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In the book Handbook of Mathematics (I. N. Bronshtein, pg 194), we have without proof. If the bases of a triangular prism are not parallel (see figure) to each other we can calculate its volume by the formula $$ V = \dfrac{(a + b + c)Q}{3} \quad (1) $$ where $Q$ is a perpendicular cut $a$, $b$ and $c$ are the lengths of the parallel edges If the bases of the prism are not parallel then its volume is $$ V = lQ \quad (2) $$ where $l$ is the length of the line segment $BC$ connecting the centers of gravity of the bases and $Q$ is the cross cut perpendicular to this line.

I struggle to prove (1) and (2) elemental form. Using the equation of the plane and double integral, can solve the problem. I would like to see a proof that uses elementary tools. Thanks for aby help.

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Step 1) show that the volume depends only on $a,b,c$ and $Q$. (see below)

Step 2) once you have 1), you know that any two prisms with the same $a,b,c,Q$ have the same volume. Now build a orthogonal prism $P$ with three equal edges of lengths $a+b+c$, and section $Q$. For such prism you easily prove the formula is true by hands.

Split the first vertical edge of $P$ in three segments of lenghts $a,b,c$. Split the second edge as $b,c,a$ and the third as $c,a,b$. In this way you divided $P$ into three prisms $P_1,P_2,P_3$ each of edges $a,b,c$. (All of theme have section $Q$). by Step 1) $V(P_1)=V(P_2)=V(P_2)$ and $V(P)$ is three times that volume.


How to prove Step 1).

Given a prism $P$ with vertical edges $a,b,c$ and horizontal section $Q$, split it in two prisms by cutting $P$ along $Q$. You reduces then to the case in which the prism has an horizontal face. But such prisms are uniquely determined by $Q,a,b,c$ so Step 1) is proved.

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  • $\begingroup$ The case of prisms with parallel faces reduces to the case of parallel, horizontal faces by cutting along $Q$ and gluing the bottom prism on top. $\endgroup$ – user126154 Jul 20 '14 at 12:57
  • $\begingroup$ Excellent solution and without the use of integrals. $\endgroup$ – Mathsource Jul 27 '14 at 2:03
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This has to do with the nontrivial fact that the center of gravity of the vertices of a triangle coincides with the center of gravity of its area. (The center of gravity of the vertices is their arithmetic mean, and the center of gravity of the area has to lie on all three medians, by physical considerations. Both approaches lead to the same point.)

Now to your problem: It is enough to consider a prism $P$ whose base triangle $Q$ lies in the $(x,y)$-plane with centroid at the origin, and the "parallel edges" of $P$ are parallel to the $z$-axis, pointing upwards.

The top facet of $P$ lies in a plane with equation $$z=\ell+ax+by\qquad\bigl((x,y)\in{\mathbb R}^2\bigr)\ .$$ The volume of $P$ is therefore given by $${\rm vol}(P)=\int_Q (\ell+ax+by)\ {\rm d}(x,y)=\ell\cdot{\rm area}(Q)+a\int_Q x\ {\rm d}(x,y)+ b\int_Q y\ {\rm d}(x,y)\ .$$ Here the last two integrals vanish because the centroid of $Q$ is at the origin.

Let $a$, $b$, $c$ be the lengths of the vertical edges of $P$. Then the top facet of $P$ is a triangle with centroid at the point $c=(0,0,\ell)$, with $\ell={a+b+c\over3}$. (We know this because the centroid is an "affine invariant".) It follows that $${\rm vol}(P)=\ell\cdot{\rm area}(Q)={a+b+c\over3}\cdot{\rm area}(Q)\ .$$

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I prefered 3D coordinates system to prove the lemma. In my opinion , this tool is the most elementary tool for that problem.

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The projection area $Q$ on $xy$ plane and bordered by $(OED)$ can be written as

$$\mathbf{M_{2}}=\begin{bmatrix}0 & 0 & 1\\x_1 & y_1 & 1\\x_2 & y_2 & 1\end{bmatrix}$$

$Q=\frac{1}{2} |det(\mathbf{M_{2}})|=\frac{1}{2} |x_2y_1-x_1y_2|$

The prism in picture can be divided in 3 tetrahedrons and the volume of prism can be written as the total volume of these 3 tetrahedrons as shown in the picture below. You can also see the paper (page 154) for extra information how to calculate the volume of a tetrahedron via 4x4 matrix.

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$$V=\frac{1}{6} |det(\begin{bmatrix}0 & 0 & a& 1\\x_2 & y_2 & z_2& 1\\x_1 & y_1 & z_1+b& 1\\x_1 & y_1 & z_1& 1\end{bmatrix})|+\frac{1}{6} |det(\begin{bmatrix}0 & 0 & a& 1\\x_2 & y_2 & z_2+c& 1\\0 & 0 & 0& 1\\x_1 & y_1 & z_1& 1\end{bmatrix})|+\frac{1}{6} |det(\begin{bmatrix}0 & 0 & 0& 1\\x_2 & y_2 & z_2+c& 1\\x_2 & y_2 & z_2& 1\\x_1 & y_1 & z_1& 1\end{bmatrix})|$$

$V=\frac{b}{6} |x_2y_1-x_1y_2|+\frac{a}{6} |x_2y_1-x_1y_2|+\frac{c}{6} |x_2y_1-x_1y_2|$

$V=\frac{a+b+c}{6} |x_2y_1-x_1y_2|=\frac{a+b+c}{3} \frac{|x_2y_1-x_1y_2|}{2}$

$V=\frac{(a+b+c)}{3} Q$

You can find the Wolframalpha links for determinant calculations above. det1,det2,det3

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