What's the area of the isosceles trapezoide below?

geometry euclidean-geometry plane-geometry

Solution 1:

enter image description here

If $AG$ and $CH$ are perp to $CD$ extend and $AB$ respectively, please notice that $\triangle ADG \cong \triangle CBH$.

So, $S_{ABCD} = S_{AHCG}$

As $\angle CAB = \frac{1}{2} \angle COB = 37^\circ, AH \approx \frac{4}{3} CH = 12$

That leads you to the answer

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