Problem 477

Let $h$ be the height of a regular tetrahedron, $S$ be the area of a face, and $V$ be the volume. Then

$$h=\frac{a \sqrt{2}}{\sqrt{3}}, S=\frac{a^{2} \sqrt{3}}{1} .$$

Therefore,

$$V=\frac{1}{3} S \cdot h=\frac{1}{3} \cdot \frac{a^{2} \sqrt{3}}{1} \cdot \frac{a \sqrt{2}}{\sqrt{3}}=\frac{a^{2} \sqrt{2}}{12} .$$

Answer

$\frac{a^{2} \sqrt{2}}{12}$.