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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 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.
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. Heron's formula is obtained by setting the smaller parallel side to zero.
The most basic area formula is the formula for the area of a ... Similar arguments can be used to find area formulas for the trapezoid [26] as well as more ...
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 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 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 area of the trapezoid can be calculated to be half the area of the square, that is 1 2 ( b + a ) 2 . {\displaystyle {\frac {1}{2}}(b+a)^{2}.} The inner square is similarly halved, and there are only two triangles so the proof proceeds as above except for a factor of 1 2 {\displaystyle {\frac {1}{2}}} , which is removed by multiplying by two ...