Date: 10/01/97 at 23:13:10 From: T.NUTTING Subject: Parallelogram The lengths of the diagonals of a parallelogram are 10 and 24. If the length of one side of the paralellogram is 13, what is the perimeter of the parallelogram? Is this right? Two sides are 13. The other two sides are 10.9. I used a2 + b2 = c2 or a2+5(25) = 12(144) a2 = 144-25, a2 = 119 a = 10.9. 13+13+10.9+10.9 = 47.8
Date: 10/10/97 at 15:43:16 From: Doctor Chita Subject: Re: Parallelogram Hi there, T. Nutting: You've made a nice try at solving this problem. However, your analysis is not quite correct. Let me try to explain why and how you might want to think about this problem. First of all, you are correct in recognizing that the diagonals of a parallelogram bisect each other. Consequently, the segments of each diagonal are 5 and 12. You know that one side of this parallelogram is 13. Therefore, one of the inner triangles has sides of 5, 12, and 13. Is this a right triangle? If so, which side is the hypotenuse? You should remember that in a right triangle the longest side is always the hypotenuse. So, if this triangle is a right triangle, 13 must be the hypotenuse. Therefore, the legs, a and b, are 5 and 12. You must first show that this triangle is a right triangle. To do this, see if a^2 + b^2 is equal to 13^2. c^2 = 5^2 + 12^2 c^2 = 25 + 144 c = sqrt (169) = 13 Therefore, the triangle is indeed a right triangle, since a^2 + b^2 = c^2. Since the triangle is right, then the angle opposite the 13-side is a right angle. Therefore, the diagonals are perpendicular to one another, and all the inner triangles are right triangles with sides of 5, 12, and 13. Since the hypotenuse of each triangle equal 13 and each hypotenuse is also a side of the parallelogram, then by definition, the parallelogram is a rhombus. Its perimeter is 4 * 13 = 52. The secret to solving this problem correctly, then, is to first show that a right triangle exists by using the converse of the Pythagorean theorem and showing that a^2 + b^2 = c^2. Second, show that the parallelogram has 4 equal sides (and is a rhombus). This was a tricky problem. You had the right ideas, but made a few wrong turns. Hope that this explanation is helpful. -Doctor Chita, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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