Lopsided trapezium

Suppose the shorter base has length $a,$ and the longer base has length $b.$ The segment connecting the midpoints of the sides has length $\frac{a+b}{2}.$ Because the ratio of the two trapezium is $\frac{5}{11},$ and they have the same height, $\frac{a+\frac{a+b}{2}}{\frac{a+b}{2}+b}=\frac{5}{11}$ After simplification, this gives $b=7a.$

Because the trapezium is circumscribed, the sums of the lengths of the opposing sides are equal (this follows from the fact that the distances from any vertex to the points where the inscribed circle touches its adjacent sides are equal).

Therefore, $a+b=30+50,$ thus $8a=80.$ So $a=10$ and $b=70,$ thus $ab=700.$