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Sagot :
When solving for the volume and height of cones, we use the formula for the volume of a cone:
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
where [tex]\( r \)[/tex] is the radius and [tex]\( h \)[/tex] is the height.
### Part A:
We are given that the height [tex]\( h \)[/tex] of the cone is 25 cm, and the radius [tex]\( r \)[/tex] is 9 cm. We need to find the volume [tex]\( V \)[/tex] of the ice cream cone.
Using [tex]\( \pi = 3.14 \)[/tex], we substitute the values into the formula:
[tex]\[ \text{Volume } (V) = \frac{1}{3} \pi r^2 h \][/tex]
[tex]\[ V = \frac{1}{3} \times 3.14 \times (9)^2 \times 25 \][/tex]
First, calculate the area of the base ( [tex]\( \pi r^2 \)[/tex] ):
[tex]\[ \pi r^2 = 3.14 \times 9^2 \][/tex]
[tex]\[ = 3.14 \times 81 \][/tex]
[tex]\[ = 254.34 \][/tex]
Next, multiply by the height and adjust for the factor [tex]\( \frac{1}{3} \)[/tex]:
[tex]\[ V = \frac{1}{3} \times 254.34 \times 25 \][/tex]
[tex]\[ = \frac{1}{3} \times 6358.5 \][/tex]
[tex]\[ = 2119.5 \, \text{cm}^3 \][/tex]
So, the volume of the ice cream that fits inside the cone is [tex]\( 2119.5 \, \text{cm}^3 \)[/tex].
### Part B:
Given the volume [tex]\( V \)[/tex] of 1272 cm³ for the smaller cone intended for kids, and keeping the radius [tex]\( r \)[/tex] at 9 cm, we need to find the new height [tex]\( h \)[/tex].
We rearrange the volume formula to solve for [tex]\( h \)[/tex]:
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
[tex]\[ 1272 = \frac{1}{3} \times 3.14 \times 81 \times h \][/tex]
First, simplify [tex]\( \frac{1}{3} \times 3.14 \times 81 \)[/tex]:
[tex]\[ \frac{1}{3} \times 3.14 \times 81 = 84.78 \][/tex]
Then solve for [tex]\( h \)[/tex]:
[tex]\[ 1272 = 84.78h \][/tex]
[tex]\[ h = \frac{1272}{84.78} \][/tex]
[tex]\[ h \approx 15.0 \][/tex]
So, the new height of the smaller cone, rounded to the nearest hundredth, is [tex]\( 15.0 \)[/tex] cm.
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
where [tex]\( r \)[/tex] is the radius and [tex]\( h \)[/tex] is the height.
### Part A:
We are given that the height [tex]\( h \)[/tex] of the cone is 25 cm, and the radius [tex]\( r \)[/tex] is 9 cm. We need to find the volume [tex]\( V \)[/tex] of the ice cream cone.
Using [tex]\( \pi = 3.14 \)[/tex], we substitute the values into the formula:
[tex]\[ \text{Volume } (V) = \frac{1}{3} \pi r^2 h \][/tex]
[tex]\[ V = \frac{1}{3} \times 3.14 \times (9)^2 \times 25 \][/tex]
First, calculate the area of the base ( [tex]\( \pi r^2 \)[/tex] ):
[tex]\[ \pi r^2 = 3.14 \times 9^2 \][/tex]
[tex]\[ = 3.14 \times 81 \][/tex]
[tex]\[ = 254.34 \][/tex]
Next, multiply by the height and adjust for the factor [tex]\( \frac{1}{3} \)[/tex]:
[tex]\[ V = \frac{1}{3} \times 254.34 \times 25 \][/tex]
[tex]\[ = \frac{1}{3} \times 6358.5 \][/tex]
[tex]\[ = 2119.5 \, \text{cm}^3 \][/tex]
So, the volume of the ice cream that fits inside the cone is [tex]\( 2119.5 \, \text{cm}^3 \)[/tex].
### Part B:
Given the volume [tex]\( V \)[/tex] of 1272 cm³ for the smaller cone intended for kids, and keeping the radius [tex]\( r \)[/tex] at 9 cm, we need to find the new height [tex]\( h \)[/tex].
We rearrange the volume formula to solve for [tex]\( h \)[/tex]:
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
[tex]\[ 1272 = \frac{1}{3} \times 3.14 \times 81 \times h \][/tex]
First, simplify [tex]\( \frac{1}{3} \times 3.14 \times 81 \)[/tex]:
[tex]\[ \frac{1}{3} \times 3.14 \times 81 = 84.78 \][/tex]
Then solve for [tex]\( h \)[/tex]:
[tex]\[ 1272 = 84.78h \][/tex]
[tex]\[ h = \frac{1272}{84.78} \][/tex]
[tex]\[ h \approx 15.0 \][/tex]
So, the new height of the smaller cone, rounded to the nearest hundredth, is [tex]\( 15.0 \)[/tex] cm.
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