Note that while the figure is made up of rectangle and parallelogram, then the area of this figure
[tex]A_{figure}=A_{rectangle}+A_{parallelogram}.[/tex]
1. The area of rectangle is
[tex]A_{rectangle\ ABCF}=AB\cdot BC.[/tex]
The vertices of rectangle are points A(-2,-2), B(0,-6), C(8,-2) and F(6,2). Then
[tex]AB=\sqrt{(-2-0)^2+(-2+6)^2}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5},\\ \\BC=\sqrt{(8-0)^2+(-2+6)^2}=\sqrt{64+16}=\sqrt{80}=4\sqrt{5}.[/tex]
Therefore,
[tex]A_{rectangle\ ABCF}=2\sqrt{5}\cdot 4\sqrt{5}=8\cdot 5=40\ un.^2[/tex]
2. The area of parallelogram can be calculated using formula
[tex]A_{parallelogram\ FCDE}=\text{base}\cdot \text{height}.[/tex]
The base of parallelogram is segment CD with length
[tex]CD=\sqrt{(8-8)^2+(3+2)^2}=\sqrt{25}=5[/tex]
and the height has length 2 un. Then
[tex]A_{parallelogram\ FCDE}=5\cdot 2=10\ un^2.[/tex]
3. Now
[tex]A_{figure}=40+10=50\ un.^2[/tex]
