## Task B-3.7.

In the figure, a parallelogram $A B C D$ is shown, with the shorter side $\overline{A B}$ being $2 \mathrm{~cm}$ long, and the height to this side is $12 \mathrm{~cm}$. A line $p$ passes through the vertex $D$ of the parallelogram, which is perpendicular to the line $C D$, and forms an angle $\alpha$ with the side $\overline{A D}$ such that $\sin \alpha=\frac{3}{5}$. Determine the volume of the solid formed by rotating the parallelogram $A B C D$

![](https://cdn.mathpix.com/cropped/2024_05_30_803c5e1b47d2c202a3a9g-22.jpg?height=408&width=369&top_left_y=1235&top_left_x=1369)
around the line $p$.
## First Solution.

By rotating the parallelogram around the line $p$, a hollow frustum of a cone is formed, that is, a frustum of a cone with a hollow part in the shape of a cone. The axial section of this frustum is shown in the sketch.

![](https://cdn.mathpix.com/cropped/2024_05_30_803c5e1b47d2c202a3a9g-22.jpg?height=577&width=676&top_left_y=1913&top_left_x=610)

Let point $E$ be the intersection of line $p$ and line $B C$, and point $S$ be the intersection of line $p$ and line $A B$.

Consider the right triangle $A D S$.

First, note that $|D S|=12 \text{ cm}$.

With the labels as shown in the sketch, we have $\sin \alpha=\frac{r}{b}=\frac{3}{5}$, or $r=\frac{3}{5} b$.

1 point

By applying the Pythagorean theorem to this triangle, we can determine $r$ and $b$.

$12^{2}=b^{2}-r^{2}=b^{2}-\left(\frac{3}{5} b\right)^{2}=\frac{16}{25} b^{2}$, so $b=15 \text{ cm}$.

1 point

It follows that $r=\frac{3}{5} b=9 \text{ cm}$.

1 point

To calculate the volume of the resulting rotational body, we need to subtract the volume of the small cone (the complement) with base radius $|C D|=2 \text{ cm}$ and the volume of the hollow part with base radius $|A S|=r=9 \text{ cm}$ from the volume of the large cone with base radius $|B S|=r+2=11 \text{ cm}$.

To calculate the volumes, we need to determine the heights of all three cones.

Note that $\angle S D A$ and $\angle D E C$ are angles with parallel sides, which means $\angle D E C=\alpha$.

In triangle $C E D$, we have $\sin \alpha=\frac{2}{|C E|}=\frac{3}{5}$, so $|C E|=\frac{10}{3} \text{ cm}$.

It follows that $|D E|=\sqrt{\left(\frac{10}{3}\right)^{2}-2^{2}}=\sqrt{\frac{64}{9}}=\frac{8}{3} \text{ cm}$ and $|S E|=12+\frac{8}{3}=\frac{44}{3} \text{ cm}$.

Let's calculate the volume of the large cone and the volume of the small cone (the complement).

$V_{\text {cone }}=\frac{1}{3} \cdot 11^{2} \pi \cdot \frac{44}{3}=\frac{5324}{9} \pi \text{ cm}^{3}$

$V_{\text {complement }}=\frac{1}{3} \cdot 2^{2} \pi \cdot \frac{8}{3}=\frac{32}{9} \pi \text{ cm}^{3}$

The volume of the hollow part is $V_{\text {hollow }}=\frac{1}{3} \cdot 9^{2} \pi \cdot 12=324 \pi \text{ cm}^{3}$.

Finally, the volume of the desired rotational body is:

$V=V_{\text {cone }}-V_{\text {complement }}-V_{\text {hollow }}=264 \pi \text{ cm}^{3}$.