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In trapezoid $ABCD$ the lengths of the bases $AB$ and $CD$ are 8 and 17 respectively. The legs of the trapezoid are extended beyond $A$ and $B$ to meet at point $E$. What is the ratio of the area of triangle $EAB$ to the area of trapezoid $ABCD$? Express your answer as a common fraction.

RektTheNoob Aug 9, 2017

#1**+2 **

Triangles EAB and EDC will be similar if AB is parallel to DC

The base of EDC = DC = 17....and the base of EAB = AB = 8

So......the scale factor of triangle EDC to triangle EAB will be 17/8

Thus......the area of triangle EDC will be (17/8)^2 = 289 / 64 that of triangle EAB

Let the area of EAB = A......so....the area of EDC will be (289 / 64) A

And the area of the trapezoid ABCD is just the area of EDC - area of EAB = [ 289 / 64 - 1 ] A =

225 / 64 A

Therefore.....the ratio of triangle EDC to that of trapezoid ABCD = A / [ (225/64) A ] = 64 / 225

As an interesting aside....note that the height of the trapezoid and the location of "E" are irrelevent....thus, AB can be any distance from DC.....the important thing is the ratio of AB to DC......

CPhill Aug 9, 2017