Hi, there-- THE PROBLEM: A square and a rectangle each has a perimeter of 48 m. The difference between the areas of two figures is 4 square meters. Find the dimensions of the rectangle. A SOLUTION: Consider the square with perimeter 48m. Let x be the length of each side. The perimeter is the distance around, or 4x. 4x = 48 x = 12 The length of each side is 12m. The area of the square is the side length times the side length, or x*x. 12*12 = 144 The area of the square is 144 square meters. We are told that the difference between the area of the square and the rectangle is 4 square meters. Then the area of the rectangle must be 4 more or 4 less. For now, assume that the rectangle has the lesser area, or 140 square meters. The formula for the area of a rectangle is length times width, or L*W. We have L*W = 140. We also know that the perimeter of the rectangle is 48. We have 2L + 2W = 48 We have two equations with two variables. Solve for L and W. Solve the first equation for L. L*W = 140 L = 140/W Substitute 140/W for L in the second equation. 2L + 2W = 48 2(140/W) + 2W = 48 Solve for W. 280/W + 2W = 48 Subtract 2W from both sides. 280/W = -2W + 48 Multiply both sides by W. 280 = -2W^2 + 48W This is a quadratic equation. Solve by factoring. 2W^2 - 48W + 280 = 0 Divide each term by 2. W^2 - 24W + 140 = 0 (W - 14)(W - 10) = 0 W = 14 or W = 10 The width of the rectangle is 10m or 14m. If the width of the rectangle is 10m, then its length is 14m because 2L + 2W = 2(10) + 2(14) = 48 The dimensions of the rectangle are 10m by 14m. You can work out the dimensions of the rectangle if it has the larger area. Then your equations will be L*W = 148 2L + 2W = 48 I'll leave that to you. Hope this helps! Feel free to email if you have any questions about the solution. Good luck with your math, Mrs. F math.in.the.vortex@gmail.com