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 the mid-point.

Hence, let the mid-point of cone = P

OP = PM = 6 cm

From ΔOPD and ΔOMN

∠POD = ∠POD    ...[Common]

∠OPD = ∠OMN   ...[Both 90°]

Hence, by the Angle-Angle similarity criterion

We have,

ΔOPD ~ ΔOMN

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 π"r"^2"h"` 

Volume of first part = `1/3 π(4)^(2)6` = 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 π"h"("r"_1^2 + "r"_2^2 + "r"_1"r"_2)`, where, h = height, r₁ and r₂ are radii, (r₁ > r₂)

Volume of second part = `1/3 π(6)[8^2 + 4^2 + 8(4)]` 

= 2π(112)

= 224π

Therefore, we get the ratio,

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