248 9. MEASUREMENTapproximating the area of a 192-sided polygon.^6 That is, he started with a hexagon
and doubled the number of sides five times. However, since the area of the polygon
with twice the number of sides is the radius of the circumscribed circle times the
perimeter of the original polygon, it was only necessary to find the perimeter of a
96-sided polygon and multiply by the radius.
Problems 31 and 32 ask for the area of a circular field of a given diameter and
circumference.^7 The method is to multiply half of the circumference by half of the
diameter, which is exactly right in terms of Euclidean geometry; equivalently, the
reader is told that one may multiply the two quantities and divide by 4. However,
in the actual data for problems the diameter given is exactly one-third of the given
circumference; in other words, the value assumed for one-dimensional ð is 3. The
assumption of that value leads to two other procedures for calculating the area:
squaring the diameter, then multiplying by 3 and dividing by 4, or squaring the
circumference and dividing by 12. An elaboration of this problem occurs in Prob-
lems 37 and 38, in which the area of an annulus (the region outside the smaller of
two concentric circles and inside the larger) is given in terms of its width and the
circumferences of the two circles.
The authors knew also how to find the volume of a pyramid. Problem 15 of
Chapter 5 asks for the volume of a pyramid whose base is a rectangle 5 chi by
7 chi and whose height is 8 chi. The answer is given as 93^ (cubic) chi. For a
frustum of a pyramid having rectangular bases the recipe is to add twice the length
of the upper base to the lower base and multiply by the width of the upper base to
get one term. A second term is obtained symmetrically as twice the length of the
lower base plus the length of the upper base, multiplied by the width of the lower
base. These two terms are then added and multiplied by the height, after which
one divides by 6. If the bases are a ÷ b and c x d (the sides of length a and c being
parallel) and the height is h, this yields what we would write (correctly) as
V = ^U2a + c)b+(2c + a)d].
6
Notice that this result is more general than the formula in the Moscow Papyrus,
which is given for a frustum with square bases.
The Pythagorean theorem. The last of the nine chapters of the Jiu Zhang Suanshu
contains 24 problems on the gougu theorem. After a few "warm-up" problems
in which two of the three sides of a right triangle are given and the third is to
be computed, the problems become more complicated. Problem 11, for example,
gives a rectangular door whose height exceeds its width by 6 chi, 8 cun and has a
diagonal of 1 zhang. One zhang is 10 chi and 1 chi is 10 cun (apparently a variant
rendering of fen). The recipe given is correct: Take half the difference of the height
and width, square it, double, subtract from the square of the diagonal, then take
the square root of half of the result. That process yields the average of the height
and width, and given their semidifference of 3 chi, 4 cun, one can easily get both
the width and the height.
3.3. The Sun Zi Suan Jing. The Sun Zi Suan Jing contains a few problems in
measurement that are unusual enough to merit some discussion. An inverse area(^6) Lam and Ang (1986) give the value as 3.14 + 169/625 = 3.142704.
(^7) All references to problem numbers and nomenclature in this section are based on the article of
Lam (1994).