2006-II-45

Note that each faces of the tetrahedron are equilateral triangle with side $3\text{ cm}$. Consider the base equilateral triangle $ABC$. Add a point $D$ on $BC$ such that $AD\perp BC$ and $BD=CD$. Note that $E$ is the centroid of $\mathrm{\Delta }ABC$.

$\begin{array}{rcl}A{C}^2& =& A{D}^2+C{D}^2\\ (3{)}^2& =& A{D}^2+(1.5{)}^2\\ AD& =& {\displaystyle \frac{3\sqrt{3}}{2}}\text{ cm}\end{array}$

Since $E$ is the centroid of $\mathrm{\Delta }ABC$, $AE:ED=2:1$. Hence, we have $AE=\sqrt{3}\text{ cm}$. In $\mathrm{\Delta }VEA$,

$\begin{array}{rcl}V{A}^2& =& V{E}^2+A{E}^2\\ V{E}^2& =& (3{)}^2–(\sqrt{3}{)}^2\\ V{E}^2& =& 6\\ VE& =& \sqrt{6}\text{ cm}\end{array}$

Therefore, the height of the tetrahedron is $\sqrt{6}\text{ cm}$.