SOLUTION: A sphere with radius 3 is inscribed in a conical frustum of slant height 10. (The sphere is tangent to both bases and the side of the frustum.) Find the volume of the frustum.
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Question 1075217: A sphere with radius 3 is inscribed in a conical frustum of slant height 10. (The sphere is tangent to both bases and the side of the frustum.) Find the volume of the frustum.
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
Let's look at a section of that frustum/cone/sphere
cut center of the sphere, and the center of the base of the cone
(and perpendicular to those bases):
The cross sections of sphere and cone are
a circle inscribed in an isosceles triangle.
The center of the circle is the incenter of the triangle,
at equal distance to all sides of the triangle,
and on the bisectors of the triangle's angles.
The height of the frustum is , the length of one side
of a 3-4-5 right triangle with side length 6, 8, and 10.
Its smaller angle is , with
and .
We are interested in right triangle AOC,
because we want to find the length of AC.
We know and we know .
The trigonometric identity
tells us that .
So, ---> .
That is , the radius of the larger base of the frustum.
Since right triangle ABC is similar to the 3-4-5 right triangle,
---> .
That is the height of the cone that the frustum came from.
The volume of that cone is
.
The similar little cone lopped off to make the frustum has
a base radius of ,
so it is a scale version of the original cone used to make the frustum.
Consequently, the volume of that little cone is ,
and the volume of the frustum is
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