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

THE REGULAR POLYGONS 327
Geom. 102 (21, 14, Heib.).
2 lb. 102 (21, 16, 17, Heib.).
In these cases (chaps. 22, 24; the Table of Chords (i e.
328 HERON OF ALEXANDRIA
now 3 + xjt
= || = i (f£)> so that the approximation used by
Heron for \/3 is here Heron. As a matter of fact, however, 6
.
ff For the side 10, the method gives
(§ + jq) = -^3-
exactly,
and only the Metrica gives the more accurate calculation.
the same result as above, for §§ . 100 = 43J.
The regular pentagon is next taken (chap. 18). Heron
The regular heptagon.
premises the following lemma.
Heron assumes (chap. 20) that, if a be the side and r the
Let ABC be a right-angled triangle, with the angle A equal
radius of the circumscribing circle, a — |r, being approximately
equal to the perpendicular from the centre of the
to §i*. Produce AC to so that CO = AC
If now AO is divided in extreme and
circle to the side of the regular hexagon inscribed in it
mean ratio, AB |
is equal to the greater
is the approximate value of \ \^3). This theorem is quoted by
segment. (For produce AB to D so that
Jordanus Nemorarius (d. 1237) as an 'Indian rule'; he probably
obtained it from Abu'l Wafa (940-98). The Metrica,
,4D = AO, and join 50, DO. Then, since
J. 2)0 is isosceles and the angle at A—^R,
shows that it is of Greek origin, and, if Archimedes really
I ADO = AA0D = ±R, and, from the
wrote a book on the heptagon in a circle, it may be clue to
equality of the triangles ABC, 0BC,
Z.A0B = LBA0 = fiS. It follows that
him. If then p is the perpendicular from the centre of the
circle on the side (a) of the inscribed heptagon, r/(-|a) = 8/3-§
the triangle J.D0 is the isosceles triangle of Eucl. IV. 10, and
or 16/7,
AD p 2 /(^a) 2 = -iw'i and 'p/\ a — (approximately)
14|/7 or 43/21. Consequently the area of the
is divided in extreme and mean ratio in B.) Therefore,
says Heron, (BA+ACf = 5 AC 2 . [This is Eucl. XIII. 1.]
heptagon = 7 . \pa — 7 .
fftr — fi^2
-
Now, since LB0C — %R, if BC be produced to E so that
CE = BC, BE subtends at an angle equal to ~R, and
The regular octagon, decagon and dodecagon.
therefore
BE is the side of a regular pentagon inscribed in the
In these cases (chaps. 21, 23, 25) Heron finds p by drawing
circle with as centre and 0B as radius. (This circle also
the perpendicular 00 from 0, the centre of the
passes through D, and BD is the side of a regular decagon in
circumscribed circle, on a side AB, and then making
the same circle.) If now B0 — AB = r, 0C = p, BE = a,
the angle OAD equal to the angle AOD.
we have from above, (r + p) — 2 5p
2
, whence, since V5 is
For the octagon,
approximately J, we obtain approximately r = %p, and
I
la = %p, so that p —
ADO = ±R, and
\a. Hence \pa —
p ^ Ja(l + V2) = Ja(l +
\o?, and f|)
the area
of the pentagon = fa
2
. Heron adds that, if we take a closer
or \a .
f§ approximately.
approximation to a/5 than |, we shall obtain the area still
For the decagon,
more exactly. In the Geometry 1 the formula is given as \^-a 2 .
ZADC = f R, and AD : DC =5:4 nearly (see preceding page)
The regular hexagon (chap. 19) is simply 6 times the
hence AD : AC =5:3, and
equilateral triangle with the same side. If A
p = \a (§ +
be the area
§
) — fa.
of the equilateral triangle with side a, Heron has proved
For the dodecagon,
that A = 2 T3
ga 4 (Metrica I. 17), hence (hexagon) 2 = $?-a\ If,
Z ADC = | £, and p = |a (2 + V3) = \a (2 + J)
= ^-a
e.g. a = 10, (hexagon) = 2
67500, and (hexagon) = 259 nearly.
approximately.
In the Geometry 2 the formula is given as ^-a', while another
'
book' is quoted as giving 6(J + T^)a 2 ; it is added that the
Accordingly A 8
= ^g-a 2 ,
A 10
= -^tt 2 , A2 = -4 4 5 '
- a
2
where a is
>
latter formula, obtained from the area of the triangle, (J + ^) a 2 ,
represents the more accurate procedure, and is fully set out by
the side in each case.
The regular enneagon and hendecagon.
1