
Imagine a prismoid container like the one pictured. I have poured water into it and I know all values of the resulting water prismoid - I know its volume, it's height, the areas of its bases, the area of the midsection and the angles of its sides.
Now I want to pour in more water for it to reach a certain volume. How much higher should the water surface (or top base) be to reach that volume?
I have really been racking my brain to make a working formula but cannot. Can any genius help me?

Let the top rectangular face have sides $A$ and $b$ and the bottom rectangular face have sides $a$ and $B$, such that $A$ and $a$ are parallel, $|A| > |a|$, and similar for the other ones.
Then consider the Stott contracted "prismatoid" with top "rectangle" with sides $C$ with length $|C|=|A|-|a|$ and $d$ with length $|d|=|b|-|b|=0$ and bottom "rectangle" with sides $c$ of length $|c|=|a|-|a|=0$ and $D$ of length $|D|=|B|-|b|$. Obviously it still has the same height and the lacing face planes all would be just shifted copies of the former, just that these lacing faces now become triangles only, whereas the top and bottom faces degenerate into line segments (of already provided lengths $C$ and $D$ respectively).
Thus you have reduced your problem in finding the height of a simplex (between an opposite pair of edges). That simplex further would be a sphenoid (having two opposite sides at orthogonal directions). But without any further informations no more can be said here. However, your problem thus has been reduced to a simple elementary geometric task.
--- rk