Question

A hollow cone is cut by a plane parallel to its base and the upper portion is removed. If csa of the remaining portion is 8/9 of that of the cone, find tge ratio in which the linesegments into which the cone's altitude is divided

Answer 1

Answer:

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Step-by-step explanation:

Let R is the radius, H is the height and L is the slant height of the original cone and

let r is the radius, h is the height and l is the slant height of the smaller cone respectively.

Now in ΔOAB and ΔOCD,

∠OAB = ∠OCD  {each 90}

∠AOB = ∠COD {common}

So, by AA similarity,

ΔOAB ≅ ΔOCD

=> OB/OD = AB/CD = OA/OC

=> l/L = r/R = h/H

Now, curved surface area of the smaller cone = curved surface area of the cone - curved surface area of the frustum

=> curved surface area of the smaller cone = (1 - 8/9) * curved surface area of the cone

=> curved surface area of the smaller cone = (1/9) * curved surface area of the cone

=> curved surface area of the smaller cone/curved surface area of the cone = 1/9

=> πrl/πRL = 1/9

=> rl/RL = 1/9

=> (r/R)*(l/L) = 1/9

=> (h/H)*(h/H) = 1/9    {using equation 1}

=> (h/H)2 = 1/9

=> (h/H) = 1/3

=> h = H/3

Now, OA/AC = h/(h - h)

=> OA/AC = (H/3)/(H - H/3)

=> OA/AC = (H/3)/(2H/3)

=> OA/AC = 1/2

=> OA : AC = 1 : 2

So, the cones altitude is divided in the ratio 1 : 2