SOLUTION: The length of a rectangle is 4yd longer than its width. If the perineter of the rectangle is 48 yd, find its area.
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-> SOLUTION: The length of a rectangle is 4yd longer than its width. If the perineter of the rectangle is 48 yd, find its area.
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Question 834441: The length of a rectangle is 4yd longer than its width. If the perineter of the rectangle is 48 yd, find its area. Answer by JulietG(1812) (Show Source):
You can put this solution on YOUR website! P = 2L + 2W (the perimeter fence takes 2 pieces of length and 2 pieces of width)
P = 48
L = W + 4
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Now substitute the value of L into the original equation
48 = 2(W+4) + 2W
Distribute the 2
48 = 2W + 8 + 2W
Subtract 8 from each side
40 = 2W + 2W
Add the Ws
40 = 4W
Divide each side by 4
10 = W
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If the width is 10, then the length is 4yd longer, or 14.
Let's add them up.
2W (20) + 2L (28) does indeed = 48.
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Now that you have the W & L, multiply them to get the area.
10 * 14 = ??
