To understand how the volume of a cone changes if its radius and height are both tripled, let's start by recalling the formula for the volume of a cone:
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
Here:
- [tex]\( V \)[/tex] is the volume of the cone,
- [tex]\( r \)[/tex] is the radius of the base of the cone,
- [tex]\( h \)[/tex] is the height of the cone,
- [tex]\( \pi \)[/tex] is a mathematical constant approximately equal to 3.14159.
Now, let's denote the initial radius by [tex]\( r \)[/tex] and the initial height by [tex]\( h \)[/tex].
If we triple both the radius and the height:
- The new radius [tex]\( R \)[/tex] is [tex]\( 3r \)[/tex],
- The new height [tex]\( H \)[/tex] is [tex]\( 3h \)[/tex].
We can now substitute the new values of the radius and height into the volume formula to find the new volume [tex]\( V_{\text{new}} \)[/tex]:
[tex]\[ V_{\text{new}} = \frac{1}{3} \pi (3r)^2 (3h) \][/tex]
Next step is to simplify this expression:
[tex]\[ \begin{align*}
V_{\text{new}} &= \frac{1}{3} \pi (9r^2) (3h) \\
&= \frac{1}{3} \pi \cdot 9r^2 \cdot 3h \\
&= \frac{1}{3} \pi \cdot 27r^2 \cdot h \\
&= 27 \left(\frac{1}{3} \pi r^2 h\right)
\end{align*} \][/tex]
So, we can see that:
[tex]\[ V_{\text{new}} = 27V \][/tex]
where [tex]\( V \)[/tex] is the initial volume. This means that the new volume is 27 times the original volume.
Therefore, when the radius and height of the cone are both tripled, the volume increases by a factor of 27.
From the given choices, the correct answer is:
D. [tex]\( V = 27 \pi r^2 h \)[/tex]
