A hollow cone is cut by a plane parallel to the base and the upper portion is removed. If the curved surface area of the remainder is 8/9th of the curved surface of the whole cone, find the ratio of the line segments into which the cone's altitude is divided by the plane.
Let R, H and L be the radius, height and slant height of the original cone respectively and r, h and l be the radius, height and slant height of the smaller cone respectively.
In ∆OAB and ∆OCD,
∠OAB = ∠OCD (90°)
∠AOB = ∠COD (Common)
∴ ∆ OAB ≅ ∆OCD (AA Similarity)
Curved surface area of the smaller cone
= Curved surface area of cone – Curved surface area of the frustum
Thus, the cones altitude is divided in the ratio 1 : 2.
Let the Height of the larger cone = H
Let the height of the smaller cone = h
Let the radius of the Larger cone = R
Let the radius of the smaller circle = r
h/H = r/R = l/L
Curved Surface area of the frustum = (8/9) Curved surface area of the cone.
pi(R + r)(L – l) = (8/9)piRL
(1 + r/R)(1 – l/L) = (8/9)
(1 + h/H)(1 – h/H ) = (8/9)
On simplifying, we get h2/H2 = 1/9
Therefore h/H = 1/3
Therefore h/(H- h) = 1/2
thumps up plzzzz