Question

A cone is inscribed in a regular square pyramid. If the pyramid has a base edge of 6" and a slant height of 9", find the volume of the cone.

Answer 1

v=(1/3)hpir²

ok, so inscribed
therefor the edge length is the diameter of the cone (the base is on bottom)
so then d=6
d/2=r
6/2=3=r


height
we need pythagoran theorem to find this
if we look at one side and draw the height then we get a right triangle with height is one leg
3 is another
and 9 is hyptonuse
a²+b²=c²
3²+h²=9²
9+h²=81
h²=72
h=6√2



h=6√2
r=3


V=(1/3)(6√2)(pi)(3²)
V=18pi√2 cubic feet

Answer 2

The volume of the cone is equal to

`V=18 \pi √(2)\ in^(3)`

Explanation:

we know that

The volume of a cone is equal to

`V=(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√(2)\ in`

Step 2

Find the volume of the cone

`V=(1)/(3) \pi r^(2)h`

we have

`r=3\ in`

`h=6√(2)\ in`

substitute the values

`V=(1)/(3) \pi 3^(2)(6√(2))`

`V=18 \pi √(2)\ in^(3)`