To solve this problem, let’s break it down step-by-step.

Step 1: Understand the Cone Dimensions

From the diagram:

• The total height of the large cone is $21 \, \text{m}$.
• The radius of the large cone’s base is $6 \, \text{m}$.
• The small cone, which is removed to create the frustum, has:
  • A height of $7 \, \text{m}$.
  • A base radius of $2 \, \text{m}$.

Part (a): Volume of the Small Cone

The formula for the volume $V$ of a cone is: $V = \frac{1}{3} \pi r^2 h$ where $r$ is the radius of the base, and $h$ is the height of the cone.

For the small cone:

• Radius $r = 2 \, \text{m}$
• Height $h = 7 \, \text{m}$

Calculating the volume: $V_{\text{small cone}} = \frac{1}{3} \pi (2)^2 (7) = \frac{1}{3} \pi \times 4 \times 7 = \frac{28}{3} \pi \approx 29.3 \, \text{m}^3$

Part (b): Volume of the Frustum

The frustum is the large cone with the small cone removed.

1. Calculate the volume of the large cone.

   • Radius $r = 6 \, \text{m}$
   • Height $h = 21 \, \text{m}$

   $V_{\text{large cone}} = \frac{1}{3} \pi (6)^2 (21) = \frac{1}{3} \pi \times 36 \times 21 = 252 \pi \approx 791.7 \, \text{m}^3$

2. Volume of the frustum is the volume of the large cone minus the volume of the small cone: $V_{\text{frustum}} = V_{\text{large cone}} - V_{\text{small cone}}$ Substituting values: $V_{\text{frustum}} = 252 \pi - \frac{28}{3} \pi = \frac{756 - 28}{3} \pi = \frac{728}{3} \pi \approx 764.0 \, \text{m}^3$

Final Answers

• Volume of the small cone: $29.3 \, \text{m}^3$
• Volume of the frustum: $764.0 \, \text{m}^3$

Would you like more details on any steps, or have further questions?

Here are some related questions for further exploration:

1. How would the volume change if the small cone’s height was increased?
2. What would be the surface area of the frustum?
3. How is the frustum’s volume formula derived from the cone volume formula?
4. What would happen to the volume if both cones were doubled in size?
5. How can we find the slant height of the frustum?

Tip: Remember that scaling a cone's dimensions (height and radius) by a factor will scale its volume by the cube of that factor.