The dimensions of the rectangle can be a length of 2ft and a width of 4ft.
Remember that for a rectangle of length L and width W, the perimeter is:
P = 2*(L + W)
And the area is:
A = L*W
In this case, we know that the area is 8 square feet and the perimeter is 12 ft, then we have a system of equations:
12ft = 2*(L + W)
8ft² = L*W
To solve this, we first need to isolate one of the variables in one of the equations, I will isolate L on the first one:
12ft/2 = L + W
6ft - W = L
Now we can replace that in the other equation to get:
8ft² = (6ft - W)*W
This is a quadratic equation:
-W^2 + 6ft*W - 8ft² = 0
The solutions are given by Bhaskara's formula:
[tex]W = \frac{-6 \pm \sqrt{(6ft)^2 - 4*(-1)*(-8)} }{2*-1} \\\\W = \frac{-6 \pm 2}{-2}[/tex]
Then we have two solutions:
W = (-6 - 2)/-2 = 4ft
W = (-6 + 2)/-2 = 2ft
If we take any of these solutions, the length will be equal to the other solution.
So the dimensions of the rectangle can be a length of 2ft and a width of 4ft.
if you want to learn more about rectangles:
https://brainly.com/question/17297081
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