In a cube with side length 6 , what is the volume of the tetrahedron formed by any vertex and the three vertices connected to that vertex by edges of the cube?
Solution: We have a tetrahedron whose base is half a face of the cube and whose height is the side length of the cube, so its volume is $\frac{1}{3} \cdot\left(\frac{1}{2} \cdot 6^{2}\right) \cdot 6=\mathbf{3 6}$.