Question

A cone of radius 8 cm and height 12 cm is divided into two parts by a plane through the mid-point of its axis parallel to its base. Find the ratio of the volumes of two parts.

Answer 1

According to the question,

Height of cone = OM = 12 cm

The cone is divided from mid-point.

Hence, let the mid-point of cone = P

OP = PM = 6 cm

From △OPD and △OMN

∠POD = ∠POD [Common]

∠OPD = ∠OMN [Both 90°]

Hence, by the Angle-Angle similarity criterion

We have,

△OPD ~ △OMN

And

Similar triangles have corresponding sides in equal ratio,

So, we have,

PD/MN = OP/OM

PD/8 = 6/12

PD = 4cm

[MN = 8 cm = radius of base of cone]

For First part i.e. cone

Base Radius, r = PD = 4 cm

Height, h = OP = 6 cm

We know that,

Volume of cone for radius r and height h, V = 1/3 πr²h

Volume of first part = 1/3 π(4)²6 = 32π

For second part, i.e. Frustum

Bottom radius, r₁ = MN = 8 cm

Top Radius, r₂ = PD = 4 cm

Height, h = PM = 6 cm

We know that,

Volume of frustum of a cone = 1/3 πh(r₁² + r₂² + r₁r₂) , where, h = height, r₁ and r₂ are radii, (r₁ > r₂₎

Volume of second part = 1/3 π(6)[8² + 4² + 8(4)]

= 2π(112) = 224π

Therefore, we get the ratio,

Volume of first part : Volume of second part = 32π : 224π = 1 : 7