In geometry, Heron's formula (sometimes called Hero's formula), named after Hero of Alexandria, gives the area of a triangle by requiring no arbitrary choice of side as base or vertex as origin, contrary to 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.
where s is the semiperimeter of the triangle; that is,
Heron's formula can also be written as
Let ΔABC be the triangle with sides a=4, b=13 and c=15. 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
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 A.D. 1247.
Trigonometric proof using the Law of cosines
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 α, β, γ the angles opposite those sides. We have
by the law of cosines. From this proof 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
The following proof is very similar to one given by Raifaizen. By the Pythagorean theorem we have and according to the figure at the right. Subtracting these yields . This equation allows us to express in terms of the sides of the triangle:
For the height of the triangle we have that . By replacing with the formula given above and applying the difference of squares identity repeatedly 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
From the first part of the Law of cotangents proof, we have that the triangle's area is both
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.
Heron's formula as given above is numerically unstable for triangles with a very small angle when using floating point arithmetic. A stable alternative involves arranging the lengths of the sides so that and computing
The brackets in the above formula are required in order to prevent numerical instability in the evaluation.
Other area formulas resembling Heron's formula
Three other area formulas 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 (ma + mb + mc)/2 as σ, we have
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 we have
Finally, denoting the semi-sum of the angles' sines as S = [(sin α) + (sin β) + (sin γ)]/2, we have
where D is the diameter of the circumcircle:
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,
Another generalization of Heron's formula to pentagons and hexagons inscribed in a circle was discovered by David P. Robbins.
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
- 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" at MathPages.com.
- A Geometric Proof of Heron's Formula
- An alternative proof of Heron's Formula without words
- Factoring Heron