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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.
Solution
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
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