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.
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