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Sagot :
The perimeter of the rectangle is 8 units
The coordinates of the rectangle is given as:
[tex]A = (-6,-4)[/tex]
[tex]B = (-4,-4)[/tex]
[tex]C = (-4,-2)[/tex]
[tex]D = (-6,-2)[/tex]
Start by calculating the distance between the vertices using the following distance formula
[tex]d=\sqrt{(x_2 -x_1)^2 + (y_2 -y_1)^2}[/tex]
So, we have:
[tex]AB=\sqrt{(-6 --4)^2 + (-4 --4)^2}[/tex]
[tex]AB=\sqrt{4}[/tex]
[tex]AB=2[/tex]
[tex]BC=\sqrt{(-4 --4)^2 + (-4 --2)^2}[/tex]
[tex]BC=\sqrt{4}[/tex]
[tex]BC=2[/tex]
[tex]CD=\sqrt{(-4 --6)^2 + (-2 --2)^2}[/tex]
[tex]CD=\sqrt{4}[/tex]
[tex]CD=2[/tex]
[tex]DA=\sqrt{(-6 --6)^2 + (-2 --4)^2}[/tex]
[tex]DA=\sqrt{4}[/tex]
[tex]DA = 2[/tex]
So, the perimeter of the rectangle is:
[tex]P =AB + BC + CD + DA[/tex]
This gives
[tex]P= 2 + 2+2+2[/tex]
[tex]P= 8[/tex]
Hence, the perimeter of the rectangle is 8 units
Read more about perimeters at:
https://brainly.com/question/397857
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