# Find the area of the parallelogram with vertices A(-3, 0), B(-1, 5), C(7, 4), and D(5, -1).

What is the area of the parallelogram with vertices A(-3, 0), B(-1, 5), C(7, 4), and D(5, -1).

You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

un4t5o4v
Jeffrey Jordon

Consider the parallelogram A=(-3;0), B(-1,5), C(7;4) and D=(5;-1)

The objective is to find the area of parallelogram

Find the are as follows:

The area of the parallelogram is the magnitude of the cross product of the adjacent edges.

That is, $Area=|\stackrel{―}{AB}×\stackrel{―}{AD}|$

Find the adjacent edges $\stackrel{―}{AB}$ and $\stackrel{―}{AD}$ as follows

$\stackrel{―}{AB}=B-A$

$=\left(-1;5\right)-\left(-3;0\right)$

$=\left(-1+3;5-0\right)$

$=\left(2;5\right)$

$\stackrel{―}{AD}=D-A$

$=\left(5;-1\right)-\left(-3;0\right)$

$=\left(8;1\right)$

Find $\stackrel{―}{AB}×\stackrel{―}{AD}$ as,

$\stackrel{―}{AB}×\stackrel{―}{AD}=\left[\begin{array}{ccc}i& j& k\\ 2& 5& 0\\ 8& -1& 0\end{array}\right]$

$=5\left(0\right)-0\left(-1\right)i-\left(2\left(0\right)-0\left(8\right)\right)j+\left(2\left(-1\right)-5\left(8\right)\right)k$

$=0i+0j-42k$

The area of the parallelogram is,

Area $=|AB×AD|$

$=|\left(0i+0j-42k|$

$=\sqrt{\left(0{\right)}^{2}+\left(0{\right)}^{2}+\left(-42{\right)}^{2}}$

$=\sqrt{\left(42{\right)}^{2}}$

= 42 square unit

Hence, the area of parallelogram is 42