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A sphere of radius \(1\) and a sphere of radius \(2\) are inscribed in a right circular cone, as shown below. Find the volume of the cone. Thank you!

May 2, 2020

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We can use  similar triangles  to  find  the height of the cone

Let x  be the distance  from the cone's apex to  the  top of the smaller circle

We  have  that

[  x  + radius of smaller circle]           [ radius of larger circle + circum of small circle + x]

_________________________  =   ___________________________________________

[ 1 + x ]         [ 2 + 2 + x]

______   =    _________    cross- multiply

1                      2

2 [ 1 + x ]  =   1 [ 2 + 2 + x ]

2 + 2x  =  4  + x

x  =  2

So....the  height of the  cone   =   4  + 2  + 2   =  8

Now...x + 1  = 3  is  the hypotenuse of a right triangle  with  one of the legs  =  the radius  of the smaller cone

Let A  be the apex of the cone....B  be  the center of the small circle  and   C  be the point where the radius of the small circle  intersects  the side of the cone

Then sin of angle ABC  =  BC/ AB  =   1 / 3

And the tangent  of  this angle  =  1/ sqrt (8)

And we can find the  radius  of the cone thusly :

tan ABC   =  r / height of cone

1/ sqrt (8)  =  r  /8

8/sqrt (8)  = r  =  sqrt (8)

tan 30°  =  r / 8

1/√ 3  =  r / 8

r  =  8 / √ 3

So....the  volume of the cone  is

(1/3)  pi ( radius)^2 (height)  =

(1/3) pi *(sqrt (8))^2(8)  =  (64/3) pi  units ^3   ≈  67.02  units ^3   May 2, 2020
edited by CPhill  May 2, 2020
edited by CPhill  May 2, 2020