Question

A hollow cone is cut by a plane parallel to the base and then the upper portion is removed. so now If the csa of 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.

Answer 1

 Let the:-

⭕️ Height of the larger cone = H

⭕️Height of the smaller cone = h

⭕️ radius of the Larger cone = R

⭕️radius of the smaller circle = r

 ⇒ h/H = r/R = l/L


It's given that the :-

⭕️CSA of the frustum = (8/9) Curved surface area of the cone.

 ⇒ π (R + r) (L – l) = (8/9) × π × R × L

 ⇒ (1 + r/R) (1 – l/L) = (8/9)

 ⇒ (1 + h/H) (1 – h/H ) = (8/9)


⭕️Simplifying, we get h²/H² = 1/9

 ∴ h/H = 1/3

 ∴ h/(H- h) = 1/2


$\tiny\bf\mathfrak \pink{ thank\: you ヅ} \:$

Answer 2

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