Parallelogram

A parallelogram.

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

1 Properties
2 Computing the area of a parallelogram
3 Proof that diagonals bisect each other
4 Derivation of the area formula
5 See also
6 External links

[ Properties

The area,

A

, of a parallelogram is

A = b h

, where

b

is the base of the parallelogram and

h

is its height.

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 an parallelogram. There are infinite affine transformations which takes given parallelogram to a square.
Opposite si

[ 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:

$V = \left| \det \begin{bmatrix} a_1 & a_2 & 1 \\ b_1 & b_2 & 1 \\ c_1 & c_2 & 1 \end{bmatrix} \right|.$

[ 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:

$\angle ABE \cong \angle CDE$

$\angle BAE \cong \angle DCE$

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, we have 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 $B D$ .

You can also prove that the diagonals bisect each other, by placing the parallelogram on a coordinate grid, and assign variables to the vertexes, you can show that the diagonals have the same midpoint.

[ Derivation of the area formula

Area of the parallelogram is in blue

The area formula,

$A_\text{parallelogram} = B \times H,\,$

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

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