SOLUTION: The length of a rectangle is 2 cm more than twice its width. If the perimeter of the rectangle is 52 cm, find the dimensions of the rectangle.

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Question 68061: The length of a rectangle is 2 cm more than twice its width. If the perimeter of the rectangle is 52 cm, find the dimensions of the rectangle.
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Let l = the length and W = the width of the rectangle. From the problem description, you can write:
L+=+2W%2B2
The perimeter, P, of a rectangle is given by:
P+=+2L+%2B+2W Substitute the L from above and P = 52 and solve for W.
52+=+2%282W%2B2%29+%2B+2W
52+=+4W%2B4+%2B+2w
52+=+6W%2B4 Subtract 4 from both sides.
48+=+6W Divide both sides by 6.
8+=+W
The width is 8 cm
L+=+2W%2B2
L+=+2%288%29%2B2
L+=+16%2B2
L+=+18
The length is 18 cm
Check:
P+=+2L%2B2W
P+=+2%2818%29%2B2%288%29
P+=+36%2B16
P+=+52