THE REGULAR POLYGONS
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now £ + | (f|), so that the approximation used by
Heron for V3 is here ff. For the side 10, the method gives
the same result as above, for f § . 100 = 43f.
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 fR. Produce AC to 0 so that C0 = AG
If now AO is divided in extreme and
mean ratio, AB is equal to the greater
segment. (For produce AB to P so that
AD = AO, and join BO, DO. Then, since
ADO is isosceles and the angle at A = ^R,
l ADO = ¿AOD = fB, and, from the
equality of the triangles ABC, OBC,
¿AOB = ZBAO = fit!. It follows that
the triangle ADO is the isosceles triangle of Fuel. IV. 10, and
AD is divided in extreme and mean ratio in B.) Therefore,
says Heron, {BA + AG) 2 — 5 AC 2 . [This is Each XIII. 1.]
Now, since ZBOG = fit!, if BO be produced to E so that
CE = BC, BE subtends at 0 an angle equal to fit!, and there
fore BE is the side of a regular pentagon inscribed in the
circle with 0 as centre and OB as radius. (This circle also
passes through D, and BD is the side of a regular decagon in
the same circle.) If now BO = AB = r, OC = p, BE = a,
we have from above, {r+p) 2 = 5p 2 , whence, since V5 is
approximately |, we obtain approximately r = |p, and
\a =■ \p, so that p — fa. Hence \pa — fa 2 , and the area
of the pentagon = fa 2 . Heron adds that, if we take a closer
approximation to V5 than f, we shall obtain the area still
more exactly. In the Geometry 1 the formula is given as \j-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 = x 3 Q-a 4 (Metrica I. 17), hence (hexagon) 2 = 2 £-oA. If,
e.g. a — 10, (hexagon) 2 = 67500, and (hexagon) = 259 nearly.
In the Geometry 2 the formula is given as V-a 2 , while ‘another
book’ is quoted as giving 6 (f-h-jV)® 2 ! ^ 48 added that the
latter formula, obtained from the area of the triangle, (f + 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.).