□ AD Ergo AH = 2AC = 1,70130161. Hence AC=0,85065080 PROBLEM VIII. Q. E. D. The Side of any regular Pentagon being given, to find its Area. Example, Suppose the given Side to be 15 Inches long, then it will be, as 1 : 1,53884176::15:22,0826264 the perpendicular Height; and by the general Rule 22,0826264×= 165,619698 the Area requir'd. Thirdly, For an Dtagon. The Side of any regular Octagon is in Proportion to the Radius of Circumfcribing Circle, As 1: to 1,30656296 &c. its Infcribd Circle, As 1: to 1,20710678 &c. BA:CA::1:1,30656296 Demonstration. Let Then Then will & BDA and A DxB be alike, by Theorem 10 and 12. b=BA=1.a=CA le = BD, and y = Bx 12a:b::e:y. viz. DA: BA: : DB : Bx That is xbb Again aa = yy. and Cx+ Bx □ DA−❑DB = □ BA. By Theorem it. 54bbaa - даауу = bbbb For CxBx CB aa 5, 7 6174bbaa-2a+b+. Or 2a-4bbaa-b+ 8C9 2 س و Io aa 2bbaa= 2bbaa+b=6+6+6+ bb=6+ 10+ bb11aa=bb++ 13 2 12a=√:bb+16=1,30656296, &c. = CA Then 13 aa bbCP, viz. CH- HP= □CP 2 14aa-bb=1,207 10678 &c. =CP. From hence 'twill be easy to find the Area of any Octagon, PROBLEMIX. The Side of any regular Octagon being given, to find its Area. Example, Suppose the Side given to be 12 Inches long; First, as 1:1,20710678:: 12: 14,48528136 = the Radius of its infcrib'd Circle; then 12 X 4 = 48 is half the Sum of its Sides, and 48 × 14,48528136=695,2935 the Area required. Fourthly, For a Decagon. The Side of any regular Decagon (viz. a Polygon of ten equal Sides) is in Proportion to the Radius of Its Circumfcribing Circle, as I : to 1,61803398 &c. Infcrib'd Circle, as 1: to 1,53884176 &c. BA:CA::1:1,61803398 Viz.BA: CP::1:1,53884176 32y:e::1:1,61803398. See Pentagon. Ie be Ie 4 1,61803398-2y=이=이 (viz. CF- PF= CP. By Theorem II. 72,61803398-0,25-1,53884176=CP 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 1:1,53884176:: 14:21,543784 = the Radius of the infcrib'd Circle; and 14 × 5 = 70 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 Circumfcribing Circle, as 1 : to 1,93185165, &c. its Inscribed Circle, as I: to 1,86632012, &c. BA:CA::1: 1,03185165 Viz.BA:CP::1: 1,86632012 11 + 13, C I 2 CB 72abb 78bb 89aa I 102abb aa I 4 Bx 2ae + ee aa2ae 2abb aaaa+bb. I aa=bb II 4bbaa - aaaa = b+ I I 4 HPF aaee-2ae aa-2abb 464 = 364 = 3 2bb = 3 = 1,7320508075 14+2bb 15 aa = 2bb+3= 3,7320508075 16 a = √ 3,7320508075 = 1,93185165 = CA Again 17 aa-bb☐ CP. viz. CF-PF CP 17, Hence 18 CP = √ aabb = 1,86632012 Q. E. D. = Con Confectary. Hence, if the Side of any regular Dodecagon be given, the Radius of its infcrib'd Circle may be easily obtain'd, and thence the Area found; as in the last 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 so, but they will also help to form a true Idea of a Circle's Periphery and Area, according to the Method which I shall lay down in the next Chapter for finding them both. CHAP. VI. A new and easy Method of finding the Circle's Periphery and Area to any assign'd Exactness (or Number of Figures) by one Æquation on'y. Also a new and facile Way of making Natural Signs and Tangents. LET us fuppofe (what is very easy to conceive) the Circle's Area to be compos'd or made up of a vast Number of plain Ifofceles Triangles, having their acutest Angles all meeting in the Circle's Center. And let us imagine the Bases of those Triangles so very small, that their Sides and their Perpendicular Heights, viz. the Radius's of their circumscrib'd and infcrib'd Circles (vide Problem 6.) may become so very near in Length to each other, as that they may be taken one for another without any sensible Error: Then will the Peripheries of their circumfcribing and inscribed Circles become (altho' not co incident, yet) so 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 Bases of those Ifofceles Triangles) to such a convenient Smallness as may be necefsary to determine and fettle the Proportion betwixt a Circle's Diameter and its Periphery (to any assign'd Exaltness) hath hitherto been a Work which requir'd great Care and much Time in its Performance; as may easily 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, Mœtius, Van Culen, &c. These proceeded with the bisecting of an Arch, and found the Value of its Chord to a convenient Number of Figures at every fingle Bifection, Yy2 Bisection, repeating their Operations until they had approach'd to the Chord design'd. And this Method is made choice of by the learned Dr. Wallis in his Treatise of Algebra, wherein, after he hath given us a large Account of the different Enquiries made by several (very eminent in Mathematical Sciences) in order to find out some easier and more expeditious Way of approaching to the Circle's Periphery, as in Chap. 82, 84, 85, 86, and several other Places, he comes to this Result, (Page 321.) "'Tis true, faith he, we might in like Manner proceed by con"tinual Trifection, Quinquisection, or other Section, if we had " for these as convenient Methods of Operation as we have for "Bisection: But because Euclid shews how to bisect an Arch 66 Geomatrically, 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 Composition, I "therefore make Choice of a continual Bisection, &c. And then he lays down these following Canons." The Subtenfe of rinto 6 ✓: 2-V3 into 12 V:2-V:2+√3&c. 24 √2-√2+√:2+√3 V:2-V:2+1:2+√:2+√3 V: 2-V:2+V2+√2+√2+√3 V2-V:21:2 V:2-V:2+√: 2+√: 2 48 96 192 384 768 &c. &c. 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, publish'd in the Alta Eruditorum at Leipfick, for the Month of February 1682, in order to find the Circle's Area, and confequently its Periphery which is this: I As I: to+-+-+音一六十一, &c. infinitely:: fo is the Square of the Diameter to the Circle's Area. But this convergeth so very flowly, that it is not worth the Time to purfue it. I shall here propose a new Method of my own, whereby the Circle's Periphery, and confequently its Area, may be obtain'd infinite |