At what distance from the base of a right circular cone must a plane be passed parallel to the base so the volume of that frustum will be 2/5 the volume of the given cone?

Question

This is from my solid mensuration class

Answer 1

If the original height of the cone is H, we have two similar cones formed by the intersecting plane. If the height of the smaller cone is h and base radius r, and the base radius of the frustum is R, then we can write p=h/H=r/R, where p is a ratio.

The ratio of the volume of the two cones is hr^2/HR^2 and the ratio of the volume of the frustum to the original cone is 2/5=1-hr^2/(HR^2)=1-p^3, so p^3=3/5 and p=(3/5)^(1/3).

If the height of the plane from the base is x, then p=h/(h+x) because H=h+x. Therefore x/h=(1-p)/p=0.291 approx.

So x=0.291h or about 29.1% of the height of the original cone.