Heron's formula: Information from Answers.com

This article is about calculating the area of a triangle. For calculating a square root, see Heron's method.

A triangle with sides a, b, and c.

In geometry, Heron's (or Hero's) formula states that the area A of a triangle whose sides have lengths a, b, and c is

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

where s is the semiperimeter of the triangle:

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

Heron's formula can also be written as:

$A={\ \sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)\ \over 16}\,}$

$A={\ \sqrt{2(a^2 b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4)\ \over 16}\,}$

$A={\ \sqrt{(a^2 + b^2 + c^2)^2 - 2(a^4 + b^4 + c^4)\,}\ \over 4}.$

1 History
2 Proof
3 Proof using the Pythagorean theorem
4 Numerical stability
5 Generalizations
- 5.1 Heron-looking formula for tetrahedra
6 See also
7 Notes
8 References
9 External links

History

The formula is credited to Heron 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, and since Metrica is a collection of the mathematical knowledge available in the ancient world, it is possible that it predates the reference given in the work. ^[1]

A formula equivalent to Heron's namely:

$A=\frac1{2}\sqrt{a^2c^2-\left(\frac{a^2+c^2-b^2}{2}\right)^2}$

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.

Proof

A modern proof, which uses algebra and trigonometry 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

$\cos(C) = \frac{a^2+b^2-c^2}{2ab}$

by the law of cosines. From this proof get the algebraic statement:

$\sin(C) = \sqrt{1-\cos^2(C)} = \frac{\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2 }}{2ab}.$

The altitude of the triangle on base a has length b sin(C), and it follows

$\begin{align} A & = \frac{1}{2} (\mbox{base}) (\mbox{altitude}) \\ & = \frac{1}{2} ab\sin(C) \\ & = \frac{1}{4}\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2} \\ & = \frac{1}{4}\sqrt{(2a b -(a^2 +b^2 -c^2))(2a b +(a^2 +b^2 -c^2))} \\ & = \frac{1}{4}\sqrt{(c^2 -(a -b)^2)((a +b)^2 -c^2)} \\ & = \frac{1}{4}\sqrt{(c -(a -b))((c +(a -b))((a +b) -c))((a +b) +c)} \\ & = \sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}. \end{align}$

The difference of two squares factorization was used in two different steps.

Proof using the Pythagorean theorem

Triangle with altitude h cutting base c into d + (c − d).

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[1]. The following argument reduces Heron's formula directly to the Pythagorean theorem using only elementary means.

In the form 4A² = 4s(s − a)(s − b)(s − c), Heron's formula reduces on the left to (ch)², or

(c b) 2 - (c d) 2

using b² − d² = h² by the Pythagorean theorem, and on the right to

(s (s - a) + (s - b)(s - c)) 2 - (s (s - a) - (s - b)(s - c)) 2

via the identity (p + q)² − (p − q)² = 4pq. It therefore suffices to show

$cb=s(s-a)+(s-b)(s-c), \,$

and

$cd = s(s-a)-(s-b)(s-c). \,$

The former follows immediately by substituting (a + b + c)/2 for s and simplifying. Doing this for the latter reduces s(s − a)(s − b)(s − c) only as far as (b² + c² − a²)/2. But if we replace b² by d² + h² and a² by (c − d)² + h², both by Pythagoras, simplification then produces cd as required.

Numerical stability

Heron's formula as given above is numerically unstable for triangles with a very small angle. A stable alternative^[2] involves arranging the lengths of the sides so that: a ≥ b ≥ c and computing

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

The parentheses in the above formula are required in order to prevent numerical instability in the evaluation.

Generalizations

Heron's formula is a special case of Brahmagupta's formula for the area of a cyclic quadrilateral; both of which are special cases of Bretschneider's formula for the area of a quadrilateral. In both cases Heron's formula is obtained by setting one of the sides of the quadrilateral to zero.

Heron's formula is also a special case of the formula of the area of the trapezoid 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,

$A = \frac{1}{4} \sqrt{- \begin{vmatrix} 0 & a^2 & b^2 & 1 \\ a^2 & 0 & c^2 & 1 \\ b^2 & c^2 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{vmatrix} }$

illustrates its similarity to Tartaglia's formula for the volume of a three-simplex.

Another generalization of Heron's formula to polygons inscribed in a circle was discovered by David P. Robbins.^{[citation needed]}

Heron-looking formula for tetrahedra

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^{[citation needed]}

$\text{Volume} = \frac{\sqrt {\,( - a + b + c + d)\,(a - b + c + d)\,(a + b - c + d)\,(a + b + c - d)}}{192\,u\,v\,w}$

where

$\begin{align} a & = \sqrt {xYZ} \\ b & = \sqrt {yZX} \\ c & = \sqrt {zXY} \\ d & = \sqrt {xyz} \\ X & = (w - U + v)\,(U + v + w) \\ x & = (U - v + w)\,(v - w + U) \\ Y & = (u - V + w)\,(V + w + u) \\ y & = (V - w + u)\,(w - u + V) \\ Z & = (v - W + u)\,(W + u + v) \\ z & = (W - u + v)\,(u - v + W). \end{align}$

Notes

^ Weisstein, Eric W., "Heron's Formula" from MathWorld.
^ W. Kahan Miscalculating Area and Angles of a Needle-like Triangle

References

Heath, Thomas L. (1921). A History of Greek Mathematics (Vol II). Oxford University Press. pp. 321–323.

External links

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	Hero's formula (mathematics)
	Bretschneider's formula
	Hadwiger–Finsler inequality

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Heron's formula

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