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Vol. IV, No. 3.] Notes on Indian Mathematics. 121 
[N.S.] 
(a) The first. rule was, of course, well known to the Greeks: 
Heron gives it in this form. An attempt to find, by practical 
means,:a circle equal in area in a given square is exhibited in the 
—— (Thibaut in Journ, Asiatic Soc. Bengal, 1875, 253). 
magupta’s rule is: “The Square of the diameter m nlti- 
lied by three is the practical area; the square root extenoheds from 
ten times the same is the real value.” (XIT., 40. 
M. ibn Musa says: “ The area of any circle will be found by 
ers meg the circumference rte y half the diameter; since in every 
polygon of equal sides and a: —— such as triangles, quadrangles, 
peittagons and so on, the area is found by multiplying half the 
circumference by half the ducnster of the middle circle that can 
thew through it.” (Rosen, 72 
(b) The formula for the volume of a sphere is not even 
approximately correct. It may be expressed thus :—Volume= 
ar/ rt, This would give =16/9. Strangely enough Ahmes 
gave the value t=(16/9)%. Suppose Ahmes’ value to be correct, 
then the area of a ade would be (16/9) #r? alti irig ogee by 
” r would give the volume of the sphere, It looks s though 
r had, by mistake, been taken as the root of (4,°)47 
Rodet writes: “La formule qu'il donne pour * ‘volume de la 
sphere Rey/ 7 n’est méme pas une approximation numérique 
- Mais elle a, pour histoire des mathéma atiques, d’antant plus 
de valeur, parce qu’elle nous démontre que si set inbrens avait nan 
quelque enseignment des Grecs, il ignorait an moins les travaux 
d’Archiméde.” I do not for a moment think that Aryabhata “had 
reed access to the woes of Archimedes, but it is quite possible 
that Aryabhata’s incorrect rule was indirectly derived from the 
great mathematician’s formula. 
It is doubtful whether Archimedes was known in Aryabhata’s 8° 
time even to the ordinary Alexandrian mathematicians. He w 
out of fashion amongst the degenerates of that age. 
8. Each of the two sides multiplied by their distance apart and 
divided by their sum gives the lines from the point of intersection. 
By multiplying nay the sum of the sides by their distance apart, 
the area of the glee 
Let ABCD be any ee having the sides AB and CD 
parallel, then the ‘triangles see and CED (E being the — of 
intersection of the diagon 
will be similar and D c ra 
AE: EC:: AB: DC 
from which we get: 
AE=A0C. AB/(AB+ DO) 
and similarly ; 
A'E' = A'0'. AB/(AB+ DO) ~~ 
where AC! is the perpendi- 
cular the lel sides 
through. the point of intersection 


