2. Let $A B C D A' B' C' D'$ be a cube with edge $a$, where $A B C D$ is one of the faces of the cube, and $A A', B B', C C'$, and $D D'$ are edges of the cube. Calculate (in terms of $a$) the height of the pyramid $A C B' D'$, dropped from the vertex $D$.
Solution. Let's note that all the edges of the pyramid $A C B' D$ are equal and have a length of $b=a \sqrt{2}$. This means that $A C B^{\prime} D^{\prime}$ is obtained from the given cube when four equal triangular pyramids $A C B^{\prime} B, A C D^{\prime} D, A B^{\prime} D^{\prime} A^{\prime}$, and $C B^{\prime} D^{\prime} C^{\prime}$, each with a volume of $\frac{a^{3}}{6}$, are removed. Therefore, the volume $V$ of $A C B^{\prime} D^{\prime}$ is $\frac{a^{3}}{3}$. On the other hand,

![](https://cdn.mathpix.com/cropped/2024_06_05_50e976f6299e4a88a6a3g-01.jpg?height=536&width=603&top_left_y=1669&top_left_x=881)
for the volume $V$, we have $V=\frac{b^{2} h \sqrt{3}}{12}$, where $h$ is the desired height. By equating the two volumes and solving, we get $h=\frac{2 a \sqrt{3}}{3}$.