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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. Expressing Heron's formula with a Cayley–Menger determinant in terms of the squares of the distances between the three given vertices,
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 ...
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.)
In calculus, the trapezoidal rule (also known as the trapezoid rule or trapezium rule) [a] is a technique for numerical integration, i.e., approximating the definite integral: (). The trapezoidal rule works by approximating the region under the graph of the function f ( x ) {\displaystyle f(x)} as a trapezoid and calculating its area.
Since and are both perpendicular to , they are parallel and so the quadrilateral is a trapezoid. The theorem is proved by computing the area of this trapezoid in two different ways. The theorem is proved by computing the area of this trapezoid in two different ways.
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 ...