To find out how the volume of a cone changes when the radius is quadrupled and the height is reduced to [tex]\(\frac{1}{5}\)[/tex] of its original size, let's follow a step-by-step approach.
### Step 1: Understand the original volume formula
The volume [tex]\(V\)[/tex] of a cone is given by:
[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.
### Step 2: Define the changes in dimensions
Let's denote the original radius by [tex]\(r\)[/tex] and the original height by [tex]\(h\)[/tex].
- New radius = [tex]\(4r\)[/tex]
- New height = [tex]\(\frac{1}{5}h\)[/tex]
### Step 3: Calculate the new volume
We need to substitute the new dimensions into the volume formula.
The new volume [tex]\(V_{\text{new}}\)[/tex] will be:
[tex]\[ V_{\text{new}} = \frac{1}{3} \pi (4r)^2 \left(\frac{1}{5}h\right) \][/tex]
### Step 4: Simplify the expression
First, calculate [tex]\((4r)^2\)[/tex]:
[tex]\[ (4r)^2 = 16r^2 \][/tex]
Now, substitute this into the formula:
[tex]\[ V_{\text{new}} = \frac{1}{3} \pi (16r^2) \left(\frac{1}{5}h\right) \][/tex]
Simplify inside the parentheses:
[tex]\[ V_{\text{new}} = \frac{1}{3} \pi \cdot 16r^2 \cdot \frac{1}{5}h \][/tex]
[tex]\[ V_{\text{new}} = \frac{1}{3} \pi \cdot \frac{16r^2h}{5} \][/tex]
[tex]\[ V_{\text{new}} = \frac{16}{15} \cdot \frac{1}{3} \pi r^2 h \][/tex]
### Step 5: Compare with the original volume
The original volume was:
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
So the new volume is:
[tex]\[ V_{\text{new}} = \frac{16}{15} V \][/tex]
### Step 6: Conclusion
The new volume of the cone is [tex]\(\frac{16}{15}\)[/tex] times the original volume.
None of the provided answer choices directly match the exact form we derived ([tex]\(\frac{16}{15} \cdot \frac{1}{3} \pi r^2 h\)[/tex]), but we did conclude that the factor change in volume is [tex]\(\frac{16}{15}\)[/tex]. Therefore, if we choose the best match of the given answers based on this simplified factor of increase, even if the options do not strictly make sense:
D. [tex]\(V = \frac{16}{15} m^2 h\)[/tex]
Appears to relate closest to the correct factor, as it reflects the same multiplicative change ( [tex]\(\frac{16}{15}\)[/tex]).
Hence, the best answer from the options provided is:
D. [tex]\(V = \frac{16}{15} m^2 h\)[/tex]
