333 PART II. The Solution of fome Problems in Plane Geometry. W Hen you have a Geometrical Queftion, or Problem, propos'd to be refolv'd by ALGEBRA, the first Thing you are to do is to illuftrate it by a Contruction, or an exact Description, if you can do it; but, if you can't, defcribe a Figure as nearly reprefenting the true one, and as fully fetting forth the feveral Parts of the Problem, as you can guefs; and fuppofe this Figure to be the true one, in which, fome of the Lines being known or given, and fome of them unknown, the fame Problem may therefore be folv'd after various Ways or Manners. Next, having confider'd the Nature of the Problem, you are thence to take a curfory View of the most obvious Ways of folving it; and, having chofen the best in your Opinion, then defign the unknown Line or Lines you have chofen along with the given Lines or Quantities (or as many of them as are fufficient to determine the Problem, and moft fit for your Purpofe.) in proper Symbols. Afterwards, if it be requifite, prepare the Figure by drawing and producing perpendicular, parallel, ftraight, &c. Lines in fuch Parts, and after fuch Manners, in and about it, as are fuitable to the Nature of the Problem, according to the Method of Solution you have before chofen, in order, by the Help of these Lines, or Mediums, to deduce a Connexion in the Operation, between the Lines and Quantities in the Figure defign'd by the Symbols: And then proceed to the Operation, wherein the foregoing Rules and Methods, in this Treatife (with a competent Knowledge in Euclid's Elements) will be your Guide. Now, as the moft fimple reduc'd Equation is beft, you are always to endeavour to attain to it in the eafieft Manner poffible: But, as it can't be reasonably expected that you can fucceed in this, in all Cafes, at the first, you may repeat your Endeavours in another Solution of the fame Problem; and fo on, till you hit upon the beft. Thefe Repetitions will, in molt Cafes, be needless to you after ufing yourself, with due Application, Application, to the Solutions of various Sorts of Problems, beginning with some of the eafiest, and proceeding gradually to more difficult ones, which (I believe) is the best Rule that can be given to a young Mathematician, in order to enable him to make a Progrefs in this moft noble Science. I will therefore now proceed to fuch Problems. Let { G O infcribe a Rhombus in a given long Square, i. e. Having the Sides of the Ld , viz. A B and BC given. To find the Segment BF or D E, which being cut off, the Remainder FC or AE will be the Side of the Rhombus fought. S│1AB=b(=DC) = 20. Suppofe 3 BF = x(DE)=? Then 4 FC=c-x(=BĆ-BF)=FA=EC=EA. 47.1 Euc. El. 5b b + x x = FAq= cc 2 cx + xx. 5, 620x = cc — bb. 6÷207x= C bb 2 C =15. Confequently 82 c. c + b :: cb. x. C b .. Conftruction. By 12. 6 Eucl. El. Make BG=2c•BH=c+ b :: BI BF⇒x= DE. Then draw the ftraight Lines AF and CE, and you will have the Rhombus AFCE infcribed in the given Ld ABCD, as was required. PROB. The Perimeter (viz. the Sum of the Sides A B, AC and BC) of any Ld A ABC, and the Perpendicular CD let fall from the Right-angle C to the oppofite Side A B, being given: Thence to find the Triangle. E 30 -y. And 4 BC =y. Then 47.1 Eucl. 6xx (= A B q) = b b — 2 b x + x x → 2 b y + 2xy +2yy (=ACq+BC q). Elem. The Ld As A B C and CBD are fimilar; confequently, 4.6 Eucl. El. 8 × 2 7x(AB). b-x-y (AC)::y (BC). c (DC). 8cx by xyyy. - 92cx=2 by — 2 x y — 2yy. 6 +9 10xx + 2 cx b b 2 bx + xx. 1126+2012 x = bi bb Confequently 13/6 + cb :: 16. x. I b 2 Construction. First, By 12. 6 Eucl. El. Make, As AE = b + c • A F =b::AG=b. AB=x. *Then on AB, as a Diameter, defcribe the Semi-circle Ac CB. At the Distance *N, B. Thus far the Problem is Simple; but in the following Part thereof (a Circle being necessarily us'd in the Construction) it is a Plane Problem. DC=c draw CC AB, which will interfect the Semicircle in the Points c and C. From either of the Points of Interfection C let fall CDL A B. Draw the Lines AC and BC, and the Triangle is conftructed. Suppose it done, and the Line (or 1) AD drawn ; then Then 5 DC b a. And fuppofe 4 BD=a= =? 47. 1 Eucl. El. 6d d-aa=ADq=cc-bb2ba-aa. zba—aa. 6, 7dd-cc + b b = 2 ba. 7268 dd-cc + b = a = 7/7. BE2 First, By 12. 6 Eucl. El. Make BE 2b. BFd+c Then BH BC a BD PROB. per In an Obtuse-angled Triangle, having the three Sides given feverally; to determine the Point where a Line let fall pendicularly from the Vertex fhall cut the Bafe produc'd, i. e. Having given AB, BC, AC, feverally in the annex'd Fig. to find CD. Suppofe it done, and AD drawn; then АВ =d=24. And 4 DCα= Then 5 BD = a + b. 47. 1 Eucl. El. 6dd-aa-zab—bb=DAq=cc-aa. 6, 7 dd-cc-bb2ab. Construction. First, By 12. 6 Eucl. El. Make CE 2b. CF=d+c :: CG=dc- CH. Then HC-BC= a DC = the Segment fought. N. B. The As A CB in thefe two Problems being defcrib'd (by 22. 1 Eucl. Elem.) the Segments fought may be Geometrically determin'd only by letting fall the LAD on the Base BC produc'd, if need be. Note likewife, There are other Methods of constructing the foregoing Problems; but I don't think it worth while to infert any more of them. *A a A |
