Parallelogram

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A parallelogram. Opposite sides are parallel and congruent. The parallelogram in this case is a Rhomboid as its angles are oblique

In geometry, a parallelogram is a quadrilateral with two sets of parallel sides. The opposite or facing sides of a parallelogram are of equal length, and the opposite angles of a parallelogram are equal. The three-dimensional counterpart of a parallelogram is a parallelepiped.

[edit] Properties

Opposite sides of a parallelogram are equal in length.
Opposite angles of a parallelogram are equal in measure.
The area, $A$ , of a parallelogram is $A = b h$ , where $b$ is the base of the parallelogram and $h$ is its height.
Opposite sides of a parallelogram will never intersect.
The area of a parallelogram is twice the area of a triangle created by one of its diagonals.
The area of a parallelogram is also equal to the magnitude of the vector cross product of two adjacent sides.
The diagonals of a parallelogram bisect each other.
Any non-degenerate affine transformation takes a parallelogram to another parallelogram.
There is an infinite number of affine transformations which take any given parallelogram to a square.

[edit] Types of parallelograms

Parallelogram - A quadrilateral whose opposite sides are parallel.
Rectangle - A parallelogram with four angles of equal size (right angles).
Rhombus - A parallelogram with four sides of equal length.
Square - A parallelogram with four sides of equal length and four angles of equal size (right angles).

[edit] Proof that diagonals bisect each other

To prove that the diagonals of a parallelogram bisect each other, first note a few pairs of equivalent angles:

$\angle ABE \cong \angle CDE$ (alternate)

$\angle BAE \cong \angle DCE$ (alternate).

Since they are angles that a transversal makes with parallel lines $A B$ and $D C$ .

Also, $\angle AEB \cong \angle CED$ since they are a pair of vertical angles.

Therefore, $\triangle ABE \sim \triangle CDE$ since they have the same angles.

From this similarity, one has the ratios

${AB \over CD} = {AE \over CE} = {BE \over DE}.$

Since $A B = D C$ , we have

${AB \over CD} = 1.$

Therefore,

A E = C E

B E = D E .

$E$ bisects the diagonals $A C$ and $B D$ .

It can also be proved that the diagonals bisect each other, by placing the parallelogram on a coordinate grid, and assigning variables to the vertices, it can be shown that the diagonals have the same midpoint.

There is yet another way to prove that the diagonals of a parallelogram bisect each other.

It is known that AB = DC, because opposite sides of a parallelogram are equal in length. We also know that segment AB is parallel to segment DC by the definition of a parallelogram. If two parallel lines are cut by a transversal, alternate interior angles are equal in measure. Therefore, angle BAE is equal to angle DCE, and angle ABE is equal to angle CDE. The ASA postulate proves triangles ABE and CDE congruent. Corresponding sides of congruent triangles are equal in length, so AE = CE and BE = DE. Thus E is the mid-point of AC, and also of BD, so these diagonals of the parallelogram bisect each other.

[edit] The area formula

Area of the parallelogram is in blue

The area formula,

$A = B \times H,\,$

can be derived as follows:

The area of the parallelogram to the right (the blue area) is the total area of the rectangle less the area of the two orange triangles. The area of the rectangle is

$A_\text{rect} = (B+A) \times H\,$

and the area of a single orange triangle is

$A_\text{tri} = \frac{1}{2} A \times H\,$ or $S_\text{tri} = \frac{1}{2} bh.$

Therefore, the area of the parallelogram is

$A = A_\text{rect} - 2 \times A_\text{tri} = \left( (B+A) \times H \right) - \left( A \times H \right) = B \times H.\,$

[edit] Computing the area of a parallelogram

Let $a,b\in\R^2$ and let $V=[a\ b]\in\R^{2\times2}$ denote the matrix with columns $a$ and $b$ . Then the area of the parallelogram generated by $a$ and $b$ is equal to $| det(V) |$

Let $a,b\in\R^n$ and let $V=[a\ b]\in\R^{n\times2}$ . Then the area of the parallelogram generated by $a$ and $b$ is equal to $\sqrt{\det(V^T V)}$

Let $a,b,c\in\R^2$ . Then the area of the parallelogram is equivalent to the absolute value of the determinant of a matrix built using a, b and c as rows with the last column padded using ones as follows: