THE REGULAR POLYGONS 
327 
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.).