Question

Hello I need help with this immediately please see attached image thank you so much I will give Brainliest

Answer 1

Note that while the figure is made up of rectangle and parallelogram, then the area of this figure

$A_{figure}=A_{rectangle}+A_{parallelogram}.$

1. The area of rectangle is

$A_{rectangle\ ABCF}=AB\cdot BC.$

The vertices of rectangle are points A(-2,-2), B(0,-6), C(8,-2) and F(6,2). Then

$AB=\sqrt{(-2-0)^2+(-2+6)^2}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5},\\ \\BC=\sqrt{(8-0)^2+(-2+6)^2}=\sqrt{64+16}=\sqrt{80}=4\sqrt{5}.$

Therefore,

$A_{rectangle\ ABCF}=2\sqrt{5}\cdot 4\sqrt{5}=8\cdot 5=40\ un.^2$

2. The area of parallelogram can be calculated using formula

$A_{parallelogram\ FCDE}=\text{base}\cdot \text{height}.$

The base of parallelogram is segment CD with length

$CD=\sqrt{(8-8)^2+(3+2)^2}=\sqrt{25}=5$

and the height has length 2 un. Then

$A_{parallelogram\ FCDE}=5\cdot 2=10\ un^2.$

3. Now

$A_{figure}=40+10=50\ un.^2$