SOLUTION: Question is Area = 35sq ft and Perimeter = 27 ft. Solve for Length in the Perimeter. I have P = 2(W) x 2(L) and that would be 27= 2(W)x 2(L) and then move 2(W) to the other side so
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-> SOLUTION: Question is Area = 35sq ft and Perimeter = 27 ft. Solve for Length in the Perimeter. I have P = 2(W) x 2(L) and that would be 27= 2(W)x 2(L) and then move 2(W) to the other side so
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Question 235487: Question is Area = 35sq ft and Perimeter = 27 ft. Solve for Length in the Perimeter. I have P = 2(W) x 2(L) and that would be 27= 2(W)x 2(L) and then move 2(W) to the other side so 27-2(W) = 2(L) then divide both by 2. 27-2(W)/2 = L. But I can't figure it out. Nothing makes 35sq ftof area and 27 feet in a perimeter. My only options as I see it are 35 and 1 or 7 and 5. Someone please help me figure this out. Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Area = 35sq ft and Perimeter = 27 ft. Solve for Length in the Perimeter.
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Area = LW
Perimeter = 2(L+W)
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Substitute and solve:
35 = LW
27 = 2(L+W)
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L = 35/W
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Substitute into the Perimeter equation and solve for W:
13.5 = (35/W)+ W
Multiply thru by W and solve:
W^2 -13.5W + 35 = 0
W = [13.5 +- sqrt(13.5^2-4*1*35)]/2
W = [13.5 +- sqrt(42.25)]/2
Positive solution:
W = 20/2 = 10 ft. (the width of the rectangle)
L+W = 13.5 so L = 3.5 ft. (length of the rectangle)
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Cheers,
Stan H.
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