( 338 ) 
p p? 
D an area D" equal to — 
q q (pg)? 
Q. So the volume of this pyramid 
is equal to 
1 B 1 2g— 
Lj Gee 
3 2 pq 3 2¢q 
In the same manner we find that the volume of a tetrahedron 
of the second group can be represented by 
1 2 1 2p— 
dee 
3 2 pq 3 2p 
Moreover we have S D'+ D"+ > D"=D"; so the volume of 
the entire prismoid is given by the formula 
1 2p — 2g — Pip 
I=? En per 
6 q pg q 
1 
For p=q=1 we reobtain the result J = En h(P 4 QH4M), 
as it ought to be. From the two formulae we deduce 
PADD =9(9—P)P+p(p—)Q+4pg - - (2) 
For p=1, g=2 we find the remarkably simple relation 
l 1 
I= MOED) ee ee 
Still in another manner the volume can be expressed by means 
of two parallel sections. By interchanging p and q in (1) we get 
/ q 
lag er PPH (pt) p+0°D,- 
Pd P 
By addition of this equation to (1) we find 
h ú 
lago P+O+ Pt DHB] @ 
For p=//3+1 and g =1/3 — 1 this relation gives finally 
1 V3+1 (/s=1 
FS hk D tS cia re ee a RAS 
2 ( Vs v ew ) 
ee a ae re 
