[ [ correct tetrahedron ]

Find the height of a regular tetrahedron with edge $a$.

#
Let $ABCD$ be a regular tetrahedron with edge $a$, and $M$ be the center of the face $ABC$. Since $DM$ is the height of the tetrahedron, triangle $AMD$ is a right triangle. By the Pythagorean theorem, we find that

$$
DM=\sqrt{AD^{2}-AM^{2}}=\sqrt{a^{2}-\left(\frac{a \sqrt{3}}{3}\right)^{2}}=\sqrt{1-\frac{1}{3}}=a \sqrt{\frac{2}{3}}
$$

## Answer  $\sqrt{\frac{2}{3}}$