A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921

iJ
MEASUREMENT OF SOLIDS 335
describing circle and the length of the path of its centre.
For, he says, since 1 4 is a radius (of the path of the centre),
28 is its diameter and 88 its circumference.
c
If then the tore
be straightened out and made into a cylinder, it will have 88
for its length, and the diameter of
the base of the cylinder is
12; so that the solid content of the cylinder is, as we have
seen, '
9956f (= 88 .
.
144K
(e) The tivo special solids of Archimedes s ' Method '.
Chaps. 14,
15 give the measurement of the two remarkable
solids of Archimedes's Method, following Archimedes's results.
(() The Jive regular solids.
In chaps. 16-18 Heron measures the content of the five
regular solids after the cube. He has of course in each case
to find the perpendicular from the centre of the circumscribing
sphere on any face. Let p be this perpendicular, a the
edge of the solid, r the radius of the circle circumscribing any
face. Then (1) for the tetrahedron
a 2 = 3r 2 ,
p = 2 a 2 — \a 2 = §a 2 .
(2) In the case of the octahedron, which is the sum of two
equal pyramids on a square base, the content is one-third
of that base multiplied by the diagonal of the figure,
i.e. J .a2 , a/2 a or J a/2, a3 ;
in the case taken a = 7, and
Heron takes 10 as an approximation to \/(2 . 7 2 ) or a/98, the
result being J. 10.49 or 163|. (3) In the case of the icosahedron
Heron merely says that
p
:
a = 93 : 127 (the real value of the ratio is \ / 7 + 3 n/ 5 \ .
(4) In the case of the dodecahedron, Heron says that
~ n ali
• , /25 + 11 \/5 . .„ / .
p:a = 9 : 8 (the true value is -
J
/ -> and, if v5 is
put equal to J, Heron's ratio is readily obtained).
Book II ends with an allusion to the method attributed to
Archimedes for measuring the contents of irregular bodies by
immersing them in water and measuring the amount of fluid
displaced.
336 HERON OF ALEXANDRIA
Book III. Divisions of figures.
This book has much in common with Euclid's book On divisions
(of figures), the problem being to divide various figures,
plane or solid, by a straight line or plane into parts having
a given ratio.
In III. 1-3 a triangle is divided into two parts
in a given ratio by a straight line (1) passing through a vertex,
(2) parallel to a side, (3) through any point on a side.
III. 4 is worth description :
'
Given a triangle ABC, to cut
out of it a triangle DEF (where D, E, F are points on the
sides respectively) given in magnitude and such that the
triangles AEF, BFD, GET) may be equal in area.' Heron
assumes that, if D, E, F divide the sides so that
AF: FB = BD: DC = CE: EA,
the latter three triangles are equal in area.
He then has to find the value of
each of the three ratios which will
result in
given area.
Join AD.
the triangle DEF having a
Since BD:CD = CE-.EA,
BC:CD= CA-.AE,
and AABC:AADC=AADC:AADE.
Also AABC: AABD = AADC: AEDC.
But (since the area of the triangle DEF is given) AEDC is
given, as well as AABC. Therefore AABD x A A DC is given.
Therefore, if AH be perpendicular to BC,
AH 2 .BD.DC is given;
therefore BD . DC is given, and, since BC is given, D is given
in position (we have to apply to BC a rectangle equal to
BD . DC and falling short by a square).
As an example Heron takes AB =13, BC =14, CA = 15,
ADEF = 24. AABC is then 84, and AH = 12.
Thus AEDC= 20, and AH 2 . BD. DC = 4 . 84
. 20 = 6720;
therefore BD .DC = 6720/144 or 46| (the text omits the §).
Therefore, says Heron, BD — 8 approximately. For 8 we