The problem provided asks to find the height of a cone given its volume $V = 32\pi \, \text{cm}^3$ and radius $r = 4 \, \text{cm}$.

The formula for the volume of a cone is:

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

where:

• $V$ is the volume,
• $r$ is the radius of the base, and
• $h$ is the height of the cone.

Step 1: Plug in the known values

We know that $V = 32\pi \, \text{cm}^3$ and $r = 4 \, \text{cm}$, so the equation becomes:

$32\pi = \frac{1}{3} \pi (4)^2 h$

Step 2: Simplify the equation

First, simplify $(4)^2$:

$32\pi = \frac{1}{3} \pi (16) h$

Next, cancel $\pi$ from both sides:

$32 = \frac{1}{3} \times 16 h$

Now, simplify further:

$32 = \frac{16}{3} h$

Step 3: Solve for $h$

Multiply both sides by 3 to eliminate the fraction:

$96 = 16 h$

Now, divide both sides by 16:

$h = \frac{96}{16} = 6 \, \text{cm}$

Final Answer:

The height of the cone is $6 \, \text{cm}$.

Would you like a more detailed explanation, or do you have any further questions?

Here are 5 related questions for practice:

1. What is the volume of a cone if the height is 9 cm and the radius is 5 cm?
2. How does changing the height of a cone affect its volume, assuming the radius is constant?
3. If the volume of a cone is $48\pi \, \text{cm}^3$ and the radius is 3 cm, what is the height?
4. Derive the formula for the height of a cone in terms of volume and radius.
5. What is the surface area of a cone with a radius of 4 cm and a height of 6 cm?

Tip: Always check if the units are consistent in geometry problems to avoid calculation mistakes.