________________
under the next sutra (IV, 13) is quite illuminating in this matter. He treats the barley-figure in two ways, namely as made of two triangles (upper and lower in FIG. 4) and as made of double segment (FIG. 3). The given dimensions are perpendicular width, NK 12, and each (curvilinear) side PNG or PKQ = 30.
While treating the yava-figure as double segment (FIG. 3), we have h = 6, and from (24) cs-h 30-6= 24 (viloma vidhi).
Hence by (13), area will be 90 for half figure, and 180 for the given barley figure.
For applying the method of triangles, the area of the upper triangle is found by the old empirical rule. 15
R
P
N
Δ
2r
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(KP PN) 2
SR
r
15+15 2
which gives the same answer 180 for the full figure. Indeed ancient methods were peculiar. 16 If one can see, rules (23) and (25) are same in the context
here!
Q P
E G
KN
')
2
(1/2)
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= 90
c/2
h
IN
*************
с FIG. 6
HF
(25)
S
FIG. 5
Now we will give two rationales for the formula (19). One is based on averaging which was frequently used in ancient days. 17 It can be seen easily that the area of the semi-circle (with л= 3) in FIG. 5 is the mean of the areas of the inscribed triangle PNQ and the circumscribed rectangle PRSQ. It is quite natural to apply the same process to the segment PNQ in FIG. 6 by analogy. So we get
A = ( PNQ+ area PRSQ)/2 = (ch/2+ch)/2 which gives (19). Second method is based on equating the area of the segment by apparently equivalent trapezium which was a pride figure in Jaina mathematics. In FIG. 6, GHh, EF c/2. Firstly, we note that the rule (13) approximates the area of the segment by the trapezium PGHQ. But this rule (13) always gives less area. So the segmental area may be approximated by the trapezium
14
Arhat Vacana, 14(1), 2002
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