To find out how the volume of a cone changes when the radius is multiplied by 3 while keeping the height constant, let's go through the calculations in detail.
1. Volume of a Cone Formula:
The volume [tex]\( V \)[/tex] of a cone can be calculated using the formula:
[tex]\[
V = \frac{1}{3} \pi r^2 h
\][/tex]
where [tex]\( r \)[/tex] is the radius of the base, and [tex]\( h \)[/tex] is the height of the cone.
2. Initial Volume:
Given that the initial radius is [tex]\( r \)[/tex] and the height is [tex]\( h \)[/tex], the initial volume [tex]\( V_{\text{initial}} \)[/tex] is:
[tex]\[
V_{\text{initial}} = \frac{1}{3} \pi r^2 h
\][/tex]
Using the given problem, let's assume:
[tex]\[
r = 1, \quad h = 1
\][/tex]
Then, the volume can be explicitly calculated as:
[tex]\[
\text{Initial Volume} = \frac{1}{3} \pi (1)^2 (1) = \frac{1}{3} \pi
\][/tex]
Numerically:
[tex]\[
\text{Initial Volume} \approx 1.047
\][/tex]
3. New Radius:
The problem states that the radius is multiplied by 3. So the new radius [tex]\( r_{\text{new}} \)[/tex] is:
[tex]\[
r_{\text{new}} = 3r
\][/tex]
4. New Volume:
The new volume [tex]\( V_{\text{new}} \)[/tex] of the cone with the new radius and the same height [tex]\( h \)[/tex] is:
[tex]\[
V_{\text{new}} = \frac{1}{3} \pi (3r)^2 h
\][/tex]
Simplifying this, we get:
[tex]\[
V_{\text{new}} = \frac{1}{3} \pi 9r^2 h = 9 \left(\frac{1}{3} \pi r^2 h\right)
\][/tex]
Therefore:
[tex]\[
\text{New Volume} = 9 \times \text{Initial Volume}
\][/tex]
Numerically:
[tex]\[
\text{New Volume} \approx 9 \times 1.047 \approx 9.425
\][/tex]
5. Volume Multiplication Factor:
The multiplication factor is the ratio of the new volume to the initial volume:
[tex]\[
\frac{V_{\text{new}}}{V_{\text{initial}}} = \frac{9 \times \text{Initial Volume}}{\text{Initial Volume}} = 9
\][/tex]
Therefore, the volume is multiplied by a factor of 9 when the radius of the cone is tripled, while keeping the height constant.
The correct answer is:
C. 9
