PROBLEM X. The Side of any regular Decagon being given, to find its Area. Example. Let the given Side be 14 Inches long; then as I: 1,53884176:: 14: 21,543784 = the Radius of the infcrib'd Circle; and 14 x 570 is half the Sum of its Sides. Lastly, 21,543784 × 70 = 1508,06488 the Area required. Fifthly, For a Dodecagon. The Side of any regular Dodecagon (viz. a Polygon of twelve equal Sides) is in Proportion to the Radius of its Viz. S Circumfcribing Circle, as 1 to 1,93185165 &c. as 1 to 1,86632012 &c. BACA: 1 : 1,93185165 11±12araa 12 C13aaaa· aa = bb HP aaee—2a0 ४ 1aa2a ✓ bb — ¦ aa ·4bbaa + 4b4 =364 = 3 1314aa2bb = √3 = 1,7320508075 14 + 2bb 15 ca=2bb + √3 = 3.7320508075 15 w2 16 a✓ 3,73205080751,93185165=CA Again 170bb CP. viz, CF-PFCP 17. Hence 18CP ✓ aabb 1,86632012. Q. E. D. Cun Confectary. Hence, if the Side of any regular Dodecagon be given, the Radius of it's infcrib'd Circle may be eafily obtain'd, and thence the Area found; as in the laft Problem. The Work of the 'foregoing Polygons, being well confider'd, will help the young Geometer to raise the like Proportions for others, if his Curiofity requires them: And not only fo, but they will alfo help to form a true Idea of a Circle's Periphery and Area, according to the Method which I fhall lay down in the next Chapter for finding them both. CHAP. VI. A new and eafy Method of finding the Circle's Periphery and Area to any affign'd Exactness (or Number of Figures) by one Equation only. Alfo a new and facile Way of making Natural Sines and Cangents. LET us fuppofe (what is very easy to conceive) the Circle's Area" to be compos'd or made up of a vaft Number of plane Ifofceles Triangles, having their acuteft Angles all meeting in the Circle's Center. And let us imagine the Bafes of thofe Triangles fo very fmall, that their Sides and their Perpendicular Heights, viz. the Radius's of their circumfcrib'd and infcrib'd Circles (vide Problem 6.) may become fo very near in Length to each other, as that they may be taken one for another without any fenfible Error: Then will the Peripheries of their circumfcribing and infcribed Circles become (altho' not co-incident, yet) fo very near to each other, as that either of them may be indifferently taken for one and the fame Circle. But how to find out the Sides of a Polygon (viz. the Bafes of those Ifofceles Triangles) to fuch a convenient Smallnets as may be neceffary to determine and fettle the Proportion betwixt a Circle's Diameter and its Periphery (to any affign'd Exactnefs) hath hitherto been a Work which requir'd great Care and much Time in its Performance; as may eafily be conceiv'd from the Nature of the Method us'd by all those who have made any confiderable Progress in it, viz. Archimedes, Snellius, Hugenius, Meetius, Van Culen, &c. Thefe proceeded with the bifecting of an Arch, and found the Value of its Chord to a convenient Number of Figures at every fingle Bifection, Y y 2 Bifection, repeating their Operations until they had approach'd to the Chord defign'd. And this Method is made choice of by the learned Dr Wallis in his Treatife of Algebra; wherein, after he hath given us a large Account of the different Enquiries made by feveral (very eminent in Mathematical Sciences) in order to find out fome easier and more expeditious Way of approaching to the Circle's Periphery, as in Chap. 82, 84, 85, 86, and feveral other Places, he comes to this Refult, (Page 321.) " 'Tis true, faith he, we might in like Manner proceed by con"tinual Trifection, Quinquifection, or other Section, if we had " for these as convenient Methods of Operation as we have for "Bifection: But because Euclid fhews how to bifect an Arch "Geometrically, but not to trifect, &c. and the one may be "done (Algebraically) by refolving a Quadratick Æquation, but "not those other, without Æquations of a higher Compofition, I "therefore make Choice of a continual Bifection, &c." And then he lays down these following Canons. 56 √:2√2+ √2+√3 48 ✓:2-✓:2+ √2+√2+ √3 96 &c. √:2−√:2+ √: 2+√: 2+√2+√3 192 √:2−√:2+ √:2+ √:2+ √:2+ √:2+√3 384 √:2~√:2+√:2+√:2+√:2+ √:2+ √:2+√3 &c. 768 Sic. How tedious and troublesome the Work of these complicated Extractions is, I leave to the Confideration of those, who either have had Experience therein, or out of Curiofity will give themselves the Trouble of making Trial. Again, in Page 347, the Doctor inferts a particular Method propos'd by Libnitius, publifh'd in the Ala Eruditorum at Leipfick, for the Month of February 1682, in order to find the Circle's Area, and confequently its Periphery, which is this: As I to} − } + { +15-13 13 +14-15, &c. infinitely fo is the Square of the Diameter to the Circle's Arca. But this convergeth fo very flowly, that it is not worth the Time to puriue it. :: I I I fhall here propofe a new Method of my own, whereby the Circle's Periphery, and confequently its Area, may be obtain'd infinitely infinitely near the Truth, with much greater Eafe and Expedition than either that of Bifection, or that of Libnitius, as above, or any other Method that I have yet feen; it being perform'd by refolving only one Equation, deduced by an eafy Procefs from the Property of a Circle, (known to every Cooper) which is this: The Radius of every Circle is equal to the Chord of one fixth Part of its Periphery. That is, ADD HHG, the Chords of one third Part of the Semicircle, are each equal to A F its Radius. Then, if the Arch AD be trifected, it will be ABBZ = ZD. R=AF = 1 Let = AD = 1 1 a = AB. D H Then R:a::a: Be A And 2 R:a::R FB: BZ:: Fe: ex = AD. A AFB, and BAe, are alike. ( Because the L at B And AB Ae=Dx, &c. 4 x &c. 53 R1a-aaa-RRc. That is, 3a-aaa=1. Here a the Chord of Part of the Circle. 3 Next, To trifect the Arch AB. Universal Aritas Let 13-3a the laft Chord. 13227y3-2735 +977 y9 = a3 1 x 3 39y-3y2 = 3a projs 2006 12200 30 Again, To trifect the Arch whereof y is the Chord. 2187a7 +5103a) +5103a' — 2835a13 + 945a15 = y7 627a 19683a9-59049a11 + 78732a13 — 161236a15 = y9 2 x 30 781043-810a5 +270a7 — 30a9 = 30y3 3 x 27 86561a5 4× (6561a5 10935a7 + 7290a? — 2430a11 + 405a13 +27a's 27ys 9919683a7 45927a9 +45927a11 — 25515a13 +8505a15=9y1 '+} = 27a-819a3 +7371a5-30888a7 +7 Here a the Chord of Part of the Circle. 162 Proceeding on in this Method of continually trifecting the Arch of every new Chord and ftill connecting the produced Equations into one, as in the two laft Trife&ions, 'twill not be difficult to obtain the Chord of any affign'd Arch, how small foever it be. Now, in order to facilitate the Work of raising these Equations to any confiderable Height, 'twill be convenient to add a few useful Obfervations concerning their Nature, and of fuch Contractions as may be fafely made in them; which, being well understood, will render the Work very easy. 1. I have obferv'd, that every Trifection will gain or advance ons Figure in the Circle's Periphery, but no more. Therefore fo many Places of Figures as are at firft defign'd to be perfect in the Periphery, fo many Trifections must be repeated to raise an Equation that will produce a Chord answerable to that Defign. 2. I have also found, that all the fut erior Powers (of a) whose Indices are greater than the Number of Trifections, (viz. whofe Indices are greater than the Number of defign'd Figures) may be wholly rejectad as infignificant. 3. When once the Number of Trifections and thence the highest Power (of a) is determin'd, the third Procefs (viz. the third Trifection) may be made a fix'd or conflant Canon; for by it, and Multiplication only, all the fucceeding Trifections (how many foever they are) may be compleated without repeating the feveral Ínvoluti gris. |