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.
Step-by-step explanation:
The volume of a cone can be calculated with this formula:
[tex]V=\frac{\pi r^2h}{3}[/tex]
Where "r" is the radius and "h" is the height.
We know that the radius is half the diameter. Then:
[tex]r=\frac{12cm}{2}=6cm[/tex]
We know the volume and the radius of the conical container, then we can find "h":
[tex]120\pi cm^3=\frac{\pi (6cm)^2h}{3}\\\\(3)(120\pi cm^3)=\pi (6cm)^2h\\\\h=\frac{3(120\pi cm^3)}{\pi (6cm)^2}\\\\h=10cm[/tex]
The diameter and height doubled are:
[tex]d=12cm*2=24cm\\h=10cm*2=20cm[/tex]
Now the radius is:
[tex]r=\frac{24cm}{2}=12cm[/tex]
And the container capacity is
[tex]V=\frac{\pi (12cm)^2(20cm)}{3}=960\pi cm^3[/tex]
Then, to compare the capacities, we can divide this new capacity by the original:
[tex]\frac{960\pi cm^3}{120\pi cm^3}=8[/tex]
Therefore, the container's capacity would be 8 times its original capacity.
