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Sagot :
To determine the volume of the icicle, which is in the shape of an inverted cone, we need to use the formula for the volume of a cone. The formula is:
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
where:
- [tex]\( V \)[/tex] is the volume
- [tex]\( r \)[/tex] is the radius of the base of the cone
- [tex]\( h \)[/tex] is the height of the cone
- [tex]\(\pi\)[/tex] (pi) is approximately 3.14.
First, we need to find the radius of the cone. The diameter is given as 9 mm, so we can find the radius by dividing the diameter by 2:
[tex]\[ r = \frac{\text{diameter}}{2} = \frac{9 \text{ mm}}{2} = 4.5 \text{ mm} \][/tex]
Next, we can substitute the values into the volume formula. The height [tex]\( h \)[/tex] is given as 27 mm.
[tex]\[ V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \cdot 3.14 \cdot (4.5)^2 \cdot 27 \][/tex]
Now, we calculate [tex]\( r^2 \)[/tex]:
[tex]\[ r^2 = (4.5)^2 = 20.25 \][/tex]
Next, we multiply [tex]\( \pi \)[/tex], [tex]\( r^2 \)[/tex], and [tex]\( h \)[/tex]:
[tex]\[ \pi \cdot r^2 \cdot h = 3.14 \cdot 20.25 \cdot 27 \][/tex]
Calculate the product inside the brackets:
[tex]\[ 3.14 \cdot 20.25 \cdot 27 = 1758.795 \][/tex]
Now, apply the [tex]\( \frac{1}{3} \)[/tex] factor:
[tex]\[ V = \frac{1}{3} \cdot 1758.795 = 586.265 \][/tex]
Lastly, we round this volume to the nearest hundredth:
[tex]\[ V \approx 572.26 \text{ mm}^3 \][/tex]
Therefore, the amount of frozen water in the icicle is approximately [tex]\( 572.26 \)[/tex] cubic millimeters.
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
where:
- [tex]\( V \)[/tex] is the volume
- [tex]\( r \)[/tex] is the radius of the base of the cone
- [tex]\( h \)[/tex] is the height of the cone
- [tex]\(\pi\)[/tex] (pi) is approximately 3.14.
First, we need to find the radius of the cone. The diameter is given as 9 mm, so we can find the radius by dividing the diameter by 2:
[tex]\[ r = \frac{\text{diameter}}{2} = \frac{9 \text{ mm}}{2} = 4.5 \text{ mm} \][/tex]
Next, we can substitute the values into the volume formula. The height [tex]\( h \)[/tex] is given as 27 mm.
[tex]\[ V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \cdot 3.14 \cdot (4.5)^2 \cdot 27 \][/tex]
Now, we calculate [tex]\( r^2 \)[/tex]:
[tex]\[ r^2 = (4.5)^2 = 20.25 \][/tex]
Next, we multiply [tex]\( \pi \)[/tex], [tex]\( r^2 \)[/tex], and [tex]\( h \)[/tex]:
[tex]\[ \pi \cdot r^2 \cdot h = 3.14 \cdot 20.25 \cdot 27 \][/tex]
Calculate the product inside the brackets:
[tex]\[ 3.14 \cdot 20.25 \cdot 27 = 1758.795 \][/tex]
Now, apply the [tex]\( \frac{1}{3} \)[/tex] factor:
[tex]\[ V = \frac{1}{3} \cdot 1758.795 = 586.265 \][/tex]
Lastly, we round this volume to the nearest hundredth:
[tex]\[ V \approx 572.26 \text{ mm}^3 \][/tex]
Therefore, the amount of frozen water in the icicle is approximately [tex]\( 572.26 \)[/tex] cubic millimeters.
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