10,11

The area of the base of the pyramid is 3, and the volume of the pyramid is also 3. Two planes parallel to the base of the pyramid have been drawn. The areas of the resulting sections are 1 and 2. Find the volume of the part of the pyramid located between the planes.
Each of the specified planes cuts off a pyramid similar to the given one. Let $k_1$ and $k_2$ be the similarity coefficients, $V_1$ and $V_2$ be the volumes of the cut-off pyramids, $V=3$ be the volume of the given pyramid, and $v$ be the desired volume. Since the ratio of the areas of similar figures is equal to the square of the similarity coefficient, then

$$
k_1=\frac{1}{\sqrt{3}}, \quad k_2=\frac{\sqrt{2}}{\sqrt{3}}
$$

Since the ratio of the volumes of similar bodies is equal to the cube of the similarity coefficient, then

$$
\begin{aligned}
& V_1=k_1^3 \cdot V=\left(\frac{1}{\sqrt{3}}\right)^3 \cdot 3=\frac{1}{\sqrt{3}}, \\
& V_2=k_2^3 \cdot V=\left(\frac{\sqrt{2}}{\sqrt{3}}\right)^3 \cdot 3=\frac{2\sqrt{2}}{\sqrt{3}}
\end{aligned}
$$

Therefore,

$$
v=V_2-V_1=\frac{2\sqrt{2}}{\sqrt{3}}-\frac{1}{\sqrt{3}}=\frac{2\sqrt{2}-1}{\sqrt{3}}=\frac{2\sqrt{6}-\sqrt{3}}{3}
$$

## Answer

$\frac{2\sqrt{6}-\sqrt{3}}{3}$