# 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

Algebra ->  Algebra  -> Rectangles -> 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      Log On

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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(57224)   (Show Source): You can put this solution on YOUR website!Area = 35sq ft and Perimeter = 27 ft. Solve for Length in the Perimeter. ----------- Area = LW Perimeter = 2(L+W) ------------------------ Substitute and solve: 35 = LW 27 = 2(L+W) ---------------- L = 35/W ------ 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) ========================================================= Cheers, Stan H.