Tetrahedron $ABCD$ has $AB=5$, $AC=3$, $BC=4$, $BD=4$, $AD=3$, and $CD=\tfrac{12}5\sqrt2$. What is the volume of the tetrahedron? $\textbf{(A) }3\sqrt2\qquad$ $\textbf{(B) }2\sqrt5\qquad$ $\textbf{(C) }\dfrac{24}5\qquad$ $\textbf{(D) }3\sqrt3\qquad$ $\textbf{(E) }\dfrac{24}5\sqrt2$
Solution
Since $3^2+4^2=5^2$, triangles $ABC$ (right angle at $C$) and $ABD$ (right angle at $D$) are right triangles with the common hypotenuse $AB=5$.
Drop perpendiculars from $C$ and from $D$ to $AB$. In each right triangle the altitude to the hypotenuse is $\frac{3\cdot4}5=\frac{12}5$, and the foot is at distance $\frac{3^2}5=\frac95$ from $A$ (because $AC=AD=3$). So both feet are the same point $P$ of $\overline{AB}$, with $PC=PD=\frac{12}5$.
Both $PC$ and $PD$ are perpendicular to $AB$. Moreover $PC^2+PD^2=2\cdot\frac{144}{25}=\frac{288}{25}=CD^2$, so $\angle CPD=90^\circ$.
Hence $AB$ is perpendicular to the plane $PCD$, and the tetrahedron is split by that plane into two pyramids with base $\triangle PCD$ and heights $AP$ and $PB$, so $V=\frac13[PCD]\cdot AB=\frac13\cdot\frac12\left(\frac{12}5\right)^2\cdot5=\frac{24}5$. The answer is (\textbf{C}).
$\textbf{(A) }3\sqrt2\qquad$ $\textbf{(B) }2\sqrt5\qquad$ $\textbf{(C) }\dfrac{24}5\qquad$ $\textbf{(D) }3\sqrt3\qquad$ $\textbf{(E) }\dfrac{24}5\sqrt2$ 