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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.

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    @jorki, Yes, I see it now, M is on the line where ADM and BCM intersect. Thanks.2012-06-22

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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 $256/15\approx17$.

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    Thanks. How do you obtain the upper limit for the volume? Could you explain it in details, please?2012-06-24