SOLUTION: A rectangle has a perimeter of 52ft and area of 120 sq ft. find the width of the rectangle.

Question 693422: A rectangle has a perimeter of 52ft and area of 120 sq ft. find the width of the rectangle.
Answer by RedemptiveMath(80) (Show Source): You can put this solution on YOUR website!
We are given the perimeter and area of a rectangle, so it would be helpful to know how to find its perimeter and area. Doing so will help us find its length and width since we know that perimeter and area deal with a rectangle's length and width. So, the formulas for perimeter and area of a rectangle (respectively) are

2l + 2w = P and lw = A.

Now we just plug the values we have from the given information into our formulas:

2l + 2w = 52 ft
lw = 120 ft^2

We can use substitution to find either variable first:

lw = 120 ft^2
w = 120/l or l = 120/w;
2l + 2(120/l) = 52 ft or 2(120/w) + 2w = 52 ft
2l + 240/l = 52 ft or 240/w + 2w = 52 ft
2l^2 + 240 = 52l or 240 + 2w^2 = 52w
2l^2 - 52l + 240 = 0 or 2w^2 - 52w + 240 = 0
2(l^2 - 26l + 120) = 0 or 2(w^2 - 26l + 120) = 0
2(l-6)(l-20) = 0 or 2(w-6)(w-20) = 0
l = 6 or 20; w = 6 or 20.

There are two things you can notice here. The first is as we went along the manipulation process, we could see that l and w looked the same throughout. The second is that we are given two answers for both measurements. This means that this problem has two sufficient answers. Most of math deals with just finding one answer for a problem, but this problem gives us two. To figure out which ones work, we need to use simple logic. We know that the length and width can't both be 6 or 20 because the area must be 120 ft^2. Therefore, one must be 6 while the other is 20. This satisfies the conditions of the area and perimeter.

l = 6 ft and w = 20 ft
OR
l = 20 ft and w = 6 ft.

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