GEOMETRY. 



PROBLEM XXVII. 



To inscribe a square in a given circle. 

 Fig. 30. Draw the diameter A C, and per- 

 pendicular thereto, the diameter B D : 

 the lines AB, BC, CD, and DA, will form 

 a correct square. 



PROBLEM XXVIII. 



To describe a circle around a square. 

 Fig. 30. In the square ABCD, draw the 

 diagonals AC, BD, their intersection at E 

 \villgive the centre of a circle, whose ra- 

 dius may be any one of the four converg- 

 ing lines ; say EA, that will enclose the 

 square. 



PROBLEM XXIX. 



To describe a circle tvithin a given square. 

 Fig. 31. Divide the square into four equal 

 parts, by the lines AC, BD, whose inter- 

 section, at E, shows the centre of a circle 

 to be drawn with any one of the converg- 

 ing lines, say EA, 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 AC, 

 BD ; draw*the tangents GH, FK, parallel 

 to AC, and GF, HK, parallel to BD ; 

 which will give the required square. 



PROBLEM XXXI. 



To make an isosceles triangle^ having 

 fach of the angles at the base double that at 

 the summit. Fig. 32. Cut any given line, 

 as AB, into extreme and mean propor- 

 tions, (as in Problem 18) ; then, from A, 

 as the centre, draw a circle BDE, with 

 the opening AB, and apply the line BD 

 within its circumference, equal to AC, 

 the greater portion of AB ; join CD, ABD 

 will be the isosceles triangle sought. 



PROBLEM XXXII. 



To describe a regular pentagon. Fig. 33. 

 Make the isosceles triangle ACD within 

 the circle ABCDE ; the base CD will give 

 the fifth part of the circumference. 



PROBLEM XXXIII. 



To describe a regular pentagon about a 

 circle. Fig. 33. This is done by drawing 

 parallels to the lines AB, BC, CD, DE, 

 EA ; making them all tangents to the cir- 

 cle ; on the same principle, a square, a 



hexagon, &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 pen-, 

 tagon, and take their point of intersec- 

 tion, G, as a centre, using either of the 

 converging lines, DG, or EG, for a radius. 

 Where a circle 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 ei- 

 ther of the converging lines serving for a 

 radius. 



PROBLEM XXXV. 



To inscribe a regular hexagon -within a 

 circle. Fig. 34. The radius of a circle 

 being equal to one-sixth of its circumfer- 

 ence, establishes a very easy mode of 

 setting off the six sides as follows : draw 

 the diameter AB, 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 complete. 



PROBLEM XXXVI. 



To form a quindecagon, or figure of 15 

 equal sides, -within circle. An equilate- 

 ral triangle being inscribed within a cir- 

 cle, 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 pentagon 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 AB equal to three 

 and one seventh diameters AD of the cir- 

 cle, and from the centre C draw CB, and 

 CA : the triangle CAB will be equal in 

 contents to the circle AD. 



PROBLEM XXXVIII. 



To change a pentagon into a triangle. 

 Fig 36. Continue the base line AB to C, 

 and from the centre D let a perpendicu- 

 lar fall on AB, bisecting it in E. Measure 

 from B a space equal to four times EB. 

 Through the centre D draw DF, paralle* 



