The width of a rectangle is 4 feet more than the length. If the length is doubled and the width is decreased by 2 feet, a new rectangle is formed whose perimeter is 68 feet. Find the dimensions of the original rectangle.

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Answer:

Step-by-step explanation:

We'll work through this by setting up a table, as follows:

               original                 new

width

length

It tells us the width of a rectangle is 4 feet more than the length. If we don't know the length, we'll call it L; that means that the width is L + 4, because the width is the length plus 4 more feet. Put that into column 1:

                  original                 new

width            L + 4

length             L

The next sentence tells us how the original rectangle is being manipulated to create a new rectangle, so that information goes into the column named "new". The original length is doubled:

                   original                new

width             L + 4

length              L                        2L

and the original width is decreased by 2:

                     original                 new

width              L + 4                   L + 4 - 2 (which simplifies to L + 2)

length                L                         2L

The perimeter of the new rectangle is 68. The formula for the perimeter is

P = 2L + 2w, so filling in our info along with the given perimeter:

68 = 2(L + 2) + 2(2L) and

68 = 2L + 4 + 4L and

68 = 6L + 4 and

64 = 6L so

L = [tex]\frac{32}{3}[/tex] and the width then is

w = [tex]\frac{32}{3}+4[/tex] so

w = [tex]\frac{44}{3}[/tex]

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