Answer:
The volume of the cone is equal to
[tex]V=18 \pi \sqrt{2}\ in^{3}[/tex]
Step-by-step explanation:
we know that
The volume of a cone is equal to
[tex]V=\frac{1}{3} \pi r^{2}h[/tex]
where
r is the radius of the base
h is the height of the cone
In this problem the length of the square base of the pyramid is equal to the diameter of the base of the cone
so
[tex]r=6/2=3\ in[/tex]
Step 1
Find the height of the cone
Applying the Pythagoras Theorem
[tex]l^{2}=r^{2}+h^{2}[/tex]
where
l is the slant height
in this problem we have
[tex]l=9\ in[/tex]
[tex]r=3\ in[/tex]
Solve for h
[tex]h^{2}=l^{2}-r^{2}[/tex]
substitute the values
[tex]h^{2}=9^{2}-3^{2}[/tex]
[tex]h^{2}=72[/tex]
[tex]h=6\sqrt{2}\ in[/tex]
Step 2
Find the volume of the cone
[tex]V=\frac{1}{3} \pi r^{2}h[/tex]
we have
[tex]r=3\ in[/tex]
[tex]h=6\sqrt{2}\ in[/tex]
substitute the values
[tex]V=\frac{1}{3} \pi 3^{2}(6\sqrt{2})[/tex]
[tex]V=18 \pi \sqrt{2}\ in^{3}[/tex]
