20880000 8072000e 1210800ee, 2464230, 25ee +125409000 + 376227008 + 3762270ee 246423025 49284605e ·+354683070 + 35468307e Viz. 213489045 + 15734402e + 87239,75ee=2741839 &c. Ꭰ And r+e= 13,7=r for a fecond Operation, with which you may proceed, as in the laft 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 publish'd. CHAP. V Practical Problems, and Kules 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 Dimenfions 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 Poles or Perches, (as in Surveying of Land, &c) then the Area will be fquare Perches, &c. Thefe 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. PROBLEM I. To find the Superficial Content, or Area of a Square, or of any Right-angled Parallelogram. Rule. { Multiply the Length into its Breadth, and the Product will be the Area required. (See Lemma 1. Page 302.) Example, Suppofe the Line AB=6 Yards, and the Breadth AC or B D 3 Yards, then ABX AC = 6 X3 18 will be the Number of Jquare Yards contain'd in the Area of the Parallelogram A B C D. This is fo evident by the Figure only, that it needs no Demonftration. A 3 PROBLEM II. 6 B To find the Area of any Oblique-angled Parallelogram, viz. either of a Rhombus or Rhomboides. Bule. { Multiply the Length into its perpendicular Height (or B That is, the Side ABX BPthe Area of the Rhombus ABCD. For if BP be drawn perpendicular to C D, and AG be made parallel to B P, then will GCPD and GP CD. Confequently ▲ AGC = ABPD, and ABGP Rhombus ABCD. But ABX BP ABGP. Therefore ABX BP, or CDX BP = the Area of the Rhombus ABCD. G P Example, Suppofe the Side A B 23 Inches, and the Perpendicular BP17,5 Inches, (being the shortest or nearest Distance between the two Sides, AB, and CD.) then ABX BP=2317,5 =402,5 Square Inches, being the Area of the Rhombus required. The like may be done for any Rhomboides whole Length and perpendicular Breadth is given. X x 2 PRO. PROBLEM III. To find the Superficial Content, or Area of any plain Triangle. Every plain 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 perpendiRule.cular Height, or half the Bafe into the whole Perpendicular, and the Product will be the Area. That is, BDX 1⁄2 C P, or 1⁄2 B D X C P A Area of ▲ B C D. B P N F D fore AA BCA BCP, and for the like Reafons A CFDA CP D. Therefore ▲ BCP+ ACPD ABCD. Con fequently BDX CP, or BDX CP will be the Area of A BG D. Example, Suppofe the Base B D = 32 Inches, and the perpendicular Height CP14 Inches. Then BDX CP = 16 × 14 224. Or B D X C P = 32 × 7 = 224. Or thus, 32 x 14 448. Then 2) 448 (224 the Area of the Triangle B CD 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 oppofite 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 PerpenKule.diculars, or half the Sum of the Perpendiculars into the Diagonal, and the Product will be the Area. That is, ACX BP + ED. Or ACX BP + ¦ E D = Area of the Trapezium A B C D. 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, Confequently, BP + EDX AC, or ; BP + E D× AG will be the Area of the Trapezium, as above. BP+EDX AC= 29 X 15,5 = 14 2 E D B P 478,5. Or ACX BP + ED = 33 X2 2 =478,5. Or thus, 29 X 33957. Then 2) 957 (478,5 any of thefe Products are the Area of the Trapezium ABC D. PROBLEM V. To find the Superficial Content or Area of any irregular Polygon or many fided Figure, which by fome Authors is call'd a Triangulate, because (as I fuppofe) it must be divided into Triangles, as in the annexed Figure ABCDFG; by which it is evident, that the Sum of the Area's of all thofe Triangles, found as in the laft Problem, &c. will be the Area of their circumferibing Polygon. B C A D F PROBLEM VI. To find the Superficies, or Area of any regular Polygon, viz. of any regular Pentagon, Heragón, Heptagon, Datagon, &c. General Kule. That is, Multiply half the Sum of its Sides into the Radius of the inferib'd Circle, or half the faid Radius into the Sum of the Sides, and the Product will be the Area required. AB+BD+DE+EF÷FG+GH+HK÷KA 2 :XCP the Area of the annexed Octagon; wherein it is evident, that its Area is compos'd of so many equal fofceles Triangles as there are Number of Sides in the Polygon, viz. of eight Ijofceles Triangles, whofe Bafes are the Sides of the Octagon, viz. AB=BD=DE, &c. And the Sides of those Triangles, CA, CB, CD, &o. are the Radius's of the circumfcribing Circle; and their perpendicular Heights, viz. PP, is the Radius of the inferibed Circle. But G But the Area of any one of those Triangles is This, being equally evident. in all regular Polygons whatfoever, makes the Rule general for finding their Area's. Now, because it is requir'd to have the Radius of the propos'd Polygon's infcrib'd Circle, I hall here infert (and demonftrate) the Proportions that are between the K AB XC P by F E Sides of feveral regular Polygons, and the Radius's both of their infcrib'd and circumfcribing Circles; the one will help to delineate or project the Polygon (if Occafion require it) and the other will help to find its Area. And First, Of an Equilateral Triangle. The Side of any Equilateral plain Triangle is in Proportion to the Radius of 0,25 0,75 = AG, confequently, ✔0,75=0,86602540=AG: B 0,80602540 &c. ་་་་ H A Then AG: AB:: AB: AH, by Theorem 13, that is, 0,8660254: 1 :: 1: 1,15470054 &c. AH, then AH= 0,57735027 = AC. Again, AG: DG:: DG: CG, that is, 0,8660254: 0,5:: 0,5 0,28867513=CG. Q. E. D. Now, by the Help of the First of these Proportions, it will be eafy to refolve the following Problem. PRO |