A history of Greek mathematics - Wilbourhall.org

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328 HERON OF ALEXANDRIA
Heron. As a matter of fact, however, 6 (§ + jq) = -^3-
exactly,
and only the Metrica gives the more accurate calculation.
The regular heptagon.
Heron assumes (chap. 20) that, if a be the side and r the
radius of the circumscribing circle, a — |r, being approximately
equal to the perpendicular from the centre of the
circle to the side of the regular hexagon inscribed in it (for |
is the approximate value of \ \^3). This theorem is quoted by
Jordanus Nemorarius (d. 1237) as an 'Indian rule'; he probably
obtained it from Abu'l Wafa (940-98). The Metrica,
shows that it is of Greek origin, and, if Archimedes really
wrote a book on the heptagon in a circle, it may be clue to
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-§
or 16/7, whence p 2 /(^a) 2 = -iw'i and 'p/\ a — (approximately)
14|/7 or 43/21. Consequently the area of the
heptagon = 7 . \pa — 7 .
fftr — fi^2
The regular octagon, decagon and dodecagon.
In these cases (chaps. 21, 23, 25) Heron finds p by drawing
For the decagon,
the perpendicular 00 from 0, the centre of the
circumscribed circle, on a side AB, and then making
the angle OAD equal to the angle AOD.
For the octagon,
I ADO = ±R, and p ^ Ja(l + V2) = Ja(l + f|)
-
or \a .
f§ approximately.
ZADC = f R, and AD :
AC =5:3, and p = \a (§ + §
) — fa.
hence AD :
For the dodecagon,
DC =5:4 nearly (see preceding page)
Z ADC = | £, and p = |a (2 + V3) = \a (2 + J)
Accordingly A 8
= ^g-a 2 ,
A
the side in each case.
10
= -^tt 2 , A2
The regular enneagon and hendecagon.
= - 4
4
5 - a
= ^-a
approximately.
'
2
>
where a is
In these cases (chaps. 22, 24; the Table of Chords (i e.