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Ratio of Areas of Triangle and Parallelogram


Date: 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
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Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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