192 



THE POPULAR EDUCATOR. 



K L T intersect in the point Q. From the centre Q, with the dis- 

 tance Q A, or Q B, describe the circle A N B, and along its circum- 

 ference set off the arcs A M, M N, N o, each equal to A B, to deter- 

 mine the angular points M, N, o of the pentagon A M N o B, which 

 is completed by joining the straight lines A M, M N, N o, o B. 



The student who is acquainted with the method of drawing 

 the equilateral triangle, square, and regular pentagon, will readily 

 see that the construction of figures, 

 the number of whose sides are mul- 

 tiples of 3, 4, and 5, depends merely on 

 the repeated bisection of arcs of 

 circles or angles, as may be seen by 

 . _].M_ an inspection of Figs. 69, 70, and 66. 

 In Fig. 69, ABC is an equilateral 

 triangle, or trigon, inscribed in the 

 circle ABC. By bisecting the arcs 

 A B, B c, c A, in the points D, E, F, and 

 joining A D, D B, B E, E c, c F, and F A, 

 we get the hexagon A D B E c F, the 

 number of whose sides is 3x2, 

 or 6. If the arcs A D, D B, B E, E c, c F, and F A, were again 

 bisected in the points G, H, K, L, M, N, we should get a dodecagon, 

 or twelve-sided figure, the number of whose sides is 3 x 2 x 2 

 = 12 ; by joining the extremities of the arcs A G, G D, etc., or 

 drawing chords subtending these arcs, and by another bisection 

 of the twelve arcs, into which the circle is divided by the angu- 

 lar points of the dodecagon, we should get a twenty-four-sided 

 figure. The same results would be obtained by bisecting in suc- 

 cession the angles at the point o, the centre of the circle, and 

 drawing chords, as before, to the arcs obtained by the successive 

 bisections of the angles at the centre, and the consequent bi- 

 sections of the arcs on which they stand. 



If the arcs A B, B c, c A, or the angles A o B, B O c, c o A, were 

 trisected, instead of being divided into two equal parts, and 

 chords were drawn subtending the arcs thus subdivided, we 

 should have a nonagon or enneagon, a polygon the number of 

 whose sides is equal to 3 x 3, or 9, and by bisecting the nonagon 

 we should obtain an eighteen-sided figure, the number of whose 

 sides is equal to 3 x 3 x 2, or 18. 



Similarly in Fig. 70, in which the square A B c D is inscribed 

 in the circle A B c D, an octagon, AEBFCGDH, is obtained 

 by bisecting the arcs A B, B c, c D, DA, 

 in the points E, F, G, H, and joining the 

 straight lines A E, E B, etc., the octagon 

 being a figure the number of whose sides 

 is 4 x 2, or 8. A further bisection of 

 the arcs A E, E B, etc., would give us a B 

 figure the number of whose sides is 

 4x2x2, or 16 ; and so on. 



In the same way, an inspection of Fig. 

 66, in Problem XLVIL, page 149, shows 

 that if we bisect the arcs A B, B c, c D, 

 T> E, E F, in the points R, s, T, P, Q, and join 

 their extremities, we get a decagon, or regular polygon, the 

 number of whose sides is 5 x 2, or 10 ; while if the arcs A B, B c, 

 etc., were trisected, we should obtain a quindecagon, or regular 

 polygon, the number of whose sides is 5 x 3, or 15. The learner 

 can readily calculate for himself the results of further subdivi- 

 sions of the arcs of the circle in the cases of the pentagon, de- 

 cagon, and quindecagon. 



The reader may have noticed already, that of all figures having 

 equal sides and equal angles there are only three kinds that 

 can be fitted together so as to cover a flat surface without 

 . .. any interstices or openings be- 

 \ tween the pieces used. These 

 are the equilateral triangle, the 

 square, and the hexagon. This 

 number of regular figures that 

 will cover a flat surface with- 

 ., / out the addition of any other 



-*~- figures, regular or irregular, may 



be further reduced to two when 

 we remember that the hexagon 

 itself is composed of six equilateral triangles, as may be seen 

 from Fig. 71, in which the thick lines show how any number of 

 equilateral triangles regularly disposed in rows may be grouped 

 into hexagons. The learner will find it a useful practice 

 to make drawings of the various regular polygons arranged in 



Fig. 71. 



order, in such patterns as his taste and fancy may dictate, 

 for models of tesselated pavement, marquetry, and parquetry. 

 An illustration of our meaning will be found in Fig. 72, a design 

 which consists of perpendicular lines of pentagons, black and 

 white in alternation, the lozenges or rhombuses (see Definition 30, 

 Vol. I., page 53) being divided by transverse perpendicular and 

 horizontal lines into right-angled triangles, which are also black 

 and white in alternation as they are r 

 contiguous to a white or black pen- 

 tagon respectively. The designs 

 made by the learner, for pavement, 

 may be done in pieces of red, black, 

 and white paper, or in paper coloured 

 to imitate various kinds of marble, 

 pasted together on cardboard. The 

 patterns for marquetry, which con- 

 sists of various kinds of ornamental 

 wood cut in small pieces and veneered 

 on a flat surface of deal or some corn- 



Fig. 72. 



mon timber, to form the top of a table or the panel of a cabinet; 

 and those for parquetry, or blocks of wood about an inch in thick- 

 ness, put together in symmetrical patterns to form floors, may 

 be cut out in paper stained in imitation of various kinds of 

 wood. It may be said that coloured paper in imitation of wood 

 may be purchased of any one who sells materials for bookbind- 

 ing, while paper in imitation of marble may be procured at the 

 same place for fine work, and from the paper-hanger for making 

 designs on a large scale. We have called the attention of the 

 reader to the method named above of making geometrical de- 

 signs in coloured paper, to show him how readily a knowledge 

 of geometry may be applied to art purposes. 



For filling up any space with small compartments all of the 

 same size and form, the hexagon is the most convenient, because 

 its shape assimilates more closely than either the equilateral tri- 

 angle or the square to that of the circle, the strongest form for 

 the arrangement of material to bear pressure, as in the case of 

 the barrel drain or circular sewer. Less material, too, would be 

 used in forming a number of hexagonal compartments than in 

 filling the same space with compartments in the form of equi- 

 lateral triangles or squares having the same depth and area of 

 base as the hexagon. A remarkable instance in nature of the 

 use of the hexagon for making the most of space for stowage, 

 with the least possible quantity of material, is found in the honey- 

 comb, of which the cells are hexagonal in form, terminating at 

 the bottom in a roof or floor, consisting of three parallelo- 

 grams, the opposite angles of which are about 110 and 70 

 respectively. 



The construction of a hexagon is easy enough, whether it be 

 required to inscribe it in a given circle, or to construct it on a 

 given straight line, because the radius of the circumscribing 

 circle is always equal to the side of the hexagon that it sur- 

 rounds. Thus, if we have to inscribe a hexagon in the circle 

 ACE (Fig. 73), it is manifest that all we have to do is to open the 

 compasses to the extent of A G, the radius of the circle ACE, 

 and divide the circumference into six equal parts by apply- 

 ing the opening of the compasses round the circumference, 

 and marking the points A, B, c, D, E, F, in succession. The 

 hexagon is then completed by joining A B, B c, c D, D E, E F, 

 F A, the chords of the six equal parts 

 into which the circumference has 

 been divided. 



PROBLEM LI. To construct a hexa- 

 gon on a given straight line. Let A B 

 (Fig. 73) be the given straight line 

 on which it is required to construct 

 a hexagon. From A and B as cen- 

 tres, with the distances A B and B A 

 respectively, draw two arcs intersect- 

 ing each other in G. The point G is 

 the centre of a circle circumscribing 

 the required hexagon. From the cen- 

 tre G with the distance G A describe the circle ACE; then from 

 A and B as centres, with the radius A B and B A respectively, 

 draw arcs cutting the circumference of the circle A c E in the 

 points F and c ; and from these points as centres, with the same 

 radius, draw arcs cutting the. circumference in the points E and 



D. Join A B, B C, C D, D E, E F, F A ; A B C D E F is the 



required, for it is constructed on the given straight line A B. 



Fig. 73. 



