THE CON STRUCT I OK OF 
350 
The demonstration whereof is evident from the last 
proposition: and in the same manner may 6, 8, or 10, 
&c. equal circles be inscribed in a given circle, to touch 
one another. 
The method of calculation in this, or any other case, 
will also be the same as in the last problem ; for in the 
triangle MNH will be given the ratio of XM to NH 
(as 2 to l) and the included angle MNH equal to 
126 3 , 120', 112 1 °, or 108°, &c. according as the num 
ber of circles is 5,6, 8, or io, &c. from which the angle 
MHN will be given; then in the triangle CMH will be 
given all the angles, and the side CH, to find CM. 
PROBLEM xxxix* 
The perimeter of a right-angled triangle, whose sides 
are in geometrical progression, being given to describe the 
triangle. 
CONSTRUCTION. 
Upon AC, equal to the given perimeter, describe the 
semi-circle ABC, and let AC be divided in D, ac 
cording to extreme and mean proportion; make DB 
perpendicular to AC, meeting the periphery of the