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.
Let ORN be the cone, when this is cone is divided into two parts by a plane through the mid-point of its axis parallel to its base, the upper and lower parts obtained are cone and a frustum respectively.
Given,
Height of cone = OM = 12 cm
As, the cone is divided from mid-point, let P be the mid-point of cone, then
OP = PM = 6 cm
Now, In △OPD and △OMN
∠POD = ∠POD [Common]
∠OPD = ∠OMN [Both 90°]
So, By Angle-Angle similarity criterion
△OPD ~ △OMN
And
[Similar triangles have corresponding sides in equal ratio]
[MN = 8 cm = radius of base of cone]
Now, For First part i.e. cone
Base Radius, r = PD = 4 cm
Height, h = OP = 6 cm
As we know, volume of cone for radius r and height is
Now, For second part, i.e. Frustum
Bottom radius, r1 = MN = 8 cm
Top Radius, r2 = PD = 4 cm
Height, h = PM = 6 cm
As we know
Volume of frustum of a cone
Where, h = height, r1 and r2 are radii
(r1 > r2)
So,
Volume of first part : Volume of second part = 32π : 224π = 1 : 7