Note. There is one Thing very remarkable respecting these Five Bodies, which is, that if an absolute Plenum takes Place in the Universe (a Doctrine held by some Philosophers), then the conftituting Particles of Matter must be in the Shape of fome one of these Solids; for there are no other Bodies, let their Figurability be what they may, but will, when combined together, leave some Vacuity or Interstice between them.

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ADDITIONAL PROBLEMS.

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O continue

Problem 1.

a Right Line to a greater Length than

can be drawn by a Ruler at one Operation.

Suppose A B be the Line given, which cannot be made longer at one Operation, by Reason of the Ruler being of the fame Length.

Operation. With the Compasses opened to the Length of the given Line A B, set one Foot in A, and with the other describe the Arch bc; upon which, from the End of the given Line at B, set off two Points, as e and f. On each of which Points alternately fet one Foot of the Compasses (opened to any Wideness) and describe the Arches interfecting each other at h; to which, from the End of the given Line, lay a Ruler, and continue the said Line at Pleasure. By this Means a Line of any determinate Length may be drawn with a very short Ruler.

Problem 2.

To find the Length of any Arch of a Circle.

Let ACD be the Arch, whose Length is required.

Operation. Divide the Chord AD into 4 equal Parts, and fet off one Part from A to b; then draw a Line from b to the End of the 3d Division on AD, and it will be nearly equal to half the Arch; which doubled, will give the Length of the whole Arch ACD required.

To find the fame more exactly in Numbers.
Rule.

Multiply the Radius of the Circle by the Number of Degrees in the given Arch, and that Product multiply again by 0174533 (a Decimal), and this last Product will be the Length of the Arch required.

Suppose the Diameter of a Circle be 22.6 Inches, and the Arc, or Part of the Circumference given, be 52 Degrees 15 Minutes, what is its Length ?

Operation. The Decimal of 15 Minutes is .25, which added to 52 Degrees is 52.25. Then 52.25 × 11.3 the Radius = 590.425, which × .01745 gives 10.30291625 Inches, the true Length of the Arch required.

Problem 3.

To divide a given Right Line into an infinite Number of Parts.

Let the Line given be EF to be divided into a Number of Parts, exceeding any finite Number.

Operation. First, set the Line EF upright between the 2 parallel Lines A Band CD, and suppose them infinitely extended to the Right Hand; then it is evident, that in the Line CD infinitely extended there may be taken an infinite Number of Points, a, b, c, d, &c. Now if to each of these Points there be drawn Right Lines from the Point A taken in the Line A B, to the Left of the Line EF, each of these Lines A a, Ab, Ac, &c. will cut off a fmall Portion of the Line EF; but because, the Points a, b, c, &c. are infinite in Number, so likewife are the Lines A a, Ab, Ac, &c. and confequently the Parts, or small Portions, they will cut off from the Line EF will be infinite in Number too. Whence it is manifest that the Line EF, however small, may be divided into an infinite Number of Parts.

Note. The smallest Particle of Matter, as well as the largeft, is capable of an infinite Divifion.

Problem 4.

To shew that an Angle, as well as a Line, may be continually diminished, and yet never be reduced to Nothing.

Operation. Let A B be a Right Line produced to an infinite Length beyond B. On this Line let there be placed an infinite Number of Equilateral Triangles, as Aab, bcd, def, &c. close to each other. Then from the Point A, draw the Lines Ac, Ae, Ag, &c. to the Tops of the 2d, 3d, 4th, &c. Triangles. Whence it is plain, that every Line drawn from A, to the Top of every succeeding Triangle, will make a less Angle with the Line A B, than the Line immediately before it. But no Right Line drawn from the Point A to the Top of any Triangle set upon the Line A B, how far off soever, could ever coincide with the Line A B; therefore the Angle at A will be continually diminishing, but can never be exhausted, or come to nothing.

Note. The Line ab of the first Triangle will never be quite cut off by any Line drawn from A to the Top of any Triangle; a Part of it towards the Bottom will still remain, which proves here, as in the laft Problem, that Matter is divisible ad infinitum.