GEOMETRY. 
PROBLEM XXIX. 
To describe a circle within a given square. 
Fig. 31. Divide the square into four equal 
parts, by the lines A C, B D, whose inter- 
section at E, shows the centre of a circle 
to be drawn with any one of the converg- 
ing lines, say E A, as a radius. 
PROBLEM XXX. 
To describe a square on a given circle. 
Fig. 31. Divide the circle into four equal 
parts, (or quadrants) by the lines A C, B D ; 
draw the tangents GH, FK, parallel to 
A C, .and G F, H K, parallel to B D ; which 
will give the required square. 
PROBLEM XXXI. 
To make an isosceles triangle, having each 
of the angles at the base double that at the 
summit. Fig. 32. Cut any given line, as 
AB, into extreme and mean proportions, 
(as in Problem 18); then, from A, as the 
centre, draw a circle B D E, with the open- 
ing A B, and apply the line B D within its 
circumference, equal to A C, the greater 
portion of AB; join CD, A B D will be 
the isosceles triangle sought. 
PROBLEM XXXII. 
To describe a regular pentagon. Fig. 33. 
Make the isosceles triangle A CD within 
the circle ABCDE; the base CD will 
give the fifth part of the circumference. 
PROBLEM XXXIII. 
To describe a regular pentagon about a cir- 
cle. Fig. 33. This is done by drawing pa- 
rallels to the lines A B, B C, CD, DE, 
EA; making them all tangents to the cir- 
cle ; on the same principle, a square, a hex- 
agon, &c. may be drawn around a circle, 
from a similar figure inscribed within it. 
PROBLEM XXXIV. 
To describe a circle around a pentagon. 
Fig. 33. Bisect any two angles of a penta- 
gon, and take their point of intersection, G, 
as a centre, using either of the converging 
lines, D G, or E G, for a radius. Where a cir- 
cle is to be described within a pentagon, 
you must bisect any two of the faces, and 
raise perpendiculars at those points, which 
will meet in the centre either of the con- 
verging lines serving for a radius. 
PROBLEM XXXV. 
To inscribe a regular hexagon within a cir- 
cle. Fig. 34. The radius of a circle being 
equal to one-sixth of its circumference, es- 
tablishes a very easy mode of setting off the 
six sides as follows: draw the diameter 
A B, set one leg of your compasses at A, 
and draw the segment DF, and from B 
draw the segment CE; thus dividing the 
circle into six equal portions ; draw lines 
joining them, and the figure will be com- 
plete. 
PROBLEM XXXVI. 
To form a quindecagon, or figure of 15 
equal sides, within a circle. An equilateral 
triangle being inscribed within a circle, by 
assuming the distance between three points 
of a hexagon, say from A to C in the last 
figure for a side, let one point of such 
triangle be applied to each angle of a penta- 
gon in succession ; its two other points will 
divide the opposite sides in three equal 
parts, as the figure changes place within the 
pentagon. 
PROBLEM XXXVII. 
To change a circle to a triangle. Fig. 35. 
Draw the tangent A B equal to 3.) diameters 
A D of the circle, and from the centre C 
draw C B, and C A : the triangle CAB 
will be equal in contents to the circle A D. 
PROBLEM XXXVIII. 
To change a pentagon into a triangle. Fig. 
36. Continue the base line A B to C, and 
from the centre D let a perpendicular fall 
on A B, bisecting it in E. Measure from 
B a space equal to four times E B. Through 
the centre D draw D F, parallel and equal 
to E C ; draw F C : the parallelogram 
contained under E C D F will equal the 
area of the pentagon. Or the pentagon 
may be changed to a triangle by adding to 
A B four times its own length, and draw- 
ing a line from the centre, to the produced 
termination of A B ; the angle at the centre 
would then be obtuse. 
PROBLEM XXXIX. 
To draw a spiral line from a given point. 
Fig. 37. Draw the line A B through the 
given point C, and from C draw the semi- 
circle D E, then shift to D for a centre, and 
make the semi-circle A E In the opposite 
side of the line : shift again from D to C 
for a centre, and draw the semi-circle F G ; 
and then continue to change the centres 
alternately, for any number of folds you 
may require ; the centre C serving for all 
above, the centre D for all below, the 
line A B. 
With respect to the application of geome- 
try to its pristine intent, namely, the mea- 
