To find the volume of the second cone, we start by considering the relationship between the dimensions of the first cone and the second cone. The volume of a cone is given by the formula:

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

where $r$ is the radius of the base and $h$ is the height.

The problem states:

1. The volume $V$ of the first cone is 200 cm³.
2. The second cone has a height that is 6 times bigger and a base radius that is half as long as the first cone.

Given these changes:

• Let $h$ and $r$ be the height and radius of the first cone.
• Then, the height of the second cone is $6h$ and the radius is $\frac{r}{2}$.

Substituting these into the volume formula for the second cone, we get: $V' = \frac{1}{3} \pi \left(\frac{r}{2}\right)^2 (6h)$ $V' = \frac{1}{3} \pi \frac{r^2}{4} (6h)$ $V' = \frac{1}{3} \pi \frac{6r^2h}{4}$ $V' = \frac{1}{3} \pi \frac{3r^2h}{2}$ $V' = \frac{3}{2} \left(\frac{1}{3} \pi r^2 h\right)$ $V' = \frac{3}{2} V$

Since $V = 200$ cm³: $V' = \frac{3}{2} \times 200$ $V' = 300$ cm³.

Therefore, the volume of the second cone is 300 cm³.

Would you like more details on any part of this calculation, or do you have any other questions?

1. Would you like to learn how the volume changes when dimensions of a cone are altered in other ways?
2. Interested in understanding the derivation of the cone's volume formula?
3. Curious about the impact of changing only the height or only the radius on the cone’s volume?
4. Would you like to explore similar problems with other geometric shapes such as cylinders or spheres?
5. Need to solve another problem involving volumes of composite shapes?

Tip: Always double-check the units in your calculations to ensure consistency and correctness in your answers.