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  2. Bretschneider's formula - Wikipedia

    en.wikipedia.org/wiki/Bretschneider's_formula

    Bretschneider's formula generalizes Brahmagupta's formula for the area of a cyclic quadrilateral, which in turn generalizes Heron's formula for the area of a triangle.. The trigonometric adjustment in Bretschneider's formula for non-cyclicality of the quadrilateral can be rewritten non-trigonometrically in terms of the sides and the diagonals e and f to give [2] [3]

  3. Quadrilateral - Wikipedia

    en.wikipedia.org/wiki/Quadrilateral

    The area can be also expressed in terms of bimedians as [16] = ⁡, where the lengths of the bimedians are m and n and the angle between them is φ. Bretschneider's formula [17] [14] expresses the area in terms of the sides and two opposite angles:

  4. Tangential quadrilateral - Wikipedia

    en.wikipedia.org/wiki/Tangential_quadrilateral

    If the incircle is tangent to the sides AB, BC, CD, DA at T 1, T 2, T 3, T 4 respectively, and if N 1, N 2, N 3, N 4 are the isotomic conjugates of these points with respect to the corresponding sides (that is, AT 1 = BN 1 and so on), then the Nagel point of the tangential quadrilateral is defined as the intersection of the lines N 1 N 3 and N ...

  5. 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.

  6. Barycentric coordinate system - Wikipedia

    en.wikipedia.org/wiki/Barycentric_coordinate_system

    Two solutions to the 8, 5 and 3 L water pouring puzzle using a barycentric plot. The yellow area denotes combinations achievable with the jugs. The solid red and dashed blue paths show pourable transitions. When a vertex lands on the dotted triangle, 4 L has been measured.

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  8. Tetrahedron - Wikipedia

    en.wikipedia.org/wiki/Tetrahedron

    Its surface area is four times the area of an equilateral triangle: = =. [7] Its volume can be ascertained similarly as the other pyramids, one-third of the base times height. Because the base is an equilateral, it is: [ 7 ] V = 1 3 ⋅ ( 3 4 a 2 ) ⋅ 6 3 a = a 3 6 2 ≈ 0.118 a 3 . {\displaystyle V={\frac {1}{3}}\cdot \left({\frac {\sqrt {3 ...

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