Lesson Rectangles
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richard1234(7193)
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BS in mathematics w/ computer science. Previous tutoring experience, and I have participated in the USAMO.
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 {{{L = 6W/(W-6)}}} 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, {{{L = (6X + 36)/X = 6 + 36/X}}} (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!
