A history of Greek mathematics - Wilbourhall.org

THE REGULAR POLYGONS 327
now 3 + xjt
= || = i (f£)> so that the approximation used by
Heron for \/3 is here .
ff For
the side 10, the method gives
the same result as above, for §§ . 100 = 43J.
The regular pentagon is next taken (chap. 18). Heron
premises the following lemma.
Let ABC be a right-angled triangle, with the angle A equal
to §i*. Produce AC to so that CO = AC
If now AO is divided in extreme and
mean ratio, AB is equal to the greater
segment.
(For produce AB to D so that
,4D = AO, and join 50, DO. Then, since
J. 2)0 is isosceles and the angle at A—^R,
I ADO = AA0D = ±R, and, from the
equality of the triangles ABC, 0BC,
Z.A0B = LBA0 = fiS. It follows that
the triangle J.D0 is the isosceles triangle of Eucl. IV. 10, and
AD is divided in extreme and mean ratio in B.) Therefore,
says Heron, (BA+ACf = 5 AC 2 . [This is Eucl. XIII. 1.]
Now, since LB0C — %R, if BC be produced to E so that
CE = BC, BE subtends at an angle equal to ~R, and therefore
BE is the side of a regular pentagon inscribed in the
circle with as centre and 0B as radius. (This circle also
passes through D, and BD is the side of a regular decagon in
the same circle.) If now B0 — AB = r, 0C = p, BE = a,
we have from above, (r + p) 2 — 5p
2
, whence, since V5 is
approximately J, we obtain approximately r = %p, and
la = %p, so that p — \a. Hence \pa — \o?, and the area
of the pentagon = fa
2
. Heron adds that, if we take a closer
approximation to a/5 than |, we shall obtain the area still
more exactly. In the Geometry 1 the formula is given as \^-a 2 .
The regular hexagon (chap. 19) is simply 6 times the
equilateral triangle with the same side. If A be the area
of the equilateral triangle with side a, Heron has proved
that A = 2 T3
ga 4 (Metrica I. 17), hence (hexagon) 2 = $?-a\ If,
e.g. a = 10, (hexagon) = 2
67500, and (hexagon) = 259 nearly.
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
latter formula, obtained from the area of the triangle, (J + ^) a 2 ,
represents the more accurate procedure, and is fully set out by
1
Geom. 102 (21, 14, Heib.).
2 lb. 102 (21, 16, 17, Heib.).