To the extent possible under law, the Text Creation Partnership has waived all copyright and related or neighboring rights to this keyboarded and encoded edition of the work described above, according to the terms of the CC0 1.0 Public Domain Dedication (http://creativecommons.org/publicdomain/zero/1.0/). This waiver does not extend to any page images or other supplementary files associated with this work, which may be protected by copyright or other license restrictions. Please go to http://www.textcreationpartnership.org/ for more information.
AS an Introduction to our Subject, at our first entry, we shall begin with the Definition of Geometry, and the Explication of those Terms that are hereafter to be made use of.
Geometry is a Science which treats of Magnitudes, and it is the principal part of the Mathematicks.
1. Magnitude is a continued Quantity, which consists either in Lines or Angles, in Surfaces or Bodies.
2. A Line is a length without breadth, whose Extremities are called Points.
3. A Point is something which cannot be divided.
4. A Line is right, crooked, or mixt: a right Line is that which goes without turnings from one•• Point to another. A crooked Line is that which goes off from its bounds, by one or more turnings: and a mixt Line is that which is part right and part crooked.
5. A Surface is length with breadth: whose bounds are called Lines.
6. These Lines are either Parallel or Oblique. Parallel Lines are those which are every way equally distant from each other. Oblique Lines are those which meeting at a Point make an Angle.
7. It is called a right Angle when one of the Lines is perpendicular upon the other. It is called an Oblique Angle when one of the Lines is not perpendicular upon the o∣ther. It is called a right-lined Angle when ••t is made up of two right Lines. It is called a Curve-lined Angle, when it is made up of two crooked Lines. It is called a mixt Angle when one of the Lines is right, and the other curv'd. It is called an A∣cute Angle when it is less than a right one.
And it is called an Obtuse Angle, when it is greater than a right Angle.
8. A Figure is a Magnitude bounded by one or more Lines and Surfaces, where∣of some are plain, some solid.
9. A Circle is a plain Figure bounded by a Circumference, the middle Point whereof is called the Center, from which all right Lines drawn to the Circumfe∣rence are equal.
10. A Diameter is a right Line, which goes thro' the Center, and touches the Cir∣cumference in two Points.
11. A Semi-Circle is the half of an En∣tire Circle.
12. An Oval is a plain Figure bounded by its own Circumference, within which no Point can be taken from which all right Lines drawn to the Circumference can be equal.
13. A Triangle may be considered ei∣ther with respect to its Sides, or its Angles: in respect to its Sides, it is eithe•• 〈…〉〈…〉 lateral, which has three equal Sides: or an Isosceles, which has two Sides only e∣qual: whose third is called the Base: or a Scalene, which has three unequal sides: in respect of its Angles, it is either right-an∣gled, which has one right Angle: or A∣cute-angled,
which has three acute Angles: or Obtuse-angled, which has one Obtuse and two Acute Angles. One of the Lines in a right-angled Triangle is called the Base: the second the Perpendicular: and the third the Hypothenusal.
14. A Parallelogram is a Figure which has four Sides, and four Angles: and it is either equilateral which has four equal sides, or in-equilateral whose sides are unequal.
15. An equilateral Parallelogram has either right-Angles, as a Square or Oblique Angles, as a Rhombus.
16. An in-equilateral Parallelogram has either right-Angles as a rectangled Paralle∣logram: which is likewise called a long∣square; or not, and then it is a Rhomboid: every other square Figure is called a Tra∣pezium.
17. A Tetraëdron is a solid Figure, or a Body which is bounded by four equal Triangles.
18. A Pyramid, is a Body bounded by several Triangles, which making one Sur∣face for a Base, meet all in one Point.
19. A Prism is a Body which has two Bases, equal and parallel each to other.
20. A Parallelepiped is a Body bounded by six Parallelograms, whereof the two
opposite ones are always equal and parallel.
21. A Trapezium is a Body bounded by plain Surfa••es, which from neither a Prism, nor a Parallelepiped.
22. A Cone is a Body made by the turn∣ing of a right-angled Triangle round a Cir∣cle, the angular Point of the right angle being fixed in the Center.
23. A Cylinder is a Body made by the turning of a right-angled Parallelogram round two Circles one at the top, the other at the bottom.
24. A Cube is a Body bounded by six Squares.
25. An Octaëdron is a Body bounded by eight equal and equilateral Triangles.
26. A Dodecaëdron is a Body bounded by twelve equal and equilateral Pentagons.
27. An Icosaëdron is a Body bounded by twenty equal and equilateral Triangles.
28. A Pentaëdral Prism is a Body bound∣ed by three right-angled Parallelograms, having two equilateral Triangles for its Bases.
29. A Rhomb is a Body bounded by six equilateral Oblique-angled Squares.
30. A Rhomboid is a Body bounded by 4 Oblique-angled Parallelograms, having two Oblique-angled equilateral Squares for their Bases.
MArk upon the given Line any two points AB where you please, and by them draw two Arches EF, in draw∣ing a right Line by EF you have your Parallel.
1. Draw an Arch from the point G, which touches the given Line only in K.
2. With the distance GK, draw ano∣ther Arch upon the given Line MN.
3. The Line drawn by GN shall be parallel to the given Line HI.
1. Take a distance from the given point (what you please) to the right, as AB.
2. Take the same distance to the lest, as AC.
3. Take BC and draw two Arches which intersect each other at D.
4. Joyn D and A together, and you will have your Perpendicular.
1. Draw a Circle to the point B, from the center C, which may be taken at plea∣sure.
2. From the point D, where the Circle cuts the given Line, draw a right Line through the Center C, till it touches the Circle on the other side, as at E.
3. Joyn E and B together, and you will have your Perpendicular.
1. Draw an Arch from the point G, which may cut the given Line in two places, as CD.
2. From CD draw two Arches under the Line which shal•• intersect each other at E.
3. Joyn E and G together, and you will have your Perpendicular GF.
1. From the points AB draw two Arches which shall intersect above the given Line at C, and below it at D.
2. Joyn C and D together, and you shall have your Line cut into two equal parts.
1. Draw CD a Line at pleasure, and mark as many parts as the given Line AB ought to have; six for Example, upon this Line as CE.
2. Take CE and draw two Arches which may intersect in F.
3. Draw all these six parts in F.
4. Take the given Line AB, and lay it upon the Lines FE and FC, as GH.
5. Joyn GH together, you will have your given Line AB divided into six equal parts.
1. Let the given Angle be BAC.
2. Draw the Arch ADE.
3. Draw a Line IL, and upon this Line an Arch with the distance AD, as FG.
4. Set the distance DE upon the Arch FG, as FH.
5. Draw a right Line from the point I to the point H, which will make an Angle e∣qual to the given Angle.
1. Let the given Angle be BAC draw the Arch ADE.
2. From the points DE draw two Ar∣ches which shall intersect each other at F.
3. The Line drawn from F to A will cut the Angle into two equal parts.
1. With the points AB, make two Ar∣ches which shall intersect at D above, and at E below the given points.
2. Draw a right Line from the point D to G.
3. Do the same thing with BC, so that they may intersect in FG.
4. Draw another right Line from F to G, and there where this Line intersects the first, as for instance at G, you will find the Cen∣ter, by which you may draw the Circle through the three given points.
1. Divide the given Circle into four equal parts ABCD, with the Center mark'd at E.
2. Take the distance AB and put it from E upon the Perpendicular, as EF, and with this distance draw a Circle, as big again as that given.
3. To make a Circle twice as big, take AF, and put it from E upon the first Per∣pendicular,
as EG, and the distance EG will be the Semi-diameter of the demanded Circle.
4. To make a Circle thrice as big, put the distance AG upon the same Perpendicu∣lar, as EH, and the distance EH will be the Semi-diameter of the demanded Circle. Thus you may enlarge a Circle for ever.
1. Draw a Line, and thereupon the Circle BAC.
2. From C draw another Circle through B, as CBD.
3. Take care that these two Circles inter∣sect each other, as in EF.
4. From the point of intersection E, draw right lines through BC, which shall touch the two Circles in GH.
5. From the point E, draw an Arch from G to the point H.
6. Keep this distance, and do the same thing from the other point of intersection F, as IK, and so your Oval will be compleat.
1. Upon a straight Line, draw the Arch ABC.
2. Take BC, and fixing one point of the Compass upon B, draw from the point C another Arch, which shall touch the same Line in D.
3. Take the first Center A, and draw a∣nother Arch from D, that shall touch the line in E.
4. Return to the point B, and from it as a Center draw from E another Arch to F.
5. Take A as a Center, and draw an Arch from F to G; and so on for ever.
1. Divide a Circle into four equal parts, as ABCD, whose Center is E.
2. From A thro' E draw the Arch FG; so also draw from B the Arch IH, and from C draw and the third KL; and at last from D draw the last Arch MN.
3. Having by these Arches divided the Circle into twelve equal parts, afterwards di∣vide every one of these twelve into three o∣ther equal parts, and then you will have thirty six equal parts.
4. Divide afterwards every one of these thirty six into ten equal parts, and your Circle will be divided into three hundred and sixty equal parts.
Take a given line AB, and from those two points draw two Arches which shall intersect at C; from C draw two right lines to A and B; these will make an Equilateral Triangle.
1. Let the given Triangle be ABC, draw a line at pleasure, and mark thereup∣on the length AB, as DE.
2. Take the distance BC, and from the point E draw an Arch.
3. Take the distance AC, and from D draw another Arch which shall intersect the first at F.
4. Joyn FE and FD together, you will have a Triangle equal to the given one A¦BC.
1. Let ABC be the Triangle to be divi∣ded into five other Triangles: first divide the longest side BC into five equal parts, so
that the fifth part, marked at D may be joyned to A.
2. Divide the longest of the two other sides into four equal parts, and let one of them marked G be joyned to D.
3. Divide the remaining part DC into three equal parts, and joyn that marked F by a right Line to G.
4. Divide the remainder GC into two e∣qual parts, and joyn that marked E by a right Line with F. So you may have five equal Triangles: viz. ABD, ADG, GDF, GFE, and EFC.
1. Take the Line AB, and raise a Perpen∣dicular upon B, of the same height as BC.
2. With the same distance make two Ar∣ches from A and C which shall intersect at D.
3. Joyn DA, and DC by right Lines: and you will have a Square.
1. Draw a long Line AB, and upon B raise a small Perpendicular BC.
2. Take the distance BC and by the point A draw an Arch.
3. Once more take the Line AB, and by
the Point C make an Arch which shall inter∣sect the other at E; joyn E to A and C and you will have this Parallelogram.
1. Draw a long Line AB, and mark upon it the distance AC.
2. Raise a Perpendicular upon C, equal to AC, as CD.
3. Divide the Line AC into two equal parts, as AE, EC.
4. Set the Point of the Compass upon E; and with the other Point draw an Arch from D, which shall touch the given Line at F.
5. By the Points A, C, and the distance AF, draw two Arches intersecting at G.
6. Take only the distance A, C, and by the Points G, A, draw two Arches intersect∣ing each other in I: so likewise, by the Points G, C, draw two other Arches which shall in∣tersect each other in K.
7. Joyn AI, IG, GK, and KC together by right Lines, and you have your Pentagon.
1. Divide the Diameter of AB the given Circle into seven equal parts.
2. Raise a Perpendicular upon B three times as long as the Diameter AB, and a se∣venth part over, as BC.
3. Draw a right Line from C to D the Center of the given Circle, and that gives you the Triangle.
This Problem must be wrought after the same manner as the foregoing one.
1. Divide the Diameter AB into seven equal parts.
2. Double this Diameter, and add a se∣venth part of itself to it, as AC.
3. Divide the first Diameter AB into two equal parts, as AD.
4. Divide the Line DC into two equal parts, and taking the middle Point E, draw from D. to C an arch, as DFC.
5. Raise a Perpendicular upon E to touch the Circle in F: this Line EF shall be a side of the required Square: the rest is wrought by the eighteenth Problem.
1. Let one of the given Squares be AB¦CD, and the other EFGH.
2. Joyn them both, so that the sides BC, FE, may make one right Line CBE.
3. Joyn A to E, and raise a Perpendicu∣lar upon E of the same length, as EK.
4. Keep the same length, and from the points KA, make two Arches to intersect in I.
5. Joyn KI and IA, and your Square will be made.
Let the given Parallelogram be ABCD.
1. Lengthen the Base DC, and add to it the other side of the Parallelogram BC, as CE.
2. Divide the distance DE into two e∣qual parts, and from the middle point F draw an Arch, from D towards E.
3. Raise a Perpendicular upon C to touch the Arch in G; this will be a side of the re∣quired Square; set it upon the first length∣ned Line, and you will have another side, as CI.
4. Keep the same distance, and from the points GI, make two Arches to intersect in H; joyn IGH, and HI together, and your Square will be made.
Let the given Square be ABCD.
1. Lengthen the side AB, and take the distance BD, and set it upon the lengthen∣ed side AB, as AE, and it will be a side of a Square double to that given at first.
2. Afterwards observe what was said a∣bove in the Eleventh Problem.
1. From the points AB let fall P••rpendi∣culars to the lower Line, as AE, and joyn it afterwards to the Perpendicular B, as BF.
2. And when you have the right angled Parallelogram AEBF, work on by the twen∣ty fifth Problem.
1. From the points BC, let fall Perpen∣diculars upon the Line, and set the distance BE, upon the other point C, as CF.
2. Joyn EF together, and the sides BC EF will make a right angled Square equal to a given Rhomb.
1. Lengthen the Base CB, and divide the same Base into two equal parts, as CD, DB.
2. From the point B let fall a Perpendi∣cular equal to BD, as BF.
3. Lengthen BF upwards, and from the point A draw a Parallel to the Base CB, which shall cut the lengthened Line BF in E.
4. Divide the distance FE into two e∣qual parts, and from G the middle point draw an Arch from E to F; and from the point H, where this Arch touches the first lengthened Line CB, to B shall be the first side of the Square, which laid upon the Line BE, will give another side BI; afterwards
from the points IH, make two Arches which shall intersect each other in K; and joyn H¦K and KI together, and your Square will be made.
1. Divide the Base AB into two equal parts, as AD, DB.
2. Divide the side CB into two equal parts, as BE, EC.
3. Draw a right Line from D to E, what length you please, and take the distance DE, and set it from the point E, to lengthen it, as EF.
4. Take DF, and from the point B make an Arch.
5. Take DB, and from F intersect this Arch in G.
6. Joyn FG and GB together, and your Parallelogram will be made.
1. Lengthen the side of the Square AB, adding double its own length to it, as BF.
2. Divide the side of the Square BC into two equal parts, as BE, EC.
3. Take BF, and from the point E make an Arch.
4. Take BE and intersect this Arch from the point F in G.
5. Afterwards joyn EG and GF toge∣ther, and you have the Parallelogram re∣quired.
Let the given Figure be ABCDE.
1. Take any one of the points, which you please, A for instance:
2. From A make as many Triangles as there are opposite Angles in the Figures A¦BC, ACD, ADE.
3. Then take a Line at pleasure, of the length of AB, as FG.
4. From the point G, with the distance BC, make an Arch.
5. With the distance AC intersect this Arch, from F in H, and joyn GH toge∣ther.
6. From the point H, with the distance CD make another Arch, and intersect it from F in I, with the distance AD, and this must be joyned to H.
7. Take AE, and from this point F make the last Arch; intersect it from I in K, with
the distance DE, and joyn it with IF: so the Figure FGHIK, will be equal to the given Figure.
1. MAke an equilateral Triangle ABC.
2. Place another equilateral Trian∣gle upon every side, as ABE, BCD, & ACF, and your Tetraëdron is made.
1. Make an equilateral right-angled Square ABCD.
2. Upon every side place another Square equal to the first, as ABEF, BCGH, DCIK, ADLM.
3. Joyn EF, GH, IK, and LM together by right Lines.
4. Place another Square equal to the rest upon one of these four, as KINO, and the Cube will be made.
1. Make a Parallelogram ABCD, and lengthen the sides AC, BD, above and be∣low.
2. Mark any distance above AB you please, as AE, BF; and mark the same under CD as CG, DH, then joyn E, F, and G, H, toge∣ther.
3. Above EF, mark the distance AC or BD, at I and K: and joyn I, K, together.
4. Lengthen EF and AB; upon the leng∣thened Lines mark the distance AC, or BD, as FL, BM, EN and AO.
5. Joyn L and M, N and O, together, and your Parallelipiped will be made.
1. Make a long Square ABCD.
2. Divide AB into twenty two equal parts.
3. Upon what Point you please raise a Per∣pendicular, of the length of seven of these twenty two parts, as F.G.
4. Upon the Line DC mark the distance FB, as DH.
5. Let fall a Perpendicular from H of the length of FG as HI.
6. Divide FG and HI into two equal parts: and from their middle Points draw Circles: and the Cylinder will be made.
1. From the Point A taken at pleasure draw an Arch.
2. Divide it into twenty two equal parts; and joyn the Extremities G, H, with A.
3. Let fall a Perpendicular where you please from the Arch, equal to seven of those parts as BC.
4. Divide BC into two equal parts: from its middle point, draw a Circle, which shall be a Base of a Cone made upon the Arch GH.
1. Make a right-angled Parallelogram ABCD.
2. Divide AB and CD into three equal parts, as AE, EF, FB, and CG, GH, HD.
3. Joyn EG and FH together by right Lines.
4. From E, and F make two Arches to in∣tersect in I, and joyn IE, and IF together.
5. Upon GH make an equilateral Tri∣angle K, and your Prism will be made.
1. Upon one Line make three equilateri∣al Triangles, as ABC, BDE, and DFG.
2. Lengthen the Line CG; and mark upon the lengthened Line the distance EG, as GH.
3. Joyn H, F together, and with the same distance from D, F, make 2 Arches to inter∣sect at K, and from C, E, two others to in∣tersect at I.
4. Joyn CI and IE together, as also DK and FK, and the Octaëdron will be finished.
1. Make a regular Pentagon ABCDE: the Center whereof may be F.
2. Take the distance AF, and therewith from every side draw two Arches mutually intersecting, as ABG, BCH, CDI, DEK, and EAL.
3. From the Points G, H, I, K, L, with the distance AB, describe five Circles, and divide each of them into five equal parts: and you will have five other equal Pentagons about your first.
4. With the distance HM or HN from M and N draw two Arches to intersect each
other in O: from this Point, by the same distance, draw another Circle thro' N and M, and divide it as you did the other into five equal parts, and as M, N, P, Q, R, and there joyn them all together.
5. Take the distance ON, and from P, and Q draw two arches to intersect each o∣ther in S: which is a Center from whence a Circle shall be drawn thro' PQ, which shall afterwards be divided into five equal parts, QP, PW, WU, UT, and TQ.
6. Repeat the same operation upon TQ, TU, UW, and WP, and your Dodecaëdron will be finished.
1. Make five equilateral Triangles upon the same Line, as ABC, BDE, DFG, FHI, and HKL.
2. From the Points AC make two arches which shall intersect each other in M.
3. Joyn AM and ML together by right Lines.
4. Upon MC raise the equilateral Trian∣gle N, upon CE, O; upon EG, P: upon GI, Q; and lastly upon IL, R.
5. Repeat the same operation upon their Bases: for instance, set the equilateral Tri∣angle S upon AB; the Triangle T upon
BD; the Triangle U upon DF; W upon FH; and at last the Triangle X upon HK: which will finish the Icosaëdron.
1. Upon the same Line make four equila∣teral Triangles. GIM, IKN, KLO, LHP.
2. Joyn M, P together by a right Line: as also GM, IN, KN, LO and H.P.
3. Add to the Line NI its own length beyond it, as IQ.
4. Add to the Line OK double its own length, as KR, RS.
5. Joyn RI and SQ together.
6•• Add to the Line IR its own length be∣yond it as RT: do the same to NK, as KU: then joyn U, T, together, and your solid Rhomb is made.
1. Make an equilateral Triangle ABC, and from BC another D.
2. Add to the Line CD twice its own length, as DE, EF.
3. Add to DB its own length, as BG.
4. Add to CB thrice its own length, as BH, HI, IK.
5. With the distance IB from the point G make an arch.
6. Make another arch with the distance BG from I which shall intersect the first at L, and joyn G, L and L, I by right Lines.
7. Add to GH thrice its own length; HM, MO, & OP, and joyn M and I together.
8. From K, and M draw two Arches to intersect at N, and joyn KN, NM, MD, OE, and PF together: and the Rhomboid will be finished.
1. Make a Square ABCD: and divide AB and CD into five equal parts, as CE, EF, FG, GA, HD; as also AI, IK, KL, LM, and MB.
2. Joyn EI, and FK, and GL, and HM, and DB together.
3. Divide one part of each of these Lines into twenty two equal parts: as FG and KL.
4. Divide one of those twenty two parts into three other equal parts.
5. Take eighteen of these twenty two parts, and 2/•• of one more, as FN, and KO; and draw two arches from K, L, as also from F, G, which shall intersect in P and Q: from P, and Q as Centers draw two Circles, which being divided into five equal parts by the distances FG and KL, your Figure will be compleated.
1. FRom the point A taken at pleasure draw an arch, which divide into three equal parts, BC, CD and DE.
2. Joyn AB, AC, AD and AE with right Lines; as also BC, CD, and DE with other right Lines.
3. From the points C, D, with the distance CD, draw two Arches to intersect in F.
4. Joyn CF and DF, and the Pyramid will be compleated.
1. From the point A draw an Arch: di∣vide it into four equal parts, BC, CD, DE, and EF, which are to be joyned together by right Lines, as also AB, AF, AC, AD, and AE.
2. From the point C, let fall a Perpendi∣cular equal to CD, as CG.
3. With the distance CG from G, D, draw two Articles to intersect each other in H.
4. Joyn HG and DH together by right lines and you have the required Pyramid.
1. From the point A draw a circle, in which first mark a larger arch, as BC, then a smaller one, as CD, then mark a third arch (DE) equal to the first, and a fourth (EF) equal to the second.
2. Joyn AB, AC, AD, AE, AF, as also BC, CD, DE, EF by right lines.
3. From C let fall a Perpendicular equal to CD, as CG.
4. Take the distance CG, and from B draw an arch.
5. Take the distance BC, and from G draw another arch to intersect the first in H; joyn BH and HG together, and the Problem is finished.
1. Draw an arch from A, which divide into five equal parts, BC, CD, DE, EF, FG.
2. Joyn them all to A, and one to ano∣ther, B to C, C to D, D to E, E to F, F to G, by right lines.
3. Divide one of these five parts, as DE, into twenty two equal parts, and one of them into three.
4. Take eighteen of the twenty two parts, and two of the three little parts, and from the points DE, make two arches to intersect each other in H, which is the Center, from whence you are to draw a circle big enough to be divided into five parts, equal to the distance DE. When you have this Penta∣gon for a Base, the work is done.
1. Draw an arch from A, and divide it into six equal parts, B, C, D, E, F, G, I.
2. Joyn them to the center A, as also B to C, C to D, D to E, E to F, F to G, G to I.
3. Take one of these six parts (DE) and from D and E make two arches to intersect in H.
4. From H draw a circle big enough to be divided into six parts equal to DE.
5. Having this Hexagon as a Base for the Pyramid, the Probl••m is wrought.
1. From A draw an arch, and divide into seven equal parts, B, ••, D, E, F, G, H.
2. Joyn them to A, and also B to C, C to D, D to E, E to H, H to G, G to F, F to I, by right lines.
3. Divide one of these parts, FG for in∣stance, into seven equal parts, and then one of these seven into eight more; then take eight of the larger parts, and the half of a little one, and with that distance from the points FG, make two arches to intersect each other in K; for which as a center draw a circle thro' FG, which must be divided in∣to seven parts equal to FG, which when joyned together, will serve for a Base of the Pyramid required.
1. From A draw an arch, divide it into eight equal parts, B, C, D, E, H, I, F, G, joyn these points one to another, and also to A by right lines.
2. Divide one of these parts, as FG, into seven equal parts, and one of these seven in∣to seven others.
3. Take nine of the larger, and one and a half of the smaller parts, and with that distance from F and G, make two arches to intersect each other in K.
4. From K thro' FG draw a circle, divide it into eight parts equal to FG, and then joyn them by right lines: so you'll have an Octagon to be a Base to the Pyramid requir'd.
1. From A draw an arch, and divide it into nine equal parts, B, C, D, E, F, G, H, I, K; joyn these together, and with A by right lines.
2. Divide any one of these parts, as KM, into seven equal parts, and one of them into seven more.
3. Take ten of the former, and two of the latter, and from the points KM make two arches to intersect each other in L, with the length of those parts, as a given distance.
4. From L draw a circle thro' KM, which must be divided into nine parts equal to K M; joyn these together by right lines: so you have the Enneagon which was wanted for a Base for the Pyramid required.
1. From A draw an arch, which divide into ten equal parts, to be joyned to each o∣ther, and to A, as B, C, D, E, F, G, H, I, K, L.
2. Divide one of these ten parts into se∣ven, and one of these seven into eight more.
3. Take eleven of the former, and two and a half of the latter, and with that di∣stance ••rom IK, draw two arches which shall intersect each other in M.
4. From M draw a circle thro' IK, which divide into ten parts equal to IK, and there will be a Decagon for a Base to the Pyramid.
1. From A draw an arch, and divide it into eleven equal parts, to be joyned each to other, and all to A, as B, C, D, E, F, G, H, I, K, L, M.
2. Divide one of them (FG) into seven equal parts, and one of these seven into se∣ven more.
3. Take twelve of the former, and one and a half of the latter, for a distance, where∣by from FG to draw arches which shall in∣tersect each other in N.
4. From N draw a circle thro' FG, which may be divided into eleven parts equal to F G•• joyn them together, and you have the Base of the required Pyramid.
1. From A draw an arch, which divide into twelve equal parts, which are to be joyned to A, and each to other by right lines, as B, C, D, E, F, G, H, I, K, L, M, N.
2. Divide any one (FG) into seven e∣qual parts, and one of these seven into se∣ven others.
3. Take thirteen of the former, and one and a half of the latter, for a distance where∣by from FG to draw two arches which shall intersect each other in O.
4. From O draw a circle thro' FG, and divide into twelve parts equal to FG, which when joyned by right lines, will give a Base of the Pyramid that was required.