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