SOLUTION: At what distance from the base of a right circular cone must a plane be passed parallel to the base in order that the volume of the frustum formed shall be three-fifths of the volu
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Question 1193372: At what distance from the base of a right circular cone must a plane be passed parallel to the base in order that the volume of the frustum formed shall be three-fifths of the volume of the given cone?
Pls. if anyone can help me? I try the volume of the frustum formula but I still didn't get it but I know that the answer is 0.26319h. Hope anyone could notice me. Thank you very much. More power! Answer by greenestamps(13200) (Show Source):
The nice thing about this problem is that you don't need to deal with the formula for the volume of a frustum, or with ANY geometric formula.
If the volume of the frustum is 3/5 of the volume of the whole cone, then the small cone that is cut off by the plane is 2/5 of the volume of the whole cone.
The two cones are similar, so the ratio of the volumes is the cube of the ratio of the heights. Knowing that the volume of the small cone is 2/5 of the volume of the whole cone, the ratio of the heights of the two cones is
= 0.73681 to 5 decimal places.
Then, since the height of the small cone is 0.73681 times the height of the whole cone, the height of the frustum is 1-0.73681 = 0.26319 times the height of the whole cone.
Clarification for the student who asked a question about my response....
The ratio of the volumes is the cube (3rd power) of the ratio of the heights; so the ratio of the heights is the CUBE ROOT (1/3 power) of the ratio of the volumes.
