1888.] Dr E. Sang on Diameter of a Circle. 351 
is the triangle J)Vd portion of a regular polygon on the base 
~Dd, having twice as many sides as before, with^ the same total 
boundary. PC is the inscribing and PO the circumscribing 
radius thereof. 
Similarly, by making PQ equal to PD, joining DQ, bi- 
secting CQJin E and drawing FE/ parallel to BZ>, we get FQ/ 
portion of a regular polygon of four times the original number of 
sides and having the same length of perimeter. And so we may 
continue until the difference between QE and QF be undistinguish- 
able. 
The geometrical construction is simple, the corresponding arith- 
metical work is not less so. This may be best shown by an 
example. 
Let it be proposed to describe a circle whose circumference shall 
be an English mile, and let an exactitude to within one-tenth of a 
foot in the radius be deemed sufficient. 
Having made AB one-eighth of a mile, "or 660 feet, and the 
perpendicular AO also of that length, the triangle BO 6 becomes the 
quarter of a square having O for its centre. The inscribing radius, 
which we shall denote by r, is OA = 660, and the circumscribing 
radius, denoted say by E, is found by adding the squares of OA 
and AB together, and extracting the square root of the amount. 
Thus 
OA = 660-0, OA2=^43 5600 
AB = 660-0, AB2 = 43 5600 
OB2 = 87 1200, OB = 933-4 
AO= 660-0 
OB= 933-4 
AP = 1593-4, CP = 796-7, CP2 = 63 4731 
DC2 = 10 8900 
PD2 = 72 3631, PD = 862-3 
and so on. 
This work may be concisely arranged as under, s being written 
for the half side. 
From this we conclude that a circle must have a diameter of 
1680-7 feet in order that its circumference may be exactly one mile. 
If we make use of the ordinary tables of square numbers, the 
above computation is done by mere inspection, and need not occupy 
VOL. XV. 30/10/88 z 
