Heron's formula

This article is about calculating the area of a triangle.

A triangle with sides a, b, and c.

Formulation

In geometry, Heron'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 formulas, there is no need to calculate other distances in the triangle first.

Heron's formula states that the area 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; that is,

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

Heron's formula can also be written as

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

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

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

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

History

The formula is credited to Heron (or Hero) of Alexandria, and a proof can be found in his book, Metrica, written c. CE 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.

A formula equivalent to Heron's, namely

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

was discovered by the Chinese independently[citation needed] of the Greeks. It was published in Shushu Jiuzhang (“Mathematical Treatise in Nine Sections”), written by Qin Jiushao and published in 1247.

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,

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

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

Another generalization of Heron's formula to pentagons and hexagons inscribed in a circle was discovered by David P. Robbins.[13]

reference article：https://en.wikipedia.org/wiki/Heron%27s_formula

Formulation

History

Generalizations

猜你喜欢