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GEOMETRY is the Art of Measu∣ring well.
2. The Subject of Geometry is Magnitude, or continued Quantity, whose parts are joyned together by a common term or limit.
3. Magnitude is either a Line, or some∣thing made of a Line or Lines.
4. A Line is a Magnitude, consisting on∣ly of length, without either breadth or thick∣ness, the term or limit whereof is a Point.
5. A Point is an indivisible sign in Magni∣tude. A Point therefore is no quantity, but the beginning of all continued quantities, which are divisible in power infinitely.
6. A Line is either considered simply by it self, or else comparatively with another Line.
7. A Line considered simply of it self, is either Right or Oblique.
8. A Right line, is that which lyeth equal∣ly between his Points.
9. An Oblique line, is either circular or mixt.
10. A Periphery, or Circular Line, is that which is equally distant from the middle of the comprehended space, which middle is called the Centre, and the distance between that Centre and the Circumference, is cal∣led the Radius.
11. Lines compared to one another are or the same or different Species.
12. Lines compared together of the same Species, are either Parallel or Angular.
13. Parallel lines, are such as are equally distant in all places, and are either Right lined Parallels, or Circular.
14. Right lined Parallels, are such as being in one and the self same plane, and infinite∣ly produced on both sides, do never meet in any part.
15. A Circular Parallel is a Circle drawn within or without another Circle.
16. Angular lines are such as inclining, or bowing to one another, touch one another, but not in a direct line.
17. An Angle is either Right or Oblique.
18. A Right Angle, is that whose legs or sides are Perpendicular to one another.
19. An Oblique Angle, is that whose legs or sides do incline to one another upon one side more than upon another.
20. An Oblique Angle is either Acute, or Obtuse.
21. An Acute Oblique Angle, is that which is less than a Right.
22. An Obtuse Oblique Angle, is that which is greater than a Right Angle.
23. The Measure of an Angle, is the Arch of a Circle described upon the Angular Point, and intersected between the sides of the Angle sufficiently prolonged; but of this Measure there can be no certain know∣ledge, unless the quantity of that Arch be ex∣pressed in Numbers.
24. Every Circle therefore is supposed to be divided into 360 equal parts, called De∣grees, and every Degree into 60 Minutes, and every Minute into 60 Seconds, and so forward; others suppose every Degree to be subdivided into 10 parts, and every one of those into 10 more, and so forward, as far as you please.
25. A Semi-circle is the half of a whole Circle, and containeth 180 Degrees.
26. A Quadrant, or fourth part of a Cir∣cle, is 90 Degrees; and seeing that a Right Line falling Perpendiculary upon a Right Line, doth make the Angles on both sides e∣qual, and cutteth a Semi-circle into two e∣qual parts, the fourth part of a Circle, or 90 Degrees, must needs be the Measure of a Right Angle.
27. Thus are Lines compared with Lines of the same Species, the comparing of Lines of different Species, is the comparing of Right Lines with those that are Oblique or Circular.
28. And Right Lines, as they have refer∣ence to, or are compared with the circum∣ference of a Circle, are either such as are in∣scribed within it, or applyed to it.
29. A Right Line, inscribed in a Circle, either passeth through the Centre, as the Diameter and Radius, or is drawn besides the Centre, as Chords and Sines.
30. A Diameter, is a Right Line inscrib∣ed through the Centre of the Circle, divi∣ding the Circle into two equal parts.
31. The Radius of a Circle is the one half of the Diameter, or a Right Line drawn from the Centre to the Circumference; thus the Right Line GBD, in Fig. 1. is a Dia∣meter, GB, or BD, the Radius.
32. A Chord or Subtense, is an inscribed Right Line drawn through or besides the Centre bounded at both ends with the Cir∣cumference.
33. A Chord or Subtense, drawn through the Centre is the same with the Diameter.
34. A Chord or Subtense, drawn besides the Centre, is a Right Line bounded at both ends with the Circumference, but always less than the Diameter.
35. Sines are either Right or Versed.
36. A Right Sine is half the Chord of the Double Arch and it is either the whole Sine, and Sine of 90 Deg. or Sine less than the whole.
37. The whole Sine is equal to the Semi∣diameter or Radius of a Circle, as the right Line BE.
38. A Sine less than whole, is half the Chord of any Arch less than a Semi-circle; as CA is the sine of CD.
39. A Versed Sine, is a part of the Diame∣ter lying between the right sine and the cir∣cumference, as the Right Line AD, which is one part of the Diameter, is the versed sine of the Arch C D, and the right line AG, which is the other part of the Dia∣meter, is the versed sine of the Arch CEG.
40. A Right line applied to a Circle, is either a Tangent or Secant.
41. A Tangent, is a right line without but touching the Circle, drawn Perpendicular to the end of the Radius or Diameter, conti∣nued to the Secant.
42. A Secant, is a right line drawn from the Centre of the Circle, through the term
of an Arch, and continued to the Tengant; thus the right line FD, is the Tangent, and the right line BF, is the Secant of the Arch CD, or of the Arch CEG, the Comple∣ment thereof to a Semi-circle.
43. These Lines thus inscribed in, or ap∣plyed to a Circle, may to any limited Radi∣us be drawn or made upon a Rule of Wood, Brass, or other Metal; or, a Table may be made, expressing the length of these lines in numbers, answering to every Degree and part of a Degree in the Quadrant or Semi-circle; That is, the lines of Chords and Ver∣sed Sines may be made to any part of a Semi-circle, and the lines of Sines, Tangents and Secants, to any part of a Quadrant: The use of such Scales and Tables is such, that no Student in Geometrie can well be without them; here therefore I will lay down such Propositions as will sufficiently demonstrate the way of making these lines upon a Scale or Ruler, but as to the construction of the Tables by which the lengths of these lines are expressed in Numbers: I refer them to my Trigonometria Britannica, and other Books of the like nature.
Let it be required to erect a Perpendicu∣lar to the line DG, from the Point B, in Fig. 2. take two equidistant Points, as D and G, open your Compasses to a convenient distance, and setting one Foot of your Com∣passes in B, draw the Arch EC, and keep∣ing your Compasses at the same distance, set one Foot in G, and with the other draw the Arch HIF, and through the Intersections of these two Arches draw a right line, as BL, which shall be perpendicular to the Point B.
But if it were required to erect a Perpen∣dicular from the end of a line, do thus, your Compasses being opened to any convenient distance, set one Foot in the Point given, as at A, in the line AB, and the other at D, or where you please, and making D the Cen∣tre, draw the Arch CAE, and from the points C and D, draw the right line CDE, then draw the line AC, which shall be Per∣pendicular to the line AB, from the point A, as was required.
Let the given line be DG, and let the point assigned be L, at the distance of LD draw the Arch DAGF, then setting one Foot of your Compasses in D, draw the Arch IK, and keeping your Compasses at the same distance, set one Foot in G, and with the o∣ther draw the Arch M, the right line LBD drawn through the Intersections of those two Arches shall be Perpendicular to DG, from the Point L, as was required.
But if it were required to let fall a Per∣pendicular from the point E, upon the line AB, draw the line EDC at pleasure, which being bisected at D, upon D as a Centre at the distance of ED, draw the Arch EAC, so shall the line EA be perpendicular to AB, as was required.
Draw the line AC, and from the points A and C erect the Perpendiculars AE and XC, and at any distance of the Compasses,
set off as many equal parts as you please up∣on the Perpendiculars AE, and XC, and draw the Parallel lines EX, FV, GT, HS, KR, LQ, MP, and NO; and let it be re∣quired to divide the right line into three e∣qual parts, open your Compasses to the length of the line given, and setting one foot in A, where the other foot shall touch the third Parallel, make a mark which is at Z, draw the line AZ, so shall the line AZ be divided into three equal parts, as was desi∣red.
And thus may that line be made, which is commonly called the Diagonal Scale.
Draw the Diameter BC, and upon the point A, describe the Circle CDBL, then draw the Diameter DAN, at Right Angles, to the Diameter CAB.
2. The Semidiameter or Radius of a Cir∣cle will divide the Circle into 6 equal parts, and so is equal to the Chord of 60 degrees, AC, therefore being set from D to F, shall mark out the Arch DF, 60 degrees.
3. The side of a Pentagon of fifth part of a Circle, is 72 degrees; now then, if you bisect the Radius AC in the point E, and
make EG = ED; then shall DG = DM, the side of a Pentagon or Chord of 72 degrees, and FM the difference between DM, 72 and DF 60, that is the Chord of 12, which by bisection shall give the Chord of 6 and 3 degrees, and so the Circle may be divided into 120 parts, as was propounded.
4. A Circle being thus divided into 120 degrees, the Arches are so equal, that the third part of the Chord of 3 degrees will subdivide it into 36, without sensible error; and your Circle being thus divided into 360 parts, lines at every degree, or half degree, drawn parallel to the Diameter, shall consti∣tute the line of Chords, & half those Chords the line of sines; and the Segments of the Diameter, the line of versed sines, and as for the Tangents and Secants, a line touching the Circle drawn perpendicular to the end of the Diameter, and continued to the seve∣ral lines drawn from the Centre, through every degree of the Quadrant, shall consti∣tute the line of Tangents, and those lines drawn from the Center to the Tangents, shall constitute the line of Secants also. And thus may a Scale be made with the lines of Sines, Tangents, Secants, and equal parts.
HItherto we have spoken of the first kind of Magnitude, that is, of Lines, as they are considered of themselves, or among themselves.
2. The second kind of Magnitude, is that which is made of Lines, that is a Fi∣gure.
3. A Figure is that which is every where bounded, whether it be with one only limit as a Circle; or with more, as a Triangle, Quadrangle, Pyramis, or Cube, &c.
4. The terms or limits of every Figure, are either Lines or Superficies.
5. A Figure, which is terminated by Lines is a Superficies.
6. A Figure, which is bounded or limited with several Superficies, is a Body or Solid.
7. A Superficies is a Magnitude, consisting of length and breadth, and is either right lined, curved lined, or composed of both.
8. A Right Lined Plane or Superficies, is that which is Terminated with right lines; and it is either a Triangle, or a Triangu∣late.
9. A Triangle, or the first right lined Fi∣gure, is that which is comprehended by
three right lines. It is distinguished from the sides, or from the Angles.
10. In respect of the sides, a Triangle is either Isopleuron, Isosceles, or Scalenum.
An Isopleuron Triangle, is that which hath three equal sides. An Isosceles, which hath two equal sides. And a Scalenum, whose three sides are all unequal.
11. In respect of the Angles, a Triangle is either Right or Oblique.
12. A Right Angled Triangle, is that which hath one right line.
13. An Oblique Angled plane Triangle, is either Acute or Obtuse
14. An Oblique and Obtuse Angled plane Triangle, hath two Acute Angles and one Obtuse; an Acute angled Triangle hath all the three Angles Acute.
15. The second sort of right lined planes is called a Triangulate, or a Plane, composed of Triangles.
16. The sides of a Triangulate, are in number more by two than the Triangles, of which it is composed.
17. A Triangulate, is either a Quadrangle, or a Multangle.
18. A Quadrangle, is a Plane comprehen∣ded, by four right lines, and is either a Pa∣rallelogram or a Trapezium.
19. A Parallelogram, is a Quadrangle, whose opposite sides are Parallel or Equidi∣stant, and it is either Right Angled or O∣blique.
20. A Right Angled Parallelogram, is that which hath every Angle Right; and it is ei∣ther a Square or an Oblong.
21. A Square is a Right Angled Parallelo∣gram, whose four sides are equal, and the Angles Right.
22. An Oblong, is a Right Angled Paral∣lelogram, whose Angles are all right, but the sides unequal.
23. An Oblique angled Parallelogram, is that whose Angles are all Oblique, and is either a Rhombus, or a Rhomboides.
24. A Rhombus, is an Oblique Angled Pa∣rallelogram, of equal sides.
25. A Rhomboides, is an Oblique angled Parallelogram of unequal sides.
26. A Trapezium, is a Quadrangle, but not a Parallelogram, and it is either Right angled, or Oblique.
27. A Right Angled Trapezium, hath two opposite sides, parallel, but unequal, and the sides between them perpendicular.
28. An Oblique Angled Trapezium, is a Quadrangle, but not a Parallelogram, hav∣ing at least two Angles thereof Oblique, and none of the lines Parallel.
29. A Right angled Multangled Plane, is that which is comprehended by more than four lines.
30. A Multangled Right lined Plane, or Polygon, is either Ordinate and Regular, or Inordinate and irregular.
31. Ordinate and Regulate Polygons, are such as are contained by equal sides and an∣gles, as a Pentagon, Hexagon, &c.
32. Inordinate or Irregular Polygons, are such as are contained by unequal sides and angles.
32. Having thus shewed what a right lin∣ed Figure is, with the several sorts of them, we will now shew, how they may be measur∣ed, both in respect of the lines by which they are bounded, and also of their Area or Superficial Content.
33. And first we will shew how the lines, and angles of all plane Figures, especially Triangles, may be measured, as being the first and chiefest of them, and into which all other may be reduced.
34. The sides of all plane Triangles, and other plane Figures, are to be Measured by the scale or line of equal Parts.
35. The Angles may be measured by the lin••s of Sines, Tangents, or Secants, as well as by the line of chords; but here it shall suffice to shew how any Angle may be pro∣tracted, or being protracted, be Measured by the line of Chords only.
Draw a line at pleasure at AB, then o∣pen your Compasses to the number of 60 de∣grees in your line of Chords, and setting one of that extent in B, with the other describe the Arch CD, and from the point B, let it be required to make an Angle of 40 degrees; open your Compasses to that extent in the line of Choads, and setting one Foot in D, with the other make a mark as at E, and draw the line EB, so shall the Angle ABE contain 40 degrees, as was required.
Let the quantity of the Angle ABE, be required; open your Compasses in the line of Chords, from the beginning thereof to 60 degrees, and setting one foot thereof in the point B, with the other describe the Arch DE, then take in your Compasses the distance between E and D, and applying that extent to the line of Chords, it will shew you the number of degrees contained
in that Angle, which in our Example will be found to be 40 degrees.
These things premised, we will now shew you how all plane Triangles may be measu∣red, in respect of their Sides and Angles, both by the Scale, and also by the Tables of Sines and Tangents.
IN the Solution of plane Triangles, the an∣gles only being given, the sides cannot be found, but the reason of the sides only; it is therefore necessary, that one of the sides be known.
2. In all plane Triangles, the three angles are equal to two Right: two Angles therefore being given, the third is also gi∣ven; and one of them being given, the sum of the other two is also given.
3. In a Right angled plane Triangle, one of the Acute Angles being given, the other is also given, it being the Complement of the other to a Quadrant or 90 degrees.
4. In a Right Angled plane Triangle, there are seven Cases, whose Solution shall be shewed in the Problems following.
5. The sides comprehending the Right angle we call the legs, and the side subtend∣ing the Right angle, we call the Hypothenuse.
In the right angled plane Triangle ABC, let there be given the legs.
Draw a line at pleasure, as AB, and upon the point A, erect the perpendicular AC, and by help of your Scale of equal parts, set off from A to B, 512, and also from A to C, 384, and draw the line BC, for the Hy∣pothenuse, which being Measured by the scale of equal Parts, will be found to be 640. and by the line of Chords, the angle at B 36.87, whose complement is the angle ACB, 53.13.
By the Tables, the Proportions are,
Draw a line at pleasure, as AB, and at Right angles the point A erect the perpen∣dicular AC, and by your scale of equal parts set off from A to B 512, and upon the point B lay down the angle ABC, 36.87. and draw the line BE, till it cut the perpendi∣cular AC, then measure the lines BC and AC, by the scale of equal parts, so shall the one, to wit, BC, be the Hypothenuse, and AC, the other leg inquired.
By the Tables, the Proportions are,
3. Rad. . AB ∷ 4 B . AC.
4. Sine C . AB ∷ Rad. . BC.
Draw a line at pleasure, as AB, and upon the point B protract one of the angles gi∣ven, suppose the lesser ABC, 36.87. and
draw the line BC, & by your scale of equal parts, number the given Hypothenuse from B to C 640. and from the point C to the line AB, let fall the Perpendicular AC, then is BA one, and CA, the other leg inquired.
By the Tables, the Proportion is,
5. Rad. . BC ∷ sB. AC.
Draw a line at pleasure, as AB, and by your scale of equal parts, number from B to A, the quantity of the given leg AB, 512. then upon the point A erect the Perpendicu∣lar AC, and opening your Compasses to the extent of your Hypothenuse BC 640, set one Foot in B, and move the other, till it touch the Perpendicular AC, and there draw BC, so shall AC be the leg inquired, and either Angle may be found by the line of Chords.
By the Tables, the Proportions are,
6. BC . Rad. ∷ AB . Sine C.
7. Rad. . BC ∷ Sine B . AC.
6. Hitherto we have spoken of Right angled plane Triangles, the Problems fol∣lowing concern such as are Oblique.
In the Oblique angled plane Triangle BCD, let there be given the side CB 632, and the Angles DCB 11.07. D. 26.37.
Draw the line CB at pleasure, and by your scale set off from C to B 632, and up∣on those points protract the given Angles DCB 11.07 CBD. 142.56, and draw the lines CD and BD, till they intersect one a∣nother, then shall the one side be CD 865, and the other DB 273.
By the Tables, the Proportion is,
1. Sine BDC . BC ∷ sDCB . DB.
In the Oblique angled plane Triangle BCD, let there be given,
Draw the line CD at pleasure, and by your scale set off from C to D, 865, and up∣on the point D protract the Angle CDB 26.37 and draw the line DB, then open your Compasses to the length of the other side CB 632, and setting one foot in C, turn the other about till it touch the line DB, which will be in two places, in the point B or point nearest to D, if the angle opposite to the side CB be Obtuse, but in the point E, or point farthest from D if Acute; accor∣ding therefore to the Species of that Angle, you must draw either the line CB or CE, and then you may measure the other angles and the third side, as hath been shewed.
By the Tables, the Proportion is,
2. CB . Sine D ∷ CD . Sine B.
3. Sine D . CB ∷ Sine C . BD.
In the Oblique angled plain Triangle BCD, let there be given,
Draw a line at pleasure, as DC 865, and by your Scale set off from C to D, 865, then protract the Angle at C 11.07, and draw the line BC, and by your Scale set off from C to B 632, and draw the line BD, and so have you constituted the Triangle BDC, in which you measure the Angles and the third side, as hath been shewed; but to resolve this Problem by the Tables, it is somewhat more troublesome.
1. To find the Angles, the proportion is, 〈 math 〉〈 math 〉.
2. To find the third Side.
Sine D . BC ∷ Sine C . BD.
Let the given sides be DC 865. BC 632. and DB 273.
Draw a line at pleasure, as DC, and by your Scale set off from C to D 865, then o∣pen your Compasses to the extent of either of the other sides, and setting one foot of your Compasses in C, with the other draw an Occult arch, then open your Compasses to the extent of your third side, and setting one foot in D, with the other foot describe another Arch cutting the former in the point B, then will the Lines BC and DB, constitute the Triangle, whose Angles may be measured, as hath been already shewed.
To resolve this Problem by numbers, the Proportions are for the Segments of the Base.
As the base is to the sum of the other sides, so is the difference of those sides to the dif∣ference of the Segments of the Base; which being subtracted from the Base, half the re∣mainer will shew where the perpendicular must fall, suppose at F, and constitute the two Right angled Trianges BDF, and FDC, in which we have given the Hypo∣thenuses BD and DC, and the Legs BF and CF, and therefore we may find the
Angles of those Triangles, as hath been shewed in the fourth Problem.
Having shewed the Mensuration of Trian∣gular planes in respect of their sides and Angles, we will now shew how the Area or Superficial content of them, and any other plane Figures may be found: And because all many-sided Figures may be best Measu∣red by reducing them first into Right an∣gled Triangles, Quadrangles, or Trapezias, we will first shew how the Area or Superfi∣cial content of these Figures may be readily found; and first of a Right or Oblique an∣gled plane Triangle.
2. To Measure the Right angled plane Triangle BDF, in Fig. 7. Multiply BF by FD, half the Product shall be the content.
3. To Measure the Oblique angled plane Triangle BDC, let fall the Perpendicular DF, then Multiply BC by DF, half the Product shall be the content.
4. To find the Area or Superficial con∣tent of any Oblique angular Trapezium, convert it into two Oblique angled Trian∣gles, by a Diagonal, as the line BD in the Trapezium ABCD, then turn the Oblique
angled Triangles into Right, by letting fall the Perpendiculars AE and CF, then Mul∣tiply BD by the sum of AE and CF, half the Product shall be the content. In like manner may any other Irregular Multangle be also measured by turning it into Trian∣gles and Trapeziums, and computing them severally, and adding all their contents to∣gether.
Vide A.
Vide B.
The Circumference of a Circle whose di∣ameter is 1, is 3.14159 and therefore,
As 1 is to 3.14159, so is any other Dia∣meter, to the Circumference answering that Diameter.
Archimedes hath Demonstrated, that the Area of a Circle is equal to the content of a Right angled plane Triangle, whose Legs
comprehending the Right angle, are one of them equal to the Semidiameter, and the o∣ther to the Circumference of a Circle. And therefore the Area or Superficial content of a Circle may be found, by Multiplying half the Circumference by half the Diameter, or the whole Diameter by the fourth part of the Circumference, they taking the Diame∣ter of a Circle to be one, and the Circum∣ference 3.14159, the Superficial content of such a Circle will be found to be 07853975.
And therefore, As 1 is to 78539, so is the Square of any other Diameter to the Super∣ficial content required.
The Chord or Subtense of the fourth of a Circle, whose Diameter is 1, is 7071067; therefore, as one, to 7071067, so is the Dia∣meter of any other Circle, to the Side required.
By the Diameter to find the Circumfe∣rence, the proportion by the tenth Problem is; As 1 to 3.14159, so the Diameter to the Circumference, and therefore putting the Circumference of a Circle to be 1.
As 3.14159 . 1 ∷ 1. 318308.
And therefore as 1 to 318308, so is any other Circumference, to the Diameter sought.
As the Square of the Circumference of a Circle given, is to the Superficial content of that Circle, so is the Square of the Circum∣ference of any other Circle, to the Superfi∣cial content of that other Circle.
And in a Circle whose Diameter is 1, the Circumference is 3.14159, and the Area 7853975, and supposing an Unite to be the Circumference of a Circle, it is, as the square of 3.14159 . 7853975 ∷ 1 . 0079578, and therefore, As 1 . 0.079578, so is the
square of any other Circumference, to the Area desired.
As the Circumference of a Circle whose Diameter is 1, viz. 3.14159, is to 707107, the side of the inscribed square of that Cir∣cle, so is the Circumference of any other Circle, to the side inquired; and putting the Circumference to be Unity, it is, as 3.14159 . 707107 ∷ 1 . 225078, there∣fore,
As 1 to 225078, so is the Circumference given, to the side inquired.
This is the Converse of the 11. Problem, the Diameter given, to find the Content, for which the Proportion is; as 1 to 7853975, so is the square of the diameter, to the content: and therefore we must say; as 7853975 is to 1 so 1 to 1.27324; and hence, as 1 to 1.27324, so is the Area, to the Square of the Diameter.
This is the Converse of the 14. Problem, the Circumference given, to find the Content.
As 1 to 079578, so Circumference square, to the Content: And therefore,
As 079578 . 1 ∷ 1 . 12.5664, and by consequence,
As 1 to 12.5664, so the Area, to the Square of the Circumference.
Vide C.
As the square of the Diameter of a Circle, which suppose 1, is to 3.14159 the Area, so is the square of the Axis given, to the Area that is required.
AFter the description of lines and planes, the Doctrine of Bodies is to be consi∣dered.
2. A Solid or Body, is that which hath Length, Breadth and Thickness, whose bounds or limits are Superficies.
3. A Solid is either Plane or Gibbous.
4. A Plane Solid, is that which is compre∣hended of Plane Superfices, and is either a Pyramide or Pyramidate.
5. A Pyramide, is a solid Figure, which is contained by Planes, set upon one Plane or Base, and meeting in one point.
6. A Pyramidate, is a solid Figure, com∣posed of Pyramides, and is either a Prisme or a mixt Polyhedron.
7. A Prisme, is a Pyramidate or solid Fi∣gure, by Planes, of which these two which are opposite, are equal, like, and parallel and all the other Planes are parallelograms.
8. A Prisme, is either a Pentahedron, a Hexahedron, or a Polyhedron.
9. A Pentahedron Prisme, is that, which comprehended of five sides, and the Base Triangle.
10. An Hexahedren Prisme, is that which is comprehended of six sides, and the Base a Quadrangle.
11. An Hexahedron Prisme, is either a Parallelipipedon, or a Trapezium.
12. A Parallelipipedon, is that whose sides or opposite planes are parallelograms.
13. A Prisme, called otherwise a Trapezi∣um, is that solid, whose opposite planes or sides are neither parallel nor equal.
14. A Parallelipipedon, is either Right an∣gled or Oblique.
15. A Right angled Parallelipipedon, is that which is comprehended of right angled sides and it is either a Cube or an Oblong.
16. A Cube, is a Right angled parallelipi∣pedon of equal sides.
17. An Oblong, is a right angled paralleli∣pipedon of unequal sides.
18. An Oblique angled Parallelipipedon, is that which is comprehended of oblique sides
19. A Polyhedron, is that which is compre∣hended of more than five sides, and the Base a Multangle.
20. A mixt Polyhedron, is that whose Ver∣tex is in the Centre, and the several sides exposed to view, and of this sort, there are only three; the Octahedron, the Icosohedron, and the Dodecahedron.
21. An Octahedron, is a solid Figure, which is contained by eight Equal and Equilateral Triangles.
22. An Icosohedron, is a solid Figure, which is contained by twenty Equal and Equilate∣ral Triangles.
23. A Dodecahedron, is a solid Figure, which is contained by twelve Pentagons, E∣quilateral and Equiangled.
24. A Gibbous solid, is that which is com∣prehended of Gibbous Superficies, and it is either a Sphere or Various.
25. A Sphere, is a Gibbous body, abso∣lutely Round and Globular.
26. A Various Gibbous Body, is that which is comprehended by various superficies and a circular base; and is either a Cone, or a Cylinder.
27. A Cone, is a Pyramidical Body, whose Base is a Circle.
28. A Cylinder, is a solid Body of equal thickness, having a Circle for its Base. The solid content of these several Bodies may be measured by the Problems following.
Multiply the Altitude by a third part of the Base, or the whole Base by a third part of the Altitude, the Product shall be the so∣lid Content required.
Multiply the Base of the Prisme or Cylin∣der given, by the Altitude, the Product shall be the solid content.
If the Aggregate of both the Bases of the Frustum, and of the mean Proportional be∣tween them, be drawn into the Altitude of the Frustum, the third part of the Product shall be equal to the solid content required.
A Sphere (as Archimedes hath shewed) is equal to two thirds of a Cylinder circum∣scribing it; now then, such a Cylinder be∣ing made; by the Area of a Circle multi∣plyed by the Diameter; and therefore the
Area of a Circle being multiplied by two thirds of the Diameter, the Product shall be the solid content of a Sphere.
The Area of a Circle whose Diameter is 1, is 7853975, which being multiplied by 666666, the two thirds of the Diameter, the Product 523598 is the solid content of such a Sphere; therefore,
As 1 to 523598, so is the Cube of any Ax∣is given, to the solid content required.