Trapezoidal pyramid Is the following problem correct? If it is correct, please present a solution. 
Given a trapezoidal pyramid ABCDM. The base $ABCD$ is trapezoid (where $AB\;  ||\; CD$). The planes $ADM$ and $BCM$ are perpendicular to the base $ABCD$. If $AB=5$, $DC=3$, $S_{\triangle BCM}=9$ and $S_{\triangle ABM}=20$, find the volume of the pyramid.

 A: The data given does not determine the volume of the pyramid.
Note that since $ADM$ and $BCM$ are perpendicular to $ABCD$, it follows that $M$ lies above the intersection $P$ of $AD$ and $BC$. Let $h$ be the height of $M$ above $P$, let $x$ be the perpendicular distance from $P$ to $AB$, and let $d$ be the perpendicular distance between $CD$ and $AB$. Then
$$S_{\triangle BCM}=\frac12 h|BC|=9$$
and
$$S_{\triangle ABM}=\frac12\sqrt{x^2+h^2}|AB|=20\;,$$
and with $|AB|=5$ this yields
$$x^2+h^2=64\;.$$
We also have
$$\frac{x-d}x=\frac35\;,$$
so $x=\frac52d$, and thus
$$\left(\frac52d\right)^2+h^2=64\;.$$
Writing $d=|BC|\cos\alpha$, with $\alpha$ the angle between $BC$ and the perpendicular on $AB$ and $CD$, we arrive at
$$\left(\frac52|BC|\cos\alpha\right)^2+\left(\frac{18}{|BC|}\right)^2=64\;.$$
Solving for $\cos\alpha$ yields
$$\cos\alpha=\frac{\sqrt{64-(18/|BC|)^2}}{\frac52|BC|}\;.$$
Setting this to zero yields $|BC|\gt\frac94$, and the value is below $1$ in that entire range with a maximum of $32/45$ at $|BC|=\frac94\sqrt2$.
With $V=\frac13hd(|AB|+|CD|)/2=\frac43hd=\frac43h|BC|\cos\alpha=24\cos\alpha$, the volume of the pyramid can take any value between $0$ and $256/15\approx17$.
