# SOLUTION: How do you find the dimensions of a rectangle whose perimeter is inches and whose area is square inches?

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 Geometry: Rectangles Solvers Lessons Answers archive Quiz In Depth

 Question 234475: How do you find the dimensions of a rectangle whose perimeter is inches and whose area is square inches?Answer by Theo(3504)   (Show Source): You can put this solution on YOUR website!P = 2L + 2W A = LW Solve for L in terms of W in the second equation. Use that value of L in the first equation to solve for W. Example: 2L + 2W = 100 LW = 400 Solve for L in second equation to get L = 400/W Use that value in the first equation to get 2(400/W) + 2W = 100 Once you find W, then you go back to the first equation and solve for L I'll do this one for you. Equations are: 2L + 2W = 100 LW = 400 I use second equation to solve for L in terms of W to get L = W/400 I replace L with 400/W in the first equation to get: 2*(400/W) + 2*W = 100 I multiply both sides of this equation by W to get: 2*400 + 2*W*W = 100*W which becomes: 800 + 2W^2 = 100W I subtract 100W from both sides of this equation and reorder the terms to get: 2W^2 - 100W + 800 = 0 I divide both sides by 2 to get: W^2 - 50W + 400 = 0 I factor this quadratic equation to get: (W-10) * (W-40) = 0 The result is that: W = 10 or: W = 40 I then go back to the first equation and use these values of W to solve for L. I get: L = 10 or: L = 40 When L = 10, W = 40 When L = 40, W = 10 I confirm these values by substituting in the original equations of: 2L + 2W = 100 LW = 400 They confirm so the answers are good. I get: L = 10 or 40 W = 10 or 40 When L = 10, W = 40 When L = 40, W = 10