# 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.

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I think that it is, but I cannot solve it. –  Adam Jun 21 '12 at 21:18
The area of a pyramid is $\frac{1}{3}A_{base}h$. Where $A_{base}$ is the area of the base (no matter the shape) and $h$ is the height. –  Tpofofn Jun 22 '12 at 1:24
Correction to my comment: The volume of a pyramid... –  Tpofofn Jun 22 '12 at 1:55
"The planes ADM and BCM are perpendicular to the base ABCD" This is physically impossible. –  Tpofofn Jun 22 '12 at 2:10
@Tpofofn: No, it isn't. –  joriki Jun 22 '12 at 3:35

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? –  Adam Jun 24 '12 at 20:17