As per the inquiry,

Tallness of cone = OM = 12 cm

The cone is separated from mid-point.

Consequently, let the mid-point of cone = P

Operation = PM = 6 cm

From △OPD and △OMN

∠POD = ∠POD [Common]

∠OPD = ∠OMN [Both 90°]

Consequently, by the Angle-Angle likeness rule

We have,

△OPD ~ △OMN

Also,

Comparable triangles have relating sides in equivalent proportion,

Thus, we have,

PD/MN = OP/OM

PD/8 = 6/12

PD = 4cm

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

For First part for example cone

Base Radius, r = PD = 4 cm

Tallness, h = OP = 6 cm

We realize that,

Volume of cone for range r and tallness h, \[V\text{ }=\text{ }1/3\text{ }\pi r2h\]

Volume of initial segment \[=\text{ }1/3\text{ }\pi \left( 4 \right)26\text{ }=\text{ }32\pi \]

For second part, for example Frustum

Base sweep, r1 = MN = 8 cm

Top Radius, r2 = PD = 4 cm

Stature, h = PM = 6 cm

We realize that,

Volume of frustum of a cone\[=\text{ }1/3\text{ }\pi h\left( r12\text{ }+\text{ }r22\text{ }+\text{ }r1r2 \right)\] , where, h = stature, r1 and r2 are radii, (r1 > r2)

Volume of second part \[=\text{ }1/3\text{ }\pi \left( 6 \right)\left[ 82\text{ }+\text{ }42\text{ }+\text{ }8\left( 4 \right) \right]\]

\[=\text{ }2\pi \left( 112 \right)\text{ }=\text{ }224\pi \]

Hence, we get the proportion,

Volume of initial segment : Volume of second part = 32π : 224π = 1 : 7