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Ergo

DAD

AH

=2AC 1,70130161. Hence AC=0,85065080

But AH-AC-CH=0,68819096, &c.

Q. E. D.

From hence it will be easy to refolve the following Problem.

PROBLEM VIII.

The Side of any regular Pentagon being given, to find its Area.

Example, Suppofe the given Side to be 15 Inches long, then it will be, as I 1,53884176:: 15:22,0826264 the perpendicular Height; and by the general Rule 22,0826264× = 165,619698 the Area requir'd.

Thirdly, For an Dragon.

The Side of any regular Octagon is in Proportion to the Radius of its { Circumfcribing Circle, As 1: to 1,30656296 &c. Inferib'd Circle, As I to 1,20710678 &c.

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S BACA:: 1: 1,30656296
BA: CP:: 1: 1,20710678

Demonftration.

Draw the Right-line D B, and from the Point B let fall the Perpendicular B x upon the Diameter DA.

Then will ▲ BDA and A D x B Δ DxB be alike, by Theorem 10 and 12.

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b=BAI

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a = CA

Let {BD, and y =Bx

Then I za be y. viz. DA: BA::DB: Bx

2ay

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=e=DB

b

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That is 4 × bb

Again

D DA-DB□ BA. By Theorem it. 5,4bbaa

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78 9

C

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·2a+b+. Or 2a4— 4 bbaa

8' aaaa - 2bbaa — — 1 b+

9 a 2bbaa + b = b+ — 1 b + = b + 10 aabb√ 1b +

10bb11aa = bb + √ !b+

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2

-b+

12 a=√:bb+b=1,30656296, &c. = CA Then 13 aa- bb CP, viz. CH- HP=□СP 14 aa- -1bb=1,20710678 &c. =CP.

13 2

From hence 'twill be eafy to find the Area of any Octagon.

PROBLEM IX.

The Side of any regular Octagon being given, to find its Area. Example, Suppofe the Side given to be 12 Inches long; First, as 1: 1,20710678: 12: 14,48528136 the Radius of its infcrib'd Circle; then 12 X 4 = 48 is half the Sum of its Sides, and 48 × 14,48528136695,2935 the Area required.

Fourthly, For a Decagon.

The Side of any regular Decagon (viz. a Polygon of ten equal
Sides) is in Proportion to the Radius of

Circumfcribing Circle, as 1 to 1,61803398 &c.
Infcrib'd Circle,

SBA: CA:: 1: 1,61803398

{BA: 1:

BA:CP::1:1,53884176

Demonftration.

b=BA=1.a=CA

c = DB,
DB, and y Bx

Its {

Viz.

Let {

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a

2ye 1: 1,61803398. See Pentagon.

32ye

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= 2y ==

=

a

a

4 1,61803398
1,61803398= a = CA

4÷Ie 5

Again 6

That is

aa bb CP.

viz. OCF-O PF□ CP. By Theorem 11.

72,61803398-0,25—1,53884176=CP
Y y

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PROBLEM X.

The Side of any regular Decagon being given, to find its Area. Example. Let the given Side be 14 Inches long; then, as 1: 1,53884176:: 14: 21,543784 Circle; and 14 X 570 is half the 21,543784 X 701508,06488 the

Laftly,

the Radius of the infcrib'd
Sum of its Sides.
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 & Circumfcribing Circle, as 1 to 1,93185165, &c.
{Infcrib'd Circle, as I to 1,86632012, &c.
BA: CA::1: 1,03185165
{BA:CP::1: 1,86632012

Viz.

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13,

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I

5 √bb-4aaee

6 aa- аа - аа 2ae +ee

О СВ

72a√ bb

8bb

9 aa

10 2a

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BxCx

[ocr errors]

aa 2ae

4

HPF

- aa - 2a√ bb — — aa ee — 2ae

4

Jaa = aa + bb.

bb

aabb

II 4bbaa — aaaa = b+

12 aaaa- 4bbaa =- .b4

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13 aaaa-4bbaa + 4b4 = 3b4 = 3

14 aa2bb√3 = 1,7320508075 14+2bb 15 aa = 2bb + √3 = 3,7320508075

15 ww

CP

16 a = √ 3,7320508075 = 1,93185165=CA Again 17 aa-bb-CP. viz. CFO PF 17, Hence 18 CP✓ aabb 1,86632012

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Q. E.D.

Con

Confectary.

Hence, if the Side of any regular Dodecagon be given, the Radius of its inferib'd Circle may be eafily obtain'd, and thence the Area found; as in the laft Problem.

The Work of the 'foregoing Polygons, being well confider'd, will help the young Geometer to raise the like Proportions for others, if his Curiofity requires them: And not only fo, but they will alfo help to form a true Idea of a Circle's Periphery and Area, according to the Method which I fhall lay down in the next Chapter for finding them both.

C H A P. VI.

A new and eafy Method of finding the Circle's Periphery and Area to any affign'd Exactness (or Number of Figures) by one Equation only. Alfo a new and facile Way of making Natural Signs and Tangents.

L

ET us fuppofe (what is very easy to conceive) the Circle's Area to be compos'd or made up of a vaft Number of plain Ifofceles Triangles, having their acuteft Angles all meeting in the Circle's Center. And let us imagine the Bafes of thofe Triangles to very fmall, that their Sides and their Perpendicular Heights, viz. the Radius's of their circumfcrib'd and infcrib'd Circles (vide Problem 6.) may become fo very near in Length to each other, as that they may be taken one for another without any fenfible Error: Then will the Peripheries of their circumfcribing and infcribed Circles become (altho' not co incident, yet) fo 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 Bafes of thofe Ifofceles Triangles) to fuch a convenient Smallnefs as may be neceffary to determine and fettle the Proportion betwixt a Circle's Diameter and its Periphery (to any assign'd Exactness) hath hitherto been a Work which requir'd great Care and much Time in its Performance; as may eafily be conceiv'd from the Nature of the Method us'd by all those who have made any confiderable Progrefs in it, viz. Archimedes, Snellius, Hugenius, Matius, Van Culen, &c. Thefe proceeded with the bifecting of an Arch, and found the Value of its Chord to a convenient Number of Figures at every single Bifection,

Y y 2

Bifection, repeating their Operations until they had approach'd to the Chord defign'd.

And this Method is made choice of by the learned Dr. Wallis in his Treatife of Algebra; wherein, after he hath given us a large Account of the different Enquiries made by feveral (very eminent in Mathematical Sciences) in order to find out fome easier and more expeditious Way of approaching to the Circle's Periphery, as in Chap. 82, 84, 85, 86, and feveral other Places, he comes to this Refult, (Page 321.)

"'Tis true faith he, we might in like Manner proceed by con"tinual Trifection, Quinquifection, or other Section, if we had "for these as convenient Methods of Operation as we have for "Bifection: But becaufe Euclid fhews how to bifect an Arch "Geomatrically, but not to trifect, &c. and the one may be "done (Algebraically) by refolving a Quadratick Æquation, but "not thofe other, without Equations of a higher Compofition, I "therefore make Choice of a continual Bifection, &c. And then he lays down these following Canons."

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How tedious and troublesome the Work of thefe complicated Extractions is, I leave to the Confideration of thofe, who either have had Experience therein, or out of Curiofity will give themselves the Trouble of making Trial.

Again, in Page 347, the Doctor inferts a particular Method propos'd by Libnitius, publifh'd in the Acta Eruditorum at Leipfick, for the Month of February 182, in order to find the Circle's Area, and confequently its Feriphery which is this:

:

::

As I to-+-+-+-+-5, &c. infinitely fo is the Square of the Diameter to the Circle's Area. But this convergeth fo very flowly, that it is not worth the Time to pursue it.

I fhall here propofe a new Method of my own, whereby the Circle's Periphery, and confequently its Area, may be obtain'd

infinite

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