Side Lengths of Isosceles Triangle
Date: 7/8/96 at 14:26:30 From: Anonymous Subject: Side Lengths of Isosceles Triangle There are two equilateral triangles on top of one another. The smaller one's sides are 35 on each side and 21 on the bottom. The larger one's sides are y + the 35 of the smaller one and the other side shows x as the whole length but you still have the other 35 of the other triangle and the bottom is 48. I am supposed to find the unknown lengths. The answer in the book is x = 80, y = 45, but I do not understand how to get the answer.
Date: 7/9/96 at 13:19:46 From: Doctor Beth Subject: Re: Side Lengths of Isosceles Triangle I'm going to assume that you mean that the triangles are isosceles, since they have 2 sides the same length but not the third (an equilateral triangle is one that has all three sides the same length). If I understand the problem correctly, the triangles share a common angle, and their bases are parallel. In this case, they are similar triangles since the angles of one triangle are congruent to the angles of the other triangle. Since the two triangles are similar, we know that the ratios of their sides are the same. For example, since their bases are of length 21 (small triangle) and length 48 (large triangle), we know that the ratios of the sides of the small triangle to the sides of the large triangle is 21/48. Since we know this, and we also know that the side of the small triangle is of length 35 and the side of the large triangle is of length x, we know that 35 21 -- = -- x 48 To solve for x, first we'll cross-multiply and get that 35*48 = 21x. Then we divide both sides by 21, and get that x = 80. Now since we know that the large triangle is isosceles, we know that the side that is of length x is the same length as the side of length y + 35, so since x = 80, we know that y + 35 = 80. Subtracting 35 from both sides, we learn that y = 45. This isn't the only way to do the problem. We could have solved for y first instead of for x, but the equations are a little more complicated if we do that. Hope this helps! -Doctor Beth, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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