Task B-3.7. Determine the volume of a regular 12-sided truncated pyramid if $R$ and $r$ $(R>r)$ are the radii of the circles circumscribed around the bases and if the lateral edges are inclined at an angle of $60^{\circ}$ to the base plane.
## Solution.

First, let's determine the areas of the bases.

The central angle of a regular dodecagon is $\frac{360^{\circ}}{12}=30^{\circ}$.

(1 point)

$$
B_{1}=12 \cdot \frac{R \cdot R \sin 30^{\circ}}{2}=3 R^{2}, \quad B_{2}=12 \cdot \frac{r \cdot r \sin 30^{\circ}}{2}=3 r^{2}
$$

Consider the diagonal section of the given truncated pyramid, which is an isosceles trapezoid with bases $2 R$ and $2 r$.

Sketch

(2 points)

![](https://cdn.mathpix.com/cropped/2024_05_30_bde1d6b20cdcb90302c4g-17.jpg?height=482&width=727&top_left_y=900&top_left_x=257)

We calculate the height of the truncated pyramid

$$
\operatorname{tg} 60^{\circ}=\frac{h}{R-r}, \quad h=(R-r) \sqrt{3}
$$

Then the desired volume is

$$
V=\frac{h}{3}\left(B_{1}+\sqrt{B_{1} B_{2}}+B_{2}\right)
$$

$$
\begin{aligned}
& V=\frac{(R-r) \sqrt{3}}{3}\left(3 R^{2}+\sqrt{3 R^{2} \cdot 3 r^{2}}+3 r^{2}\right) \\
& V=\frac{\sqrt{3}}{3}(R-r)\left(3 R^{2}+3 R r+3 r^{2}\right) \\
& V=\sqrt{3}(R-r)\left(R^{2}+R r+r^{2}\right)=\sqrt{3}\left(R^{3}-r^{3}\right)
\end{aligned}
$$