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  2. Area of a triangle - Wikipedia

    en.wikipedia.org/wiki/Area_of_a_triangle

    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.

  3. Heron's formula - Wikipedia

    en.wikipedia.org/wiki/Heron's_formula

    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]

  4. Triangle - Wikipedia

    en.wikipedia.org/wiki/Triangle

    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.

  5. Shoelace formula - Wikipedia

    en.wikipedia.org/wiki/Shoelace_formula

    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.

  6. Brahmagupta's formula - Wikipedia

    en.wikipedia.org/wiki/Brahmagupta's_formula

    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.

  7. Pick's theorem - Wikipedia

    en.wikipedia.org/wiki/Pick's_theorem

    Any triangle subdivides its bounding box into the triangle itself and additional right triangles, and the areas of both the bounding box and the right triangles are easy to compute. Combining these area computations gives Pick's formula for triangles, and combining triangles gives Pick's formula for arbitrary polygons. [7] [8] [13]

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