________________
IJ = E + EF + FJ
= al.10 + 121.421.10 + .01.10
= 61.61.10 GH = GA + AB + BH
= 121.10 + 121.401.10 + 1.21.10
= 61.61.10 Therefore,
the quadratic length (bhuja) GH (or IJ) = 61.61.10
and
the quadratic breadth (koti) GI (or HJ) = 11.1. The quadratic area of the rectangle = GH.HJ
= 61.61.10.11.1.
= 6.6.10.L.L.I.I. square yojanas. anther (UTCHT HIER if 796:1 (TLS v.122 second half, p. 136)
The (result of the) multiplication or division of quadratic quantities always becomes quadratic,
By this rule, the required area is equal to 6/101.1 square yojanas.
Madhavacandra's present method for the quadrature of a circular annulus is, no doubt, excessively lengthy. It can be easily made short at the following two points - (a) The quadratic quantities should be replaced by simple ones. (b) There is no need to transform an isosceles trapezium into a rectangle. Moreover, the formula for finding the area A of an isosceles trapezium,
A = 1/2 (m + b)u was known to Nemicandra and to Madhavacandra as well (cf. TLS v.114, p. 110), where m, b and u are its face (mukha), base (bhumi) and altitude (udaya). Here m = 11/10, b = 51 10 and u = 21.
Hence,
A = 1/2 (11/10 + 5110).21
= 6101.1. square yojanas. The possible reason behind why Madhavacandra has transformed the isosceles trapezium into a rectangle is that a rectangle is some more down
Arhat Vacana, October 2000
