The image shows a mathematical problem involving a solid formed by removing a smaller cone from a larger cone, where both cones are mathematically similar. Let's break down and solve the problem step by step.

Problem Breakdown:

1. Given Information:

   • The height of the large cone $H_L = 21$ cm.
   • The diameter of the base of the large cone $D_L = 18$ cm, so the radius $R_L = 9$ cm.
   • The height of the small cone $H_S = 14$ cm.

2. Similarity Ratio: Since the cones are similar, the ratio of their corresponding dimensions (height and radius) will be the same: $\frac{H_S}{H_L} = \frac{R_S}{R_L}$ Substituting the values: $\frac{14}{21} = \frac{R_S}{9}$ Simplifying: $\frac{2}{3} = \frac{R_S}{9}$ Solving for $R_S$: $R_S = \frac{2}{3} \times 9 = 6 \text{ cm}$

3. Volumes of Cones: The formula for the volume of a cone is: $V = \frac{1}{3} \pi r^2 h$

   • Volume of the large cone $V_L$: $V_L = \frac{1}{3} \pi (9)^2 (21) = \frac{1}{3} \pi \times 81 \times 21 = \frac{1}{3} \pi \times 1701 = 567 \pi \text{ cubic cm}$
   • Volume of the small cone $V_S$: $V_S = \frac{1}{3} \pi (6)^2 (14) = \frac{1}{3} \pi \times 36 \times 14 = \frac{1}{3} \pi \times 504 = 168 \pi \text{ cubic cm}$

4. Volume of the solid (frustum): The volume of the solid is the difference between the volume of the large cone and the volume of the small cone: $V_{\text{solid}} = V_L - V_S = 567 \pi - 168 \pi = 399 \pi \text{ cubic cm}$

5. Conclusion: The volume of the solid is indeed $399 \pi$ cubic cm, as required.

Would you like more details or have any further questions?

Related Questions:

1. How would the volume change if the height of the small cone was altered?
2. What would happen to the volume if the cones were not similar?
3. Can you derive the formula for the volume of a frustum directly from the cones' properties?
4. How would you calculate the surface area of the solid instead of the volume?
5. If the small cone's base were instead parallel to the large cone's base, what would the new solid's volume be?

Tip:

Always verify the similarity ratio when dealing with similar figures, as this relationship is key to solving problems involving similar shapes.