# Lesson Rectangles

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 Geometry: Rectangles Solvers Lessons Answers archive Quiz In Depth
 This Lesson (Rectangles) was created by by richard1234(5390)  : View Source, ShowAbout richard1234: MIT undergrad next year, part time mathematics tutor. Participated in USAMO and ARML. So far on this site I've seen probably 20 or 30 problems saying something like, The perimeter of a rectangle is x units. If the length is z more than the width, find the area of the rectangle. The truth is, they're not that hard! Once this is changed into algebraic form it becomes a simple algebra problem. I'll show a few examples of such problems: Example 1: The perimeter of a rectangle is 68. If the length is 4 more than twice the width, find the dimensions and the area of a rectangle. Solution: Let the dimensions of the rectangle be L and W. Since the perimeter is 68, then 2(L+W) = 68 --> L + W = 34. Also, the length is 4 more than twice the width, i.e. L = 2W + 4. Therefore we have the system of equations L + W = 34 L = 2W + 4 Plugging L = 2W + 4 = 34 into the first equation we have: 3W + 4 = 34 --> W = 10. From this, we obtain L = 24, so the area is 240 sq. units. Example 2: Find all rectangles with integer dimensions such that the area is numerically three times the perimeter. Solution: Again, just another straightforward algebra problem. If the dimensions are L and W, then LW = 6(L + W) = 6L + 6W Moving all L terms to one side, LW - 6L = 6W Now, it suffices to find all positive integers W such that L is also a positive integer. To do this, suppose W = X + 6. Then, (X must also be positive) Hence, X can equal any positive factor of 36, namely {1, 2, 3, 4, 6, 9, 12, 18, 36}. Adding 6 to each of these terms to obtain W, we have W = {7, 8, 9, 10, 12, 15, 18, 24, 42} Solving for L, we obtain: L = {42, 24, 18, 15, 12, 10, 9, 8, 7} (which is obviously symmetrical to W) Therefore, all ordered pairs (L, W) assuming L>=W, that satisfy are (42, 7), (24, 8), (18, 9), (15, 10), and (12, 12). The moral of this is, when you see a problem like this (or most any other problem), turn it into an algebra problem that you can easily solve! This lesson has been accessed 1393 times.