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Height of Tetrahedral Pyramid


Date: 09/12/2001 at 23:07:17
From: Bob
Subject: Height of tetrahedral pyramid

Dr. Math:

I'm looking for a simple formula (and derivation) of the height of a 
tetrahedral pyramid with an equilateral triangle as a base.  The 
"sides" making up the pyramid are isosceles triangles with 45-degree 
angles - essentially the pyramid is a portion of a standard cube.  I 
realize that the volume of the pyramid is defined by V=bh/3, but is 
there a simple formula that relates the height, h, to the sides of 
the base?

Thanks.


Date: 09/13/2001 at 11:25:41
From: Doctor Rick
Subject: Re: Height of tetrahedral pyramid

Hi, Bob.

Yes, there is a simple formula, and I'll show you how to derive it 
yourself - in two different ways.

By mentioning the volume, you've given me an idea for an interesting 
way to find the height of this pyramid. Since it is the corner of a 
cube, we can regard one of the isosceles triangles as the base, and 
the other edge of the cube is perpendicular to the base, so it is the 
altitude. Thus the volume of the pyramid is

  V = (1/3)bh
    = (1/3)(s^2/2)s
    = s^3/6

What is s? It is the leg of an isosceles right triangle with 
hypotenuse equal to the side of the equilateral triangle, which I'll 
call a. Use the Pythagorean theorem to find s. Plug this expression 
for s into the equation for V to get an expression for V in terms 
of a.

Now, turning the pyramid back so the equilateral triangle is the base, 
you can find a second equation for V in terms of a and h. This will 
require finding the area of an equilateral triangle, which in turn 
involves finding the length of the altitude of the equilateral 
triangle. If you don't know this, you can find it using the fact that 
the altitude of an equilateral triangle is also its median, so it 
bisects the side it is perpendicular to. This allows you to use the 
Pythagorean theorem.

The volume must be the same however we calculate it, so you can set 
the two expressions for V equal to each other. Solve for h to get a 
formula for h in terms of a.

You can also calculate the altitude h as follows. Find the side length 
s, as above. Find the median (or altitude) of the equilateral 
triangle, as above. Due to symmetry, the altitude h meets the base at 
the centroid of the triangle, which divides each median in 1/3:2/3 
proportion. Therefore we have a right triangle consisting of a side of 
length s, the altitude, and 2/3 of a median of the base. You can use 
the Pythagorean theorem to find h, and you'll get the same answer as 
by the volume method.

I'm not sure whether you really needed the formula for the altitude 
given the base side, or you just needed the volume of the pyramid, and 
you thought that this required finding the altitude. If you only 
needed the volume, then I've given you more than you needed, but I 
hope it's interesting anyway.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   


Date: 09/13/2001 at 12:22:44
From: Doctor Jubal
Subject: Re: Height of tetrahedral pyramid

Hi Bob,

The height of a pyramid depends on its shape, so unlike the volume 
formula V=bh/3 that works for all pyramids, there's no succinct 
formula that will give the height of any pyramid. However, you can 
derive formulas specific to pyramids of specific shapes, including 
your pyramid.

To derive a formula for the height of your pyramid, all you need to do 
is apply the Pythagorean theorem again and again.

Consider one of the triangles we could expose if we made a vertical 
slice cutting the pyramid in half:

    A              
       
    |\
    | \
    |  \
    |   \ s
  h |    \
    |     \
    |      \
    |_      \
    |_|______\
   C     c     B
 
A is the apex of the pyramid, B is one of the corners of the base, and 
C is the center of the equilateral triangle that is the base. That 
makes s the length of one of the non-base edges of the pyramid's 
sides, h the pyramid's height, and c the distance from the center of 
the base to one of its vertices. If we can find s and c, we can use 
the Pythagorean theorem to find h. 

Let's start with s. The sides of the pyramid are isoceles right 
triangles. I'll use a for the length of an edge along the base of the 
pyramid. a is the hypoteneuse of one of the side faces, and s is the 
length of the other two sides, so by the Pythagorean theorem:  

  s^2 + s^2 = a^2

        s^2 = (a^2)/2

          s = a * sqrt(1/2) 

Now for c. The base of the pyramid is an equilateral triangle:

                B
                ^
               /|\
              / | \
             /  |c \
          a /   |   \
           /    |C   \
          /  __/|     \
         /__/c  |i     \
        //______|_______\
       D        H 
           a/2

Points B and C are the same as before, H is the midpoint of the side
opposite B, D is another corner of the triangle. c is the length of 
segment BC, as before. Since the triangle is equilateral, c is also 
the length of CD. i is the length of CH. Applying the Pythagorean 
theorem to triangle BDH:

  (a/2)^2 + (c + i)^2 = a^2

  (a^2)/4 + (c + i)^2 = a^2

            (c + i)^2 = (3/4)a^2
        
              (c + i) = a * sqrt(3/4)

      c^2 + 2ic + i^2 = (3/4)a^2
                
We can get rid of i^2 by applying the Pythagorean theorem to triangle 
CDH:

  i^2 + (a/2)^2 = c^2

            i^2 = c^2 - (a^2)/4

Substituting this back into the previous expression

  2*c^2 + 2ic - (a^2)/4 = (3/4)a^2

            2*c^2 + 2ic = a^2              

           2c * (c + i) = a^2

   2c * [a * sqrt(3/4)] = a^2

                      c = a/(2sqrt(3/4)) = a/sqrt(3)

So now we have c and s, and one final application the Pythagorean 
theorem gives us h.

  h^2 = s^2 - c^2

  h^2 = (a^2)/2 - (a^2)/3

  h^2 = (a^2)/6

    h = a/sqrt(6)

Does this help?  Write back if you'd like to talk about this some
more, or if you have any other questions.

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

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