The aim of this problem is to get us familiar with the area of a very common quadrilateral known as a parallelogram. If we recall, a parallelogram is a pretty simple quadrilateral with two couples of parallel-faced sides.

The opposite lengths of a parallelogram are of equal dimensions and the opposing angles of a parallelogram are of equal magnitude.

## Expert Answer

Since a parallelogram is a tilted rectangle, all of the area formulas for known quadrilaterals can be used for parallelograms.

A parallelogram with one base $b$ and height $h$ can be separated into a trapezoid and a triangle with a right-angled side and can be shuffled into a rectangle. This implies that the area of a parallelogram is identical to that of a rectangle which has the same base and height.

We can define the area of a parallelogram as the absolute magnitude of the cross product of its adjacent angles, that is:

\[Area = |\overline{AB} \times \overline{AD}|\]

Finding the adjacent edges $\overline{AB}$ and $\overline{AD}$ and substituting back in the equation as follows:

\[\overline{AB} = B – A \]

Point $A$ and $B$ are given as:

\[\overline{AB} = (-1, 5) – (-3, 0) \]

\[= (-1+3) , (5 – 0)\]

\[\overline{AB} = (2, 5)\]

Now solving $\overline{AD}$:

\[\overline{AD} = D – A\]

Point $A$ and $D$ are given as:

\[\overline{AD} = (5, -1) – (-3, 0)\]

\[= (5+3) , (-1 + 0)\]

\[\overline{AD} = (8, -1)\]

Finding the cross product of $\overline{AB}$ and $\overline{AD}$ as:

\[ \overline{AB} \times \overline{AD} = \begin{bmatrix} i & j & k \\ 2 & 5 & 0\\8 & -1 & 0 \end{bmatrix}\]

\[= [5(0) – 0(-1)]i – [2(0)-0(8)]j +[2(-1)-5(8)]\]

\[= 0i +0j -42k\]

Taking the magnitude of $\overline{AB}$ and $\overline{AD}$, as the formula states:

\[Area = |\overline{AB} \times \overline{AD}|\]

\[= |0i+ 0j -42k|\]

\[= \sqrt{0^2 + 0^2 + 42^2}\]

\[= \sqrt{42^2}\]

\[Area= 42\]

## Numerical Result

The area of the parallelogram with its vertices $A(-3,0)$, $B(-1,5)$, $C(7,4)$ and $D(5,-1)$ is $42$ Square Unit.

## Example

Find the area of the parallelogram given the vertices $A(-3,0)$, $B(-1,4)$, $C(6,3)$ and $D(4,-1)$

Inserting the values into the formula of parallelogram, which is given as:

\[Area = |\overline{AB} \times \overline{AD}|\]

Finding the $\overline{AB}$

\[\overline{AB} = B – A\]

Point $A$ and $B$ are given as:

\[\overline{AB} = (-1, 4) – (-3, 0) \]

\[= (-1+3) , (4 – 0) \]

\[\overline{AB} = (2, 4)\]

Now solving $\overline{AD}$:

\[\overline{AD} = D – A\]

Point $A$ and $D$ are given as:

\[\overline{AD} = (4, -1) – (-3, 0) \]

\[= (4+3) , (-1 + 0) \]

\[\overline{AD} = (7, -1)\]

Finding the cross product of $\overline{AB}$ and $\overline{AD}$ as:

\[\overline{AB} \times \overline{AD} = \begin{bmatrix} i & j & k \\ 2 & 4 & 0\\7 & -1 & 0 \end{bmatrix}\]

\[= [5(0) – 0(-1)]i – [2(0)-0(8)]j +[2(-1)-4(7)]\]

\[ = 0i +0j -30k \]

Taking the magnitude of $\overline{AB}$ and $\overline{AD}$, as the formula states:

\[Area = |\overline{AB} \times \overline{AD}|\]

\[= |0i+ 0j -30k|\]

\[ = \sqrt{0^2 + 0^2 + 30^2}\]

\[ = \sqrt{30^2}\]

\[ = 30\]

The area of the parallelogram with vertices $A(-3,0)$, $B(-1,4)$, $C(6,3)$ and $D(4,-1)$ is $30$ Square Unit.