r, : 



MATHEMATICS. -PRACTICAL OHOMETRT. 



[PROBLEM nr. va 



pas* it round the pin at C, carrying it up in the direction 

 of the line of new road to D. Bring the line back to 

 the pin at E, and holding the end of the tape line fast 

 at E, stretch the line tight When stretched, this will 

 give a triangle, having tides three (E C), four (C D) and 

 five (E D) feet respectively in length, of which D C will 

 be perpendicular to E C, and form one side of the new 

 road required. 



This method of setting off a perpendicular is taken 

 from that beautiful theorem in Euclid (Book I., Prop. 

 XI. VII.), where in this case, ED, the hypothenuse of 

 the triangle, is equal to 6 feet, and the two sides E C 

 and C D, 3 and 4 feet respectively ; therefore 6" - 3 1 + 

 4 1 or 23 9 + 16. The sum of the three sides being 

 13 or the whole length of line used ; but any numbers 

 may be used (consistent with the length of the tape line), 

 provided they fulfil the required value. It may be 

 easily remembered that any multiple of the numbers 

 used, viz., 3, 4, 6, will suffice. 



PROBLEM IIL 



To draw a straight line parallel to a given straight 

 line (A B), and through any given point (C). (Euclid, 

 Book I., Prop. XXXI.) 



IST METHOD. Take any two points, D and E (Fig. 

 14), in the given line Fig. 14. 



AB, the further apart c 



from each other the 

 better; with the centres 

 D and E and radii equal 

 to the distance between 

 the point C and the line 

 A B * describe two arcs, 

 and draw a straight line to touch or be a tangent to 

 both arcs, then the line G F is parallel to A B, 



2so METHOD. From Fig. is. 



the point C (Fig. 15) 

 draw the line C D per- 

 pendicular to A B ; at 

 another point anywhere 

 in the line A B, but as 

 far from D as conve- 

 nient, draw E H per- 

 pendicular to A B. 

 Make EH equal to CD, 

 and draw through the 

 points C and H the straight line F G, which will be 

 parallel to A B. 



3RD METHOD. Take any point D (Fig. 16), the further 

 from C the better, in Fig. is. 



the line A B ; join D C, 



and with the centre D, 



and radius U C (or any 



other convenient radius 



which will cut D C), 



describe an arc cutting 



A B in E ; also with 



the centre C, and the game radius, describe an arc D H 



again with the centre D, and a radius equal to the dis- 

 tance between C and E, Fig. 17. 



describe an arc cutting 



the hitter arc in H ; 



draw the line F G pass- 

 ing through the points 



C and H ; the line F G 



will be parallel to A B. 

 4ra METHOD. Take I 



any point D (Fig. 17) 



in the line A B ; join 



DC and produce it, - v 



making C O equal to A >s - y 



C D ; with the centre O 



and radii O C and O D, 



describe two arcs, the hitter cutting A B in E ; join O E, 



Tb dtetaae* between C and the line ABU euily got by placing one 

 -' opening them until the other point jus 

 . rallel to A B at any distance apart, by 

 the radii of the area, as in Fig. 14. 



\ 



cutting the lesser arc in the point H ; draw the line 



? G passing through the points C and H, then F G will 



>e parallel to A B. 



STH METHOD. Take any point D (Fig. 18) in the line 

 A B ; join D C and Fig. 18. 



produce it ; with the 



lentre C, and radius 



1), describe a circle, 



cutting A B in E, and, 



DC produced, in O ; 

 with the centres E and 



3, and the same radius, m , *r~^T 



describe arcs intersect- / 



ng in the point 1 L ; 

 draw F G passing 

 through the points C 

 and H ; then F G 

 will be parallel to A B. 



PROBLEM IV. 



To divide a given straight line (AB) into any given number 



(say 6) of equal parts. 



Through A (Fig. 19) draw a line A D at any angle with 

 A B, and through B Fig. 19. 



draw B C parallel to 

 A D; from A set off 

 six equal parts on the 

 line A D (those equal 

 parts should be as nearly 

 the length of the re- 

 quired equal parti of A B 

 as can be estimated), 

 and from B on the line 

 B C set off the same number (six) equal parts of th 

 same length as before ; join B and o, 1 and 1, 2 and 2, 

 3 and 3, 4 and 4, 5 and 5, A and 6 ; then the points 

 where those lines cut A B, viz., in a, b, c, d, e, will divide 

 the line A B into the required number (six) of equal 

 parts. 



PROBLEM V. 



To find a mean or mean proportional between two given 

 Kn<a(ABandBC). 



Draw a straight line ABC (Fig. 20) equal in length 

 Fig. 20. to AB and BC together ; 



bisect AC in O, and with 

 the ceutre O and radius 

 O A, or O C, describe a 

 semicircle ; from the 

 point B = B C, the small 

 line measured from C 

 A O C, draw a line 

 AC, 



point of Uw eonpuM at C, and openi 

 bock**, ro may be drawn parallel 

 enljr aatag that diitanc* aa the radii 



AB 



/ 



A." 



"o B" 



4 perpendicular to 

 and meeting the 



circle in the point D, then B D is a mean or mean pro- 

 portional between A B and B C. 



PROBLEM VL 



To describe a square on a given straight line (A B). (Euclid, 



Book I., Prop. XL VI.) 

 From the point A (Fig. 21) draw A E perpendicular 

 to AB (Problem II.), and from AE 

 cut off A D equal to A B ; draw D C 

 and C B parallel to A B and A D re- 

 spectively, intersecting at the point 

 C ; or, with B and D as centres, and 

 radii equal to A B, describe arcs in- 

 tersecting at the point C, and join 

 D C and B C, then A B C D is the 

 square required, t 



Fig. 21. 



PROBLEM VII. 

 To describe a square which shall be equal to any number 



of given squares (1, 2, 3, 4, 5). 

 Draw any two lines, A B and A C (Fig. 22), at right 



+ Of couree, any rectangle whoae length and breadth are Riven, can be 

 constructed in a aimilar manner, by making the perpendicular equal to 

 the breadth, and either drawing parallels or using the length of the aide* 

 ai radii. 



