Tetrahedron Volume

Since CD is perpendicular to AB and EF, CD is perpendicular to the plane (triangular) AFB which separate the tetrahedron into 2 smaller tetrahedrons.

The area of the triangle AFB is $\frac {1} {2} * EF * AB = \frac {1} {2} * 10 * 12 = 60$

The volume for tetrahedron ADFB is $\frac {1} {3} * [AFB] * DF$ and the volume of the tetrahedron CAFB is $\frac {1} {3} * [AFB] * CF$

So, the volume for the big tetrahedron is $\frac {1} {3} * [AFB] * (CF + DF) = \frac {1} {3} * 60 * 12 = 240$ .

=240

$EF=10, FD=6$ so $ED=2\sqrt{34}$ . This gives us the area of $ABD$ is $12\sqrt{34}$ . We just need to find the altitude from $C$ to $ED$ . Well, the area of triangle $ECD$ is $6*10=60$ , so the altitude from $C$ to $ED$ is $\frac{30\sqrt{34}}{17}$ . And finally, the volume is $12\sqrt{34}*\frac{30\sqrt{34}}{17}*\frac{1}{3}=240$ .

I personally believe the hardest part about this question is understanding the description and visualizing this tetrahedron. Once you have done that, the following description should make sense. [The hard part is to realize (as shown below) that we can decompose the tetrahedron into 2 figures which we can easily calculate the volume of. Working with $ABCD$ directly is next to impossible. - Calvin]

On a plane, we are able to see $AB$ as the baseline of the tetrahedron, with length $12$ and a vertical running upwards from its midpoint $E$ for length $10$ . This intersects the topline of the tetrahedron, $CD$ , where $C$ points perpendicularly into the plane of the paper and $D$ points perpendicularly out of the plane of the paper, each running length $6$ .

Once we have gotten this, we can split the calculation of volume into 2 identical pyramids, $CDEA$ and $CDEB$ . Let the base be $CDE$ , then the height would be $EC$ (perpendicular). Now, each pyramid has a volume of $\frac{1}{3}(base)(height) = \frac{1}{3} (\frac{1}{2} \times 12 \times 10) (6) = 120$ . Hence total volume is $240$ .

Let us take coordinates in such a way $A=(0,0,0), B=(0,12,0), C=(-6,6,10), D=(6,6,10).$ Then the volume $V$ of the tetrahedron is $\frac{1}{6}$ of the absolute value of determinant of the vectors $\vec{AB}, \vec{AC}, \vec{AD}$ . Since $\frac{1}{6}\det\left(\begin{matrix}0&12&0\\-6&6&10\\6&6&10\end{matrix}\right)=240,$ we get the solution $V=\boxed{240}$ .

In general,if in a tetrahedron,we are given the lengths of two skew sides (say a and b),the angle between them (say x),and the shortest distance between them (say d),then the volume of tetrahedron is given by (abd*sinx)/6

We can split the tetrahedron into AEFD, AEFC, BEFD, BEFD. Then the formula for each of these is 1/3 bh= 1/3 6 30 = 60, so the total volume is 240.