Exercise 293

What is the volume of tetrahedron $ABCD$ with edge lengths $AB=3$, $AC=4$, $AD=6$, $BC=5$, $BD=3\sqrt5$, and $CD=2\sqrt{13}$?

Solution

$AB^2+AC^2=3^2+4^2=25=BC^2$, so $\triangle ABC$ is right at $A$ and has area $\frac12(3)(4)=6$.
$AB^2+AD^2=3^2+6^2=45=BD^2$ and $AC^2+AD^2=4^2+6^2=52=CD^2$. Hence $AD$ is perpendicular to the plane of $ABC$, and the height is $6$.
The tetrahedron's volume is $\frac13(6)(6)=\boxed{12}$ cubic units.