The area of a rectangle is 2 square feet. the perimeter of the rectangle is 8 ft. Find the length,l, and width, w, of the rectangle. What is l^2 and w^2?

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Because both the area and perimeter are related to the length and width, this can be represented as a system of equations. If the area of a rectangle is 2 feet, and its perimeter is 8 feet, we want to use the perimeter and area equations, as follows:

Area(a)= length(l)...

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Because both the area and perimeter are related to the length and width, this can be represented as a system of equations. If the area of a rectangle is 2 feet, and its perimeter is 8 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 = 8; l=(8-2w)/2

Thus, l*w=((8-2w)/2)*w = 2

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

-2w^2+8w-4 =0. This yields solutions (using the "quadratic formula," explained at the link below), w =.586, and w = 3.414. Whichever answer we select for the width (w) w will determine that the other answer is the length (l). Thus, if width (w) = .586, then length (l) = 3.414. It follows that w^2 = .343396, and l^2=11.655396.

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