364 
Problem of the Arbelon. 
[Nor. 
IV . — Problem of the Arbelon. 
To the Editor of the Gleanings in Science. 
Sir, 
7 
Having lately sent you a specimen of Hindfi proficiency in mathematics, it 
may be a little interesting to send one of Moliamedan intellect too. Happening 
to mention to Hakim Abdul Mojid (a well known literary character in Calcutta) 
Archimedes’ problem of the Arbelon, (of which the reader will find an account in 
the Quarterly Review for February 1810, p. 103,) he took it home, and in adayor 
two brought me the accompanying solution, which I accompany with a translation. 
Your very obedient Servant, 
E, 
(Plate XXII. fig. 21.) 
f ) jJtij L - -0-^3-i <uic I ItL-o jfjbrO 
T CIS * } Ujd y ^ ^Lc Lfljt j 
I > r ^ V 
J 
T J** ^ ^ 
_ ^ fj ^ r 
{j? ^ y XJ! ! 
r CL? * ) J ££yJt>j 1*3 LS LaS u Jo J I iaXyb 
«* ♦♦ 
CL? ^ 1 1 . 1.? ? I lXLz> 1 ) t j )) l „ <Lf 3 j f 
1 
^5^ > j * I ^ j £.<0 1$ L~J ' > k,'j) 
J 9 
t 
f j r O « I ■* J 4-> f ^ I 
L <Cyj UiXA^.1 1 c j f j * i f j 
♦♦ ^ v_ ^ - 
e _5 l — f ^ I ; J ^ ^ * 
g f ^ ^ vj'*^ *) T * TL ^ 
* S- 5 ^ S C-T L-> ^ J < 9 ^ ° ^ 
(Plate XXII. fig. 22.) 
On a given line, as AB, let there be a semicircle AHDKB, and also let there jc 
semicircles on the segments of the (given) line, viz. AETC and CPRMB. Br.m ^ 
(the point) C of AB the perpendicular CD to the circumference, and the cm e 
scribed upon CD is equal to the figure included between the semicircle on 1 f ‘ 
AB, and the two semicircles on its segments, that is the figure AETCPRMB • 
Join AD, BD, so as to produce the triangle ADB, then the angle ABO is rig » > 
desci’ibe upon AD and BD two semicircles, viz. AID, BLD, an l B |e ‘ 
AETC, and CEMB, and CFEHD, and DUROC, being similar, are equal to > 
similar arcs AID, BLD, that is equal to the arc AHDKB, and the two « lics 
CRD, that is the circle CIIDO, are equal to the figure AETCPRMBK 0 0. 
Q. E. B* 
