Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

enter image description here

share|improve this question
    
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
1  
@Tpofofn: No, it isn't. –  joriki Jun 22 '12 at 3:35

1 Answer 1

up vote 1 down vote accepted

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 $256/15\approx17$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.