Weekly challenge 30 solution: The volume of a tetrahedron
There are several ways of getting to the volume of a tetrahedron, although
most proceed along similar lines. Here we give two different approaches, the
first given by Diamor Marke (who has created some awesome diagrams in
Word to help us follow his ideas), and the second by Alma Crasmaru.
Diamor’s solution
base of tetrahedron
𝑝
𝑝√3
2
60°
𝑝
2
First we need to find an expression for the area of the base of the tetrahedron, which is an
equilateral triangle. The area of the base is therefore:
12×p×p32
=p234
Then, we need to find the perpendicular height of the tetrahedron, which, for regular tetrahedral, is
the distance from the centroid of the base to the apex.
base of tetrahedron
𝑝
60°
30°
Section of the base
of the tetrahedron
centroid of base
60°
30°
p2
This triangle is a 30 60 90 triangle, so the ratio of the side lengths is 1:3 :2, the “2” being the
hypotenuse. Since the p2 represents the second longest side, it is the “3” in the ratio. This makes
the length of the hypotenuse:
p2÷3×2
=p3
Edge of
tetrahedron
𝑝
ℎ
ℎ
𝑝
Centroid of base
The length of the perpendicular height of the tetrahedron is:
h=p2-p32
=p2-p23
=p23
The volume of any pyramid is:
=13×A×h, where A is the area of the base and h is the perpendicular height
We can use this to find the volume of the tetrahedron, which is:
13×p234×p23
=p332123
=p3212
√3
Alma’s Solution