A rectangle with sides of $15 \mathrm{~cm}$ and $20 \mathrm{~cm}$ is rotated about one of its diagonals. What is the volume of the resulting solid of revolution?
Solution. Let's denote the vertices of the rectangle as $A, B, C$, and $D$ as shown in the figure, with sides $AB=20$ and $BC=15$. Rotate the rectangle around the diagonal $BD$. The $180^{\circ}$ rotation of $A$ is $A'$. Construct a perpendicular to the diagonal $BD$ at its midpoint $F$ within the plane of the rectangle. Let this line intersect the side $AB$ at $M$ and the side $CD$ at $M'$. Clearly, the distance $FM$ is equal to the distance $FM'$, so $M$ is the reflection of $M'$ across the diagonal $BD$. The plane perpendicular to $BD$ at $F$ is a symmetry plane of the solid of revolution, so it is sufficient to examine the solid of revolution formed on one side of this plane. The total volume is twice the volume of this solid of revolution.

![](https://cdn.mathpix.com/cropped/2024_05_02_5751ac9cb3cf7202a990g-1.jpg?height=558&width=572&top_left_y=431&top_left_x=771)

During the rotation, a cone with radius $\frac{AA'}{2}=r$ and height $DO=m$ is formed (where $O$ is the intersection of $AA'$ and $BD$), and a frustum of a cone with base radii $FM=r_1$ and $AO=r_2$, and height $FO=m'$. Let's calculate the necessary data.

The diagonal of the rectangle $BD=\sqrt{400+225}=25$. The length of $DO$ can be calculated using the leg theorem in the right triangle $ABD$:

$$
DA=\sqrt{DO \cdot DB}, \quad \text { i.e., } \quad 15=\sqrt{DO \cdot 25}, \quad \text { hence } \quad DO=\frac{225}{25}=9
$$

which is the height $m$ of the cone.

The radius of the base of the cone $r=AO=\sqrt{15^{2}-9^{2}}=12$. The volume of the cone is:

$$
V_{\text {cone}}=\frac{m r^{2} \pi}{3}=\frac{9 \cdot 12^{2} \pi}{3}=432 \pi
$$

The height of the frustum of the cone is: $m'=FO=\frac{25}{2}-9=\frac{7}{2}$. The radius $r_1$ can be determined from the similar triangles $BFM$ and $BOA$: $\frac{FM}{AO}=\frac{BF}{BO}$, hence

$$
r_1=FM=AO \cdot \frac{BF}{BO}=12 \cdot \frac{12.5}{12.5+3.5}=12 \cdot \frac{12.5}{16}=\frac{3}{4} \cdot 12.5=9.375
$$

and $r_2=AO=12$.

The volume of the frustum of the cone is:

$$
\begin{aligned}
& V_{\text {frustum}}=\frac{m' \pi}{3}\left(r_1^{2}+r_1 r_2+r_2^{2}\right)=\frac{\frac{7}{2} \pi}{3}\left(9.375^{2}+9.375 \cdot 12+12^{2}\right) \\
& V_{\text {solid of revolution}}=\pi \cdot\left[864+\frac{7}{3}\left(9.375^{2}+9.375 \cdot 12+12^{2}\right)\right] \approx 5238.85
\end{aligned}
$$