PROBLEM VII. The Side of any Equilateral plain Triangle being given, to find its Area. Example, Suppose the side of the propos'd Triangle A B C to, be 25 Inches, viz. AB= BC= CA = 25 B First 1:0,866254: : AB= 25:21,650635 = BP by Theorem 13. Then AP (=iC A) X BP = the Area of A ABC by Rule to Problem 3, that is, 12,5 X 21, 630635 = 270,6329 the Area in square Inches. Or chis Problem may be otherwise resolv’d, A P C thus: Let b = AP= AC. Then 2b A B. But OAB - AP=OBP. By Theorem 11. That is, 4bb - bb = 3bb = BP. Consequently, ✓3bb = B P. Then b v 3bb = BPX A C. viz. v 3bbbb 3 = the Area of the Triangle. Secondly, For a Pentagon. 50,85065080 &c. its Infcrib'd Circle, As 1 : To 0,68819096 &c, Perpendicular Height, 1,53884176 &c. B } G Demonftration. X H н F Ergo O AD Ergo AH =2AC =1,70130161. Hence AC=0,85 65080 But AH - AC=CH=0,68819096, &c. Q. E. D. From hence it will be easy to resolve the following Problem. PROBLEM VIII. 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 22,0826264x 4 = 1653619698 the Area requir'd. В Thirdly, For an Ddagon. The Side of any regular OEtagon is in Proportion to the Radius of its { Circumscribing Circle, As I : to 1,30656296&c. Viz.{ BA:CP :: 1:1,20710678 Demontration. D A Draw the Right-line D B, and from the Point B let fall the Perpendicular B z upon the Diameter DA. Then will A BDA and ADXB H be alike, by Theorem 10 and 12. Let { Sb BA=1.0=CA = BD, and y =Bx DB 4aayy G may 2 2 bb But = bb That is 4aayy 4 400 bb ODA - ODB=BA. By Theorem it. 5 4bbaa - 4aayy = bbbb aa = yy. For Cx = Bx Cx + = OCB 4 Xbb 5, 6 بس 9 Іо! аа 2 52 61 714bbaa --2a+=ht. Or 2a+ - 4bbaa wbt 21 8 Jaaaa abbaa = – bb =Vbt Io + bb naa= bb+vb+ 12 a=v:bbtv1b=1,30656296, &c. = CA Then 13 aa--- bb=CP, viz. CH-HP- OCP 13 w 141V aa-10b=1,20710678 &c. =CP. From hence 'cwill be easy to find the Area of any Octagon. PROBLEM IX. Example, Suppose the Side given to be 12 Inches long; First, as 1:1,20710678::12:14,48528136 = the Radius of its inscrib'd Circle; then 12 X 4 = 48 is half the Sum of its Sides, and 48 x 14,48528136 = 695,2935 the Area required. Fourtbly, For a Decagon. Sides) is in Proportion to the Radius of Circle, as 1 : to 1,53884176 &c. B b=BA=I.a=CA D Then 20:b::e: That is, DA:BA::DB:Bx 2ay be 1 2 P and 2y= But 1e a 3 Again be H = 2y = == aa — *lb = OCP. 6 viz. O CF-PF=O CP. By Theorem Ir. 71 V 2,61803398—0,25=1,53884176=CP PRO Уу P R O B L E M X. The Side of any regular Decagon being given, to find its Area. Example. Let the given Side be 14 Inches long; then, 29 1:1,53884176:: 14: 21,543784 = the Radius of the inscribd Circle ; and 14 X 5 = 70 is half the Sum of its Sides. Lastly, 21,543784 X 70 = 1508,06 488 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 { , 1 Circumscribing Circle, as I : to 1,93185165, &c. , c. (B 4: CP:::1,86632012 B 2 2 HPF Demonliration. bb-Bx=ee D By Figure. 3 o Bx = aa 0 Bx = 0 Cx 5 Х aa = 2ae 4 7 8. bb I aa – 2a v bb - a= l - 2ae 72 8 aa = aa + bb-1aa - 2a v bb Laa 9 + 10 20 v bb aa = bb IO II/4bbaa ht II 12 aaaa 4bbaa 13, C 0 13 aaaa – 4bbaa + 46+ = 3b+ = 3 I. w? 14 aa - 2bb=V 3= 1,7320508075 14 + 2bb 15 aa = 2bb tv 3= 3,7320508075 15 w 16a = V 3,7320508075 = 1,93185165 = CA Again 17 aa-bb=0CP. viz. O CF-PF=QCP 17, Hence 18 CPSV aa-bb= 1,86632012 ОСВ 72 av bb 2a 9 aa аааа - 6+ 2 Q. E. D. Con Confetary. The Work of the 'foregoing Polygons, being well consider'd, will help the young Geometer to raise the like Proportions for others, if his Curiosity 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. C H A P. VI. A new and easy Method of finding the Circle's fperiphery and area to any asign’d Exactness (or Number of Figures) by one Æquation only. Also a new and facile Way of making Natural Signs and Tangents. LET us suppose (what is very easy to conceive ) the Circle's Area to be compos’d or made up of a vast Number of plain Isosceles 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 inscrib'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 circumscribing 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 Tsosceles Triangles) to such a convenient Smallness as may be necessary to determine and settle the Proportion betwixt a Circle's Diameter and its Periphery (to any align’d Exaélness) 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 considerable Progress in it, viz. Archimedes, Snellius, Hugenius, Mætius, 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 single Y y 2 Bilection, |