The area of a rectangle is 252 square feet. If the perimeter is 66 feet, find the length and width of a rectangle. What is l^2 and w^2?

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First, identify the geometric equations at hand.

First, Area: A = L * W, where L represents length and W represents width.

Perimeter: P = 2*L+ 2*W

Thus, in this example, L * W = A = 256,

and 2L + 2W = 66.

As a general rule, we know...

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First, identify the geometric equations at hand.

First, Area: A = L * W, where L represents length and W represents width.

Perimeter: P = 2*L+ 2*W

Thus, in this example, L * W = A = 256,

and 2L + 2W = 66.

As a general rule, we know that we can solve for as many equations as we have variables. Because here we have two variables and two equations, we can simply solve for one in terms of the other using one equation, and plug that solution into the other equation, in order to eliminate one variable, as follows:

Using the perimeter equation, we can rearrange to solve for L: L = (66-2W)/2

We can plug this into the area equation: ((66-2W)/2) * W = 66

This second equation we can reconfigure as a quadratic equation in standard form, as follows:

-2W² + 66W - 504 = 0

Using the quadratic formula or a graphing utility, we have W = 12, 21. Whichever is not the width is going to be the length. Thus, if we let W = 12, then L = 21. It follows that w² =144 and l²=441.

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