The height of a right circular cone is 20 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume be 1/8 of the volume of the given cone, at what height above the base is the section made?
Let the cutting plane be passing through points B and C as shown
Height of cone = AD = H = 20 cm
Height of small cone which we get after cutting = AB = hc
Let ‘r’ be the radius of small cone ∴ we have BC = r
‘R’ be radius of original cone which is to be cut ∴ we have DE = R
From figure consider ΔABC and ΔADE
∠ABC = ∠ADE = 90˚
∠BAC = ∠DAE …(common angle)
as two angles are equal by AA criteria we can say that
ΔABC∼ΔADE
Let V1 be the volume of cone to be cut
Let V2 be the volume of small cone which we get after cutting
Volume of cone = (1/3)π(radius)2(height) cm3
∴ V1 = (1/3) × π × R2 × hc
∴ V2 = (1/3) × π × r2 × 20
Given is that the volume of small cone is (1/8) times the original cone
∴ V2 = (1/8) V1
∴ (1/3) × π × r2 × 20 = (1/8) × (1/3) × π × R2 × hc
Using equation (i) we get
∴ hc3 = 203/8 cm
∴ hc = 20/2 cm
∴ hc = 10 cm
But we have to find the height from base i.e. we have to find BD from figure
∴ 20 = BD + hc
∴ 20 = BD + 10
∴ BD = 10 cm
∴ 10 cm above base the section is made.