Full text: From Aristarchus to Diophantus (Volume 2)

332 
HERON OF ALEXANDRIA 
view of their surprising character, referred to Archimedes by 
certain writers who give the traditional account of their 
origin. But whether they belong to Archimedes or another, 
it is necessary to give a sketch of these methods as well.’ 
The Book begins with generalities about figures all the 
sections of which parallel to the base are equal to the base 
and similarly situated, while the centres of the sections are on 
a straight line through the centre of the base, which may be 
either obliquely inclined or perpendicular to the base ; whether 
the said straight line (‘ the axis ’) is or is not perpendicular to 
the base, the volume is equal to the product of the area of the 
base and the 'perpendicular height of the top of the figure 
from the base. The term ‘ height ’ is thenceforward restricted 
to the length of the perpendicular from the top of the figure 
on the base. 
(a) Cone, cylinder, parallelepiped {prism), pyramid, and 
frustum. 
II. 1-7 deal with a cone, a cylinder, a ‘parallelepiped’ (the 
base of which is not restricted to the parallelogram but is in 
the illustration given a regular hexagon, so that the figure is 
more properly a prism with polygonal bases), a triangular 
prism, a pyramid with base of any form, a frustum of a 
triangular pyramid; the figures are in general oblique. 
(J3) Wedge-shaped solid ((3co/xtcn<os or crtprjvicrKos). 
II. 8 is a case which is perhaps worth giving. It is that of 
a rectilineal solid, the base of which is a rectangle ABCD and 
has opposite to it another rectangle EFGH, the sides of which 
are respectively parallel but not necessarily proportional to 
those of ABCD. Take AK equal to EF, and BL equal to FG. 
Bisect BK, CL in V, W, and draw KRPU, VQOM parallel to 
AD, and LQRN, WORT parallel to AB. Join FK, GR, LG, 
GU, HN. 
Then the solid is divided into (1) the parallelepiped with 
AR, EG as opposite faces, (2) the prism with KL as base and 
FG as the opposite edge, (3) the prism with NU as base and 
GH as opposite edge, and (4) the pyramid with RLGU as base 
and G as vertex. Let h be the ‘height’ of the figure. Now
	        
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