Problem: Find the volume of a regular tetrahedron with side a.
a
a
a
a
a
a
Solution: First some preliminaries.
An equilateral triangle with side s has height
3
3
k = ------- s and area A = ------- s 2 .
2
4
s
k
--s2
Also, the center of the triangle, C, (where the 3 medians
meet) is a trisection point of the medians, so that
k1
k = k1 + k2
2
3
and k 1 = --- k = ------- s
3
3
C
k2
1
3
and k 2 = --- k = ------- s .
3
6
In a regular tetrahedron with side s, let P and Q be vertices,
and let C be the center of the face (triangle) opposite P.
ΔPCQ is a right triangle with PQ = s and
P
3
QC = k 1 = ------- s . Then the height, h, of the tetrahedron is
3
h =
3 2
s 2 – ⎛ ------- s⎞ =
⎝ 3 ⎠
2--- 2
6
s = ------- s .
3
3
s
h
6
And solving for s, s = ------- h .
2
Then the area of one face of a regular tetrahedron
can be written in terms of the height of the tetrahedron
3
3 6 2
3 3
A = ------- s 2 = ------- ⎛ ------- h⎞ = ---------- h 2 .
⎝
⎠
4
4 2
8
Q
C
To find the volume of a regular tetrahedron with side a, orient the tetrahedron so that one vertex is at the origin,
O, and the positive x-axis goes through the center, C, of the opposite face.
O
C
x
6
6
OC is the height of the tetrahedron so OC = ------- a . Let x be a point on the x-axis in the interval [0,------- a] and
3
3
slice the tetrahedron at x perpendicular to the x-axis.
x
O
C
x
The section of the original tetrahedron that lies between the origin, O, and the slice at x is another regular
tetrahedron with height x, so the cross-section at x is an equilateral triangle and the cross-sectional area at x, A(x),
is
3 3
A ( x ) = ---------- x 2 .
8
Then the volume, V, of a regular tetrahedron with side a is
V =
∫
------6- a
3 3
------6- a
3
0
0
3
---------3- x 2 dx = 3---------3- x----8
8 3
2
= ------- a 3 .
12