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
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1+0 Answers



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......




cool cool cool

CPhill  Aug 9, 2017
edited by CPhill  Aug 9, 2017
edited by CPhill  Aug 9, 2017

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