# 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.

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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/H= 1/9

Therefore h/H = 1/3

Therefore h/(H- h) = 1/2

thumps up plzzzz

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