# Find the area of the parallelogram whose vertices are listed. (0,0), (5,2), (6,4), (11,6)

This article aims to find the area of the parallelogram. This article uses the concept of the area of the parallelogram. A parallelogram bounds a parallelogram‘s area in a given two-dimensional space. As a reminder, a parallelogram is a particular type of quadrilateral with four sides, and the pairs of opposite sides are parallel. In parallelogram, opposite sides have the same length, and opposite angles have equal measures. Since a rectangle and a parallelogram have similar properties, the area of the rectangle is equal to the area of a parallelogram.

To find area of a parallelogram, multiply the perpendicular base by its height. It should be noted that the base and altitude of a parallelogram are perpendicular to each other, while the lateral side of a parallelogram is not perpendicular to the base.

$Area = b \times h$

Where $b$ is the base and $h$ is the height of the parallelogram.

A parallelogram can be described by $4$ vertices or $2$ vectors. Since we have $4$ vertices $(ABCD)$, we find the vectors $u$, $v$ that describe the parallelogram.

$A = ( 0 , 0 )$

$B = ( 5 , 2 )$

$C = ( 6 , 4 )$

$D = ( 11 , 6 )$

$u = AB = \begin{bmatrix} 5 \\ 2 \end{bmatrix}$

$v = AC = \begin{bmatrix} 6 \\ 4 \end{bmatrix}$

Area of parallelogram is the absolute value of the determinant.

$\begin{bmatrix} u _ { 1 } & v _ { 1 } \\ u _ { 2 } & v _ { 2 } \end{bmatrix} = det \begin{bmatrix} 5 & 6 \\ 2 & 4 \end{bmatrix}= 20 \: – \: 12 = 8$

The area of the parallelogram is $8$.

## Numerical Result

The area of the parallelogram is $8$.

## Example

Find area of the parallelogram whose vertices are given. $( 0 , 0 )$, $( 5 , 2 )$, $( 6 , 4 )$ , $( 11 , 6 )$

Solution

A parallelogram can be described by $4$ vertices or $2$ vectors. Since we have $4$ vertices $( ABCD )$, we find the vectors $u$, $v$ that describe the parallelogram.

$A = ( 0 , 0 )$

$B = ( 6 , 8 )$

$C = ( 5 , 4 )$

$D = ( 11 , 6 )$

$u = AB = \begin{bmatrix} 6\\ 8 \end{bmatrix}$

$v = AC = \begin{bmatrix} 5\\ 4 \end{bmatrix}$

Area of parallelogram is the absolute value of the determinant.

$\begin{bmatrix} u _ { 1 } & v _ { 1 } \\ u _ { 2 } & v _ { 2 } \end{bmatrix} = det \begin{bmatrix} 6 & 5 \\ 8 & 4 \end{bmatrix}= 24 \: – \: 40 = 16$

The area of the parallelogram is $16$.

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