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Sagot :
To determine how much cement is needed to fill each cone-shaped top placed on the bridge supports, we need to calculate the volume of one cone. The formula for the volume [tex]\( V \)[/tex] of a cone is given by:
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
where:
- [tex]\( \pi \)[/tex] (pi) is approximately 3.14,
- [tex]\( r \)[/tex] is the radius of the base of the cone,
- [tex]\( h \)[/tex] is the height of the cone.
Assuming we have the values for radius [tex]\( r \)[/tex] and height [tex]\( h \)[/tex] as follows:
- Radius [tex]\( r = 5 \, \text{cm} \)[/tex],
- Height [tex]\( h = 9 \, \text{cm} \)[/tex].
Let's now substitute these values into the volume formula:
[tex]\[ V = \frac{1}{3} \times 3.14 \times (5)^2 \times 9 \][/tex]
First, calculate the square of the radius:
[tex]\[ 5^2 = 25 \][/tex]
Next, multiply this result by the height [tex]\( h \)[/tex] and [tex]\( \pi \)[/tex]:
[tex]\[ 25 \times 9 = 225 \][/tex]
[tex]\[ 225 \times 3.14 = 706.5 \][/tex]
Finally, multiply by [tex]\(\frac{1}{3}\)[/tex]:
[tex]\[ V = \frac{1}{3} \times 706.5 = 235.5 \, \text{cm}^3 \][/tex]
Therefore, the volume of the cement needed to fill each cone is approximately:
[tex]\[ 235.5 \, \text{cm}^3 \][/tex]
Given the options:
A) 314 cm³
B) 271.2 cm³
C) 942 cm³
D) 282.6 cm³
The result closest to our calculated volume is not exactly among the provided options, but the true answer obtained after calculation is:
[tex]\[ 235.5 \, \text{cm}^3 \][/tex]
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
where:
- [tex]\( \pi \)[/tex] (pi) is approximately 3.14,
- [tex]\( r \)[/tex] is the radius of the base of the cone,
- [tex]\( h \)[/tex] is the height of the cone.
Assuming we have the values for radius [tex]\( r \)[/tex] and height [tex]\( h \)[/tex] as follows:
- Radius [tex]\( r = 5 \, \text{cm} \)[/tex],
- Height [tex]\( h = 9 \, \text{cm} \)[/tex].
Let's now substitute these values into the volume formula:
[tex]\[ V = \frac{1}{3} \times 3.14 \times (5)^2 \times 9 \][/tex]
First, calculate the square of the radius:
[tex]\[ 5^2 = 25 \][/tex]
Next, multiply this result by the height [tex]\( h \)[/tex] and [tex]\( \pi \)[/tex]:
[tex]\[ 25 \times 9 = 225 \][/tex]
[tex]\[ 225 \times 3.14 = 706.5 \][/tex]
Finally, multiply by [tex]\(\frac{1}{3}\)[/tex]:
[tex]\[ V = \frac{1}{3} \times 706.5 = 235.5 \, \text{cm}^3 \][/tex]
Therefore, the volume of the cement needed to fill each cone is approximately:
[tex]\[ 235.5 \, \text{cm}^3 \][/tex]
Given the options:
A) 314 cm³
B) 271.2 cm³
C) 942 cm³
D) 282.6 cm³
The result closest to our calculated volume is not exactly among the provided options, but the true answer obtained after calculation is:
[tex]\[ 235.5 \, \text{cm}^3 \][/tex]
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