Answer this :

Three identical cones each have a radius of 50 and a height of 120. The cones are placed so that their circular bases are touching each other. A sphere is placed so that it rests in the space created by the three cones, as shown. If the top of the sphere is level with the tops of the cones, then the radius of the sphere is closest to

(A) 38.9          (B) 38.7          (C) 38.1          (D) 38.5          (E) 38.3

Dear student,

Let us represent the given information using a diagram.

The centers of 3 circular bases form an equilateral triangle with side length 100.Let the circumcircle of this triangle have radius D. Its radius is 1003.O is the center of the sphere with radius r and Q is the point of contact of the sphere with one of the cones.The line PT is parallel to the base and forms a right triangle.AC has slope -125 so QOhas slope 512OT= 5r13QT= 12r13The triangles ABC, APQ and QTO are similar.AP = S= S+OT= r + 5r13 = 18r13PQAP = 50120=512PQ= 512*AP = 512*18r13 = 15r26PT= PQ+QT = 15r26+12r13 = 3r2PT= BD. So,3r2=1003r = 38.4900179 38.5So, radius of the sphere is approximately 38.5 units 


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