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This yields as a special case the well-known formula for the area of a triangle, by considering a triangle as a degenerate trapezoid in which one of the parallel sides has shrunk to a point. The 7th-century Indian mathematician Bhāskara I derived the following formula for the area of a trapezoid with consecutive sides a, c, b, d:
The area of an isosceles (or any) trapezoid is equal to the average of the ... is the semi-perimeter of the trapezoid. This formula is analogous to Heron's formula to ...
The formula for the surface area of a sphere is more difficult to derive: because a sphere has nonzero Gaussian curvature, it cannot be flattened out. The formula for the surface area of a sphere was first obtained by Archimedes in his work On the Sphere and Cylinder. The formula is: [6] A = 4πr 2 (sphere), where r is the radius of the sphere.
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 ...
Perimeter#Formulas – Path that surrounds an area; List of second moments of area; List of surface-area-to-volume ratios – Surface area per unit volume; List of surface area formulas – Measure of a two-dimensional surface; List of trigonometric identities; List of volume formulas – Quantity of three-dimensional space
The formula for the area of a trapezoid can be simplified using Pitot's theorem to get a formula for the area of a tangential trapezoid. If the bases have lengths a, b, and any one of the other two sides has length c, then the area K is given by the formula [2] (This formula can be used only in cases where the bases are parallel.)
The formula was described by Albrecht Ludwig Friedrich Meister (1724–1788) in 1769 [4] and is based on the trapezoid formula which was described by Carl Friedrich Gauss and C.G.J. Jacobi. [5] The triangle form of the area formula can be considered to be a special case of Green's theorem.
where K is the area of a convex quadrilateral with perimeter L. Equality holds if and only if the quadrilateral is a square. The dual theorem states that of all quadrilaterals with a given area, the square has the shortest perimeter. The quadrilateral with given side lengths that has the maximum area is the cyclic quadrilateral. [43]