Trapezoid Area

We note that $\triangle ABE$ and $\triangle CDE$ are similar. Then $\dfrac {AE}{BE} = \dfrac {CE}{DE} \implies AE = \dfrac {CE}{DE} \times BE = \dfrac 54 \times 3 = \dfrac {15}4$ . The area of trapezoid $ABCD$ is

$\begin{aligned} [ABCD] & = \frac 12 AE \cdot BE \sin \angle AEB + \frac 12 DE \cdot AE \sin \angle DEA + \frac 12 CE \cdot DE \sin \angle CED + \frac 12 BE \cdot CE \sin \angle BEC \\ & = \frac 12 \cdot \frac {15}4 \cdot 3 \cdot \sin 150^\circ + \frac 12 \cdot 4 \cdot \frac {15}4 \cdot \sin 30^\circ + \frac 12 \cdot 5 \cdot 4 \cdot \sin 150^\circ + \frac 12 \cdot 3 \cdot 5 \cdot \sin 30^\circ \\ & = \frac {45}{16} + \frac {15}4 + 5 + \frac {15}4 \\ & = \frac {245}{16} \end{aligned}$

Therefore $a+b = 245+16 = \boxed{261}$ .

From the given informations we have

$|\overline {AB}|=\dfrac 34 |\overline {CD}|$ ,

$\tan \angle {BAC}=\dfrac {2}{5+2\sqrt 3}$

Using these we get the height of the trapezoid as $h=\dfrac {35}{2\sqrt {41+20\sqrt 3}}$ ,

and $|\overline {AB}|+|\overline {CD}|=\dfrac 74 \times \sqrt {41+20\sqrt 3}$

So, the required area of the trapezoid is

$\dfrac 12 \times \dfrac 74 \sqrt {41+20\sqrt 3}\times \dfrac {35}{2\sqrt {41+20\sqrt 3}}$

$=\dfrac {245}{16}$

So, $a=245,b=16$ , and $a+b=245+16=\boxed {261}$ .