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The area of a triangle can be demonstrated, for example by means of the congruence of triangles, as half of the area of a parallelogram that has the same base length and height. A graphic derivation of the formula T = h 2 b {\displaystyle T={\frac {h}{2}}b} that avoids the usual procedure of doubling the area of the triangle and then halving it.
A triangle with sides a, b, and c. In geometry, Heron's formula (or Hero's formula) gives the area of a triangle in terms of the three side lengths , , . Letting be the semiperimeter of the triangle, = (+ +), the area is [1]
The area formula for a triangle can be proven by cutting two copies of the triangle into pieces and rearranging them into a rectangle. In the Euclidean plane, area is defined by comparison with a square of side length , which has area 1. There are several ways to calculate the area of an arbitrary triangle.
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.
Regular polygons; Description Figure Second moment of area Comment A filled regular (equiliteral) triangle with a side length of a = = [6] The result is valid for both a horizontal and a vertical axis through the centroid, and therefore is also valid for an axis with arbitrary direction that passes through the origin.
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.
The area of a polygon can be calculated from individual quadrangles of the above type, from (analogously) individual triangle bounded by a segment of the polygon and two meridians, [15] by a line integral with Green's theorem, [16] or via an equal-area projection as commonly done in GIS.
One may prove these ratio formulas based on the facts that a triangle is half of a parallelogram, and the area of a parallelogram is easy to compute using a determinant. Specifically, let D = − A + B + C . {\displaystyle D=-A+B+C.}