SOLUTION: Suppose that the perimeter of a rectangle is 66 and the lenght is twice the width. what is the dimensions of the rectangle and what is the area
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Question 828950: Suppose that the perimeter of a rectangle is 66 and the lenght is twice the width. what is the dimensions of the rectangle and what is the area Answer by JulietG(1812) (Show Source):
You can put this solution on YOUR website! A perimeter is a fence around something. To fence a rectangle, you need 2 pieces of the length and 2 pieces of the width.
2L + 2W = P
The question gives you the following:
P = 66
L = 2W
Plug in the numbers:
2(2W) + 2W = 66
Multiply:
4W + 2W = 66
Add the Ws:
6W = 66
Divide by 6
W = 11
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Now that you have the width, you can find the length. "the length is twice the width" L = 2*11, or 22.
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Let's prove this correct.
2L + 2W = 66
2(22) + 2(11) = 66
44 + 22 does indeed equal 66.
