SOLUTION: what is the dimentions of a rectangle that has a perimeter 42 in. and an area of 104 sq in.

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Question 504121: what is the dimentions of a rectangle that has a perimeter 42 in. and an area of 104 sq in.
Answer by oberobic(2304) About Me  (Show Source):
You can put this solution on YOUR website!
P = perimeter
P = 2(L+W)
P = 42 in.
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A = area
A = L*W
A = 104 sq. in.
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L*W = 104
L = 104/W
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substitute
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2(104/W + W) = 42
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divide both sides by 2
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104/W + W = 21
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multiply both sides by W to eliminate fraction
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W(104/W + W) = 21W
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104 + W^2 = 21W
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subtract 21W from both sides
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104 + W^2 -21W = 0
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rearrange
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W^2 -21W + 104 = 0
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factor, if possible
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(W -8)(W -13) = 0
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Check with FOIL
W^2 -13W -8W + 104
OK
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W = 8 or -13.
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But a negative width is nonsense.
So, our only solution is W = 8.
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Substitute W=8 to find L.
L*W = 104
L*8 = 104
L = 13
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Check your work using the perimeter.
2(L+W) = 42 ?
2(8+13) = 2(21) = 42
Correct.
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Answer: The dimensions of the rectangle are 8 in. and 13 in.
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Done.