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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 ...
Three 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 m a, m b, and m c and their semi-sum (m a + m b + m c)/2 as σ, we have [10]
A method analogous to piece-wise linear approximation but using only arithmetic instead of algebraic equations, uses the multiplication tables in reverse: the square root of a number between 1 and 100 is between 1 and 10, so if we know 25 is a perfect square (5 × 5), and 36 is a perfect square (6 × 6), then the square root of a number greater than or equal to 25 but less than 36, begins with ...
This formula generalizes Heron's formula for the area of a triangle. A triangle may be regarded as a quadrilateral with one side of length zero. From this perspective, as d approaches zero, a cyclic quadrilateral converges into a cyclic triangle (all triangles are cyclic), and Brahmagupta's formula simplifies to Heron's formula.
Bretschneider's formula generalizes Brahmagupta's formula for the area of a cyclic quadrilateral, which in turn generalizes Heron's formula for the area of a triangle.. The trigonometric adjustment in Bretschneider's formula for non-cyclicality of the quadrilateral can be rewritten non-trigonometrically in terms of the sides and the diagonals e and f to give [2] [3]
In geometry, a Heronian triangle (or Heron triangle) is a triangle whose side lengths a, b, and c and area A are all positive integers. [ 1 ] [ 2 ] Heronian triangles are named after Heron of Alexandria , based on their relation to Heron's formula which Heron demonstrated with the example triangle of sides 13, 14, 15 and area 84 .
If also d = 0, the cyclic quadrilateral becomes a triangle and the formula is reduced to Heron's formula. The cyclic quadrilateral has maximal area among all quadrilaterals having the same side lengths (regardless of sequence). This is another corollary to Bretschneider's formula. It can also be proved using calculus. [12]
Additional proofs involve arguments based on symmetry, calculations in exterior algebra, or algebraic manipulation of Heron's formula (for which see § Soddy circles of a triangle). [22] [23] The result also follows from the observation that the Cayley–Menger determinant of the four coplanar circle centers is zero. [24]