26 
TRANSACTIONS OF THE TEXAS ACADEMY OF SCIENCE. 
But S—B S = GHPQ+QNEH —ONEF+POFG; 
B t -B s = ABGH+ADEH- CDEF+BCFG; 
m-b 2 =jkgii+jmeh—LMEF+KLFG. 
Combining with these the relation (2) and the three corresponding to 
it gives 
S—B 3 = (B 1 — B 2 )(a 2 —a 1 )a 8 /a s + (M—B 2 )4a 1 a 2 /a s , 
or, since a=a 1 -\-a a , 
a s S=(a 2 —a 1 a 2 )B 1 -f-(a®—a 1 a 2 )B 2 +4a 1 a 2 M . . . (3); 
or, also, 
S=B 1 —(3B t +B 2 —4M)a 1 /a+(2B 1 +2B 8 —4M)a*/a 8 , 
which determination clearly holds also for every other prismoid. 
Therefore in the prismoid the area of a section parallel to the bases is 
a linear function of the bases and the midsection, and a quadratic func¬ 
tion of its distance from a base. 
Plotting on a plane the altitudes a x as abscissas and the areas S as 
ordinates in a system of rectangular coordinates, the extremity of the 
ordinate has for locus a parabola with axis parallel to the axis of or¬ 
dinates. (Fig. 2.) [Not reproduced in this translation.] 
This parabola may meet the axis of abscissas at no point, or touch it at 
one point or cut it in two points. 
The first is the case when no section is null, as, say, in the hyperboloid 
of one nappe. 
The second is the case when only one section is null, as in the pyr¬ 
amid. 
Finally the third occurs if two sections are null, as is the case for the 
tetrahedron when the sections are taken parallel to two opposite edges. 
In this case the sections lying between the two vanishing sections are 
positive, if those lying without are taken negative, and inversely. How¬ 
ever, such negative surfaces will not be further considered here, as, 
indeed, they present no difficulties. 
The volume of the prismoid between the bases B x and B 2 can be found 
by various methods. It is, for example, also given by the surface cov¬ 
ered by the ordinates of the just mentioned parabola which lie between 
B, and B„. 
Very elegant is the deduction by Professor Steiner, who divides the 
prismoid into pyramids from any point in the mid section. 
It can be deduced from the last given value of S by multiplying it by 
dx and integrating with respect to x between the limits o and a. 
This gives easily Y=-|a(B 1 +B 2 +4M), as is known. To express the 
volume, instead of by M, by any chosen section S distant a x , a 2 from 
B„ B„ (3) gives 
4M=B 1 +B,+2S+(S-B 1 )o,/o I +(S—B,)a,/o,, 
