Chapter 7.8, Problem 9E

To determine

To find the area of the figure.

Answer to Problem 9E

The area of the figure is obtained as,

  $A=24\text{ square units}$

Explanation of Solution

Given information :

The vertices are given as, $\left(-3,2\right)\left(1,2\right)\left(-1,-4\right)\left(-5,-4\right)$ .

Calculation :

Area of the figure from the given vertices can be obtained by evaluating the determinant.

Using the first three matrices $\left(-3,2\right)\left(1,2\right)\left(-1,-4\right)$ , the value of $A_1$ can be obtained as,

  $\begin{array}{l}{A}_1=\pm \frac{1}{2}\left|\begin{array}{ccc}-3& 2& 1\\ 1& 2& 1\\ -1& -4& 1\end{array}\right|\\ =\pm \frac{1}{2}\left[-3\left(2+4\right)-2\left(1+1\right)+1\left(-4+2\right)\right]\\ =\pm \frac{1}{2}\left[-18-4-2\right]\\ =\pm \frac{1}{2}\cdot 24\\ =12\end{array}$

Using the other three matrices $\left(-3,2\right)\left(-1,-4\right)\left(-5,-4\right)$ , the value of $A_2$ can be obtained as,

  $\begin{array}{l}{A}_2=\pm \frac{1}{2}\left|\begin{array}{ccc}-3& 2& 1\\ -1& -4& 1\\ -5& -4& 1\end{array}\right|\\ =\pm \frac{1}{2}\left[-3\left(-4+4\right)-2\left(-1+5\right)+1\left(4-20\right)\right]\\ =\pm \frac{1}{2}\left[-8-16\right]\\ =\pm \frac{1}{2}\cdot \left(-24\right)\\ =12\end{array}$

The total area $A$ is the sum of the $A_1+A_2$ as follows,

  $\begin{array}{l}A=A_1+A_2\\ A=12+12\\ A=24\end{array}$

Therefore,

The area of the figure is obtained as,

  $A=24\text{ square units}$