________________
of the face (mukha) and the square of the diameter (8 yojanas) of the base (bhūmi) are to be multiplied by 10 which yield 16.10 = 160 yojanas and 64.10 = 640 yojanas respectively. If the square root (vargamula) is obtained through the area-factor-multiplication (kşetrakhandananugunana), the circumferences of the face and base are 12
yojanas respectively.
to 16 are 12 and 24 54
16
If the remainder in the square root of the face (mukhamülasesa) is reduced through dividing by 8 (asta) into its lowest term, then 2/3 is obtained. In the same way if the remainder it in the square root of the base (bhūmimülasesa) is reduced through dividing by 16 (sodaśa) into its lowest term, then is arrived at.
In this way the subtle values of the circumferences (suksma paridhis) of the face and base become 12 yojanas and 24 yojanas respectively.
-
D
Figure 9 If the thickness most of this field (Fig. 8a) (ksetra - bahalya) (8 yojanas) is cut up to the middle (madhya) (4 yojanas) and expanded (prasāra) then we have like Fig. 9.
Here we would like to make clear that it is the thickness (bahalya) 'b' which has been called the middle result and which has been calculated through halving the sum of the dimension 'd' and face 'm'. Therefore, we may infer that d+m
(5a]
Now, an open question arises before us as to how the above formula [5a) is obtained from the Fig. 7.
Here it is certain that Madhavacandra calls 'bahalya' to the middle diameter of a complete drum - like figure (Fig. 7). We have seen that Yativrsabha
42
Arhat Vacana, 14(1), 2002
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