The formula for the volume of a right circular cone is:

$V = \frac{1}{3} \pi r^2 h$

where:

• $r = 5.1$ inches (radius of the base)
• $h = 11.4$ inches (height)
• $\pi \approx 3.1416$

Step 1: Plug in the given values

$V = \frac{1}{3} \pi (5.1)^2 (11.4)$

$V = \frac{1}{3} \pi (26.01) (11.4)$

$V = \frac{1}{3} \pi (296.514)$

$V = \frac{296.514\pi}{3}$

$V \approx \frac{296.514 \times 3.1416}{3}$

Step 2: Compute the volume

$V \approx \frac{931.49}{3}$

$V \approx 310.5 \text{ cubic inches}$

Final Answer:

$\boxed{310.5} \text{ cubic inches}$

Would you like a step-by-step breakdown of how the calculations were performed?

Here are five related questions to explore further:

1. How does the volume formula change if the cone were truncated (frustum of a cone)?
2. How would the volume change if the height were doubled while keeping the same radius?
3. What is the lateral surface area of this cone?
4. If the cone were filled with water, how much would it weigh (given the density of water)?
5. How does the volume compare to a cylinder with the same base and height?

Tip: The volume of a cone is always one-third of the volume of a cylinder with the same base and height.