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Question
The height of a cone is 30 cm. From its topside a small cone is cut by a plane parallel to its base. If volume of smaller cone is `1/27` of the given cone, then at what height it is cut from its base?
Solution
Let the radius and height of the bigger cone be R and H, respectively.
Given: H = 30 cm
Let the radius and height of the smaller cone be r and h, respectively.
Now, in ∆AFE and ∆AGC,
∠AEF = ∠ACG (Corresponding angles)
∠AFE = ∠AGC (90° each)
∴ ∆AFE~∆AGC (AA similarity)
`=> (AF)/(AG) = (FE)/(GC)`
`=> h/H = r/R` ....(1)
it is given that
Volume of the smaller cone = `1/27 xx "Volume of the bigger cone"`
`=> 1/3 pir^2h = 1/27 xx 1/3 piR^2H`
`=>(r/R)^2 xx h/H = 1/27`
`=> (h/H)^2 xx h/H = 1/27` [Using (1)]
`=> (h/H)^3 = 1/27`
`=> h/H = 1/3`
`:. h = 1/3 xx H = 1/3 xx 30` = 10 cm
Now
FG = AG − AF = 30 cm − 10 cm = 20 cm
Thus, the small cone is cut at the height of 20 cm from the base.
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