see the attached figure to better understand the problem
Let
x------> the length side of a rectangle
y-------> the width side of a rectangle
we know that
the perimeter of a rectangle is equal to the formula
[tex]P=2x+2y[/tex]
in this problem
[tex]AB=DC=x[/tex]
[tex]AD=BC=y[/tex]
Step 1
Find the distance AB
[tex]A(-6,4)\\B(2,8)[/tex]
we know that
the distance's formula between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}[/tex]
substitute the values
[tex]d=\sqrt{(8-4)^{2} +(2+6)^{2}}[/tex]
[tex]d=\sqrt{(4)^{2} +(8)^{2}}[/tex]
[tex]dAB=\sqrt{80}\ units[/tex]
Step 2
Find the distance BC
tex]B(2,8)\\C(4,4)[/tex]
we know that
the distance's formula between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}[/tex]
substitute the values
[tex]d=\sqrt{(4-8)^{2} +(4-2)^{2}}[/tex]
[tex]d=\sqrt{(-4)^{2} +(2)^{2}}[/tex]
[tex]dBC=\sqrt{20}\ units[/tex]
Step 3
Find the perimeter
we know that
the perimeter of a rectangle is equal to the formula
[tex]P=2x+2y[/tex]
[tex]P=2AB+2BC[/tex]
we have
[tex]dAB=\sqrt{80}\ units=8.9\ units[/tex]
[tex]dBC=\sqrt{20}\ units=4.5\ units[/tex]
substitute the values of the distance in the formula
[tex]P=2*8.9+2*4.5=26.8\ units[/tex]
therefore
the answer is
The perimeter of the rectangle is equal to [tex]26.8\ units[/tex]
