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# The area of a rectangle is 252 square feet. If the perimeter is 66 feet, find the length and width of a rectangle?

Hello, Jesse --

You need to use your information about area and perimeter to set up two equations with two variables representing length and width of the rectangle. Then you will set up a quadratic equation. Let's use L for the length and W for the width of the rectangle:

For the perimeter:

2L + 2W = 66  or
L + W  = 33  or
W  = 33 - L

For the area:

LW =  252

Now substitute W = 33 - L from the first equation into the second one:

L(33 - L) = 252  or

33L - L2 = 252  or

L2 - 33L + 252 = 0

This can be solved by factoring, or by using the quadratic equation.

By factoring:

(L - 12)(L - 21) = 0  or

L = 12, or L = 21

In this case, let's use 21 because length is usually the longer dimension. Now we substitute L = 21 into the first equation to find W.

W = 33 - 21 or
W = 12

L+W = P/2 = 66/2 = 33

LW = A = 252

So L and W are the two roots of x2 - 33x + 252 = 0

which factors as (x - 12)(x - 21) = 0

So it is a 12 by 21 rectangle.

A = 252 sq. ft = l*w

2l + 2w = 66, 2(l+w) = 66, so l+w = 33

w = 252/l

l+(252/l) = 33

l^2 + 252 = 33l

l^2 - 33l - 252 = 0

You're at a quadriatic equation, which you can solve. You will get two answers - toss out the unreasonable one (hint - negative answer, won't fit with the width, etc.).

Mental math approach:

Half of the perimeter = length + width = 33 ft

Factor out 252 into two numbers such that the sum is 33: 252 = 21*12.

Therefore, the length is 21 ft and the width is 12 ft.

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