5. A cone is given with the radius of the base $r=3 dm$. Above the same base and on the same side, a cylinder is constructed whose surface area is equal to the surface area of the cone, and the volume is equal to the volume of the cone. Calculate the surface area and volume of the body which is the intersection of the cone and the cylinder.
Solution. From the equality of volumes we get $\pi r^{2} h=\frac{1}{3} \pi r^{2} H$, i.e., $H=3 h$. From the equality of areas we get

$$
2 \pi r^{2}+2 \pi r h=\pi r^{2}+\pi r s \text {, i.e., } r+2 h=s
$$

if we substitute $r=30 \text{ cm}$ we get $s=30+2 h$. Now, from the Pythagorean theorem applied to the right triangle $S O A$ we get $s^{2}=r^{2}+H^{2}$, i.e.,

![](https://cdn.mathpix.com/cropped/2024_06_05_fcce00877a1c83df80a0g-5.jpg?height=461&width=407&top_left_y=767&top_left_x=1075)

$$
(30+2 h)^{2}=900+9 h^{2}
$$

The last equation is equivalent to the equation $5 h(h-24)=0$, from which we find $h=24 \text{ cm}$, so $H=72 \text{ cm}$ and $s=78 \text{ cm}$. We still need to determine the dimensions of the cone with height $S O^{\prime}=\frac{2}{3} H=48 \text{ cm}$. Triangles $S O A$ and $S O^{\prime} A^{\prime}$ are similar, so $O^{\prime} A^{\prime}=\frac{2}{3} r$ and $S A^{\prime}=\frac{2}{3} S$. The desired volume is equal to the difference in volumes of the two cones

$$
V=\frac{1}{3} \pi r^{2} H-\frac{1}{3} \pi\left(\frac{2}{3} r\right)^{2} \cdot \frac{2}{3} H=28000 \pi \text{ cm}^{3}
$$

The area is equal to the sum of the areas of two circles, increased by the difference in the lateral areas of the two cones:

$$
P=\pi r^{2}+\pi\left(\frac{2}{3} r\right)^{2}+\pi r s-\frac{2}{3} \pi r \cdot \frac{2}{3} s=2600 \pi \text{ cm}^{2}
$$