where s is the semiperimeter of the triangle:
Heron's formula can also be written as:
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.
- 1 Example
- 2 History
- 3 Proof
- 4 Proof using the Pythagorean theorem
- 5 Proof using the Law of cotangents and the triple cotangent identity
- 6 Numerical stability
- 7 Other area formulas resembling Heron's formula
- 8 Generalizations
- 9 See also
- 10 Notes
- 11 References
- 12 External links
Let ΔABC be the triangle with sides a=7, b=4 and c=5. The semiperimeter is , and the area is
The formula is credited to Heron (or Hero) of Alexandria, and a proof can be found in his book, Metrica, written c. A.D. 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.
A formula equivalent to Heron's namely:
- , where
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 angles opposite those sides. We have
by the law of cosines. From this proof get the algebraic statement:
The altitude of the triangle on base a has length b·sin(C), and it follows
The difference of two squares factorization was used in two different steps.
Proof using the Pythagorean theorem
Heron's original proof made use of cyclic quadrilaterals, while other arguments appeal to trigonometry as above, or to the incenter and one excircle of the triangle . The following argument reduces Heron's formula directly to the Pythagorean theorem using only elementary means.
We wish to prove The left-hand side equals
while the right-hand side equals
via the identity It therefore suffices to show
Substituting into the former,
as desired. Similarly, the latter expression becomes
Using the Pythagorean theorem twice, and allows us to simplify the expression to
The result follows.
Proof using the Law of cotangents and the triple cotangent identity
but, since the sum of the half-angles is , the triple cotangent identity applies, so the first of these is
Combining the two, we get
from which the result follows.
The brackets in the above formula are required in order to prevent numerical instability in the evaluation.
Other area formulas resembling Heron's formula
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.
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.
Heron-type formula for the volume of a tetrahedron
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
- "Fórmula de Herón para calcular el área de cualquier triángulo" (in Spanish). Retrieved 30 June 2012.
- Weisstein, Eric W., "Heron's Formula", MathWorld.
- The second part of the Law of cotangents proof depends on Heron's formula itself, but this article depends only on the first part.
- P. Sterbenz (1973). Floating-Point Computation, Prentice-Hall.
- W. Kahan (24 March 2000). "Miscalculating Area and Angles of a Needle-like Triangle".
- D. P. Robbins, "Areas of Polygons Inscribed in a Circle", Discr. Comput. Geom. 12, 223-236, 1994.
- W. Kahan, "What has the Volume of a Tetrahedron to do with Computer Programming Languages?", , pp. 16-17.
- Heath, Thomas L. (1921). A History of Greek Mathematics (Vol II). Oxford University Press. pp. 321–323.