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The rhombus has a square as a special case, and is a special case of a kite and parallelogram.. In plane Euclidean geometry, a rhombus (pl.: rhombi or rhombuses) is a quadrilateral whose four sides all have the same length.
A square is a special case of a rhombus (equal sides, opposite equal angles), a kite (two pairs of adjacent equal sides), a trapezoid (one pair of opposite sides parallel), a parallelogram (all opposite sides parallel), a quadrilateral or tetragon (four-sided polygon), and a rectangle (opposite sides equal, right-angles), and therefore has all ...
Rhombus – A parallelogram with four sides of equal length. Any parallelogram that is neither a rectangle nor a rhombus was traditionally called a rhomboid but this term is not used in modern mathematics. [1] Square – A parallelogram with four sides of equal length and angles of equal size (right angles).
Its four vertices lie at the three corners and one of the side midpoints of the Reuleaux triangle. [19] [20] When an equidiagonal kite has side lengths less than or equal to its diagonals, like this one or the square, it is one of the quadrilaterals with the greatest ratio of area to diameter. [21]
An equivalent condition is that opposite sides are parallel (a square is a parallelogram), and that the diagonals perpendicularly bisect each other and are of equal length. A quadrilateral is a square if and only if it is both a rhombus and a rectangle (i.e., four equal sides and four equal angles).
Many results about plane figures are proved, for example, "In any triangle, two angles taken together in any manner are less than two right angles." (Book I proposition 17) and the Pythagorean theorem "In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle ...
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A square is a limiting case of both a kite and a rhombus. Orthodiagonal equidiagonal quadrilaterals in which the diagonals are at least as long as all of the quadrilateral's sides have the maximum area for their diameter among all quadrilaterals, solving the n = 4 case of the biggest little polygon problem.