What is the area of a rectangle whose length is six less than four times its width and whose perimeter is 48 millimeters?

Answer 1

You've got a nice little algebra problem there, but you have to know a bit about geometry, too. The area of a rectangle is the product of the length and the width, so it is given by the formula A = LW. The formula for the perimeter is two times the sum of the length and width, so we write P = 2(L + W). For this particular problem, the length, L, is six less than four times the width, W. So, we can write a formula for L in terms of W: L = 4W - 6. Call this equation (1). We also know that the perimeter is 48 (millimeters). So let's use that information to start figuring things out. Since P = 2(L + W), we can write 48 = 2(L + W). If we divide both sides by 2, we get 24 = L + W. If we subtract W from both sides, we get 24 - W = L. Let's write that as L = 24 - W. Call that equation (2). So we now have two equations for L! If we equate equation (1) and equation (2), we can figure out the value for L. Let's do it: 4W - 6 = 24 - W. Subtract W from both sides, and add 6 to both, which yields: 5W = 30. Solving for W gives you W = 6 (millimeters). Plugging that into equation (2), we get L = 24 - 6 = 18 (millimeters) Now we can calculate the area: A = LW = 18 x 6 = 108 square millimeters.