Question

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Question : 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 $\dfrac{8}{9}$ of the curved surface area of the whole cone, then find the ratio of the line segment into which cone Altitude is divided by the plane.



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Answer 1

Answer:

Step-by-step explanation:

C.S.A  of frustum = 8/9    C.S.A of cone

OA/AB = h / (H-h) =?

=> π (R+r)(L-l) = 8/9 × π R L

( π will be cancle )

(L- l / L ) = ( R / r- R ) 8/9     = D

l / L = r /R -----------------------(1) = AO/OC = h/H

(1/l-L) = (R/r+R) 8/9

(1-l/L) = 1 / (r/R+1) 8/9

(1-h/H) =1/ (1+h/H) 8/9

(1-h/H)(1+h/H) = 8/9

1 -  h^2/H^2 = 8/9

1 - 8/9 = h^2/H^2

1/9=h^2/H^2

h/H=1/3

h/H-h = h/H/(1-h/H) =1/3/1-1/3 = 1/2 => ration

OA/AB =  h / (H-h) = 1/2

the ration is 1/2

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Answer 2

Answer:

Assume that the ratio of the altitude of the bigger and the smaller cone be k:1.

Let R and r be the radii of the bigger and the smaller cone respectively.

Let H and h be the height of the bigger and the smaller cone respectively.

Consider the similar triangles △ AGC & △ AFE ,

By the property of similarity, we have,

AF/AG=FE/GC

H/h=R/r=k/1, where k is some constant.

Curved surface area of bigger cone = πRL, where L is the slant height of the bigger cone.

Curved surface area of smaller cone = πrl, where l is the slant height of the smaller cone.

Again by the property of similarity, we have,

L/l=R/r=k/1

Given that the ratio of the curved surface area of the frustum of the cone to the whole cone is 8/9.

The ratio of the curved surface area of the smaller cone to the bigger cone is 1/9.

πRL/πrl = 1/k2 = 91

k=3

H/h=31

Therefore, h/H−h = 1/3−1 = 1/2

Hence, the required ratio is 1:2.