Question
A cone has a volume of 9π in3 and a diameter of 6 in. Wilson states that a cylinder with the same height and diameter has the same volume. Which statement explains whether or not Wilson is correct?
A cylinder in which h = 1 and d = 6 has a volume of 27π in3; therefore, Wilson is incorrect
A cylinder in which h = 3 and d = 6 has a volume of 27π in3; therefore, Wilson is incorrect
A cylinder in which h = 1 and d = 6 has a volume of 9π in3; therefore, Wilson is correct
A cylinder in which h = 3 and d = 6 has a volume of 9π in3; therefore, Wilson is correct
A cylinder in which h = 1 and d = 6 has a volume of 27π in3; therefore, Wilson is incorrect
A cylinder in which h = 3 and d = 6 has a volume of 27π in3; therefore, Wilson is incorrect
A cylinder in which h = 1 and d = 6 has a volume of 9π in3; therefore, Wilson is correct
A cylinder in which h = 3 and d = 6 has a volume of 9π in3; therefore, Wilson is correct
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Answered by
GPT-4o mini
To determine whether Wilson is correct, we need to calculate the height of the cone based on its given volume and diameter.
First, we know that the volume \( V \) of a cone is given by the formula:
\[
V = \frac{1}{3} \pi r^2 h
\]
Where:
- \( r \) is the radius
- \( h \) is the height
Given that the diameter of the cone is 6 inches, the radius \( r \) is:
\[
r = \frac{d}{2} = \frac{6}{2} = 3 \text{ in}
\]
The volume of the cone is given as \( 9\pi \) in³. So we set up the equation:
\[
9\pi = \frac{1}{3} \pi (3)^2 h
\]
This simplifies to:
\[
9\pi = \frac{1}{3} \pi (9) h
\]
\[
9\pi = 3\pi h
\]
Dividing both sides by \( \pi \) (assuming \( \pi \neq 0 \)):
\[
9 = 3h
\]
Solving for \( h \):
\[
h = \frac{9}{3} = 3 \text{ in}
\]
Now, we have established that the height of the cone is 3 inches.
Next, we need to find the volume of a cylinder with the same height and diameter. The volume \( V \) of a cylinder is given by the formula:
\[
V = \pi r^2 h
\]
Using the same radius \( r = 3 \) in and height \( h = 3 \) in, the volume becomes:
\[
V = \pi (3)^2 (3) = \pi (9)(3) = 27\pi \text{ in}^3
\]
Now we compare the volumes:
- Volume of the cone: \( 9\pi \) in³
- Volume of the cylinder: \( 27\pi \) in³
Since the volumes are not the same (the cylinder has a volume three times that of the cone), Wilson's statement that a cylinder with the same height and diameter has the same volume as the cone is incorrect.
Thus, the correct statement explaining whether Wilson is correct is:
**A cylinder in which h = 3 and d = 6 has a volume of 27π in³; therefore, Wilson is incorrect.**
First, we know that the volume \( V \) of a cone is given by the formula:
\[
V = \frac{1}{3} \pi r^2 h
\]
Where:
- \( r \) is the radius
- \( h \) is the height
Given that the diameter of the cone is 6 inches, the radius \( r \) is:
\[
r = \frac{d}{2} = \frac{6}{2} = 3 \text{ in}
\]
The volume of the cone is given as \( 9\pi \) in³. So we set up the equation:
\[
9\pi = \frac{1}{3} \pi (3)^2 h
\]
This simplifies to:
\[
9\pi = \frac{1}{3} \pi (9) h
\]
\[
9\pi = 3\pi h
\]
Dividing both sides by \( \pi \) (assuming \( \pi \neq 0 \)):
\[
9 = 3h
\]
Solving for \( h \):
\[
h = \frac{9}{3} = 3 \text{ in}
\]
Now, we have established that the height of the cone is 3 inches.
Next, we need to find the volume of a cylinder with the same height and diameter. The volume \( V \) of a cylinder is given by the formula:
\[
V = \pi r^2 h
\]
Using the same radius \( r = 3 \) in and height \( h = 3 \) in, the volume becomes:
\[
V = \pi (3)^2 (3) = \pi (9)(3) = 27\pi \text{ in}^3
\]
Now we compare the volumes:
- Volume of the cone: \( 9\pi \) in³
- Volume of the cylinder: \( 27\pi \) in³
Since the volumes are not the same (the cylinder has a volume three times that of the cone), Wilson's statement that a cylinder with the same height and diameter has the same volume as the cone is incorrect.
Thus, the correct statement explaining whether Wilson is correct is:
**A cylinder in which h = 3 and d = 6 has a volume of 27π in³; therefore, Wilson is incorrect.**
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