212 



THE POPULAE EDUCATOE. 



point N. From the point B as centre, with the distance B N, 

 draw arcs cutting the circumference of the circle ACE 

 in c and H. Join c B, B u. These straight lines are sides 

 of a heptagon inscribed in the circle ACE, and the heptagon 

 itself may be completed by applying 

 the compasses, with an opening equal 

 to B N or N M, round the circum- 

 ference of the circle ACE, as may 

 be needful. The triangle A B L is an 

 equilateral triangle (see Problem 

 XVII., Vol. I., page 209), and A B is 

 the side of a hexagon that may be 

 inscribed in the circle A c E, in which 

 the heptagon B c D E F G H has already 

 been inscribed. 



PROBLEM LIII. To construct a 

 heptagon on any given straight line. 



Fig. 74. 



Fig. 75. 



Let A B (Fig. 75) be the given straight line on which it is 

 required to construct a heptagon. Produce A B indefinitely both 

 ways to x and T. Bisect A B in c, and again bisect c B in D. From 

 B along the straight line B T set off B E equal to five times B D, 

 and from A along the straight line A F set off A x equal to B E, 

 or five times B D. Then from the points A and B as centres, 

 with the distances A E, B F respectively, describe the arcs E G, 

 V 6, cutting one another in the point G ; and from G as centre, 



with a radius 

 equal to A B, de- 

 scribe the arc H K, 

 cutting the arcs 

 P G, E G, in the 

 points H and K. 

 Join G H and G K, 

 and bisect them 

 respectively in the 

 points, L, M; and 

 join AM, B L, 



intersecting each 



other in the point 

 N, which is the 

 centre of the circle 



circumscribing the required heptagon. From the centre N, at 

 the distance N A or N B, describe the circle A B K G H. Bisect 

 the arcs A H, B K, in the points o, P, and join H o, o A, B p, p K. 

 The figure ABPKGHoisa heptagon, and it is described on 

 the given straight line A B, as required. 



PROBLEM LIV. To construct an octagon on a given straight 

 line. 



As it has been remarked in a former lesson (see page 192), it 

 is easy to inscribe a hexagon in a given circle when we can place 

 an equilateral triangle within it, as the process is merely to 

 bisect the arcs intercepted between the ends of the sides of the 

 triangle, and to form the hexagon by joining the six points thus 

 obtained in the circumference of the circle. By a similar pro- 

 cess of bisection, an octagon may be inscribed in a given circle 

 when we have once placed a square within it ; while the bisec- 

 tion of the arcs intercepted between the ends of the sides of a 

 pentagon and hexagon will similarly produce a decagon and a 



dodecagon. There are, how- 

 ever, one or two processes of 

 constructing an octagon on a 

 given straight line which we 

 give here, as they may be of 

 use to the learner in cutting 

 out an octagon in timber, or an 

 octagonal flower-bed in turf. 

 First, let A B (Fig. 76) be the 

 given straight line on which it 

 is required to construct an octa- 

 gon. Produce A B indefinitely 

 both ways to x and T, and at the 

 points A and B draw the straight 

 fines A O, B p, perpendicular to A B or x Y. From A as centre, 

 with the distance A B, describe the arc BCD, cutting A o in c, 

 and XT in D ; and from B, as centre, with the distance B A, 

 describe the arc A E F, cutting B p in E, and x Y in F. Join 

 c D, E F, and bisect them in G and H respectively. Join A G, 

 B H, and produce A G to meet the arc B c D in K, and B H to 

 meet the arc A E F in L. Through the points K. \, draw the 



Fig. 76. 



Fig. 77. 



[ straight lines K M, L N, parallel to A o or B P, and along K M 

 set off K Q, equal to A B, and along L N set off L R, also equal 

 to A B. Then from the points Q and R as centres, with a 

 radius equal to A B, draw arcs cutting the perpendiculars A o, 

 B P, in s and T. Join Q s, s T, 

 T R. The figure ABLRTSQK 

 is an octagon, and it is described 

 on the given straight line A B as 

 required. 



Next, let A B (Fig. 77) be the 

 given straight line on which it is 

 required to construct an octagon. 

 Produce A B indefinitely both 

 ways to x and Y, and through 

 the points A and B draw the 

 straight lines A o, B P, perpen- 

 dicular to A B or x Y, and set off 

 along A o and B p the straight 

 lines Ac, B D, each equal to 

 A B. Join A D, B c, and produce them indefinitely to Q and B 

 respectively; and along D Q, c R, set off D E, c F, each equal to 

 A B. Through E and F draw the straight lines E G, F H, meeting 

 x Y in G and H ; and along E G, F H, set off E L and F K, each equal 

 to A B. Through K draw K M parallel to A Q, and cutting B p in 

 M ; and through L draw L N parallel to B R, and cutting A o in N. 

 Join A K, B L, E M, M N, N F. The figure ABLEMNFKisan 

 octagon, and it is described on the given straight line A B, as 

 required. If it be required to cut off the corners of a square 

 piece of wood or pasteboard as A B c D in Fig. 78, so as to form 

 an octagon, first draw the diagonals A c, 

 B D, intersecting each other at right 

 angles in E, and then from the points 

 A, B, c, D in succession, with the distances 

 A E, B E, c E, D E, describe the arcs 

 F E G, H E K, L E M, and N E o, having 

 their terminations in the sides of the 

 square. Then join OH, G L, K N, and 

 M F. The figure MFOHGLKNisan 

 octagon. 



PROBLEM LV. To inscribe a nona- ' 

 gon in a given circle. 



Let ABC (Fig. 79) be the given circle 



in which it is required to inscribe a nonagon. Draw any diameter, 

 c E passing through the centre D of the circle ABC, and produce 

 it indefinitely towards F. From the point E as centre, with the 

 distance E D, describe the arc A D B, cutting the circumference of 

 the circle A B c in the points A and B. Join A B, and pro- 

 duce it indefinitely both ways towards G and H, and let it 

 cut c F at right angles in the point K. Then from K 

 as centre, with a radius equal to D E, describe the semicircle 

 L M N, having its terminations L, N, in the straight line G H ; 

 and from L and N as centres, with the radii L K, N K respec- 

 tively, describe the arcs K o, K p, meeting the semicircle L if N 

 in the points o, P. Join D o, D p, cutting the circumference of 

 the circle A B c in the points Q, R. Join A Q, Q R, R B. These 

 three straight lines are the sides of a nonagon inscribed in the 

 circle ABC, which may be 

 completed by following the 

 same process with the arcs 

 AC, c B, or, applying the 

 compasses round the cir- 

 cumference of the circle 

 with an opening equal to 

 A Q, Q R, or R B, and join- 

 ing the points thus ob- 

 tained. The straight line 

 A B is manifestly the side 

 of an equilateral triangle 

 inscribed in the circle ABC, 

 and the process which has 

 been gone through is simply 

 the trisection of the arc A B, or, what is virtually the same thing, 

 the trisection of the angle A D B. 



PROBLEM LVI. To construct a nonagon on any given straight 

 line. 



Let A B (Fig. 80) be the given straight line on which it is 

 required to construct a nonagon. Produce A B indefinitely 

 both ways to x and Y. and on the straight line x Y, with 



Fig. 78. 



Fig. 79. 



