The diameter of one base circle of a frustum is $100 \mathrm{~mm}$. If the diameter is increased by $21 \%$, while the height and the size of the other base circle remain unchanged, the volume of the frustum also increases by $21 \%$. What is the diameter of the other base circle?
Solution. The radius of the given base circle is $50 \mathrm{~mm}$. Let the radius of the other base circle be $r$, and the height of the frustum be $m$. Then the volume of the frustum is

$$
\frac{\pi \cdot m}{3} \cdot\left(r^{2}+50 r+50^{2}\right)
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_c0fe5e1a75a28ee55adfg-1.jpg?height=350&width=453&top_left_y=356&top_left_x=825)

According to the condition:

$$
1.21 \cdot \frac{\pi \cdot m}{3} \cdot\left(r^{2}+50 r+50^{2}\right)=\frac{\pi \cdot m}{3} \cdot\left(r^{2}+50 \cdot 1.21 r+(1.21 \cdot 50)^{2}\right)
$$

Dividing both sides of the equation by $\frac{\pi \cdot m}{3}$ and rearranging, we get: $0.21 r^{2}=635.25$, from which $r=55 \mathrm{~mm}$. The diameter of the other base circle is $2 r=110 \mathrm{~mm}$.