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Question 1153218: A sphere of radius 1 and a sphere of radius 2 are inscribed in a right circular cone so that the larger sphere is on the base and the smaller sphere is right above it. Find the volume of the cone.
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
Picture slicing the cone down the middle to get a 2-dimensional picture. The picture is an isosceles triangle with two inscribed circles of radius 2 and 1.
Let AB be the base of the isosceles triangle and C be the vertex.
Let O be the center of the base of the cone, P the center of the circle of radius 2, and Q the center of the circle of radius 1.
Let R be the point of tangency of the two circles, and let S be the "top" of the smaller circle.
Finally, let T be the point of tangency of the larger circle and side BC of the triangle, and let U be the point of tangency of the smaller circle with side BC of the triangle.
The volume of the cone is one-third base times height; we need to determine the measures of OB (radius of the base) and CO (height).
Triangles CUQ, CTP, and COB are all similar, because they are all right triangles that share angle OCB.
Similar triangles CUQ and CTP are similar with ratio 1:2 (the radii of the two spheres, UQ and TP). That makes CQ half of CP; since PQ=3 (the sum of the two radii), CQ is also 3, and then the height of the cone is 3+3+2 = 8.
To find the radius of the cone, we can use similar triangles CUQ and COB.
QU=1 and CQ=3, so CU = 2*sqrt(2). Then
And then the volume of the cone is
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