The area of a rectangle is 2 square feet. The perimeter of the rectangle is 9 ft. Find the length, l, and width, w, of the rectangle. What is l^2+w^2

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If the area of a rectangle is 2 feet, and its perimeter is 9 feet, we want to use the perimeter and area equations, as follows:

Area(a)= length(l) * width(w)

Perimeter(p) = 2*length(l)+2*width(w)

Because we have two variables (l and w) and two equations, we can use one equation to...

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If the area of a rectangle is 2 feet, and its perimeter is 9 feet, we want to use the perimeter and area equations, as follows:

Area(a)= length(l) * width(w)

Perimeter(p) = 2*length(l)+2*width(w)

Because we have two variables (l and w) and two equations, we can use one equation to solve for l in terms of w, and then plug in that equation to the second equation, in order that we end up with one equation and one variable, as follows:

A=l*w =2

P=2l +2w = 9; l=(9-2w)/2

Thus, l*w=((9-2w)/2)*w = 2

We can re-write this equation in standard form quadratic:

-2w^2+9w-4 =0. This yields solutions (using the "quadratic formula," explained at the link below), w =.5, and w = 4. Whichever answer we select for the width (w) w will determine that the other answer is the length (l). Thus, if width (w) = .5, then length (l) = 4. It follows that w^2 = 1/4, and l^2=16.

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