How to find the height of a tetrahedron?

Question

A regular tetrahedron has edges of 5cm each. How high is it?

 

Please explain how to find the answer to this question. Thanks :)

Answer 1

In a regular tetrahedron, the height will be perpendicular to the base(and in it's center).

Think about the triangle formed by the height, a line drawn from where the height meets the base to one side, and then the altitude of that side of the tetrahedron.   This is a right triangle. 

 

The hypotenuse of our triangle is the altitude of one side.  The length of the altitude of any side can be found by looking at the right triangle it forms as part of the side.  We can find it's length using a 30-60-90 triangle or Pythagoras(a^2+b^2=c^2). In this case the hypotenuse would be 5cm, the side on the bottom would be 1/2 of the 5cm or 2.5 cm.  Can you find the third side?  I would call that length L.

 

Now we have one side of our triangle that includes the height of the tetrahedron (the hypotenuse).

The side on the bottom should be 1/2 of the altitude of a side or 1/2 L.

Correction: The side on the bottom is not 1/2L - the center is in the center of the base of the triangle, but that is not at 1/2 the altitude.  Using a 30-60-90 triangle formed by connecting the base of the height to a corner of the base and to the midpoint of a side of the base (pictures would be so much easier), you know 1/2 of the side is 5/2 so you can use tan 30 *5/2=√3/3 * 5/2=5√3/6

 

We have 2 sides of our triangle now.  Because it is a right triangle you can use Pythagoras again to get the 3rd side, which is the height we were looking. L^2 = (1/2L)^2+H^2.  Or H^2=L^2-(1/2L)^2.

Correction: H^2=L^2-(5√3/6)^2

 

In this case L^2=5^2-(5/2)^2

L^2=75/4

L=5√3/2

h^2=75/4-75/16   correction: H^2=75/4-75/36=50/3, H=5√6/3

h^2=225/16 ignore this now

h=15/4 ignore - see corrections

 

 

 

 

Answer 2

the formula is height h = (√6/3)a where a = length of an edge

 

h = (5√6)/3

 

How do we get this answer. We will form two right triangles. The first one lies flat in the base of the pyramid with one leg = 1/2 the edge and one leg drawn from the midpoint of the edge to the center of the triangular base. The hypotenuse is drawn from the corner(vertex) to the center. The angle at the vertex is 30 degrees so this is a 30-60-90 triangle.

 

The hypotenuse is x = (1/2)a / cos 30 = √3a/3

 

No we draw another right angle with the hypotenuse above as one of the legs and the hypotenuse is the edge coming up from the same corner(vertex). Then the height of the pyramid h is the other leg.

 

Using the Pythagorean theorem (√3a/3)² + h² = a²

 

a²/3 + h² = a²

h² = 2a²/3

h = √(2a²/3) = √2a/√3 = √2√3a / √3√3 = √6a/3

 

So in this example a = 5 and we again have (5√6)/3