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Sagot :
To determine how the volume of a cone changes when its radius is quadrupled and its height is reduced to [tex]\(\frac{1}{5}\)[/tex] of its original size, let's follow these steps:
1. 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.
2. New Radius and Height:
- The radius is quadrupled: [tex]\( r_{\text{new}} = 4r \)[/tex]
- The height is reduced to [tex]\(\frac{1}{5}\)[/tex] of its original size: [tex]\( h_{\text{new}} = \frac{h}{5} \)[/tex]
3. New Volume Calculation:
Substitute the new radius and height into the volume formula:
[tex]\[ V_{\text{new}} = \frac{1}{3} \pi (r_{\text{new}})^2 h_{\text{new}} \][/tex]
[tex]\[ = \frac{1}{3} \pi (4r)^2 \left(\frac{h}{5}\right) \][/tex]
[tex]\[ = \frac{1}{3} \pi (16r^2) \left(\frac{h}{5}\right) \][/tex]
[tex]\[ = \frac{1}{3} \pi (16r^2) \cdot \frac{h}{5} \][/tex]
[tex]\[ = \frac{16}{15} \pi r^2 h \][/tex]
4. Compare the New and Original Volumes:
The new volume [tex]\( V_{\text{new}} \)[/tex] is [tex]\(\frac{16}{15} \pi r^2 h\)[/tex], compared to the original volume [tex]\( V \)[/tex] of [tex]\(\frac{1}{3} \pi r^2 h\)[/tex].
Therefore, the volume of the cone changes to [tex]\(\frac{16}{15}\)[/tex] times the original volume when the radius is quadrupled and the height is reduced to [tex]\(\frac{1}{5}\)[/tex] of its original size.
Thus, the correct answer from the given options is:
[tex]\[ \boxed{D. \frac{16}{15} \pi^2 h} \][/tex]
1. 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.
2. New Radius and Height:
- The radius is quadrupled: [tex]\( r_{\text{new}} = 4r \)[/tex]
- The height is reduced to [tex]\(\frac{1}{5}\)[/tex] of its original size: [tex]\( h_{\text{new}} = \frac{h}{5} \)[/tex]
3. New Volume Calculation:
Substitute the new radius and height into the volume formula:
[tex]\[ V_{\text{new}} = \frac{1}{3} \pi (r_{\text{new}})^2 h_{\text{new}} \][/tex]
[tex]\[ = \frac{1}{3} \pi (4r)^2 \left(\frac{h}{5}\right) \][/tex]
[tex]\[ = \frac{1}{3} \pi (16r^2) \left(\frac{h}{5}\right) \][/tex]
[tex]\[ = \frac{1}{3} \pi (16r^2) \cdot \frac{h}{5} \][/tex]
[tex]\[ = \frac{16}{15} \pi r^2 h \][/tex]
4. Compare the New and Original Volumes:
The new volume [tex]\( V_{\text{new}} \)[/tex] is [tex]\(\frac{16}{15} \pi r^2 h\)[/tex], compared to the original volume [tex]\( V \)[/tex] of [tex]\(\frac{1}{3} \pi r^2 h\)[/tex].
Therefore, the volume of the cone changes to [tex]\(\frac{16}{15}\)[/tex] times the original volume when the radius is quadrupled and the height is reduced to [tex]\(\frac{1}{5}\)[/tex] of its original size.
Thus, the correct answer from the given options is:
[tex]\[ \boxed{D. \frac{16}{15} \pi^2 h} \][/tex]
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