Greeks,Practical andTheoretical 73
context is remote and offers few opportunities for identifying role models. An assessment of her life
and works, if any can be reliably ascribed to her, while sympathizing with her difficult, ultimately
tragic situation, is not dependent either on approving her obvious ability and charisma, or on
disapproving of the élitism, and that belief that the state would be better off if run by philosophers
which she shared with other Neoplatonists.
Appendix A. From Heron’sMetrics
[Introductory note. The standard translation (e.g. the one which you will find in Fauvel and Gray)
has been changed so that there is as little as possible ‘modernization’ of Heron’s language, and
no explicit algebra. This is more difficult than one might imagine, since (a) Heron does think
of lengths as numbers, and multiply them—this happens in the first part, (b) in the geometric
proof the kind of straightforward and perhaps over-simple statements I have made about Euclid
(areas of rectangles are just areas, not products of the lengths of sides etc.) are no longer clearly
true, and it is possible that something like algebra, of an embryonic form, was in Heron’s mind
even if you cannot see it on the page. Prepositions like ‘on’ and ‘by’ indicate areas of rectangles or
multiplication, and it is unclear which is being used. Most unexpectedly, ‘the onABC’ means the
productABtimesBC, not the area of a triangle.]
There is a general method for finding, without drawing a perpendicular, the area of any triangle
whose three sides are given. For example, let the sides of the triangle be 7, 8, 9 units. Add together
the 7 and the 8 and the 9; the result is 24. Take half of this; the result is 12. Takeaway the 7units;
the remainder is 5. Again takeawayfrom the 12 the 8; the remainder is 4. And then the 9; the
remainder is 3. Multiply the 12 by the 5; the result is 60. and this by 4; the result is 240. And this
by 3; the result is 720. Take the side [= square root] of this, and it will be the area of the triangle.
Since 720 does not have a rational square root, we shall reach a different [number] close to the root
as follows. Since the square nearest to 720 is 729 and it has a root 27, divide the 27 into the 720;
the result is 26 and two thirds. Add the 27; the result is 53 and two thirds. Take half of this; the
result is 26^1213 [This is the ‘Egyptian way’ of writing 26^56 .] Therefore the square root of 720 will
be very near to 26^1213. For 26^1213 multiplied by itself gives 720 361 ; so that the difference is a 36th
part of a unit. If we wish to make the difference less than the 36th part, instead of 729 we shall
take the number now found 720 361 , and by the same method we shall find a difference much less
than 361.
The geometrical proof of this is this (Fig. 9):In a triangle whose sides are given to find the area.
Now it is possible to draw a perpendicular and calculate its magnitude and so find the area of the
triangle, but let it be required to calculate the area without drawing the perpendicular.
Let the given triangle beABCand let each ofAB,BC,CAbe given; to find the area. Let the circle
DEZbe inscribed in the triangle with centreH[Euclid IV.4], and letAH,BH,CH,DH,EH,ZH
be joined. Then the [rectangle]BCtimesEHis twice the triangleBHC[Euclid I.41], andCAtimes
ZHis twice the triangleAHC, andABtimesDHis twice the triangleABH. So the perimeter of the
triangleABCtimesEH, that is, the [radius] of the circleDEZ, is twice the triangleABC.
LetCBbe produced, and letBFbe made equal toAD; thenCBFis half the perimeter of the
triangleABCbecauseADis equal toAZandDBtoBEandZCtoCE.SoCFtimesEHis equal to the
triangleABC. ButCFtimesEHis the side of the [square] onCFtimes the one onEH.