Heron's formula
![](http://upload.wikimedia.org/wikipedia/commons/thumb/e/eb/Triangle_with_notations_2_without_points.svg/198px-Triangle_with_notations_2_without_points.svg.png)
In geometry, Heron's formula (sometimes called Hero's formula), named after Hero of Alexandria,[1] gives the area of a triangle when the length of all three sides are known. Unlike other triangle area formulae, there is no need to calculate angles or other distances in the triangle first.
Formulation[]
Heron's formula states that the area of a triangle whose sides have lengths a, b, and c is
where s is the semi-perimeter of the triangle; that is,
Heron's formula can also be written as
Example[]
Let △ABC be the triangle with sides a = 4, b = 13 and c = 15. This triangle’s semiperimeter is
s = 1/2(a + b + c) = 1/2(4 + 13 + 15) = 16, and the area is
In this example, the side lengths and area are integers, making it a Heronian triangle. However, Heron's formula works equally well in cases where one or all of these numbers is not an integer.
History[]
The formula is credited to Heron (or Hero) of Alexandria, and a proof can be found in his book Metrica, written around AD 60. It has been suggested that Archimedes knew the formula over two centuries earlier,[3] 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.[4]
A formula equivalent to Heron's, namely,
was discovered by the Chinese independently[citation needed] of the Greeks. It was published in Mathematical Treatise in Nine Sections (Qin Jiushao, 1247).[5]
Proofs[]
Heron's original proof made use of cyclic quadrilaterals.[citation needed] Other arguments appeal to trigonometry as below, or to the incenter and one excircle of the triangle,[6] or to De Gua's theorem (for the particular case of acute triangles).[7]
Trigonometric proof using the law of cosines[]
A modern proof, which uses algebra and is quite different from the one provided by Heron (in his book Metrica), follows.[8] Let a, b, c be the sides of the triangle and α, β, γ the angles opposite those sides. Applying the law of cosines we get
From this proof, we get the algebraic statement that
The altitude of the triangle on base a has length b sin γ, and it follows
The difference of two squares factorization was used in two different steps.
Algebraic proof using the Pythagorean theorem[]
![](http://upload.wikimedia.org/wikipedia/commons/thumb/6/6a/Triangle_with_notations_3.svg/270px-Triangle_with_notations_3.svg.png)
The following proof is very similar to one given by Raifaizen.[9] By the Pythagorean theorem we have b2 = h2 + d2 and a2 = h2 + (c − d)2 according to the figure at the right. Subtracting these yields a2 − b2 = c2 − 2cd. This equation allows us to express d in terms of the sides of the triangle:
For the height of the triangle we have that h2 = b2 − d2. By replacing d with the formula given above and applying the difference of squares identity we get
We now apply this result to the formula that calculates the area of a triangle from its height:
Trigonometric proof using the law of cotangents[]
![](http://upload.wikimedia.org/wikipedia/commons/thumb/7/7e/Herontriangle2greek.svg/270px-Herontriangle2greek.svg.png)
From the first part of the law of cotangents proof,[10] we have that the triangle's area is both
and A = rs, but, since the sum of the half-angles is π/2, the triple cotangent identity applies, so the first of these is
Combining the two, we get
from which the result follows.
Numerical stability[]
Heron's formula as given above is numerically unstable for triangles with a very small angle when using floating-point arithmetic. A stable alternative[11][12] involves arranging the lengths of the sides so that a ≥ b ≥ c and computing
The brackets in the above formula are required in order to prevent numerical instability in the evaluation.
Other area formulae resembling Heron's formula[]
Three other area formulae have the same structure as Heron's formula but are expressed in terms of different variables. First, denoting the medians from sides a, b, and c respectively as ma, mb, and mc and their semi-sum 1/2(ma + mb + mc) as σ, we have[13]
Next, denoting the altitudes from sides a, b, and c respectively as ha, hb, and hc, and denoting the semi-sum of the reciprocals of the altitudes as H = 1/2(h−1
a + h−1
b + h−1
c) we have[14]
Finally, denoting the semi-sum of the angles' sines as S = 1/2(sin α + sin β + sin γ), we have[15]
where D is the diameter of the circumcircle: D = a/sin α = b/sin β = c/sin γ.
Generalizations[]
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.
Expressing Heron's formula with a Cayley–Menger determinant in terms of the squares of the distances between the three given vertices,
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.[16]
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[17]
where
See also[]
References[]
- ^ "Fórmula de Herón para calcular el área de cualquier triángulo" (in Spanish). Spain: Ministerio de Educación, Cultura y Deporte. 2004. Retrieved 30 June 2012.
- ^ Kendig, Keith (2000). "Is a 2000-Year-Old Formula Still Keeping Some Secrets?". Amer. Math. Monthly. 107: 402–415. doi:10.2307/2695295.
- ^ Heath, Thomas L. (1921). A History of Greek Mathematics. II. Oxford University Press. pp. 321–323.
- ^ Weisstein, Eric W. "Heron's Formula". MathWorld.
- ^ 秦, 九韶 (1773). "卷三上, 三斜求积". 數學九章 (四庫全書本) (in Chinese).
- ^ "Personal email communication between mathematicians John Conway and Peter Doyle". 15 December 1997. Retrieved 25 September 2020.
- ^ Lévy-Leblond, Jean-Marc (2020-09-14). "A Symmetric 3D Proof of Heron's Formula". The Mathematical Intelligencer. doi:10.1007/s00283-020-09996-8. ISSN 0343-6993.
- ^ Niven, Ivan (1981). Maxima and Minima Without Calculus. The Mathematical Association of America. pp. 7–8.
- ^ Raifaizen, Claude H. (1971). "A Simpler Proof of Heron's Formula". Mathematics Magazine. 44 (1): 27–28.
- ^ The second part of the Law of cotangents proof depends on Heron's formula itself, but this article depends only on the first part.
- ^ Sterbenz, Pat H. (1974-05-01). Floating-Point Computation. Prentice-Hall Series in Automatic Computation (1st ed.). Englewood Cliffs, New Jersey, USA: Prentice Hall. ISBN 0-13-322495-3.
- ^ William M. Kahan (24 March 2000). "Miscalculating Area and Angles of a Needle-like Triangle" (PDF).
- ^ Benyi, Arpad, "A Heron-type formula for the triangle," Mathematical Gazette" 87, July 2003, 324–326.
- ^ Mitchell, Douglas W., "A Heron-type formula for the reciprocal area of a triangle," Mathematical Gazette 89, November 2005, 494.
- ^ Mitchell, Douglas W., "A Heron-type area formula in terms of sines," Mathematical Gazette 93, March 2009, 108–109.
- ^ 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?", [1], pp. 16–17.
External links[]
- 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
- An alternative proof of Heron's Formula without words
- Factoring Heron
- Theorems about triangles
- Area