Answer: B

Small Hint:

Assign one tangent length to each vertex, so adjacent pairs sum to the four side lengths

Big Hint:

For supplementary opposite angles, the products of the tangent lengths at opposite vertices are equal

Solution:

Let the tangent lengths from the four consecutive vertices be $u,v,w,z.$ Then $$\begin{aligned} u+v&=70, &v+w&=90,\\ w+z&=130, &z+u&=110. \end{aligned}$$ Thus $v=70-u,$ $w=20+u,$ and $z=110-u.$ If the inradius is $r,$ a vertex with angle $A$ has tangent length $r\cot(\frac{A}{2}).$ Opposite angles are supplementary, so $uw=vz.$ Therefore $$u(20+u)=(70-u)(110-u),$$ giving $u=38.5.$ Hence the two segments of the $130$-side are $w=58.5$ and $z=71.5,$ whose difference is $13.$

Thus the correct answer is B.