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Volume of a Cone


Date: 5/9/96 at 19:4:26
From: Anonymous
Subject: Volume of a Cone

Hello Dr. Math, 

I was wondering if you could help me with a geometry problem that I 
have to do:

A coffee pot with a circular bottom tapers uniformly to a circular 
top having radius half that of the base. A mark halfway up (by height) 
says 2 cups. If the pot could be filled completely to the rim, how 
much coffee would it hold?

Any information on how to solve this problem would be greatly 
appreciated! Thanks.

Joe, 9th grade, Lake Braddock Secondary School.


Date: 12/11/96 at 01:02:47
From: Doctor Rob
Subject: Re: Volume of a Cone

Good problem!

Start by cutting a cross-section through the centers A and C of the 
circular top and bottom of the pot. Then extend the side lines DF and 
its counterpart until they meet at P. Connect the two centers A and C 
with another line, and extend it until it meets the extensions of the 
side lines at the same point P:

           P
          /|\
         / | \
        /  |  \
       /  A|   \
      /----|----\D  Top
     /     |     \
    /      |      \
   /      B|-------\E 2-Cup Line
  /        |        \
 /         |         \
-----------------------  Bottom
           C          F

Then CF = 2*AD, and AB = BC, so DE = EF, using similar or congruent
triangles. Now you can show that PC = 2*AC = 4*BC using similar 
triangles. You can compute the volume of the cones with vertex at P 
and bases with centers at A, B, and C, using the formula for the 
volume of a cone, V = (Pi*h*r^2)/3, where h is the height and r the 
radius of the base. The volume of the large cone V(L) minus the volume 
of the medium-sized cone V(M) will give the volume of 2 cups as marked 
on the pot. The volume of the large cone V(L) minus the volume of the 
small cone V(S) will be the volume you seek, call it x, measured in 
cups. The equation is:

   V(L) - V(S)   x
   ----------- = -
   V(L) - V(M)   2

which you will then solve for x.  All the lengths such as BC and CF 
from the diagram will cancel from the top and bottom of the fraction 
on the left in this equation, so they need not be known.

I hope this helps.  If you still need more help, write back and we'll 
give it another try.

-Doctor Rob,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

    
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry
Middle School Geometry
Middle School Higher-Dimensional Geometry

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