## Task B-3.4.

A clay pot for flowers, 4 dm high, has the shape of a geometric body that is formed by the rotation of a geometric figure from the image around its axis of symmetry, line $p$. The intersection of lines $A F$ and $B C$ is at the midpoint of segment $\overline{E D}$, and the intersection of lines $E F$ and $C D$ is at the midpoint of segment $\overline{A B}$. What is the volume of the pot (neglect the thickness of the wall)?

![](https://cdn.mathpix.com/cropped/2024_05_30_4b05efd680c81b2b8cbfg-17.jpg?height=583&width=597&top_left_y=1142&top_left_x=684)
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_4b05efd680c81b2b8cbfg-17.jpg?height=596&width=642&top_left_y=1895&top_left_x=707)

We will calculate the volume of the container by summing the volumes of two cones ($V_{1}$ is the volume of the cone with base diameter $\overline{E D}$ and height $h=4 \mathrm{dm}$, $V_{2}$ is the volume of the cone with base diameter $\overline{A B}$ and the same height) and then subtracting the common part ($V_{3}$ and $V_{4}$ are the volumes of the cones with base diameter $\overline{F C}$, and the sum of their heights is $h=4 \mathrm{dm}$).

$$
V=V_{1}+V_{2}-V_{3}-V_{4}=\frac{r_{1}^{2} \pi h}{3}+\frac{r_{2}^{2} \pi h}{3}-\frac{r_{3}^{2} \pi h}{3}=\frac{\pi h}{3}\left(r_{1}^{2}+r_{2}^{2}-r_{3}^{2}\right)
$$

By applying trigonometric ratios to triangles $G H E$ and $A G H$, we will determine the radii $r_{1}$ and $r_{2}$. Since $\operatorname{tg} 60^{\circ}=\sqrt{3}=\frac{h}{r_{1}}$, we get $r_{1}=\frac{4}{\sqrt{3}}$.

Furthermore,

$$
\operatorname{tg} 75^{\circ}=\operatorname{tg}\left(45^{\circ}+30^{\circ}\right)=\frac{1+\frac{\sqrt{3}}{3}}{1-\frac{\sqrt{3}}{3}}=\frac{3+\sqrt{3}}{3-\sqrt{3}}=2+\sqrt{3}=\frac{h}{r_{2}}
$$

so $r_{2}=\frac{4}{2+\sqrt{3}}$.

By applying the sine rule to triangle $A G F$, we will determine the radius $r_{3}$. We get

$$
\frac{x}{\sin 75^{\circ}}=\frac{r_{2}}{\sin 45^{\circ}}
$$

Furthermore,

$$
\sin 75^{\circ}=\sin \left(45^{\circ}+30^{\circ}\right)=\sin 45^{\circ} \cos 30^{\circ}+\cos 45^{\circ} \sin 30^{\circ}=\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2} \cdot \frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}
$$

Now,

$$
\frac{x}{\frac{\sqrt{6}+\sqrt{2}}{4}}=\frac{\frac{4}{2+\sqrt{3}}}{\frac{\sqrt{2}}{2}}
$$

so $x=2(\sqrt{3}-1)$, or $r_{3}=\frac{x}{2}=\sqrt{3}-1$.

Substituting into the initial formula, we calculate the volume

$$
V=\frac{4 \pi}{3}\left[\left(\frac{4}{\sqrt{3}}\right)^{2}+\left(\frac{4}{2+\sqrt{3}}\right)^{2}-(\sqrt{3}-1)^{2}\right]=\left(\frac{1360}{9}-\frac{248}{3} \sqrt{3}\right) \pi \mathrm{dm}^{3}
$$