SOLUTION: Suppose that the perimeter of a rectangle is 24 inches and the length is twice the width. What are the dimensions of this rectangle and what is the area?

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Question 315541: Suppose that the perimeter of a rectangle is 24 inches and the length is twice the width. What are the dimensions of this rectangle and what is the area?
Answer by moshiz08(60) About Me  (Show Source):
You can put this solution on YOUR website!
Perimeter is the sum of the four sides. Since we have a rectangle, opposite sides have the same length. So two of the sides have length L and the other two are the width W. Perimeter is +P+=+2L+%2B+2W+=24+.
Length is twice the width, that is +L+=+2+W. Substitute this into the perimeter equation (replace the 2W with L) to get
+2L+%2B+2W+=+2L+%2B+L+=+3L+=+24+. Thus, dividing by 3 gives L+=+8. The length is twice the width, so the width must be 4.
We have a 8 x 4 rectangle which has an area of 32 square inches.
Check that this solution works. The perimeter does indeed add up to 24 and the length is twice the width.