4. Around a sphere with radius $R$ is described a frustum of a cone where the area of one base is four times larger than the area of the other base. Calculate the volume of the frustum of the cone.
Solution. From the condition $P_{1}=4 P_{2}$, we get $r_{1}=2 r_{2}$. The triangles $A B C$ and $E F C$ are similar, so we have $r_{1}: r_{2}=(h+2 R): R$. Similarly, the triangles $C E F$ and $C D O$ are similar, so we have $r_{2}: R=h: \overline{C D}$. From triangle $\triangle C D O$, we get $\overline{C D}=\sqrt{2 R h+h^{2}}$. Therefore, we obtain the equations:

$r_{1}=2 r_{2}, r_{1} R=r_{2}(h+2 R), \quad r_{2} \sqrt{2 R h+h^{2}}=h R$, from which it follows that: $h=2 R, r_{2}=\frac{R \sqrt{2}}{2}$,

![](https://cdn.mathpix.com/cropped/2024_06_05_13ae78702b3aed5f03b4g-5.jpg?height=538&width=476&top_left_y=1442&top_left_x=990)
$r_{1}=R \sqrt{2}$

Therefore, for the volume $V$ of the truncated cone, we get:

$$
V=\frac{2 R \pi}{3}\left(r_{1}^{2}+r_{1} r_{2}+r_{2}^{2}\right)=\frac{7 R^{3} \pi}{3}
$$

## 4th year