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X. The Solidity of every Prism is compos'd of an infinite Series of equal Plains, parallel and alike to that of its Base.

XI. A Pyramid is a Solid bounded or included within several plain Triangles set upon any Polygonouş Base, having their vertical Angles all meeting together in a Point, called the Vertex, and takes its Name from the Figure of its Base, viz. if it has a square Base, 'tis call'd a square Pyramid; if a triangular Base, 'tis callid a triangular Pyramid, &c.

XII. A Cone is only a round Pyramid, which hath been already defined in Page 355, &c.

XIII. The Solidity of every Pyramid is compos'd or constituted of an infinite Series of Plains, parallel and alike to that of its Base, equally decreasing until they terminate in a Point at the Vertex.

XIV. A Sphere or Globe (viz. a Ball) is a Solid bounded or included within one regular Superficies, being form’d or genera ted by the Rotation of a Semi-circle about its Diameter (callid the Axis of a Sphere) and its Solidity is compos’d or constituted of an infinite Series of concentrick Circles, whose Diameters are the Chords of that Circle by which it was form’d.

XV. A Spheroid (or Egg.like Figure) is a Solid bounded with one regular Superficies, form’d by the Rotation of a Semi-ellipsis about its Transverse Diameter, (call’d the Axis of the Spheroid) and its Solidity is constituted of an infinite Series of concentrick Circles, whose Diameters are the Ordinates of that Ellipsis by which it was form’d.

XVI. There is another Sort of Solid callid an Oblate Sphereid, being form’d by the Rotation of an Ellipsis about its Conjugate Diameter, and it is like a flat Turnep.

· XVII. If a Semi-parabola be turn'd about its Axis, 'twill form a Solid call'd a Parabolick Conoid, being compos'd or constituted of an infinite Series of Circles, whose Diameters are the Ordinates of a Parabola.

XVIII. If a Parabola be turn'd about its Base, or greatest Ordinate, 'twill form a Solid call'd a Pyramidoid, but most commonly a Parabolick Spindle which will be constituted of an infinite Series of Circles, whose Diameters are Right Lines parallel to the Parabola's. Axis.

XIX. If an Hyperbola be turn'd about its Axis, 'twill form a Solid call'd an Hyperbolick Congid, being constituted of an infinite Series of Circles whose Diameters are the Ordinates of the Hyperbola.

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XX. The

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XX. The curve Superficies of all circular Solids, viz. Cylínders, Cones, Spheres, &c. are compos'd' of an infinite Series of the Peripheries of those Circles which conftitute their Solidities.

Upon these Definitions are grounded all the following Thesrems; and therefore, if they were diligently compar'd with their respective Figures, it must needs be of great Help to the Learner, and would render all that follows very easy; wherein I fhall begin with what hath been already demonstrated, by way of introducing the rest.

THEOREM I. The Area of every Right-angled Parallelogram is obtain’d by multiplying thé Length into its

Breadth. That is, BDXFB = the Area of the Parallelogram BD FG

by Lemma 1, compar'd with De-
finition 1.

Example.
Suppose BD=26, and FB=9,

then 26 x 9=234 the Area,
See Prob. 1, Pag. 339.

B

THEOREM II. The Area of every plain Triangle is equal to half the Area of its

BDXCA circumscribing Parallelogram. That is,

=the Areasf ABCD, in the following Figurla.

2

Demonftration. Suppose the Perpendicular CA to be divided into an infinite Number of equal Parts, as at the Points a, a, a, &c. and through F C those Points there were drawn

b/ a Right Lines parallel to the Base

b

d BD; (viz. bad, bad, bad, &c.)

bi then will those Lines be a Series

d of Terms in Arithmetick Progresfion, beginning at the PointC,(viz.

B A

D o, bd, 2 bd, 3bd, &c. as is evident by the Figure, wherein BD this greatest serm = L, and CA the Number of Terms = N.

But

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But NL=S, by Lemma 2. And S=the Triangle's Area by Definition 2. Q. E. D.

Example. Let B D = 26, and CA = 9 as above, then

26 X9=117, or 99 x 9 = 117. Or thus 26 x = 117 # the Area requir’d. [See Problem 3, Page 330.]

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THEOREM UI.

The Peripheries of Circles are in Proportion one to another as their

Diameters are.

Demonftration. Let the Periphery of a Circle be divided into any Number of equal Arches by Right Lines drawn from the Center (viz. Radii) suppose 'em 8, as in the annexed Figure, wherein AB is one of them; then, if thro' any Point

a in the Radius there be drawn a concen- D

1 trick or parallel Circle, its Periphery will also be divided into 8 equal Arches by those Radii, one whereof will be a b,

B and the A Cab will be like to A CAB. Therefore Ca:ab::CA: A B, or Ca:C1::ab: AB, consequently 2 Ca:

2 CA:: 8a6:8 A B. But 2 Ca=da the Diameter of the Circle, whose Periphery is 8 ab; and 2 GA

= D A, the Diameter of the Circle, whose Periphery is 8 A B. Therefore, &c. as by the Theorem.

Q. E. D. Example. In Chapter 6, Part III, it was found, that, if the Diameter of a Circle be 2, its Periphery will be 6,2831853, &c. Therefore, 2:6,2831853, &c. ::1:3,14159265, &c. the Periphery of the Circle whose Diameter is 1.

Corollary. Hence it follows, that because Unity, or 1, may be made the first Term in the Proportion, therefore 3,14159265, &cmay be made a constant or settled Factor ; which, being multiply'd into any propos'd Diameter, will produce the Periphery of that Circle.

Note, Instead of 3,14159265, &c. it may be sufficient to take only 3,1416.

Or,

Or, in whole Numbers the Proportion may be, As 7 : 22 :: Diam . : Periphery US these Numbers may serve, Or113:355 :: Diam. : Periphery}{ and are often used in com

mon Practice. THEOREM IV. The Area of any Sector of a Circle is equal to half the Rectangle

CAXA'B of the Radius into its Arch. That is,

= the Area

2 of ACP.

Demonttration. Suppose the Radius CA to be divided into an infinite Series of equidistant Points, as a, l, y, &c. and through those Points there were drawn concentrick or parallel Arches, as ab, e d, yf, &c. then they will be a Series of Arches in Arithmetick Progression, beginning at the Point C (viz. 0, 1, 2, 3, &c.) as plainly appears by the Figure, wherein the greatest Term is AB=L, and Number of Terms is CA=N. But NL=S the Sum of all the Series, by Lemma 2, and S=the Sector's Area, by Definition 3. Q. E. D.

Example Let the Radius CA=12, and the Arch AB=8, then 12 x

2 = 48. or x 8 = 48. or X 12 = 48, the Arca of the Sector ACB.

THEOREM V. The Area of every Circle is equal to half the Rectangle of the Radias

into its Periphery. That is, according to Archimedes, a Circle is equal to a Right-angled Triangle, whose Sides containing the Right-angle are equal, one to the Radius, and the other to ibe Perimeter of that Circle. Pro. 1. de Dimensione Circuli.

The Truth of this Theorem may be easily deduced from the last thus ; If we suppose the last Sector to be one Eighth-part of a Circle, then it follows, that

8 ABXCA

=4 ABX CA will be the Area of the whole Circle. But 4 AB=half the Circle's Periphery, and CA= half its Diameter; Therefore, &c. As per Theorem.

Q. E.D.
Example

2

2

Example. If the Diameter be Unity, or 1, the Periphery will be 3,14159265, &c. by Theorem 3. Then

3,14159265

X}=0,78539816, . & c. (or 0,7854 for common Use) will be the Area of that Circle.

Scholium. From hence naturally flows the following Proportion between the Square and its inscrib'd Circle.

As the Perimeter (viz. the Sum of the four Sides) Proportion. of any Square : is to its Area : : so is the Peri

(phery of the inscrib’d Circle: to its Area. That is, supposing AB=D= the side of the Square, and the Diameter of its inscrib’d Circle ; then 4 D= the Perimeter, DD = the Area of the Square, and 3,1416 D= the Periphery of the Circle, by Theorem 3. But 4D :DD:: 3,1416D:0,7854 DD = the Circle's Area. And if D=1;

D then 4D=4, and DD=IXI=1, and the Periphery will be 3,1416. Then 4:1::1:0,7854 &c. as in the Example above. And from hence may be ca

В. fily deduced the following Theorems.

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THEOREM VI. The Area's of all Circles are in Proportion one to another as the

Squares of their Diameters. (2. e. 12.) For if D=the Diameter of one Circle, and d=the Diameter of another Circle, then will 0,7854 DD be the Area of one Circle, and 0,7854 dd will be the area of the other Circle; as above. But 0,7854 DD: 0,7854 dd::DD:dd. Or thus, let D= the Diameter, and P= the Periphery of one Circle; d= the Diameter, and p= the Periphery of another Circle; Then/ i11DXi P=;DP = A, the Area of one Circle.

And 210 Xip=dp=a, the Area of the other Circle. I X4 3 DP=4 A

(per lait Theorem. 2 x 4 4 dp=4a

4A 37D5P=

40

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