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In Euclidean geometry, an orthodiagonal quadrilateral is a quadrilateral in which the diagonals cross at right angles. In other words, it is a four-sided figure in which the line segments between non-adjacent vertices are orthogonal (perpendicular) to each other.
As is true more generally for any orthodiagonal quadrilateral, the area of a kite may be calculated as half the product of the lengths of the diagonals and : [10] =. Alternatively, the area can be calculated by dividing the kite into two congruent triangles and applying the SAS formula for their area.
In the case of an orthodiagonal quadrilateral (e.g. rhombus, square, and kite), this formula reduces to = since θ is 90°. 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.
The direct theorem was Proposition 22 in Book 3 of Euclid's Elements. [3] Equivalently, a convex quadrilateral is cyclic if and only if each exterior angle is equal to the opposite interior angle . In 1836 Duncan Gregory generalized this result as follows: Given any convex cyclic 2 n -gon, then the two sums of alternate interior angles are each ...
A conformal map acting on a rectangular grid. Note that the orthogonality of the curved grid is retained. While vector operations and physical laws are normally easiest to derive in Cartesian coordinates, non-Cartesian orthogonal coordinates are often used instead for the solution of various problems, especially boundary value problems, such as those arising in field theories of quantum ...
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.
There are 673 6-uniform tilings of the Euclidean plane. Brian Galebach's search reproduced Krotenheerdt's list of 10 6-uniform tilings with 6 distinct vertex types, as well as finding 92 of them with 5 vertex types, 187 of them with 4 vertex types, 284 of them with 3 vertex types, and 100 with 2 vertex types.
A direct formula for the conversion from a quaternion to Euler angles in any of the 12 possible sequences exists. [2] For the rest of this section, the formula for the sequence Body 3-2-1 will be shown. If the quaternion is properly normalized, the Euler angles can be obtained from the quaternions via the relations: