Heron's theorem

A triangle with sides a, b and c

The Heron's Formula is a theorem of elementary geometry , which, after the ancient mathematician Heron of Alexandria is named. The sentence describes a mathematical formula with the help of which the area of a triangle can be calculated from the three side lengths. The formula is also called Heron's formula or Heron's formula or also the Heron's formula .

Formulation of the sentence

The area of a triangle of the Euclidean plane with the side lengths , , and half circumference ${\displaystyle A}$ ${\displaystyle a}$ ${\displaystyle b}$ ${\displaystyle c}$

${\displaystyle s\,=\,{\frac {a+b+c}{2}}}$

is

${\displaystyle A={\sqrt {s\cdot (s-a)\cdot (s-b)\cdot (s-c)}}}$

Other representations

This formula can also be expressed like this:

(V1)    ${\displaystyle A={\frac {1}{4}}\cdot {\sqrt {(a+b+c)\cdot (-a+b+c)\cdot (a-b+c)\cdot (a+b-c)}}}$

When multiplied one obtains:

(V2)    ${\displaystyle A={\frac {1}{4}}\cdot {\sqrt {2\cdot a^{2}\cdot b^{2}+2\cdot b^{2}\cdot c^{2}+2\cdot c^{2}\cdot a^{2}-a^{4}-b^{4}-c^{4}}}}$

As a further representation of the Heronic formula, the following is also common:

(V3)    , ${\displaystyle A={\frac {1}{4}}\cdot {\sqrt {4\cdot a^{2}\cdot b^{2}-(a^{2}+b^{2}-c^{2})^{2}}}}$

which can be obtained from version (V1) by regrouping and using the binomial formulas with the following equations :

${\displaystyle {\begin{aligned}16\cdot A^{2}&={\bigl (}((a+b)+c)\cdot ((a+b)-c){\bigr )}\cdot {\bigl (}(c+(a-b))\cdot (c-(a-b)){\bigr )}\\&={\bigl (}(a+b)^{2}-c^{2}{\bigr )}\cdot {\bigl (}c^{2}-(a-b)^{2}{\bigr )}\\&={\bigl (}a^{2}+2\cdot a\cdot b+b^{2}-c^{2}{\bigr )}\cdot {\bigl (}c^{2}-a^{2}+2\cdot a\cdot b-b^{2}{\bigr )}\\&={\bigl (}2\cdot a\cdot b+(a^{2}+b^{2}-c^{2}){\bigr )}\cdot {\bigl (}2\cdot a\cdot b-(a^{2}+b^{2}-c^{2}){\bigr )}\\&=4\cdot a^{2}\cdot b^{2}-(a^{2}+b^{2}-c^{2})^{2}\end{aligned}}}$

Finally, a representation with a determinant can be derived from the version (V3) :

(V4)    ${\displaystyle A={\frac {1}{4}}\cdot {\sqrt {-\det \left({\begin{matrix}0&1&1&1\\1&0&a^{2}&b^{2}\\1&a^{2}&0&c^{2}\\1&b^{2}&c^{2}&0\end{matrix}}\right)}}}$

This is a special case of the Cayley-Menger determinant , with which one can calculate the volume of a simplex , the generalization of triangles to any dimensions , for example a tetrahedron in three dimensions.

(V4) is obtained from (V3) using Laplace's expansion theorem and elementary matrix transformations as follows:

${\displaystyle {\begin{aligned}4\cdot a^{2}\cdot b^{2}-(a^{2}+b^{2}-c^{2})^{2}&=4\cdot a^{2}\cdot b^{2}-(c^{2}-a^{2}-b^{2})^{2}\\&=\det \left({\begin{matrix}-2\cdot a^{2}&c^{2}-a^{2}-b^{2}\\c^{2}-a^{2}-b^{2}&-2\cdot b^{2}\end{matrix}}\right)\\&=\det \left({\begin{matrix}1&a^{2}&b^{2}\\0&-2\cdot a^{2}&c^{2}-a^{2}-b^{2}\\0&c^{2}-a^{2}-b^{2}&-2\cdot b^{2}\end{matrix}}\right)\\&=\det \left({\begin{matrix}1&a^{2}&b^{2}\\1&-a^{2}&c^{2}-a^{2}\\1&c^{2}-b^{2}&-b^{2}\end{matrix}}\right)\\&=-\det \left({\begin{matrix}0&1&0&0\\1&0&a^{2}&b^{2}\\1&a^{2}&-a^{2}&c^{2}-a^{2}\\1&b^{2}&c^{2}-b^{2}&-b^{2}\end{matrix}}\right)\\&=-\det \left({\begin{matrix}0&1&1&1\\1&0&a^{2}&b^{2}\\1&a^{2}&0&c^{2}\\1&b^{2}&c^{2}&0\end{matrix}}\right)\end{aligned}}}$

Numerical example

A triangle with the sides , and half the circumference . Inserted into the formula you get the area ${\displaystyle a=4}$ ${\displaystyle b=13}$ ${\displaystyle c=15}$ ${\displaystyle s={\frac {a+b+c}{2}}={\frac {4+13+15}{2}}=16}$

${\displaystyle A={\sqrt {s\cdot (s-a)\cdot (s-b)\cdot (s-c)}}={\sqrt {16\cdot (16-4)\cdot (16-13)\cdot (16-15)}}={\sqrt {16\cdot 12\cdot 3\cdot 1}}={\sqrt {576}}=24}$

Another representation of the formula gives

${\displaystyle A={\frac {1}{4}}\cdot {\sqrt {(a+b+c)\cdot (-a+b+c)\cdot (a-b+c)\cdot (a+b-c)}}={\frac {1}{4}}\cdot {\sqrt {(4+13+15)\cdot (-4+13+15)\cdot (4-13+15)\cdot (4+13-15)}}={\frac {1}{4}}\cdot {\sqrt {32\cdot 24\cdot 6\cdot 2}}={\frac {1}{4}}\cdot {\sqrt {9216}}={\frac {1}{4}}\cdot 96=24}$In this example, the side lengths and the area are whole numbers . Therefore a triangle with sides 4, 13 and 15 is a Heronian triangle .

Connection with quadrilateral tendons

As a borderline case , the formula can be obtained from the formula for the area of a chordal quadrilateral if two of the corner points merge so that one of the sides of the quadrilateral has a length of zero. The Brahmagupta formula applies to the area of ​​a quadrilateral tendon

${\displaystyle A={\sqrt {(s-a)\cdot (s-b)\cdot (s-c)\cdot (s-d)}}}$,

here half the circumference

${\displaystyle s={\frac {a+b+c+d}{2}}}$

is.

proof

Proof with the Pythagorean theorem

According to the Pythagorean theorem , and (see figure). Subtraction gives , so ${\displaystyle b^{2}=h^{2}+d^{2}}$ ${\displaystyle a^{2}=h^{2}+(c-d)^{2}}$ ${\displaystyle a^{2}-b^{2}=c^{2}-2\cdot c\cdot d}$

${\displaystyle d={\frac {-a^{2}+b^{2}+c^{2}}{2\cdot c}}}$

The following applies to the height of the triangle . Substituting the last equation yields ${\displaystyle h}$ ${\displaystyle h^{2}=b^{2}-d^{2}}$

${\displaystyle {\begin{aligned}h^{2}&=b^{2}-\left({\frac {-a^{2}+b^{2}+c^{2}}{2\cdot c}}\right)^{2}\\&=\left({\frac {2\cdot b\cdot c}{2\cdot c}}\right)^{2}-\left({\frac {-a^{2}+b^{2}+c^{2}}{2\cdot c}}\right)^{2}\\&={\frac {(2\cdot b\cdot c+(-a^{2}+b^{2}+c^{2}))\cdot (2\cdot b\cdot c-(-a^{2}+b^{2}+c^{2}))}{4\cdot c^{2}}}\\&={\frac {((b+c)^{2}-a^{2})\cdot (a^{2}-(b-c)^{2})}{4\cdot c^{2}}}\\&={\frac {((b+c)+a)\cdot ((b+c)-a)\cdot (a+(b-c))\cdot (a-(b-c))}{4\cdot c^{2}}}\\&={\frac {2\cdot s\cdot 2\cdot (s-a)\cdot 2\cdot (s-c)\cdot 2\cdot (s-b)}{4\cdot c^{2}}}\\&={\frac {4\cdot s\cdot (s-a)\cdot (s-b)\cdot (s-c)}{c^{2}}}\end{aligned}}}$

Applying the square root on both sides yields

${\displaystyle h={\frac {2}{c}}\cdot {\sqrt {s\cdot (s-a)\cdot (s-b)\cdot (s-c)}}}$

From this follows for the area of the triangle

${\displaystyle {\begin{aligned}A&={\frac {c\cdot h}{2}}\\&={\frac {c}{2}}\cdot {\frac {2}{c}}\cdot {\sqrt {s\cdot (s-a)\cdot (s-b)\cdot (s-c)}}\\&={\sqrt {s\cdot (s-a)\cdot (s-b)\cdot (s-c)}}\end{aligned}}}$

Proof with the law of cosines

According to the cosine law ,

${\displaystyle \cos \gamma ={\frac {a^{2}+b^{2}-c^{2}}{2\cdot a\cdot b}}}$

Inserted into the trigonometric Pythagoras it follows

${\displaystyle \sin \gamma ={\sqrt {1-\cos ^{2}\gamma }}={\sqrt {\left({\frac {2\cdot a\cdot b}{2\cdot a\cdot b}}\right)^{2}-\left({\frac {a^{2}+b^{2}-c^{2}}{2\cdot a\cdot b}}\right)^{2}}}={\frac {1}{2\cdot a\cdot b}}\cdot {\sqrt {(2\cdot a\cdot b)^{2}-(a^{2}+b^{2}-c^{2})^{2}}}}$

The height of the triangle on the side is the length . Substituting the last equation yields ${\displaystyle a}$ ${\displaystyle b\cdot \sin \gamma }$

${\displaystyle {\begin{aligned}A&={\frac {a\cdot b\cdot \sin \gamma }{2}}\\&={\frac {1}{4}}\cdot {\sqrt {(2\cdot a\cdot b)^{2}-(a^{2}+b^{2}-c^{2})^{2}}}\\&={\frac {1}{4}}\cdot {\sqrt {(2\cdot a\cdot b+(a^{2}+b^{2}-c^{2}))\cdot (2\cdot a\cdot b-(a^{2}+b^{2}-c^{2}))}}\\&={\frac {1}{4}}\cdot {\sqrt {((a+b)^{2}-c^{2})\cdot (c^{2}-(a-b)^{2})}}\\&={\frac {1}{4}}\cdot {\sqrt {((a+b)+c)\cdot ((a+b)-c)\cdot (c+(a-b))\cdot (c-(a-b))}}\\&={\frac {1}{4}}\cdot {\sqrt {(a+b+c)\cdot (-a+b+c)\cdot (a-b+c)\cdot (a+b-c)}}\\&={\sqrt {s\cdot (s-a)\cdot (s-b)\cdot (s-c)}}\end{aligned}}}$

Proof with the cotangent theorem

The Inkreisradius of the triangle is . With the help of the cotangent theorem can be obtained for the area ${\displaystyle r}$

${\displaystyle {\begin{aligned}A&=r\cdot ((s-a)+(s-b)+(s-c))\\&=r^{2}\cdot \left({\frac {s-a}{r}}+{\frac {s-b}{r}}+{\frac {s-c}{r}}\right)\\&=r^{2}\cdot \left(\cot {\frac {\alpha }{2}}+\cot {\frac {\beta }{2}}+\cot {\frac {\gamma }{2}}\right)\end{aligned}}}$

With the equation for triangles (see Trigonometry formula collection ) it follows ${\displaystyle \cot {\frac {\alpha }{2}}+\cot {\frac {\beta }{2}}+\cot {\frac {\gamma }{2}}=\cot {\frac {\alpha }{2}}\cdot \cot {\frac {\beta }{2}}\cdot \cot {\frac {\gamma }{2}}}$

${\displaystyle {\begin{aligned}A&=r^{2}\cdot \left(\cot {\frac {\alpha }{2}}\cdot \cot {\frac {\beta }{2}}\cdot \cot {\frac {\gamma }{2}}\right)\\&=r^{2}\cdot \left({\frac {s-a}{r}}\cdot {\frac {s-b}{r}}\cdot {\frac {s-c}{r}}\right)\\&={\frac {(s-a)\cdot (s-b)\cdot (s-c)}{r}}\end{aligned}}}$

In addition, (see illustration). The multiplication of these equations gives ${\displaystyle A={\frac {r\cdot (a+b+c)}{2}}=r\cdot s}$

${\displaystyle A^{2}=s\cdot (s-a)\cdot (s-b)\cdot (s-c)}$

and from it Heron's theorem.

literature

• Hermann Athens, Jörn Bruhn (ed.): Lexicon of school mathematics and related areas . tape 2 , F-K. Aulis Verlag Deubner, Cologne 1977, ISBN 3-7614-0242-2 .
• Anna Maria Fraedrich: The sentence group of the Pythagoras (=  textbooks and monographs on didactics of mathematics . Volume 29 ). BI-Wissenschaftsverlag, Mannheim / Leipzig / Vienna / Zurich 1994, ISBN 3-411-17321-1 .
• György Hajós : Introduction to Geometry . BG Teubner Verlag, Leipzig (Hungarian: Bevezetés A Geometriába . Translated by G. Eisenreich [Leipzig, also editing]).
• Max Koecher , Aloys Krieg : level geometry . 3rd, revised and expanded edition. Springer Verlag, Berlin (inter alia) 2007, ISBN 978-3-540-49327-3 .
• Theophil Lambacher , Wilhelm Schweizer (Ed.): Lambacher-Schweizer . Mathematical teaching material for higher schools. Geometry. Edition E. Part 2 . 13th edition. Ernst Klett Verlag, Stuttgart 1965.

Web links

Wikibooks: Archive of Evidence: Geometry: Planimetry: Triangle: Heron's Theorem  - Learning and teaching materials

• Eric W. Weisstein : Theorem of Heron . In: MathWorld (English).
• Elementary proof
• Proof using the cosine theorem (German) (PDF; 88 kB)
• Walter Fendt: The Heronian formula for the triangular surface (PDF; 82 kB) - proof and conclusions

Individual evidence

1. ↑ For detailed evidence see also Wikibooks evidence archive .
2. ↑ Please note that the roles of the side lengths can be interchanged as desired. ${\displaystyle a,b,c}$
3. ^ György Hajós : Introduction to Geometry . BG Teubner Verlag, Leipzig, p. 380–381 (Hungarian: Bevezetés A Geometriába . Translated by G. Eisenreich [Leipzig, also editorial]).
4. ↑ Max Koecher , Aloys Krieg : level geometry . 3rd, revised and expanded edition. Springer Verlag, Berlin (among others) 2007, ISBN 978-3-540-49327-3 , p. 111 .
5. ↑ Here, too, the roles of the side lengths can be swapped, which leads to an equivalent, but correspondingly modified representation. ${\displaystyle a,b,c}$