Question 1153218
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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.<br>
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.<br>
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).<br>
Triangles CUQ, CTP, and COB are all similar, because they are all right triangles that share angle OCB.<br>
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.<br>
To find the radius of the cone, we can use similar triangles CUQ and COB.<br>
QU=1 and CQ=3, so CU = 2*sqrt(2).  Then<br>
{{{OB/OC = QU/CU}}}
{{{OB/8 = 1/(2*sqrt(2))}}}
{{{OB = 8/(2*sqrt(2)) = 2*sqrt(2)}}}<br>
And then the volume of the cone is<br>
{{{(1/3)(pi)(r^2)(h) = (1/3)(pi)((2*sqrt(2))^2)(8) = (1/3)(pi)(8)(8) = (64/3)(pi)}}}<br>