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

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

$V=\frac{1}{3} \pi 3^{2}(6\sqrt{2})$

$V=18 \pi \sqrt{2}\ in^{3}$

Answer 2

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