Ratio of Areas of Triangle and ParallelogramDate: 7/17/96 at 1:55:27 From: Anonymous Subject: Ratio of Triangle, Parallelogram Areas Hi, Can you give me some help with this problem? In the diagram, we have BC||DF, DE||AC, BE=5, and EC=3 /\A / \ D/-------\F / \ \ / \ \ / \ \ B---------E--------C Compute the Fraction: Area of triangle ADF -------------------------- Area of parallelogram DECF Date: 7/17/96 at 10:37:53 From: Doctor Anthony Subject: Re: Ratio of Triangle, Parallelogram Areas We start with ADF/DBE = 3^2/5^2 = 9/25 This is because areas of similar triangles are proportional to the squares of corresponding sides. DBE/FEC = 5/3 This is because these triangles have equal heights, so the ratio of the areas is the same as the ratio of their bases. Now DECF = 2*FEC so DBE/DECF = 5/6 We now have the chain of ratios ADF/DBE * DBE/DECF = 9/25 * 5/6 = 3/10 i.e. ADF/DECF = 3/10 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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