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# parallelogram

[par-uh-lel-uh-gram]
parallelogram, closed plane figure bounded by four line segments, or sides, with opposite pairs of sides parallel and equal in length. The rhombus, rectangle, and square are special types of parallelograms. Any side of a parallelogram is a base; an altitude is the perpendicular distance from a base to the opposite parallel side. The area of a parallelogram is equal to the product of the lengths of its base and altitude. The diagonals of a parallelogram, connecting opposite vertices, bisect one another; either diagonal divides the parallelogram into two congruent triangles.

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.

## Properties

• The area, $A$, of a parallelogram is $A = BH$, 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.
• Opposite sides of a parallelogram are equal.
• Opposite angles of a parallelogram are equal.
• Each diagonal bisects the parallelogram into two congruent triangles.
• It is possible to create a tessellation of a plane with any parallelogram.

The properties of having equal opposite sides and opposite angles are shared with the antiparallelogram, a type of non-convex quadrilateral in which the two longer edges cross each other.

## Computing the area of a parallelogram

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

Let $a,binR^n$ and let $V=\left[a b\right]inR^\left\{ntimes2\right\}$. Then the area of the parallelogram generated by $a$ and $b$ is equal to $sqrt\left\{det\left(V^T V\right)\right\}$

Let $a,b,cinR^2$. 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:

$V = left| det begin\left\{bmatrix\right\}$
`       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 $AB$ and $DC$.

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

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

Since $AB = DC$, we have

$\left\{AB over CD\right\} = 1$.

Therefore,

$AE = CE$
$BE = DE$

$E$ bisects the diagonals $AC$ and $BD$.

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

The area formula,

$A_text\left\{parallelogram\right\} = B times H,,$

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

$A_text\left\{rect\right\} = \left(B+A\right) times H,$

and the area of a single orange triangle is

$A_text\left\{tri\right\} = frac\left\{1\right\}\left\{2\right\} A times H,$ or $S_text\left\{tri\right\} = frac\left\{1\right\}\left\{2\right\} bh,$

Therefore, the area of the parallelogram is

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

### Alternate method

An alternative, less mathematically sophisticated method, to show the area is by rearrangement of the perimeter. First, take the two ends of the parallelogram and chop them off to form two more triangles. Each of these two new triangles are equal in every way with the orange triangles. This first step is shown to the right.

The second step is merely swap the left orange triangle with the right blue triangle. Clearly, the two blue triangles plus the blue rectangle have an area equivalent to $B H$.

To further demonstrate this, the first image on the right could be printed off and cut up along the lines:

1. Cut along the lines between the orange triangles and the blue parallelogram
2. Cut along the vertical lines on the end to form the two blue triangles and the blue rectangle
3. Rearrange all five pieces as shown in the second image