Suppofe the Line D F Then B parallel to AB; CA being produced to F And 6 BC: CA:: DC: CF But 7 BCbb + aa. Let AF e, and FD = y ✓: bb + aa =a+e 51055-25a+aa= CD Per Fig. 11ss-25a+aaaa+2ae+ee+yy=□ CF+ FD aa 12 ss 2sa=2ae+ee+yy II But 13 ddee+yy=□AF+□ FD Let 15 2x=ss-dd 14, 1516x-sa—ae This Equation being brought out of the Fractions, and into Numbers, will become 2018a+ + 125409a3 2464230,25a2 + 35468307a= 274183922,25; which being divided by 2018, the Co efficient of the higheft Power of a, will be-a4+62,145 bai - 1221,1254 +17575,9697a=135869,138875 &c. And And from hence the Value of a may be found, as in the aft Problem, due Regard being had to the Signs of every Term. This Work of reducing, or preparing Equations, for a Solution by Divifion, hath always been taught both by ancient and modern Wri-. ters of Algebra, as a Work fo neceffary to be done, that they do not fo much as give a Hint at the Solution of any adfected Æqua tion without it. Now it very often happens, that, in dividing all the Terms of an Equation, fome of their Quotients will not only run into a long Series, but also into imperfect Fractions (as in this Equation above) which renders the Solution both tedious and imperfect. To remedy that Imperfection, I fhall here fhew how this Equation (and confequently any other) may be refolv'd without fuch Divifion, or Reduction. Let b 2018. c=125409. d = 2464230,25 br4 4brrre-6brree = bat' Then willers + 3erre + 3cree=+ca3 -drr2dre +fr+fë.. G dee == daa This is plain and eafily conceived. The next Thing will be, how to estimate the firft Value of r; and to perform that, let G be divided by b, only fo far as to determine how many Places of whole Numbers there will be in the Quotient; confequently, how many Points there must be (according to the Height of the Equation.) Thus b 2018) G = 274183922,25 (130000 7238, &c. Now from hence one may as eafily guess at the Value of r, as if all the Terms had been divided. That is, I suppose r = 10, which being involved, &c. as the Letters above direct, will be 20880000- 8072000€ 1210800ee + 125409000 + 37622700e + 3762270ee 246423025-49284605e- 2464230,25ee +354683070 + 35468307€ Viz. 213489045 + 15734402e + 87239,75ee=2741839 &c. And Ꭰ D· &c. Firftr 10 r+e = 13,7⋅r for a fecond Operation, with which you may proceed, as in the last Problem, and fo on to a third Operation, if Occafion require fuch Exactness. But this may be fufficient to fhew the Method of refolving any adfected Equation, without reducing it; which is not only very exact, but also very ready in Practice, as will fully appear in the laft Chapter of this Part, concerning the Periphery and Area of the Circle, &c. wherein you will find a farther Improvement in the Numerical Solution of High Equations than hath hitherto been publifh'd. CHAP. V. Practical Problems, and Rules for finding the Superficial Contents, or Area's of Right-lin❜d Figures. Before I proceed to the following Problems, it may be convenient to acquaint the Learner, that the Superficies or Area of any Figure, whether it be Right-lin'd or Circular, is compofed or made up of Squares, either greater or lefs, according to the different Meafures by which the Dimensions of the Figure are taken or meafur'd. That is, if the Dimenfions are taken in Inches, the Area will be compofed of fquare Inches; if the Dimenfions are taken in Feet, the Area will be compofed of fquare Feet; if in Yards, the Area will be fquare Yards; and if the Dimenfions are taken by Pales or Perches, (as in Surveying of Land, &c.) then the Area will be Square Perches, &c. These Things being understood, and the the Definitions in the 283 and 284 Pages well confider'd, will help to render the following Rules very easy. ! To find the Superficial Content, or Area of a Square; or of any Right-angled Parallelogram. Rule. {be the Area required. (See Lemma 1. Page 302.) Multiply the Length into its Breadth, and the Product will Example. Suppofe the Line AB=6 Yards, and the Breadth AC or BD 3 Yards, then AB x AC = 6 X3 18 will be the Number of Square Yards contain'd in the Area of the Parallelogram ABCD. This is fo evident by the Figure only, that it needs no Demonftration. 3 PROBLEM II. 6 B To find the Area of any Oblique-angled Parallelogram, viz. either of a Rhombus or Rhomboides. Multiply the Length into its perpendicular Height (or · Rule. { Breadth) and the Product will be the Area requir'd. B That is, the Side ABX BP = the Area of the Rhombus ABCD. For if BP be drawn perpendicular to CD, and AG be made pa-" rallel to B.P, then will GCPD and GP CD, Confequently A AGC = ABPD, and AB GP Rhombus ABCD. But A B x BP ABGP. Therefore AB x BP, or CD x BP = the Area of the Rhombus ABCD. Example. Suppofe the Side AB 23 Inches, and the Perpendicular BP = 17,5 Inches, (being the forteft or nearest Diflance between the two Sides, A B, and C D.) then ABX BP = 23 × 17,5 =402,5 Square Inches, being the Area of the Rhombus required. The like may be done for any Rhomboides whofe Length and perpendicular Breadth is given. X x 2 PRO To find the Superficial Content, or Area of any plane Triangle, Every plane Triangle is equal to half its circumfcribing Parallelsgram, (41. e. 1.) which affords the following Rule. Multiply the Bafe of the given Triangle into half its perpenRule.dicular Height, or half the Bafe into the whole Perpendicular, and the Product will be the Area. A C F That is, BDx CP, or BDx CP Area of ABCD. For AC BP, A B=CP, and BC is common to both AA; therefore A ABCA BCP, and for the like Reafons ACFDA CPD. Therefore A BCP + ACPD ABFD. Confequently BD × CP, or B D x BCD. B P D CP will be the Area of A Example. Suppofe the Base B D = 32 Inches, and the perpendicular Height CP 14 Inches. Then BDX CP 16 x 14 = 224. Or BD × ÷ C P = 32x7224. Or thus, 32 x 14 = 448. Then 2) 448 (224 the Area of the Triangle BCD in fquare Inches. PROBLEM IV. To find the Superficies, or Area of any Trapezium. Firft, divide the given Trapezium into two Triangles, by drawing a Diagonal from one of its acute Angles to the opposite Angle; and let fall two Perpendiculars (from the other two Angles) upon the Diagonal, as in the following Figure. Then Multiply half the Diagonal into the Sum of the two PerpenRule.diculars, or half the Sum of the Perpendiculars into the Diagonal, and the Product will be the Area.. That is, AC BP+ED. Or AC x 1⁄2 BP + ¦ ED= For the AABC is Half its circumfcribing Parallelogram; and -the AACD is alfo Half of its circumfcribing Parallelogram, as hath been prov'd at the laft Problem. Confequently, |