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Task: calculate the area of the triangle $\text{AEB}$.

I could calculate the height of the trapezoid, but I can not see how that would be a help.

Please help me? Any hints or solutions?

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Trancing a segment from $C$ perpendicular to $AB$ and using Pythagorean theorem, we get $h=3\sqrt{3}$ for the height of this trapezoid. $\triangle ABE$ and $\triangle CDE$ are similar, since their corresponding angles are congruent. The ratios of the lengths of their sides is $12:6=2$. This ratio is the same for their heights. Then, theses heights are $2\sqrt{3}$ and $\sqrt{3}$. Therefore, the area of $\triangle ABE$ is $12\sqrt{3}$.

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    'Then, theses heights are $2\sqrt{3}$ and $\sqrt{3}$'. I don't understand where you get $\sqrt{3}$ from. Does it exist a connection between $h$ and the hights of $\triangle \text{ABE}$ and $\triangle \text{CDE}$? Or how did you calculate the last mentioned triangles heights? – 2017-02-04
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    Yes, $h$ is the sum of the heights $h_1$ and $h_2$ of $\triangle ABE$ and $\triangle CDE$, respectively, relative to the bases $AB$ and $CD$. So, we get $h_1+h_2=3\sqrt{3}$ and $h_1:h_2=2$. – 2017-02-04
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    Another question, though. This caught my interest: how did you find the nice numbers you are using? For example, if I would calculate $h$, I would write the answer as $h=\sqrt{6^{2}-3^{2}}$, while you write this as $3\sqrt{3}$. – 2017-02-04
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    $\sqrt{6^2-3^2}=\sqrt{27}=\sqrt{3^2\cdot 3}=3\sqrt{3}$. – 2017-02-04