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Volume of a Tetrahedron

Date: 02/27/2003 at 15:48:16
From: Haze
Subject: Tetrahedron

How do I find the volume of a tetrahedron?


Date: 03/02/2003 at 18:43:35
From: Doctor Ian
Subject: Re: Tetrahedron

Hi Haze,

The most straightforward way to find it is to apply the Pythagorean
theorem several times. 

Here is an equilateral triangle, with vertices A, B, and C; center D;
and midpoint of base E.  It's just like the one at the base of your
tetrahedron. 
         
             A
            . .
           . . .
          .  .  .
         .   .   .  
        .    .    .
       .     D     .
      .   .     .   .
     .  .          . .
    C . . . .E. . . . B

For brevity, we'll use 

    a = AB = BC = CA
    
    h = AE
    
    d = DE
    
    r = CD
    
The Pythagorean theorem tells us that

    h^2 + (a/2)^2 = a^2
    
              h^2 = a^2 - (a/2)^2
              
              h^2 = 3a^2/4
                 
                h = a sqrt(3)/2 

So the area of the whole triangle is 

    area = (1/2) * base * height              

         = (1/2) * a * a sqrt(3)/2
              
         = a^2 sqrt(3)/4
              
The area of triangle DBC is 1/3 the area of the whole triangle, so 

       (1/3) a^2 sqrt(3)/4 = (1/2) * base * height
  
                           = (1/2) a d
                      
  (2/a)(1/3) a^2 sqrt(3)/4 = d                      

               a sqrt(3)/6 = d
               

The Pythagorean theorem tells us that 

        r^2 = (a/2)^2 + d^2               
        
            = (a/2)^2 + [a sqrt(3)/6]^2
            
            = a^2/4 + 3a^2/36
            
            = a^2/4 + a^2/12
            
            = (3a^2 + a^2)/12
            
            = 4a^2/12
            
            = a^2/3
            
          r = a/sqrt(3)
          
So, if we make a tetrahedron from four of these triangles, and drop an
altitude from the apex of the tetrahedron to the center of the base,
we get a right triangle hypotenuse a, base r, and height H.  The
Pythagorean theorem tells us that

     r^2 + H^2 = a^2
     
           H^2 = a^2 - r^2
           
               = a^2 - a^2/3
               
               = 2a^2/3
               
             H = a sqrt(2/3)
             
Since the volume of any pyramid is 1/3 the area of the base times
the height, 

   Volume of a Pyramid
   http://mathforum.org/library/drmath/view/55041.html 

the volume is

    volume = (1/3) * area of base * height
    
           = (1/3) * [a^2 sqrt(3)/4] * [a sqrt(2/3)]
           
           = a^3 (1/3) [sqrt(3)/4] [sqrt(2/3)]
           
           = a^3 sqrt(2)/12

Happily, this agrees with the formula in our Dr. Math Geometric 
Formulas FAQ:

   Pyramid & Frustum Formulas   
   http://mathforum.org/dr.math/faq/formulas/faq.pyramid.html 

I hope this helps!

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Polyhedra

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