The Tetrahedron
Surface area
The regular tetrahedron has four equivalent equilateral triangles as faces. The interior angle of the triangle is `pi`, and so the each of the interior angles in an equilateral triangle is `pi/3`
| `L^2=d^2+L^2/4` `d=sqrt(3)/2L` |
The area of the right angle triangle is `A=1/2bh`, and so
`A_(right)=1/2(L/2sqrt(3)/2L)=L^2sqrt(3)/8`
Which is half the area of the equilateral triangle, and so the area of one face of the tetrahedron is `A_(face)=L^2sqrt(3)/4` and the total surface area is `A_("total")=L^2sqrt(3)`
Calculation of the volume
The volume can be found from calculus by integrating the area of a face as a function of z with the height. Ie.,
`V=int_0^hA(z)dz`.
To pursue this approach, it is necessary to calculate the height with one face of the tetrahedron confined to the xy-plane.
To do this, we find the orthocentre of the base, equilateral triangle.
|
`c =L/(2cos(pi/6))=L/sqrt(3)` `b=1/2Ltan(pi/6)` |
We can now construct a right angle triangle across the base and rising from the orthocentre, which has an altitude equal the height of the tetrahedron
`h^2=L^2-L^2/3` `h=sqrt(2/3)L` |
Now we must express the area of a face as a function of z, and we note that the side length, L, varies with z,
`A(z)=L(z)^2sqrt(3)/4`
| The two triangles are similar, which means to say that
`h/L=z/(L(z))` Hence `L(z)=zL/h` |
So the height-dependant face area is
`A(z)=L^2/h^2sqrt(3)/4z^2`
We now can integrate this function with respect to h,
`V=L^2/h^2sqrt(3)/4int_0^hz^2dz=L^2/h^2sqrt(3)/12[z^3]_0^h=L^2sqrt(3)/12h`
We replace the height with its definition in terms of L,
`V=L^3sqrt(3)/12sqrt(2)/sqrt(3)=L^3/12sqrt(2)=L^3/(6sqrt(2)`
The Tetrahedral bond angle
The tetrahedron has a special place in Chemistry, being the geometry of sp3 hybridised species
To investigate some of the properties of the tetrahedron, it is useful to orient one inside a cube
We note that the central atom, carbon in this case, is located in the absolute centre of the cube, where the sides of the cube being of length `a`, then the central atom is located at `(a/2,a/2,a/2)`.
| `d^2=a^2+a^2` `d=asqrt(2)` |
| `tantheta="opposite"/"adjacent"=(asqrt(2))/2 2/a=sqrt(2)` |
We note that the angle `theta` is one-half the tetrahedral bond angle, and so `theta_"tet"=2arctansqrt(2)=109.47`