Math Is Fun Forum

MoyMarian

Member

Registered: 2009-01-10

Posts: 29

Cone II

A cone of altitude h is to be cut into three parts of equal weight. How far from the vertex must the cuts parallel to the base be made?

JurassicPlank1

Guest

Re: Cone II

equal weight = equal volume

The volume of a Cone is:

                    (1/3)*PI*R²*H

      Radius of base = R
      Height = H

Lets make H = 1 and R = 1; so H = R = 1

Now we have the volume as: (1/3)*PI

To get 3 equal volumes divide this result by 3

hence (1/3)*(1/3)*PI = (1/9)*PI

Lets call the distance from the top of the cone to the bottom of the first segment 'h' and the radius 'r' so that:

          (1/3)*PI*r²h = (1/9)*PI = (1/3)*(1/3)*PI      councel like terms

          r²h = (1/3)

since H = R then h = r

          h³ = (1/3)

          h =  0.693 (since H = 1 this is 69.3% of the height)

Lets call the distance from the top of the cone to the bottom of the second segment 'h' and the radius 'r', so that
the first volume + the second volume = 2/3 of the total volume.

          (1/3)*PI*r²h = (2/3)*(1/3)*PI

          r²h = (2/3)

Since H = R then h = r

          h³ = 2/3

          h =  0.874 H (since H is uinty)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Cone II

@MoyMarian

If you are confused by JurassicPlank1’s inconsistent and misleading use of symbols in the above solution, I don’t blame you. Not all of JurassicPlank1’s solutions are reliable anyway. For example, JurassicPlank1’s solution to this problem

http://www.mathisfunforum.com/viewtopic.php?id=11551

is incorrect.

Copy JurassicPlank1 at your peril!