Answer:
[tex]240 \pi\ in^{3}[/tex]
Step-by-step explanation:
we know that
If two figures are similar, then the ratio of its volumes is equal to the scale factor elevated to the cube
Let
z------> the scale factor
x-----> the volume of the larger cone
y-----> the volume of the original cone
[tex]z^{3}=\frac{x}{y}[/tex]
In this problem we have
[tex]z=2[/tex] -----> scale factor
substitute
[tex]2^{3}=\frac{x}{y}[/tex]
[tex]8=\frac{x}{y}[/tex]
[tex]x=8y[/tex]
That means-----> The volume of the larger cone is 8 times the volume of the original cone
Find the volume of the original cone
[tex]V=\frac{1}{3}\pi r^{2} h[/tex]
we have
[tex]r=3\ in[/tex]
[tex]h=10\ in[/tex]
substitute
[tex]V=\frac{1}{3}\pi (3^{2})(10)=30 \pi\ in^{3}[/tex]
therefore
The volume of the larger cone is equal to
[tex]8*30 \pi\ in^{3}=240 \pi\ in^{3}[/tex]
