Question

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a conical container can hold 120π cubic centimeters of water. the diameter of the base of the container is 12 centimeters.
the height of the container is centimeters. if its diameter and height were both doubled, the containers capacity would be times its original capacity.

Explanation:

Step1: Recall the volume formula for a cone

The volume \( V \) of a cone is given by the formula \( V=\frac{1}{3}\pi r^{2}h \), where \( r \) is the radius of the base and \( h \) is the height.
We know that the diameter \( d = 12\) cm, so the radius \( r=\frac{d}{2}=\frac{12}{2} = 6\) cm, and the volume \( V = 120\pi\) cubic cm.
Substitute these values into the volume formula:
\(120\pi=\frac{1}{3}\pi\times(6)^{2}\times h\)

Step2: Solve for the height \( h \)

First, simplify the right - hand side of the equation:
\(\frac{1}{3}\pi\times(6)^{2}\times h=\frac{1}{3}\pi\times36\times h = 12\pi h\)
So our equation becomes \(120\pi=12\pi h\)
Divide both sides of the equation by \(12\pi\):
\(h=\frac{120\pi}{12\pi}=10\) cm.

Step3: Analyze the effect of doubling diameter and height on volume

If the diameter is doubled, the new radius \( r_{new}=\frac{2d}{2}=d = 12\) cm (since original \( d = 12\) cm, new \( d'=24\) cm, so \( r'=\frac{24}{2}=12\) cm). The original radius \( r = 6\) cm, so the new radius is \(2r\).
If the height is doubled, the new height \( h_{new}=2h\) (original \( h = 10\) cm, new \( h'=20\) cm).
The new volume \( V_{new}=\frac{1}{3}\pi(r_{new})^{2}h_{new}\)
Substitute \( r_{new}=2r\) and \( h_{new}=2h\) into the formula:
\(V_{new}=\frac{1}{3}\pi\times(2r)^{2}\times(2h)=\frac{1}{3}\pi\times4r^{2}\times2h = 8\times\frac{1}{3}\pi r^{2}h\)
Since the original volume \( V=\frac{1}{3}\pi r^{2}h\), then \( V_{new}=8V\). So the new capacity is 8 times the original capacity.

Answer:

The height of the container is \(10\) centimeters. If its diameter and height were both doubled, the container's capacity would be \(8\) times its original capacity.