344 
HERON OF ALEXANDRIA 
Quadratic equations solved in Heron. 
We have already met with one such equation (in Metrica 
HI. 4), namely x 2 — 14« + 46| = 0, the result only (x = 8-|) 
being given. There are others in the Geometrica where the 
process of solution is shown. 
(1) Geometrica 24, 3 (Heib.). ‘Given a square such that the 
sum of its area and perimeter is 896 feet: to separate the area 
from the perimeter ’ : i.e. x‘ 2 + ‘ix= 896. Heron takes half of 
4 and adds its square, completing the square on the left side. 
(2) Geometrica 21, 9 and 24, 46 (Heib.) give one and the same 
equation, Geom. 24, 47 another like it. ‘Given the sum of 
the diameter, perimeter and area of a circle, to find each 
of them/ 
The two equations are 
\\di l + ^~d =212, 
and ii d 2 + - 2 T 9 - d = 67\. 
Our usual method is to begin by dividing by throughout, 
so as to leave d' 1 as the first term. Heron’s is to multiply by 
such a number as will leave a square as the first term. In this 
case he multiplies by 154, giving 1 l 2 cZ 2 -h58 . lid = 212 . 154 
or 67-|.154 as the case may be. Completing the square, 
he obtains (11 d + 29) 2 = 32648 + 841 or 10395 + 841. Thus 
11 ¿¿ + 29=: V(33489) or \/(11236), that is, 183 or 106. 
Thus 11 d — 154 or 77, and cl = 14 or 7, as the case may be. 
Indeterminate problems in the Geometrica. 
Some very interesting indeterminate problems are now 
included by Heiberg in the Geometrica? Two of them (chap. 
24, 1-2) were included in the Geeponicus in Hultsch’s edition 
(sections 78, 79^; the rest are new, having been found in the 
Constantinople manuscript from which Scheme edited the 
Metrica. As, however, these problems, to whatever period 
they belong, are more akin to algebra than to mensuration, 
they will be more properly described in a later chapter on 
Algebra. , 
3 Heronis Alexandrini opera, voi. iv, p. 414. 28 sq.