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The centers of four squares all constructed either internally or externally on the sides of a parallelogram are the vertices of a square. [8] If two lines parallel to sides of a parallelogram are constructed concurrent to a diagonal, then the parallelograms formed on opposite sides of that diagonal are equal in area. [8]
The square has Dih 4 symmetry, order 8. There are 2 dihedral subgroups: Dih 2, Dih 1, and 3 cyclic subgroups: Z 4, Z 2, and Z 1. A square is a special case of many lower symmetry quadrilaterals: A rectangle with two adjacent equal sides; A quadrilateral with four equal sides and four right angles; A parallelogram with one right angle and two ...
Informally: "a box or oblong" (including a square). Square (regular quadrilateral): all four sides are of equal length (equilateral), and all four angles are right angles. 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.
Vectors involved in the parallelogram law. In a normed space, the statement of the parallelogram law is an equation relating norms: ‖ ‖ + ‖ ‖ = ‖ + ‖ + ‖ ‖,.. The parallelogram law is equivalent to the seemingly weaker statement: ‖ ‖ + ‖ ‖ ‖ + ‖ + ‖ ‖, because the reverse inequality can be obtained from it by substituting (+) for , and () for , and then simplifying.
The elements of a polytope can be considered according to either their own dimensionality or how many dimensions "down" they are from the body.
A rhombus is an orthodiagonal quadrilateral with two pairs of parallel sides (that is, an orthodiagonal quadrilateral that is also a parallelogram). A square is a limiting case of both a kite and a rhombus. Orthodiagonal quadrilaterals that are also equidiagonal quadrilaterals are called midsquare quadrilaterals. [2]
It's a classic tale: You have last-minute guests coming over for dinner or a bake sale fundraiser you didn't find out about until the night before—and now you need to concoct some tasty treats ...
Splitting the thin parallelogram area (yellow) into little parts, and building a single unit square with them. The key to the puzzle is the fact that neither of the 13×5 "triangles" is truly a triangle, nor would either truly be 13x5 if it were, because what appears to be the hypotenuse is bent.