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• Level: GCSE
• Subject: Maths
• Word count: 1275

# The Tetrahedron.

Extracts from this document...

Introduction

THE  TETRAHEDRON

Figure 1

Let each side of the tetrahedron (in blue) = s.
The tetrahedron has 6 sides, 4 faces and 4 vertices.

Here the base is marked out in gray: the triangle ABC.
From
The Equilateral Triangle  we know that:
Area ABC =  (\/¯3 / 4)s².

Now we need to get the height of the tetra = ED.
E
|\
|  \
|    \
----\
D      A
figure 2

From  The Equilateral Triangle we know that:
DA  = (1 / \/¯3)s.

Now that we have DA, we can find  ED, the height of the tetrahedron. We will call that  h.
h² =  ED = EA² - DA²  =  s² - (1/3)s² = (2/3)s²
=  ED = (\/¯2 / \/¯3)s

So V(tetra) = 1/3 * area ABC * height(tetra) = 1/3 *  (\/¯3 / 4)s² *  (\/¯2 / \/¯3)s,
V(tetra)  =   (\/¯2 / 12) s³  =    (1 / 6\/¯2 )s³      = 0.11785113s³

What is the surface area of the tetrahedron?  It is just the sum of the areas of its 4 faces.
We know from above that the area of a face is
(\/¯3 / 4)s².
So  the total surface area of the tetrahedron =
\/¯3 s².

We have found the volume of the tetra in relation to it's side.

Middle

Figure 4
So FC is the side of the tetrahedron, DC is the side of the cube, FD is the diagonal of the cube and the diameter of the sphere enclosing the cube. By Pythagorean Theorem, FC = \/¯2 * FG  and also FD²  = FC² + DC² . Assuming (temporarily) the side of the cube DC is 1, then FC = \/¯2 and
FD = \/¯(1 + 2) = \/¯3.
So FG(side of cube), FC(side of tetrahedron) , and FD(diameter of enclosing sphere)  have a 1, \/¯2, \/¯3 relationship.

We know OF = 1, so FD = 2. We also know FC / FD  = \/¯2 / \/¯3, or
FC =  (\/¯2 / \/¯3 ) * diameter of enclosing sphere.  The radius OF is 1/2 the diameter FD.
So FC = (\/¯2 / \/¯3) * 1/2 radius
FC = [(2\/¯2) / \/¯3] * radius.
Now we have the side of the tetrahedron in terms of the radius of the enclosing sphere.

So we write s =[(2\/¯2) / \/¯3] ror, s = 1.632993162r.
s  is also called the chord length.

What about the centroid of the tetrahedron?
The centroid is at O in Figure 3. The distance OF is the distance from O to any of the 4 vertices of the tetrahedron, which we have already seen is the radius (r) of the enclosing sphere = 1.

Conclusion

OJ = distance from centroid to middle of a face, in this case, face ACF,
OI = distance from centroid to middle of a side, in this case, the side FH.
The 4 faces are CFH, CFA,HFA, HCA as before. The centroid is at O. The middle of face CFA is J. The midpoint of the side of the tetra FH is at I.
We have already figured out OF = OC = OA = OH, the distance from the centroid to any vertex. This is, as you recall,  [\/¯3 / (2\/¯2)]s
Let's get OI first. Go back to Figure 5. The distance OI on that figure is the same as the distance OI in Figure 7.
OI² = OF² - IF² =  [\/¯3 / (2\/¯2)]² - 1/4 = 3/8 - 1/4 = 1/8.
So OI = [1 / (2\/¯2)]s.

It turns out that the triangle FJO in Figure 7 is right. This will enable us to get the distance OJ.
OF = [\/¯3 / (2\/¯2)]s.
From
The Equilateral Triangle  we know FJ = 1 / \/¯3.
So OJ² = OF² - FJ² =  [\/¯3 / (2\/¯2)]² - (1 / \/¯3)² = 3/8 - 1/3 = 9/24 - 8/24  = 1/24.
OJ = 1/\/¯24 = 1 / (2\/¯6).
OJ = [1 / (2\/¯6)]s.

To get a good idea of relative distances, re write OF  as [3 / (2\/¯6)]s and OI as [\/¯3 / (2\/¯6)]s
So the relationship between OJ, OI and OF is 1, \/¯3, 3.

Special characters:
\/¯  ° ¹ ² ³ × ½ ¼ Ø  \/¯(Ø² + 1)

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