XVIII. SECTION The Application of Algebra to the Solution of Geometrical Problems. W refolved by algebra, you are, in the first place, HEN a geometrical problem is proposed to be to defcribe a figure that fhall reprefent, or exhibit the feveral parts or conditions thereof, and look upon that figure as the true one; then, having confidered attentively the nature of the problem, you are next to prepare the figure for a folution (if need be) by producing, and drawing, fuch lines therein, as appear moft conducive to that end. This done, let the unknown line, or lines which you think will be the eafieft found (whether required or not) together with the known ones (or as many of them as are requifite) be denoted by proper fymbols; then proceed to the operation, by obferving the relation that the feveral parts of the figure have to each other; in order to which, a competent knowledge in the elements of geometry is abfolutely neceffary. As no general rule can be given for the drawing of lines, and electing the most proper quantities to fubfti. tute for, so as to, always, bring out the moft fimple conclufions (becaufe different problems require different methods of folution); the best way, therefore, to gain experience in this matter, is to attempt the folution of the fame problem feveral ways, and then apply that which fucceeds beft, to other cafes of the fame kind, when they afterwards occur. I fhall, however, fubjoin a few general directions, which will be found of use. 1o. In preparing the figure, by drawing lines, let them be either parallel or perpendicular to other lines in the figure, or fo as to form fimilar triangles; and, if an angle be given, let the perpendicular be oppofite to that angle, and alfo fall from the end of a given line, if poffible. 2o. In electing proper quantities to fubftitute for, let thofe be chofen (whether required or not) which lie nearest neareft the known, or given parts of the figure, and by help whereof the next adjacent parts may be expreffed, without the intervention of furds, by addition and subtraction only. Thus, if the problem were to find the perpendicular of a plane triangle, from the three fides. given, it will be much better to fubftitute for one of the fegments of the bafe, than for the perpendicular, though the quantity required; because the whole bafe being given, the other fegment will be given, or expreffed, by fubtraction only, and fo the final equation come out a fimple one; from whence the fegments being known, the perpendicular is eafily found by common arithmetic : whereas, if the perpendicular were to be firft fought, both the fegments would be furd quantities, and the final equation an ugly quadratic one. 3°. When, in any problem, there are two lines or quantities alike related to other parts of the figure, or problem, the best way is to make ufe of neither of them, but to fubftitute for their fum, their rectangle, or. the fum of their alternate quotients, or for fome line or lines, in the figure, to which they have both the fame relation. This rule is exemplified in Prob. 22, 23, 24, and 27: 4. If the area, or the perimeter of a figure be given, or fuch parts thereof as have but a remote relation to the parts required, it will, fometimes. be of ufe to affume another figure fimilar to the propofed one, whereof one fide is unity, or fome other known quantity; from whence the other parts of this figure, by the known proportions of the homologous fides, or parts, may be found, and an equation obtained, as is exemplified in Prob. 25 and 32. Thefe are the most general obfervations I have been able to collect;. which I fhall now proceed to illuftrate by proper examples. The bafe (b), and the fum of the hypothenufe and perpendicular (a) of a right-angled triangle ABC, being given; to find the perpendicular. Let Let the perpendicular BC be denoted by x; then C the hypothenufe AC will 2 be expreffed by a-x: but (by Euc. 47. 1.) AB 1+ BC2 AC; that is, b + x2 = a2 2ax + xx; The diagonal, and the perimeter of a rectangle, ABCD, being given; to find the fides. D Put the diagonal BD-a, half the perimeter (DA C + AB) = b, and AB x; then will AD = b a; which, folved, gives The area of a right-angled triangle ABC, and the fides of a rectangle EBDF infcribed therein, being given; to determine the fides of the triangle. Put DF a, DE = b, BC = x, and the measure C of the given area ABC d: then, by fimilar triangles, we fhall have x b(CF): a (DF):: x (BC): AB = D F ax Bd, and consequently ax2 2bd : which, folved, a a Having the lengths of the three perpendiculars PF, PG, PH, drawn from a certain point P within an equilateral triangle ABC, to the three fides thereof; from thence to determine the fides. Let lines be drawn from P to the three angles of the triangle; and let CD be perpendicular to AB: call PF, a; PG,b; PH, c; and AD: then will AC (AB)=2x, and CD (= VACAD2) = √3xx= *V3; and confequently the area of the whole triangle ABC (= CD × AD) = xxy3. But this triangle is xx√3. composed of the three triangles APB, BPC, and APC; A H G P DF B whereof the respective areas are ax, bx, and cx. Therefore we have xx√3 = ax + bx + cx ; and from thence, by divifion, x= a+b+c √3 V. PROBLEM Having the area of a rectangle DEFG, infcribed in a given triangle ABC; to determine the fides of the rectangle. Let CI be perpendi cular to AB, cutting DG in H; and let CI = a, AB = b, DG =x, and the given area CC: then it will be, as b Let the perpendicular BC be denoted by x; then The diagonal, and the perimeter of a rectangle, ABCD, being given; to find the fides. D Α' Put the diagonal Ba, half the perimeter (DA + AB) b, and AB Cx; then will AD = b x; and therefore, AB2 + AD being BD, we' have x2+ b2 2bx + x1 a; which, folved, gives √2a2 b2+b B x= 2 PROBLEM III. The area of a right-angled triangle ABC, and the fides of a rectangle EBDF infcribed therein, being given; to determine the fides of the triangle. |