Not enough areas!

$\triangle AOB\sim\triangle COD \text{ since }AB\parallel CD.$ The sides are proportional to the square roots of the areas of similar triangles so let $AB=5a \text{ and }CD=11a$ . Draw a line through O that is perpendicular to AB and CD, meeting AB at E and CD at F. Let $OE=h_1 \text{ and }OF=h_2$ (Heights of the 2 triangles). Let height of trapezium ABCD be $h=h_1+h_2$ . Then $\begin{aligned} \text{area of }\triangle AOB=& \frac{5a\cdot h_1}{2}=25 & \implies ah_1=10 \\ \text{area of }\triangle COD=& \frac{11a\cdot h_2}{2}=121 & \implies ah_2=22 \end{aligned}$ Add the two equations to get $ah_1+ah_2=32 \\a(h_1+h_2)=32 \\ah=32\\ \text{area of } ABCD=\dfrac{5a+11a}{2}\times{h}\\=8ah=8\times{32}=\boxed{256}$