Answer:
The volume of the cone is equal to
![V=18 \pi \sqrt{2}\ in^{3}](https://tex.z-dn.net/?f=V%3D18%20%5Cpi%20%5Csqrt%7B2%7D%5C%20in%5E%7B3%7D)
Step-by-step explanation:
we know that
The volume of a cone is equal to
![V=\frac{1}{3} \pi r^{2}h](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B1%7D%7B3%7D%20%5Cpi%20r%5E%7B2%7Dh)
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](https://tex.z-dn.net/?f=r%3D6%2F2%3D3%5C%20in)
Step 1
Find the height of the cone
Applying the Pythagoras Theorem
![l^{2}=r^{2}+h^{2}](https://tex.z-dn.net/?f=l%5E%7B2%7D%3Dr%5E%7B2%7D%2Bh%5E%7B2%7D)
where
l is the slant height
in this problem we have
![l=9\ in](https://tex.z-dn.net/?f=l%3D9%5C%20in)
![r=3\ in](https://tex.z-dn.net/?f=r%3D3%5C%20in)
Solve for h
![h^{2}=l^{2}-r^{2}](https://tex.z-dn.net/?f=h%5E%7B2%7D%3Dl%5E%7B2%7D-r%5E%7B2%7D)
substitute the values
![h^{2}=9^{2}-3^{2}](https://tex.z-dn.net/?f=h%5E%7B2%7D%3D9%5E%7B2%7D-3%5E%7B2%7D)
![h^{2}=72](https://tex.z-dn.net/?f=h%5E%7B2%7D%3D72)
![h=6\sqrt{2}\ in](https://tex.z-dn.net/?f=h%3D6%5Csqrt%7B2%7D%5C%20in)
Step 2
Find the volume of the cone
![V=\frac{1}{3} \pi r^{2}h](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B1%7D%7B3%7D%20%5Cpi%20r%5E%7B2%7Dh)
we have
![r=3\ in](https://tex.z-dn.net/?f=r%3D3%5C%20in)
![h=6\sqrt{2}\ in](https://tex.z-dn.net/?f=h%3D6%5Csqrt%7B2%7D%5C%20in)
substitute the values
![V=\frac{1}{3} \pi 3^{2}(6\sqrt{2})](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B1%7D%7B3%7D%20%5Cpi%203%5E%7B2%7D%286%5Csqrt%7B2%7D%29)
![V=18 \pi \sqrt{2}\ in^{3}](https://tex.z-dn.net/?f=V%3D18%20%5Cpi%20%5Csqrt%7B2%7D%5C%20in%5E%7B3%7D)