A cone of radius 8cm and hieght 12cm is divided by into two parts by plane through the mid point of its axis parallel to its base.find the ratio of the volume of two parts
The radius of the original cone is 8cm and the height is 12 cm.
Then, the volume of the original cone = 1\3(pi*r^2*h)=1\3*pi*(8)^2*(12)= 256pi
The cone is divided into two equal parts by drawing a plane through the mid points of its axis and parallel to the base.
Then, height of the top part will be half of the original height.
Then, the height of the small cone = {12}/{2}=6
And, the radius of the small cone = {8}/{2}=4
Volume of the small cone=1/3*pi*(4)^2*(6) =32pi
Therefore, volume of the frustum
=Volume of the original cone – Volume of the small cone
=256pi 32pi = 224pi
Compare the volume of the two part:
Volume of the frustum : Volume of the small cone = 224pi : 32pi
Volume of the frustum : Volume of the small cone=7:1
Answer : Required ratio of the two parts is 7:1
