Q. 174.2( 14 Votes )

A solid cone of base radius 10 cm is cut into two parts through the midpoint of its height, by a plane parallel to its base. Find the ratio of the volumes of the two parts of the cone.

Answer :

Let ‘H’ be the height of cone ‘R’ be the radius of base of cone.

R = 10 cm

When the cone is cut at midpoint of H by a plane parallel to its base we get a cone of height H/2 and a frustum also of height H/2

Let the radius of the base of the cone which we got after cutting and the radius of upper circle of frustum be ‘r’ as shown in figure

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


r = 5 cm

Let Vc be volume of the cone and Vf be the volume of frustum

Volume of cone = (1/3)π(radius)2(height) cm3

Vc = (1/3) × π × r2 × (H/2) cm3

= (1/3) × π × 52 × (H/2) cm3

= (1/3) × π × 25 × (H/2) cm3

Volume of frustum = (1/3)πh(R2 + r2 + Rr) cm3

Vf = (1/3) × π × (H/2)(102 + 52 + 10 × 5) cm3

= (1/3) × π × (H/2) × 175 cm3

The ratio of volumes of two parts after cutting = Vc:Vf = 1:7

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