A polyhedron $ABCDEF$ is depicted in the image above. The coordinates of the vertices are known. They are $A(25, -15, 0), B(25, 15, 0), C(-25, 15, 0), D(-25, -15, 0), E(-18, -5, 30), F(18, 5, 30) $
Find the volume of this polyhedron.
Based on a comment by ajotatxe below, and a hint by Gribouillis below, I've computed the volumes of tetrahedrons $ADCE, ACEF, ABCF$. For tetrahedron $ADCE$ we have
$AD = D - A = (-25, -15, 0) - (25, -15, 0) = (-50, 0, 0) $
$AC = C - A = (-25, 15, 0) - (25, -15, 0) = (-50, 30, 0) $
$AE = E - A = (-18, -5, 30) - (25, -15, 0) = (- 43, 10 , 30 ) $
So the volume of tetrahedron $ADCE$ is
$ V_1 = \frac{1}{6} | AD \cdot (AC \times AE) | = 7500 $
And this is equal to the volume of $ABCF$
Now
$ AF = (18, 5, 30) - (25, -15, 0) = (-7, 20, 30) $
So the volume of $ABCF$ is
$ V_2 = \frac{1}{6} | AE \cdot (AC \times AF) | = 7900 $
Hence, the required volume is
$ 2 \times 7500 + 7900 = 15000 + 7900 = \boxed{22900} $
Hint 1 Add the edge AC to the image. The polyhedron is the union of 3 tetrahedrons. Add the volumes of the 3 tetrahedrons.
Hint 2 The intersection of the polyhedron and the horizontal plane at altitude $z$ is an irregular hexagon which you can compute the vertices. Compute the area of this hexagon as a function of $z$ and integrate in $z$ from $0$ to $30$.