Question

We need a rock garden with a perimeter of 12 feet and an area of 8 square feet what is the length and width

Answer 1

The dimensions of the rectangle can be a length of 2ft and a width of 4ft.

How to find the dimensions of the garden?

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:

$W = \frac{-6 \pm \sqrt{(6ft)^2 - 4*(-1)*(-8)} }{2*-1} \\\\W = \frac{-6 \pm 2}{-2}$

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:

brainly.com/question/17297081

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