Turning a Perimeter into a Scale FactorDate: 02/17/2003 at 17:42:16 From: Krissy A. Subject: Turning a perimeter into a scale factor Perimeter and area ratios of similar figures are given. Find each scale factor. perimeter ratio = 81 scale factor = ? Date: 02/17/2003 at 19:50:17 From: Doctor Rick Subject: Re: Turning a perimeter into a scale factor Hi, Krissy. Let's get at the idea by working backward. Suppose we know the scale factor; what will the ratio of perimeters be? For instance, suppose we have two triangles; one has sides of 3, 4, and 5 inches; the other has sides of 33, 44, and 55 inches. The scale factor is 11: you multiply each side length of the first triangle to get the corresponding side length of the second triangle. Now look at the perimeters. The perimeter of the first triangle is 3+4+5 = 12 inches. The perimeter of the second triangle is 33+44+55 = 132 inches. The ratio of perimeters is 132/12 = 11. Do you notice that it's the same as the scale factor? This will always be true! Here is why. We can write the sides of the second triangle as 3*11, 4*11, and 5*11. Then the perimeter is 3*11 + 4*11 + 5*11 = (3 + 4 + 5)*11 using the distributive property. To find the ratio of perimeters, divide this by the perimeter of the first triangle: (3+4+5)*11 ---------- = 11 3+4+5 Let's continue and think about the ratio of areas. The triangles I chose happen to be right triangles (do you know how to show this?) so the area is half the product of the two shorter sides. Thus the area of the first triangle is (3*4)/2 = 6 square inches. The area of the second triangle is (33*44)/6 = 726 square inches. The ratio of areas is 726/6 = 121. This ratio happens to be 11 squared. It will always be true that the ratio of areas is the square of the scale factor. Again, we can see why this is true. Writing the sides of the second triangle as 3*11 and 4*11, the area is 3*11 * 4*11 = (3*4)*(11*11) Divide this by the area of the first triangle to find the ratio of areas: (3*4)*(11*11) ------------- = 11*11 = 11^2 3*4 Do you see how it works now? What is the answer to your problem? - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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