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

THE REGULAR POLYGONS 329
presumably Hipparehus's Table) is appealed to. If AB be the
side (a) of an enneagon or hendecagon inscribed in a circle, AC
the diameter through A, we are told that the Table of Chords
gives § and ^ as the respective approximate values of the
ratio AB /AC. The angles subtended at the centre by the
side AB are 40° and 32 X
8
T °
respectively,
and Ptolemy's Table gives,
as the chords subtended by angles of
40° and 33° respectively, 41? 2' 33"
and 34P 4' 55" (expressed in
120th
parts of the diameter) ; Heron's
figures correspond to 40^ and 33^
36' respectively. For the enneagon
AG = 9AB 2 2 , whence BC 2 =, SAB 2
or approximately 2££-AB 2 ,
and
BC = --§-a\ therefore (area of
enneagon)^ . AABC^-%
1 ^. For
the hendecagon AC = 2 %5
-AB 2 and BC = 2 ^f-AB 2 , so that
2
BC = %r a >
and area of hendecagon = .
^ AABC = --f-a
.
formula for the ratio between the side of any
An ancient
regular polygon and the diameter of the circumscribing circle
is preserved in Geepon. 147 sq. (— Pseudo-Dioph. 23-41),
namely d n
— n—. Now the ratio na n /d n tends to it as the
o
number (
n) of sides increases, and the formula indicates a time
when it was generally taken as = 3.
(e)
The Circle.
Coming to the circle (Metrica I. 26) Heron uses Archimedes's
value for 7r, namely - 2 2
T-, making the circumference of
a circle *f-r and the area ^d 2 ,
where r is the radius and d the
diameter. It is here that he gives the more exact limits
for 77- which he says that Archimedes found in his work On
Plinthides and Cylinders, but which are
not convenient for
calculations. The limits, as we have seen, are given in the
text as toVt^ 71 < "
-w&stT' an 3 h, he adopts another procedure.
This is exactly modelled on Archimedes's quadrature of
a segment of a parabola. Heron proves (Metrlca I. 27-29, 32)
that, if ADB be a segment of a circle, and D the middle point
of the arc, and if
similarly bisected at E, F,
the arcs AD, DB be
A ADB < 4 (A AED + A DFB).
Similarly, if the same construction be
made for the segments AED, BFD, each
of them is less than 4 times the sum of the two small triangles
in the segments left over.
1
It follows that
(area of segmt. ADB) > A ADB { 1 h £ + (£)
2
+ ...}
> %AADB.
If therefore we measure the triangle, and add one-third of
it, we shall obtain the area of the segment as nearly as
possible.' That is, for segments in which b > Zh, Heron
takes the area to be equal to that of the parabolic segment
with the same base and height, or f bh.
In addition to these three formulae for S, the area of
a segment, there are yet others, namely
S = J (b + h) h (1 + 2t)> Mensurae 29,
X = i(b + h)h(l+ T\), „ 31.