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In Euclidean geometry, Brahmagupta's formula, named after the 7th century Indian mathematician, is used to find the area of any convex cyclic quadrilateral (one that can be inscribed in a circle) given the lengths of the sides. Its generalized version, Bretschneider's formula, can be used with non-cyclic quadrilateral.
An arbitrary quadrilateral and its diagonals. Bases of similar triangles are parallel to the blue diagonal. Ditto for the red diagonal. The base pairs form a parallelogram with half the area of the quadrilateral, A q, as the sum of the areas of the four large triangles, A l is 2 A q (each of the two pairs reconstructs the quadrilateral) while that of the small triangles, A s is a quarter of A ...
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. Brahmagupta's formula gives the area K {\displaystyle K} of a cyclic quadrilateral whose sides have lengths a , {\displaystyle a,} b , {\displaystyle b,} c , {\displaystyle c ...
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]
A Watt quadrilateral is a quadrilateral with a pair of opposite sides of equal length. [6] A quadric quadrilateral is a convex quadrilateral whose four vertices all lie on the perimeter of a square. [7] A diametric quadrilateral is a cyclic quadrilateral having one of its sides as a diameter of the circumcircle. [8]
Euler's quadrilateral theorem or Euler's law on quadrilaterals, named after Leonhard Euler (1707–1783), describes a relation between the sides of a convex quadrilateral and its diagonals. It is a generalisation of the parallelogram law which in turn can be seen as generalisation of the Pythagorean theorem .
If the lengths of the three sides are known then Heron's formula can be used: () () where a, b, c are the sides of the triangle, and = (+ +) is half of its perimeter. [2] If an angle and its two included sides are given, the area is 1 2 a b sin ( C ) {\displaystyle {\tfrac {1}{2}}ab\sin(C)} where C is the given angle and a and b are its ...
In particular, to find the quadrilateral, or the triangle, or another particular figure, with the largest area amongst those with the same shape having a given perimeter. The solution to the quadrilateral isoperimetric problem is the square, and the solution to the triangle problem is the equilateral triangle. In general, the polygon with n ...