The problem given involves comparing the effects of changing the height and radius of a right cone on its volume. Let’s break down the solution step-by-step.
1. Initial Volume Calculation:
The initial cone has a radius of 2 units and a height of 6 units. The volume [tex]\( V \)[/tex] of a cone is given by:
[tex]\[
V = \frac{1}{3} \pi r^2 h
\][/tex]
Substituting the values:
[tex]\[
V = \frac{1}{3} \pi (2)^2 6 = \frac{1}{3} \pi \cdot 4 \cdot 6 = 8 \pi \text{ units}^3
\][/tex]
So, the initial volume [tex]\( V \)[/tex] is [tex]\( 8 \pi \)[/tex] units[tex]\(^3\)[/tex].
2. Effect of Changing the Height to 3 Units:
When the height is changed to 3 units, while keeping the radius unchanged (2 units), the new volume [tex]\( V' \)[/tex] is:
[tex]\[
V' = \frac{1}{3} \pi (2)^2 3 = \frac{1}{3} \pi \cdot 4 \cdot 3 = 4 \pi \text{ units}^3
\][/tex]
This new volume is [tex]\( 4 \pi \)[/tex] units[tex]\(^3\)[/tex].
3. Effect of Changing the Radius to 1 Unit:
When the radius is changed to 1 unit, while keeping the height unchanged (6 units), the new volume [tex]\( V'' \)[/tex] is:
[tex]\[
V'' = \frac{1}{3} \pi (1)^2 6 = \frac{1}{3} \pi \cdot 1 \cdot 6 = 2 \pi \text{ units}^3
\][/tex]
This new volume is [tex]\( 2 \pi \)[/tex] units[tex]\(^3\)[/tex].
4. Comparison of Changes:
- Halving the height from 6 units to 3 units results in the volume being [tex]\( 4 \pi \)[/tex] units[tex]\(^3\)[/tex], which is indeed half of the initial volume [tex]\( 8 \pi \text{ units}^3 \)[/tex].
- Halving the radius from 2 units to 1 unit results in the volume being [tex]\( 2 \pi \)[/tex] units[tex]\(^3\)[/tex], which is one-fourth of the initial volume [tex]\( 8 \pi \text{ units}^3\)[/tex].
5. Final Statements:
Therefore, changing the height to half of its original value and changing the radius to half of its original value do not have the same effect on the volume. Halving the height of the cone halves the volume, while halving the radius of the cone results in one-fourth the volume.
Filling in the blanks in the given statements:
If the height is now 3, then the new volume is [tex]\(\mathbf{4 \pi}\)[/tex] units[tex]\(^3\)[/tex].
If the radius is now 1, then the new volume is [tex]\(\mathbf{2 \pi}\)[/tex] units[tex]\(^3\)[/tex].
Therefore, changing the height to half of its original value and changing the radius to half of its original value does [tex]\(\mathbf{not}\)[/tex] have the same effect on the volume. Halving the height of the cone [tex]\(\mathbf{halves}\)[/tex] the volume, while halving the radius of the cone results in [tex]\(\mathbf{one-fourth}\)[/tex] the volume.
