Question

A hollow cone is cut by a plane parallel to the base and the upper portion is removed. If the CSA of the remainder is 8/9 of the CSA of the whole cone, find the ratio of the line segments into which altitude of the cone is divided by the plane.

Answer 1

CSA of frustum = 8/9
λ (R + r) (L-l) = 8/9λ RL

(R+r/R) (L-l/L) = 8/9
(1 + r/R) (1- l/L) = 8/9
(1 + h/H) (1- h/H) = 8/9
1- h²/H² = 8/9
h²/H²= 1/9
h/H = 1/3
H= 3h

Answer 2

Solution:-

❖Let OAB be the given hollow cone cut by the plane CD parallel to base AB and let OCD be removed .Then ,the remainder is the frustum CABD of the given cone

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Let

•OE= h units

•OF= H units

•OD= l units

•OB= L units

•ED= r units

•FB= R units

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In triangles OED and OFB, we have;

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$\sf \: \: \: \angle EOD=\angle FOB \: \: (common)$

$\: \: \: \sf \angle OED=\angle OFB = {90}^{0}$

⠀⠀

$\sf\therefore \triangle OED \sim \triangle OFB$

$\sf\implies \dfrac{OE}{OF}=\dfrac{OD}{OB}=\dfrac{ED}{FB}$

$\sf\implies \dfrac{h}{H}=\dfrac{l}{L}=\dfrac{r}{R}.....(1) \: \: \: \: |by \: thales \: theorum|$

Now, (curved surface area of the frustum CABD)=8/9 (curved surface area of the cone OAB)

⠀⠀ ⠀⠀ ⠀

⇨(curved surface area of the cone OCD)=(curved surface area of the cone OAB)-(curved surface area of the frustum CABD)

⠀⠀⠀=(curved surface area of the cone OAB)-8/9 (curved surface area of the cone OAB)

⠀⠀⠀=1/9 (curved surface area of the cone OAB)

⠀⠀

$\sf \implies \pi rl = \dfrac{1}{9} \pi RL$

$\sf\implies \bigg(\dfrac{r}{R} \bigg) \bigg( \dfrac{l}{L} \bigg) = \dfrac{1}{9}$

$\sf\implies \bigg(\dfrac{h}{H} \times \dfrac{h}{H} \bigg) = \dfrac{1}{9}$

$\sf \implies \dfrac{h}{H} = \dfrac{1}{3} \: \: \: |using \: (1)|$

$\sf \implies {H} = 3{h}$

Now, EF=(OF-OE)=(H-h)=(3h-h)=2h

$\sf\implies \dfrac{OE}{OF}=\dfrac{h}{2{h}} = \dfrac{1}{2}$

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Hence, (OE:EF)=1:2