A rectangle has a length of 2 more than seven times the width. If the perimeter is 52 units,find the width and length.

Question

Perimeter of a rectangle word problems

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Answer 1

Let the length be L and the width be W. The wording of the problem tells you L is 2 more than 7 times W. As an equation that says: L=2+7W. The perimeter, P, of a rectangle is the sum of all 4 sides, so it's P = 2L+2W.

You then substitute L=2+7W into the perimeter equation: P = 2L+2W = 2(2+7W)+2W. Then, you distrubute and combine like terms:

2(2+7W)+2W = 4+14W+2W = 4+16W.

Finally, you are told the perimeter is 52, so therefore: P = 4+16W = 52. Subtract 4 from both sides to get: 16W = 48. Divide both sides by 16 to get: W=3. So, you have the width, then plug that in to find the length:

L=2+7W = 2+7(3) = 2+21=23. W=3, L=23.