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Find the dimensions of a rectangle if its perimeter is 52 inches when the width is 9 less than that length ?

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Final answer:

To find the dimensions of a rectangle with a 52-inch perimeter and the width 9 inches less than the length, we establish two equations: w = l - 9 and 52 = 2l + 2w. Solving these yields a length of 17.5 inches and a width of 8.5 inches.

Step-by-step explanation:

To find the dimensions of a rectangle when given the perimeter and a relationship between its length and width, we use algebra. The formula for the perimeter (P) of a rectangle is P = 2l + 2w, where l is the length and w is the width.

In this case, we are given that the perimeter (P) is 52 inches and the width is 9 inches less than the length (l). Therefore, we can set up the following equations: w = l - 9 and 52 = 2l + 2w. Substituting the first equation into the second gives us 52 = 2l + 2(l - 9).

Now, we will solve the equation:

52 = 2l + 2l - 18

52 + 18 = 4l

70 = 4l

l = 17.5 inches (length)

w = l - 9 = 17.5 - 9 = 8.5 inches (width)

Therefore, the dimensions of the rectangle are 17.5 inches in length and 8.5 inches in width.

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