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