squares]] factorization was used in two different steps.
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| ==Proof using the Pythagorean theorem==
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| [[Image:Triangle with notations 3.svg|thumb|270px|Triangle with altitude ''h'' cutting base ''c'' into ''d'' + (''c'' − ''d'').]]Heron's original proof made use of [[cyclic quadrilateral]]s, while other arguments appeal to [[trigonometry]] as above, or to the [[incenter]] and one [[excircle]] of the triangle [http://www.math.dartmouth.edu/~doyle/docs/heron/heron.txt]. The following argument reduces Heron's formula directly to the [[Pythagorean theorem]] using only elementary means.
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| We wish to prove <math>4T^2=4s(s-a)(s-b)(s-c).</math> The left-hand side equals
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| :<math>4 T^2 = (c h)^2 = c^2(b^2-d^2) = (c b)^2 - (c d)^2</math>
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| while the right-hand side equals
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| :<math>4s(s-a)(s-b)(s-c) = [s(s-a)+(s-b)(s-c)]^2 - [s(s-a)-(s-b)(s-c)]^2</math>
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| via the identity <math>(p+q)^2-(p-q)^2=4pq.</math> It therefore suffices to show
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| :<math>cb=s(s-a)+(s-b)(s-c)</math>
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| and
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| :<math>cd=s(s-a)-(s-b)(s-c).</math>
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| Substituting <math>2s=(a+b+c)</math> into the former,
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| :<math>s(s-a)+(s-b)(s-c)=\frac{1}{4}(a+b+c)(-a+b+c) + \frac{1}{4}(a-b+c)(a+b-c) = \frac{1}{4}[(b+c)^2-a^2] + \frac{1}{4}[a^2-(b-c)^2] = \frac{1}{4}[(b+c)^2 - (b-c)^2] = cb</math>
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| as desired. Similarly, the latter expression becomes
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| :<math>s(s-a)-(s-b)(s-c)=\frac{1}{4}[(b+c)^2-a^2] - \frac{1}{4}[a^2-(b-c)^2] = \frac{1}{2}(b^2+c^2-a^2).</math>
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| Using the Pythagorean theorem twice, <math>b^2=d^2+h^2</math> and <math>a^2=(c-d)^2+h^2,</math> allows us to simplify the expression to
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| :<math>\frac{1}{2}(b^2+c^2-a^2) = \frac{1}{2}[d^2+c^2-(c-d)^2] = cd.</math>
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| The result follows.
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| ==Proof using the [[Law of cotangents]] and the [[Proofs of trigonometric identities#Miscellaneous -- the triple cotangent identity|triple cotangent identity]]==
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| [[File:Herontriangle2.svg|thumb|270px|right||Geometrical significance of ''s-a'', ''s-b'', and ''s-c''. See the [[Law of cotangents]] for the reasoning behind this.]]
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| 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
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| :<math>
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| \begin{align}
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| T &= 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) \\[8pt]
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| &= r^2\big(\cot(A/2) + \cot(B/2) + \cot(C/2)\big) \\[8pt]
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| \end{align}
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| </math>
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| and
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| :<math>T = rs</math>
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| but, since the sum of the half-angles is <math>\tfrac{\pi}{2}</math>, the [[Proofs of trigonometric identities#Miscellaneous -- the triple cotangent identity|triple cotangent identity]] applies, so the first of these is
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| :<math>
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| \begin{align}
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| T &= r^2\left(\cot(A/2)\ \cot(B/2)\ \cot(C/2)\right) = r^2\ \frac{s-a}{r}\ \frac{s-b}{r}\ \frac{s-c}{r} \\[8pt]
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| &= \frac{(s-a) (s-b) (s-c)}{r} \\[8pt]
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| \end{align}
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| </math>
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| <br /> | |
| Combining the two, we get
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| :<math>T^2 = s (s-a) (s-b) (s-c)</math>
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| from which the result follows.
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| == Numerical stability ==
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| Heron's formula as given above is [[Numerical stability|numerically unstable]] for triangles with a very small angle. A stable alternative <ref>{{cite book|title=Floating-Point Computation, Prentice-Hall|author=P. Sterbenz|year=1973}}</ref>
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| <ref>{{cite web|url=http://www.cs.berkeley.edu/~wkahan/Triangle.pdf|title=Miscalculating Area and Angles of a Needle-like Triangle|author=W. Kahan|date=24 March 2000}}</ref> involves arranging the lengths of the sides so that <math>a \ge b \ge c</math> and computing
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| :<math>T = \frac{1}{4}\sqrt{(a+(b+c)) (c-(a-b)) (c+(a-b)) (a+(b-c))}.</math>
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| The brackets in the above formula are required in order to prevent numerical instability in the evaluation.
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| ==Other area formulas resembling Heron's formula==
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| {{Main|Triangle#Formulas resembling Heron's formula}}
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| Several other triangle area formulas have the same functional form as Heron's formula but with the sides replaced by either the [[median (geometry)|medians]], the reciprocals of the [[Altitude (triangle)|altitudes]], or the [[sine]]s of the angles.
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| == 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.
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| 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.
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| Expressing Heron's formula with a [[Cayley–Menger determinant]] in terms of the squares of the [[distance]]s between the three given vertices,
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| :<math> T = \frac{1}{4} \sqrt{- \begin{vmatrix}
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| 0 & a^2 & b^2 & 1 \\
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| a^2 & 0 & c^2 & 1 \\
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| b^2 & c^2 & 0 & 1 \\
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| 1 & 1 & 1 & 0
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| \end{vmatrix} } </math>
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| illustrates its similarity to [[Tartaglia's formula]] for the [[volume]] of a [[Simplex|three-simplex]].
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| 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>
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| === Heron-type formula for the volume of a tetrahedron ===
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| If ''U'', ''V'', ''W'', ''u'', ''v'', ''w'' are lengths of edges of the tetrahedron (first three form a triangle; ''u'' opposite to ''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. 16-17.</ref>
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| :<math>
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| \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>
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| where
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| : <math>
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| \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}
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| </math>
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| ==See also==
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| * [[Heronian triangle]]
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| == Notes ==
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| <references/>
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| == References ==
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| * {{cite book
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| | author=Heath, Thomas L.
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| | title=A History of Greek Mathematics (Vol II)
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| | publisher=Oxford University Press
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| | year=1921
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| | pages=321–323}}
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| == External links ==
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| *[http://www.cut-the-knot.org/pythagoras/herons.shtml A Proof of the Pythagorean Theorem From Heron's Formula] at [[cut-the-knot]]
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| *[http://www.mathopenref.com/heronsformula.html Interactive applet and area calculator using Heron's Formula]
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| *[http://www.math.dartmouth.edu/~doyle/docs/heron/heron.txt J.H. Conway discussion on Heron's Formula]
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| *{{MathPages|id=home/kmath196/kmath196|title=Heron's Formula and Brahmagupta's Generalization}}
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| *[http://jwilson.coe.uga.edu/EMT668/EMAT6680.2000/Umberger/MATH7200/HeronFormulaProject/GeometricProof/geoproof.html A Geometric Proof of Heron's Formula]
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| {{DEFAULTSORT:Heron's Formula}}
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| [[Category:Triangle geometry]]
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| [[Category:Articles containing proofs]]
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| [[Category:Area]]
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| [[Category:Theorems in plane geometry]]
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