4. Two pyramids have a common base - a square with side $a$. The pyramids are located on the same side of the square. The heights of the two pyramids pass through the midpoints of two opposite sides of the square and have length $b$. Calculate the volume of the common part of the two pyramids.
![](https://cdn.mathpix.com/cropped/2024_06_05_e8e81a1ae01081700bfcg-5.jpg?height=538&width=1296&top_left_y=284&top_left_x=248)
Solution. Let $A B C D E$ and $A B C D F$ be the given pyramids. From the rectangle $A B E F$, it follows that $M$ is the midpoint of $A E$. Similarly, $N$ is the midpoint of $C F$. According to this, $\overline{A B}=a$ and the height of $\triangle A B M^{\prime}$ dropped from $M^{\prime}$ is $\frac{b}{2}$. On the other hand, $M N$ is the midline in $\triangle A D E$, so it follows that $\overline{M N}=\frac{a}{2}$. According to this, $\overline{M M^{\prime}}=\overline{M N^{\prime}}=\frac{a}{4}$ and

$$
V_{A B C D M N}=V_{A B C D M^{\prime} N^{\prime}}=\frac{1}{2} a \frac{b}{2} a-2 \frac{1}{3} \frac{1}{2} a \frac{b}{2} \frac{a}{4}=\frac{a^{2} b}{4}-\frac{a^{2} b}{24}=\frac{5}{24} a^{2} b
$$

## 4th year