PROBLEM VII. The Side of any Equilateral plain Triangle being given, to find its Area. Example, Suppofe the Side of the propos'd Triangle A B C to be 25 Inches, viz. AB BC=CA=25 First 1: 0,866254:: AB 25: 21,650635 BP by Theorem 13. Then AP (=C A) XBP the Area of AABC by Rule to Problem 3, that is, 12,5 X 21, 650635= 270,6329 the Area in Square Inches. Or this Problem may be otherwife refolv'd, A thus: Let b⇒ A PAC. A B. But □ AB-AP Then 26 = B P C BP. By Theorem 11. That is, 4bb-bb3bb□ BP. Confequently, ✔ 3bb = B·P. Then b3bb BPX AC. viz. 3bbbb X√3 = the Area of the Triangle. Secondly, For a Pentagon. The Side of any Pentagon is in Proportion to the Radius of AD X GF 1 (because AB AG=DF, and AD=AF=DG) hence it will be AD 1,61803398, then □ AD-DH= = AH by Theor. 11. But DH=AB, therefore AD-AB =AH=1,5388417h. Again, AH: A D:; AD: AX=2AC. For AHD and A ADX are alike. Ergo =2AC=1,70130161. Hence AC=0,85~65080 Q. E. D. But AH-AC-CH=0,68819096, &c. From hence it will be easy to refolve the following Problem. PROBLEM VIII. The Side of any regular Pentagon being given, to find its Area. Example, Suppofe the given Side to be 15 Inches long, then it will be, as 11,53884176:: 15:22,0826264 the perpendicular Height; and by the general Rule 22,0826264×165,619698 the Area requir'd. its Thirdly, For an Ddagon. 1: The Side of any regular Octagon is in Proportion to the Radius of I D Let {b = BA=1; a= CA {BD, y=Bx eBD, and y = Bx Then I 2a be y. viz. DA: BA::DB: Bx That is 4 X bb Again 4 4aa bb ODA ODB=□ BA. By Theorem it. 5 4bbaa4aayy = bbbb 6aa = yy. Saayy. For CxBx and CxBx□ CB=aa -b+ b+ 2 - 2bbaa —— 1 b+ 9 to 2 ᄆ 8 aaaa 9/a4 IO aa Or 2a4. • 2bbaa + b+= b+ — {b+ = {b+ bb11aa=bb +✔ b+ 12a=√:bb+✔b=1,30656296, &c. = CA Then 13 aa 13 w2 bb CP, viz. CH-HP-OCP 14√aa-1bb=1,20710678 &c. =CP. From hence 'twill be eafy to find the Area of any Octagon. PROBLEM IX. The Side of any regular Octagon being given, to find its Area. Example, Suppofe 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 448 is half the Sum of its Sides, and 48 X 14,48528136695,2935 the Area required. Fourthly, For a Decagon. The Side of any regular Decagon (viz. a Polygon of ten equal. Its {Circumfcribing Circle, as 1 : to 1,61803398 &c. Circle, as 1 to 1,53884176 &c. Viz. {BA: CA:: 1: 1,61803398 aa DA:BA::DB:Bx Szay = be and 2y= be a P H 32y:e1: 1,61803398. See Pentagon. 4 1,61803398 bb□ CP. viz. CF-O PF□ CP. By Theorem 11. ✔2,61803398-0,25=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 X 570 is half the Sum of its Sides. Laftly, 21,543784 X 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. Let b s Circumfcribing Circle, as I to 1,93185165, &c. as I to 1,86632012, &c. BA: CA: 1: 1,03185165 Demonstration. BA=1. a=CA as before And exA; then a — e = Cx 11 13, + C 78 5√bbaa ee 2ae +ee Bx=Cx 72a√ bb aa = 2ae 8bb 9 aa 10 2a II 4bbaa I bb aa2abb. aa aabb. aa = bb aaaa = b+ 12 aaaa- 4bbaa == -b4 HPF 13 aaaa- -4bbaa + 4b+ = 36+ = 3. 1, 2 14 aa2bb 3 = 1,7320508075 142bb 15 aa = 2bb + √3 = 3,7320508075 aa 15 w 16 a = √ 3,7320508075 1,93185165=CA Again 17 aabb=CP. viz. □ CF — □ PF = Q 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 inferib'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 raife 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 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 vaft Number of plain Iffceles 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 Smallness as may be neceffary to determine and fettle the Proportion betwixt a Circle's Diameter and its Periphery (to any affign'd Exactness) 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 thofe who have made any confiderable Progrefs in it, viz. Archimedes, Snellius, Hugenius, Matius, 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 single Y y 2 Bifection, |