# NCERT Exemplar Solutions Class 10 Mathematics Solutions for Surface Areas and Volumes - Exercise 12.3 in Chapter 12 - Surface Areas and Volumes

Question 2 Surface Areas and Volumes - Exercise 12.3

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.

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

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