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**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:

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 πr^{2}h

Volume of first part = 1/3 π(4)^{2}6 = 32π

For second part, i.e. Frustum

Bottom radius, r_{1} = MN = 8 cm

Top Radius, r_{2} = PD = 4 cm

Height, h = PM = 6 cm

We know that,

Volume of frustum of a cone = 1/3 πh(r_{1}^{2} + r_{2}^{2} + r_{1}r_{2}) , where, h = height, r_{1} and r_{2} are radii, (r_{1} > r_{2})

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**

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