Task B-3.2. The length of the side of the rhombus $A B C D$ is $a$. The rhombus first rotates around the line on which the side $\overline{A B}$ lies, and then around the line on which the longer diagonal $\overline{A C}$ lies. By these rotations, we obtain two rotational bodies. The ratio of their volumes is $V_{1}: V_{2}=9: \sqrt{3}$. Determine the acute angle of the rhombus and the ratio of the surface areas of the resulting rotational bodies.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_38fd7d226823545bc90fg-12.jpg?height=368&width=1068&top_left_y=825&top_left_x=496)

By rotating around the line $A B$, a rotational body is formed, consisting of a cone and a cylinder from which the same cone is missing inside. The volume of this body is

$$
\begin{aligned}
& V_{1}=V_{\text {cone }}+V_{\text {cylinder }}-V_{\text {cone }}=V_{\text {cylinder }} \\
& V_{1}=r^{2} \pi v, \quad r_{\text {cylinder }}=v_{\text {rhombus }}=a \sin \alpha, \quad v_{\text {cylinder }}=a
\end{aligned}
$$

Thus, $V_{1}=a^{3} \pi \sin ^{2} \alpha$.

If the rhombus rotates around the line $A C$, a rotational body is formed, consisting of two identical cones, and its volume is

$$
V_{2}=2 V_{\text {cone }}=\frac{2}{3} r^{2} \pi v
$$

Since $r_{\text {cone }}=\frac{e}{2}=a \sin \frac{\alpha}{2}$, and $v_{\text {cone }}=\frac{f}{2}=a \cos \frac{\alpha}{2}$.

The volume is

$$
V_{2}=\frac{2}{3} a^{3} \pi \sin ^{2} \frac{\alpha}{2} \cos \frac{\alpha}{2}
$$

From the given ratio $V_{1}: V_{2}=9: \sqrt{3}$, it follows that

$$
\begin{aligned}
& \frac{\sin ^{2} \alpha}{\frac{2}{3} \sin ^{2} \frac{\alpha}{2} \cos \frac{\alpha}{2}}=\frac{9}{\sqrt{3}} \\
& \frac{4 \sin ^{2} \frac{\alpha}{2} \cos ^{2} \frac{\alpha}{2}}{\frac{2}{3} \sin ^{2} \frac{\alpha}{2} \cos \frac{\alpha}{2}}=\frac{9}{\sqrt{3}}
\end{aligned}
$$

From this, $\cos \frac{\alpha}{2}=\frac{\sqrt{3}}{2}$, so $\frac{\alpha}{2}=\frac{\pi}{6}$, and $\alpha=\frac{\pi}{3}$.

Let's determine the ratio of the surface areas. The surface area of the first body is

$$
\begin{aligned}
O_{1} & =2 \cdot \mathrm{Pl}_{\text {cone }}+\mathrm{Pl}_{\text {cylinder }}=2 \cdot v_{\text {rhombus }} \cdot \pi \cdot a+2 \cdot v_{\text {rhombus }} \cdot \pi \cdot a \\
& =4 v_{\text {rhombus }} a \pi=4 a \sin \alpha \cdot a \pi=4 \pi a^{2} \cdot \frac{\sqrt{3}}{2}=2 \pi \sqrt{3} a^{2}
\end{aligned}
$$

The surface area of the second body is $O_{2}=2 r_{\text {cone }} \pi a=2 a \sin \frac{\alpha}{2} \pi a=a^{2} \pi$.

The ratio is $O_{1}: O_{2}=2 \sqrt{3}: 1$.