The base of a pentahedron is a $10\times6$ rectangle. Its lateral faces are two isosceles triangles with base of length $6$ and congruent sides of length $7$, and two isosceles trapezoids with bases of length $6$ and $10$ and nonparallel sides of length $7$. What is the volume of the pentahedron?

1. Let the base be $ABCD$ with $AB=CD=10$, $BC=DA=6$, and let the top edge be $\overline{PQ}$ with $PQ=6$, parallel to $\overline{AB}$, centered over the base, with $PA=PD=QB=QC=7$.
2. $P$ lies above a point at distance $\frac{10-6}2=2$ from $\overline{AD}$ (along $\overline{AB}$) and at distance $3$ from $\overline{AB}$, so its height is $h=\sqrt{7^2-2^2-3^2}=6$.
3. Cutting by the vertical planes through $P$ and $Q$ perpendicular to $\overline{AB}$ gives a middle triangular prism of length $6$ with cross-section a triangle of base $6$ and height $6$ (area $18$), so its volume is $108$.
4. Each end is a pyramid with apex $P$ (or $Q$), height $6$ and a rectangular base $2\times6$, of volume $\frac13\cdot12\cdot6=24$.
5. The total volume is $108+2\cdot24=\boxed{156}$.