jmc

Let the trapezium have diagonal legs of length $x$ and a shorter base of length $y$. Drop altitudes from the endpoints of the shorter base to the longer base to form two right-angled triangles, which are congruent since the trapezium is isosceles. Thus using the base angle of $\arcsin (0.8)$ gives the vertical side of these triangles as $0.8x$ and the horizontal side as $0.6x$. Now notice that the sides of the trapezium can be seen as being made up of tangents to the circle, and thus using the fact that "the tangents from a point to a circle are equal in length" gives $2y+0.6x+0.6x=2x$. Also, using the given length of the longer base tells us that $y+0.6x+0.6x=16$. Solving these equations simultaneously gives $x=10$ and $y=4$, so the height of the trapezium is $0.8\times 10=8$. Thus the area is $\frac{1}{2}(4+16)(8)=\boxed{80}$.