# SOLUTION: What are the dimensions of a rectangle that has a perimeter of 38 units an an area of 60 square units?

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 Question 623639: What are the dimensions of a rectangle that has a perimeter of 38 units an an area of 60 square units?Found 2 solutions by ankor@dixie-net.com, math-vortex:Answer by ankor@dixie-net.com(15667)   (Show Source): You can put this solution on YOUR website!what are the dimension of a rectangle that has a perimeter of 38 units an an area of 60 square units. : Perimeter equation 2L + 2W = 38 simplify divide by 2 L + W = 19 W = (19-L) : The area equation L * W = 60 Replace W with (19-L) L(19-L) = 60 19L - L^2 = 60 0 = L^2 - 19L + 60 Factors to (L-4)(L-15) = 0 Two solutions, but assume L is the larger value L = 15 units Find W W = 19 - 15 W = 4 units : Check the perimeter and the area with these values Answer by math-vortex(472)   (Show Source): You can put this solution on YOUR website!```Hi, there-- The Problem: What are the dimensions of a rectangle that has a perimeter of 38 units an an area of 60 square units? A Solution: Let W be the width of the rectangle. Let L be its length. Now we need to write two equations that use information in the problem about the length and width of the rectangle. {1} The rectangle has a perimeter of 38 units. Perimeter means the distance around the outside of the rectangle. We calculate the perimeter P by adding the two widths and two lengths. In an equation, we write this relationship as 2W + 2L = 38 {2} The area of the rectangle is 60 square units. The formula for the area of a rectangle is length times width. We can express with relationship by the equation, L * W = 60 Now we have a system of two equations with two variables. We will use the substitution method to find the length and width. Divide every term in the first equation by 2. 2W + 2L = 38 W + L = 19 Rewrite this equation in "W=..." form. (Subtract L from both sides.) W = 19 - L We see that W and 19-L are equivalent. Make this substitution in the second equation. L * W = 60 L * (L - 19) = 60 Multiply to clear the parentheses. L^2 - 19L = 60 Subtract 60 from both sides of the equation. L^2 - 19L - 60 = 0 We will solve this equation by translating the it to factored form. We need two numbers whose product is 60 and whose sum is -19. The numbers are -15 and -4. The equation in factored form is (L - 15)(L - 4) = 0 The solutions to this equation are L - 15 = 0 OR L - 4 = 0 L = 15 OR L = 4 If L=15, then the W is 4 units since the area is 60 square units and 15*4=60. By similar reasoning, if L=4, then the width is 15 units. We need to check that these values for length and with give the correct perimeter. 15 + 15 + 4 + 4 = 38, so all is well. THe dimensions of the rectangle are 4 units and 15 units. Feel free to email me if you have questions about the solution. Ms.Figgy math.in.the.vortex@gmail.com ```