110 
Mr. Whewell on calculating ^ 
§ 4. Direct symmetrical Tetrahedron and rectangular Octahedron. 
so. Axy Zy fig. 13, be a tetrahedron, and let all its 
edges be bisected, and the bisections joined by lines drawn 
in the faces. We shall thus have an octahedron DEFGHK. 
If we consider EFHK as the common base of the two pyra- 
mids of which the octahedron is composed, when EFHK is 
a rectangle, the octahedron is called rectangular ; and when 
EFHK is a square, the octahedron is called square. 
Let EFHK be a rectangle, the octahedron being a right 
one. Then all the faces of the octahedron will be isosceles 
triangles, of which DEF, DHK, GFE, GHK will be equal 
to each other, and the other four also equal to each other. 
Also, it is easily seen that the triangle Ay z has its sides 
double of those of EFG, and is similar to it ; and similarly 
xy z has its sides double of KHG. Therefore the two tri- 
angles Ay Zy xy z are both isosceles, [y z being the base,) 
and are equal in every respect ; and similarly y Ax and z Ax 
are isosceles triangles equal in every respect. 
Hence the solid angles at y and z are equal in every re- 
spect, and also those at A and x. And a plane passing 
through A x and through the middle oi y z would divide the 
tetrahedron symmetrically into two equal portions. Hence 
we have called this the direct symmetrical tetrahedron. 
We may suppose the solid angle A to be filled with paral- 
lelepipeds, the planes of which are parallel to the planes 
Axy, AxZy AyZy in the same manner as the solid angle 
A, fig. 1. And by removing these parallelepipeds according 
to any law, as in fig. 1 , we obtain a secondary plane, of 
which the symbol and the equation may be known from the 
law. 
