CHA P. II. The First Rudiments, or Leading and Preparatory IN N order to perform the following Problems, the young Geometer ought to be provided with a thin freight Ruler, made either of Brafs or Box-wood, and two Pair of very good Compaffes, viz. one Pair called Three-pointed Compaffes, being very useful for drawing of Figures or Schemes, either with Black Lead or Ink; and one Pair of plain Compaffes with very fine Points, to measure and fet off Distances; alfo he should have a very good Steel Drawing Pen: And then he may proceed to the Work with this Caution; that he ought to nake himself Master of one Problem before he undertakes the next: That is, he ought to understand the Defign, and, as far as he can, the Reafon of every Problem, as well as how to do it; and then a little Practice will render them very easy, they being all grounded upon thefe following Poftulates. Poftulates or Petitions. 1. That a Right-line may be drawn from any one given Point to another. 2. That a Right-line may be produced, encreased, or made longer from either of its Ends. 3. That upon any given Point (or Center) and with any given Distance (viz. with any Kadius) a Circle may be defcribed. Two Right-lines being given, to find their Sum and PROBLEM II. To bifect, or divide a Right-line given (as AB) into two equal Parts (10. e. 1.) .D." From both Ends of the given Line (viz. A and B) with any Radius greater than half its Length, defcribe Two Arches that may cross each other in two Points, as at D and F; then join those Points D F with a Right-line, and it will bifect the Line AB in the Middle at C; viz. it will make AC=CB; as was required. A PROBLEM III. To bifet a Right-lin'd Angle given, into two equal Angles.. (9. e. 1.) Upon the Angular Point, as at C, with any convenient Radius, defcribe an Arch as AB; and from those Points A and B, defcribe two equal Arches croffing each other, as at D; then join the Points C and D with a Right-line, and it will bifect the Arch AB, and confequently the Angle; as was requir'd. PROBLEM IV. A D B At a Point A, in a Right-line given AB, to make a Right-lin'd Angle equal to a Right-lin'd Angle given C. (23. e. 1.) Upon the given Angular Point C defcribe an Arch, as FD, (making C D any Radius at pleasure) and with the fame Radius defcribe the like Arch upon the given Point A, as fd; that is, make the Arch fd equal to the Arch FD; Then join the Points A and f with a Right-line, and it will form the Angle requir'd. C F D PRO PROBLEM V. To draw a Right-line, as FD, parallel to a given Right-line AB, that fhall pass thro' any affign'd Point, as at x, viz. at any DiStance requir'd. (31. e. 1.) M...... F A H -D -B N Take any convenient Point in the given Line, as at C, (the farther off x the better;) make Cx Radius, and with it upon the Point C, defcribe a Semicircle, as HMx N; then make the Arch HM equal to the Arch x N; thro' the Points M and x draw the Right-line FD, and it will be parallel to the Line A C, as was requir❜d. PROBLEM VI. C To let fall a Perpendicular, as C x, upon a given Right-line AB, from any affign'd Point that is not in it, as from C. (12. e. 14) Upon the given Point C defcribe fuch an Arch of a Circle as will cross the given Line AB in two Points, as at d and f; Then bifect the Distance between those two Points df (per Probl. 2.) as at x. Draw the Right-line C x, and it will be the Perpendicular requir'd. PROBLEM VII. L B To erect or raise a Perpendicular upon the End of any given Right-Line, as at B; or upon any other Point affign'd in it. (II. e. 1.) Upon any Point (taken at an Adventure) out of the given Line, as at C, defcribe fuch a Circle as will pass through the Point from whence the Perpendicular muft be raifed, as at B, (viz. make CB Radius): And from the Point where the Circle cuts the given Line, as at A, draw the Circle's Diameter AC D; then from the Point D draw the Right-line D B, and it will be the Perpendicular as was requir'd. PRO PROBLEM VIII. To divide any given Right-line, as A B, into any proposed Number of equal Parts. (10. e. 6.) At the extream Points (or Ends) of the given Line, as at A and B, make two equal An gles (by Prob. 4.) continuing their Sides AD and B C to any fufficient Length; then upon thofe Sides, beginning at the Points A and B, fet off the propofed Number of equal Parts (Suppofe 'em 5.) If Right-lines be drawn (cross the given Line) from one Point to the other, as in the annexed Figure, thofe Lines will divide the given Line A B into the Number of equal Parts required. PROBLEM IX. To defcribe a Circle that shall pass (or cut) thro' any Three Points given, not lying in a Right-line, as at the Points A B D. Join the Points A B and B D with Right-lines; then bisect both thofe Lines (per Problem 2.) the Point where the bifecting Lines meet, as at C, will be the Center of the Circle required. The Work of this Problem being well understood, 'twill be easy to perform the two following, without any Scheme, viz. B D 1. To find the Center of any Circle given. (1. e. 3.) By the last Problem 'tis plain, that if three Points be any where taken in the given Circle's Periphery, as at A, B, D, the Center of that Circle may be found as before. 2. If a Segment of any Circle be given, to compleat or defcribe the whole Circle. This may be done by taking any three Peints in the given Seg ment's Arch, and then proceed as before. PRO PROBLEM X. Upon a Right-line given, as A B, to defcribe an Equilateral Triangle. (1. e. 1.) Make the given Line Radius, and with it, upon each of its extream Points or Ends, as at A and B, describe an Arch, viz. A C and B C; then join the Points AC and B C with Rightlines, and they will make the Triangle requir'd. A PROBLEM XI. Three Right-lines being given, to form them into a Triangle, (provided any two of them, taken together, be longer than the Third) (22. e. 1.) Let the given Lines be Make either of the shorter Lines (as AC) Radius, and upon either End of the longest Line (as at A) defcribe an Arch; then make the other Line CB Radius, and upon the other End of the longest Side (as at B) defcribe another Arch, to crofs the First Arch (as. at C): Join the Points C A and C B with Right-lines, and they will form the Triangle required. PROBLEM XII. Upon a given Right-line, as A B, to form a Square. (46. e. 1.) D Upon one End of the given Line, as at B, erect the Perpendicular B D, equal in Length with the given Line, viz. make BD = A B ; that being done, make the given Line Radius, and upon the Points A and D defcribe equal Arches to cross each other, as at C; then join the Points CA and CD with Right-lines, and they will form the Square required. A PRO |