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!
+128475 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]
_________________________ = ___________________________________________
radius of smaller circle radius of large circle
[ 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
