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PROBLEM VII.

The Side of any Equilateral plain Triangle being given, to find its

Area. Example, Suppose the side of the propos'd Triangle A B C to, be 25 Inches, viz. AB= BC= CA = 25 B First 1:0,866254: : AB= 25:21,650635 = BP by Theorem 13. Then AP (=iC A) X BP = the Area of A ABC by Rule to Problem 3, that is, 12,5 X 21, 630635 = 270,6329 the Area in square Inches.

Or chis Problem may be otherwise resolv’d, A P C thus: Let b = AP= AC. Then 2b A B. But OAB - AP=OBP. By Theorem 11. That is, 4bb - bb = 3bb = BP. Consequently, ✓3bb = B P. Then b v 3bb = BPX A C. viz. v 3bbbb 3 = the Area of the Triangle.

Secondly, For a Pentagon.
The Side of any Pentagon is in Proportion to the Radius of
Circumscribing Circle,

50,85065080 &c. its Infcrib'd Circle, As 1 : To 0,68819096 &c, Perpendicular Height,

1,53884176 &c.
ABAC::1:0,85065080
Viz. A B C H:::o,688 Igo96

B
LAB : AH::1:1,53884176

}

G

Demonftration.

X
Let A B = I. And draw the

H н
Diagonals AD, AF, and DG, which
will be equal to one another. Then will

F
AGX DF + AD XGF= AFX DG
by Theorem 19. Consequently, AGX
DF-AFX DG: -- AD XGF, that is, O AB=OAD:-
AD XGFI (because AB=AG=DF, and AD=AF=DG)
hence it will be AD = 1,61803398, then AD- ODH=
DAHhy Theor. Il. But DH=AB, therefore v DAD-TOAB
=AH = 1,5388417. Again, AH: AD:: AD: AX=2AC.
For A AHD and A ADX are alike,

Ergo

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O AD

Ergo

AH

=2AC =1,70130161. Hence AC=0,85 65080 But AH - AC=CH=0,68819096, &c.

Q. E. D. From hence it will be easy to resolve the following Problem.

PROBLEM VIII.

The Side of any regular Pentagon being given, to find its Area. Example, Suppose the given Side to be 15 Inches long, then it will be, as 1 : 1,53884176::15:22,0826264 the perpendicular Height; and by the general Rule 22,0826264x 4 = 1653619698 the Area requir'd.

В

Thirdly, For an Ddagon. The Side of any regular OEtagon is in Proportion to the Radius of its

{ Circumscribing Circle, As I : to 1,30656296&c.
Inscribed Circle, As I : to 1,20710678 &c.

Viz.{ BA:CP :: 1:1,20710678

Demontration. D

A Draw the Right-line D B, and

from the Point B let fall the Perpendicular B z upon the Diameter DA.

Then will A BDA and ADXB H

be alike, by Theorem 10 and 12. Let { Sb BA=1.0=CA

= BD, and y =Bx
Then
I 20:b::e: y. viz. DA:BA :: DB:B:

DB
6

4aayy
3 = e = ODB

G

may

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2

2

bb

But

= bb

That is

4aayy 4 400

bb ODA - ODB=BA. By Theorem it. 5 4bbaa - 4aayy = bbbb aa = yy. For Cx = Bx

Cx + = OCB

4 Xbb

5, 6

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بس 9

Іо! аа

2

52 61 714bbaa --2a+=ht. Or 2a+ - 4bbaa wbt

21 8 Jaaaa abbaa =
09 at
9 at 2bbaa+b+=b4-6 = 16+

bb =Vbt Io + bb naa= bb+vb+

12 a=v:bbtv1b=1,30656296, &c. = CA Then 13 aa--- bb=CP, viz. CH-HP- OCP 13 w 141V aa-10b=1,20710678 &c. =CP. From hence 'cwill be easy to find the Area of any Octagon.

PROBLEM IX.
The Side of any regular O&agon being given, to find its Area.

Example, Suppose the Side given to be 12 Inches long; First, as 1:1,20710678::12:14,48528136 = the Radius of its inscrib'd Circle; then 12 X 4 = 48 is half the Sum of its Sides, and 48 x 14,48528136 = 695,2935 the Area required.

Fourtbly, For a Decagon.
The Side of any regular Decagon (viz. a Polygon of ten equal

Sides) is in Proportion to the Radius of
Circumscribing Circle, as 1 : to 1,61803398 &c.

Circle, as 1 : to 1,53884176 &c.
Viz. { BA:CA:: 1:1,61803398

B
Demonftration.

b=BA=I.a=CA
Let
{c= DB, and y = Bx

D Then

20:b::e: That is,

DA:BA::DB:Bx

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2ay be

1

2

P

and 2y=

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But

1e

a

3
4 IC

Again
That is

be

H
3 | 2y:8::1:1,61803398. See Pentagon,
4 1,61803398

= 2y = ==
5 | 1,61803398 =a=CA

aa *lb = OCP. 6

viz. O CF-PF=O CP. By Theorem Ir. 71 V 2,61803398—0,25=1,53884176=CP

PRO

Уу

P R O B L E M X. The Side of any regular Decagon being given, to find its Area.

Example. Let the given Side be 14 Inches long; then, 29 1:1,53884176:: 14: 21,543784 = the Radius of the inscribd Circle ; and 14 X 5 = 70 is half the Sum of its Sides. Lastly, 21,543784 X 70 = 1508,06 488 the 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 { , 1

Circumscribing Circle, as I : to 1,93185165, &c.
Inscrib’d Circle

, c.
SBA:CA::1:1,03185165
Viz.

(B 4: CP:::1,86632012

B

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2

2

HPF

Demonliration.
Let b=BA=I.a=CA as before
And e = xA; then a -= Cx
First

bb-Bx=ee D
{

By Figure.
But 2 Bx=CA=a

3 o Bx = aa
I, 3) 4 b6-aa = ee
4 W
? 5 V bbczau=ee
Again 6 aa-aa = aa 2ae tee
Viz.

0 Bx = 0 Cx 5 Х

aa = 2ae 4

7

8. bb I aa 2a v bb - a= l - 2ae 72 8 aa = aa + bb-1aa

- 2a v bb

Laa 9

+ 10 20 v bb aa = bb IO II/4bbaa

ht II 12 aaaa

4bbaa 13, C 0 13 aaaa 4bbaa + 46+ = 3b+ = 3 I. w? 14 aa - 2bb=V 3= 1,7320508075 14 + 2bb 15 aa = 2bb tv 3= 3,7320508075 15 w 16a = V 3,7320508075 = 1,93185165 = CA

Again 17 aa-bb=0CP. viz. O CF-PF=QCP 17, Hence 18 CPSV aa-bb= 1,86632012

ОСВ 72 av bb

2a

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9 aa

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аааа

- 6+

2

Q. E. D.

Con

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Confetary.
Hence, if the side of any regular Dodecagon be given, the Ra-
dius of its infcrib'd Circle inay be easily obtain'd, and thence the
Area found; as in the last Problem.

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

C H A P. VI.

A new and easy Method of finding the Circle's fperiphery

and area to any asign’d Exactness (or Number of Figures) by one Æquation only. Also a new and facile Way of

making Natural Signs and Tangents. LET us suppose (what is very easy to conceive ) the Circle's Area

to be compos’d or made up of a vast Number of plain Isosceles Triangles, having their acutest Angles all meeting in the Circle's Center. And let us imagine the Bases of those Triangles so very small, that their Sides and their Perpendicular Heights, viz. the Radius's of their circumscrib'd and inscrib'd Circles (vide Problem 6.) may become so very near in Length to each other, as that they may be taken one for another without any sensible Error: Then will the Peripheries of their circumscribing and inscribed Circles become (althonot co incident, yet) so 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 Bases of those Tsosceles Triangles) to such a convenient Smallness as may be necessary to determine and settle the Proportion betwixt a Circle's Diameter and its Periphery (to any align’d Exaélness) hath hitherto been a Work which requir'd great Care and much Time in its Performance; as may easily be conceiv'd from the Nature of the Method us'd by all those who have made any considerable Progress in it, viz. Archimedes, Snellius, Hugenius, Mætius, Van Culen, &c. These proceeded with the bisecting of an Arch, and found the Value of its Chord to a convenient Number of Figures at every single

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