Chapter 7.8, Problem 22E

To find : The area of the parallelogram with the given vertices by using the determinant.

To determine

The area of the parallelogram is 64 square units.

Given information:

The vertices of the parallelogram are $\left(0,0\right),\left(0,8\right),\left(8,-6\right),\left(8,2\right)$ .

Formula used:

The area of the parallelogram with the vertices (0,0),( a , b ),( c , d ) and ( a + c , b + d ) is given by,

  $A=\left|\det \left(A\right)\right|$

Here, $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

Calculation:

It is given that the vertices of the parallelogram are $\left(0,0\right),\left(0,8\right),\left(8,-6\right),\left(8,2\right)$ .

Determine the area of the parallelogram as follows,

Consider $\left(0,8\right)=\left(a,b\right)\text{ and }\left(8,-6\right)=\left(c,d\right)$ .

  $A=\left[\begin{array}{cc}0& 8\\ 8& -6\end{array}\right]$

So, the area of the parallelogram is computed as,

  $\begin{array}{c}A=\left|\det \left[\begin{array}{cc}0& 8\\ 8& -6\end{array}\right]\right|\\ =\left|-6\left(0\right)-8\left(8\right)\right|\\ =64\end{array}$

Therefore, the area of the parallelogram is 64 square units.

Answer to Problem 22E

The area of the parallelogram is 64 square units.

Explanation of Solution

Given information:

The vertices of the parallelogram are $\left(0,0\right),\left(0,8\right),\left(8,-6\right),\left(8,2\right)$ .

Formula used:

The area of the parallelogram with the vertices (0,0),( a , b ),( c , d ) and ( a + c , b + d ) is given by,

  $A=\left|\det \left(A\right)\right|$

Here, $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

Calculation:

It is given that the vertices of the parallelogram are $\left(0,0\right),\left(0,8\right),\left(8,-6\right),\left(8,2\right)$ .

Determine the area of the parallelogram as follows,

Consider $\left(0,8\right)=\left(a,b\right)\text{ and }\left(8,-6\right)=\left(c,d\right)$ .

  $A=\left[\begin{array}{cc}0& 8\\ 8& -6\end{array}\right]$

So, the area of the parallelogram is computed as,

  $\begin{array}{c}A=\left|\det \left[\begin{array}{cc}0& 8\\ 8& -6\end{array}\right]\right|\\ =\left|-6\left(0\right)-8\left(8\right)\right|\\ =64\end{array}$

Therefore, the area of the parallelogram is 64 square units.