In the figure, $ABCD$ in a trapezoid. Given that $AD$ is parallel to $EC$ and $EB=2(AE)$ Find the area of $ABCD$ and $AEFD $

enter image description here

I know the area of a trapezoid $A=\frac{(b_1+b_2)}{2}h$ and I think that base $AB$ is equal to $AE+EB$ where so that's $AB=\frac{EB}{2}+2AE$ and i also think that triangle $EFB$ is similar to triangle $DFC$? and now I'm stuck and it says "Hint" In two similar triangles, the ratio of their areas is the square of the ratio of their sides.


Let's call $a=CD$ the base lenght of the figure.

The we have $A(ABCD)=(AB+CD)\times \dfrac h2$, with $h$ being the height of the figure.

$AB=AE+EB=CD+2 \times CD$

Hence $A(ABCD)=2ah$


$A(ABD)=AB\times \dfrac h2=\dfrac{3ah}{2}$

Since we know that $AE=2EB$, that is $EB=\dfrac{2AB}{3}$, then Thales is whispering to us that the triangle $EFB$ has an area $\dfrac 49$ times smaller than triangle $ABD$.

Thus $A(AEFD)=\dfrac 59 \times A(ABD)=\dfrac 59 \times \dfrac{3ah}{2}$


EDITED: I thought initially (looking at the picture, error error...) that $AE=2EB$ whereas it is the contrary in the problem text, $EB=2AE$. I changed the result accordingly.

  • $\begingroup$ wait how did EB=AB/3? isn't over 2? $\endgroup$
    – Mickey
    Jan 16 '15 at 14:06
  • 1
    $\begingroup$ @Mickey Thank you, I misread the problem by looking at the error inducing picture. $\endgroup$
    – Martigan
    Jan 16 '15 at 14:15
  • $\begingroup$ Hi, I tried to answer it again and I think I made a mistake with my drawing at first, can you see it now? I also don't get AB= AE + EB = CD + CD/2, EB says its 2times AE right? thank you $\endgroup$
    – Mickey
    Jan 17 '15 at 0:09
  • $\begingroup$ I don't understand $A(ABCD)$ = $2ah$ isn't it $A(ABCD) = \frac{3CD}{4}h$? because I need to perform lcd with CD + CD/2 then $\frac{2CD+CD}{2}\times\frac{h}{2}$? $\endgroup$
    – Mickey
    Jan 17 '15 at 0:21
  • $\begingroup$ The area of the trapezoid is $\dfrac{(b_1+b_2)h}{2}$. Here $b_1+b_2=CD+3 \times CD=4a$ (since $CD=a$ in my definition). I forgot to change the previous lines when I edited the answer. $\endgroup$
    – Martigan
    Jan 19 '15 at 8:03

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