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e

20880000 80720001 1210800ee, + 125409000 + 376227001 + 3762270ee

=G 246423025 49284605 - 2464230, 25ee + 354683070 + 354683074 Viz. 213489045 + 15734402e + 87239975e2=2741839 &c. Hence 157344022 + 87239,75ee = 60694877,25 Consequently, 180,ze tee = 695,72 =D

And

180,3te Operation 180,3) 695.72 (3,7 = 1

tes 3,7 549 1 Divifor = 183 146,72

Firit r = 10 2 Divisor = 184,0 128,80

te= 3,7

&c. ate= 13,7 = for a second Operation, with which you may proceed, as in the laft Problem, and so on to a third Operation, if Occasion require such Exactness. But this may be sufficient to thew the Method of refolv. ing any adfe£ted Æquation, without reducing it ; which is not only very exact, but also very ready in Practice, as will fully appear in the last Chapter of this part, concerning the Periphery and Area of the Circle, &c. wherein you will find a farther Improvement in the Numerical Solution of High Æquations than hath hitherto been publish’d.

CHA P. V

Pra£tical Problems, and Kules for finding the Superficial

Contents, or area's of Right-lind Figures. BEfore I proceed to the following Problems, it may be conver

nient to acquaint the Learner, that the Superficies or Area of any Figure, whether it be Right-lin'd or Circular, is composed or made up of Squares, either greater or less, according to the different Measures by which the Dimensions of the Figure are taken or mea. fur'd.

That is, if the Dimensions are taken in Inches, the Area will be composed of square Inches; if the Dimensions are taken in Feet, the Area, will be composed of square Feet ; if in Yards, the Area will be square Yards ; and if the Dimensions are taken by Peles or Perches, (as in Surveying of Land, &c ) then the area will be square Perches, &c. These Things being understood, and

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the Definitions in the 283 and 284 Pages well consider’d, will help to render the following Rules very easy.

PRO B L E M I.

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To find the Superficial Content, or Area of a Square ; or of

any Rigbl-angled Parallelogram. Rule.

Multiply the Length into its Breadth, and the Product will be

the Area required. (See Lemma 1. Page 302.) Example, Suppose the Line AB=6 Yards, and the Breadth AC or BD

6 A

B = 3 Yards, then A B X AC = 6 X3= 18 will be the Number of

3 Square Yards contain' in the Area of the Parallelogram A B C D. This is so evident by the Figure only, that it needs no Demonstration.

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PRO B L E M II.

To find the Area of anv Oblique-angled Parallelogram, viz.

either of a Khombus or Rhomboides. Rule. {

Multiply the Length into its perpendicular Height (or

Breadth) and the Product will be the Area required. That is, the Side AB X BP = the Area of the Rhombus ABCD. For if B P be drawn per

B pendicular to CD, and AG be made parallel to BP, then will GC=PD and GP=CD. Consequently A AGC= A BPD, and ABGP= Rhombus ABCD. But AB X BP=O ABGP.

P Therefore A B X BP, or CDXBP= the Area of the Rhombus ABCD.

Example, Suppose the Side A B = 23 Inches, and the Perpendicular B P=17,5 Inches, (being the fortest or nearest Distance between the two Sides, A B, and CD.) then A B X BP=23% 17,5 = 40295 Square Inches, being the Area of the Rhombus required.

The like may be done for any Rhomboides whole Length and perpendicular Breadth is given.

X x 2

PRO

PROBLEM III.

To find the Superficial Content, or Area of any plain Sriangle.

Every plain Triangle is equal to half its circumscribing Parallelo. gram, (41. 6. 1.) which affords the following Rule.

Multiply the Base of the given Triangle into half its perpendiKule. cular Height, or half the Base into the whole Perpendicular,

and the Product will be the Area.

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That is, BDX;CP, or BDXCP= Area of ABCD. For AC=BP, AB=CP, and

А

F F. BC is common to both A A; there. fore A ABC=ABCP, and for the like Reasons A CFD=A CPD. Therefore A BCP+ B

P

D ACPD=ID ABCD, Confequently BDXCP, or BDX ICP will be the Area of A BCD.

Example, Suppose the Base B D = 32 Inches, and the perpendiCular Height CP=14 Inches.

Then BDXCP = 16 X 14 = 224. Or BDXCP = 32 * 7 = 224. Or thus, 32 * 14 = 448. Then 2) 448 (224 = the Area of the Triangle B C D in square Inches.

PROBLEM IV.

To find the Superficies, or Area of any Trapezium. First, divide the given Trapezium into two Triangles, by draw. ing a Diagonal from one of its acute Angles to the opposite Angle; and let fall two Perpendiculars (from the other two Angles) upon the Diagonal, as in the following Figure. Then

Multiply half the Diagonal into the Sum of the two PerpenKule. diculars, or half the Sum of the Perpendiculars into the Dia

gonal, and the Product will be the Area. That is, į AC X BPF ED. Or A CXBP +;ED= Area of the Trapezium Ą B C D.

For the A Á B C is Half its circumscribing Parallelogram; and the A AC Dis also Half of its circumscribing Parallelogram, as hath been prov'd at the last Problem.

Consequently,

Consequently, B P + EDX AC, or BP+it DX AG will be the Area of the Trapezium,

B as above.

E

Example, Suppose the Diagonal

A AC= 33 feet, the Perpendicu

P lar BP = 15 Feet, and the Perpendicular ED=14 Feet. Then

D BP + ED = 29 Feet, and BP+EDX AC= 29 X 15,5 = 478,5. Or AC XBP titi=33X?? = 478,5. Or thus, 29 X 33 = 957: Then 2) 957 (478,5 any of these Products are the Area of the Trapezium A B C D.

PROBLEM V. To find the Superficial Content or Area of any irregular Polygon or many fided Figure, which by some

B Author's is calld a Triangulate, because

C (as I suppoje) it must be divided into Triangles, as in the annexed Figure

A

D ABCDFG; by which it is evident, that the Sum of the Area's of all those Triangles, found as in the last Pro

F blem, &c. will be the Area of their circumferibing Polygon.

PROBLEM VI. To find the Superficies, or Area of any regular Polygon, viz. of ang regular Pentagon, beragon, Deptagon, Daagon, &c.

Multiply half the Sum of its Sides into the Radius General Kule.

of the inscrib'd Circle, or half the said Radius into
the Sum of the Sides, and the Produkt will be the

Area required.
That is,
AB+BD+DE+EF+FG+GH+HK+KA

:XCP the Area of the annexed Ołtagon ; wherein it is evident, that its Area is compos'd of so many equal isosceles Triangles as there are Number of Sides in the Polygon, viz. of eight ljosceles Triangles, whose Bases are the sides of the OEtagon, viz. AB=BD=DE, &c. And the sides of those Triangles, CA, CB, CD, &o are the Radius's of the circumscribing Circle ; and their perpendicular Heights, viz. PP, is the Radius of the inscribed Circle.

But

2

E

· But the Area of any one of those Triangles is A B XC P by Problem 3. Consequently the Sum of all their Area's will be CP into half the Sum of all their Bafes, as above.

IB This, being equally evident. in all regular Polygons whatsoever, makes the Rule general for finding their Area's.

Now, because it is requir'd to have the Radius of the propos'd Polygon's inscribd Circle, I Ihall here insert (and demonstrate) the Proportions that are between the Sides of several regular Polygons, and the Radius's both of their in fcribd and circumscribing Circles; the one will help to delineate or project the Polygon (if Occasion require it) and the other will help to find its Area.

K

}

And First, Of an Equilateral Triangle. The Side of any Equilateral plain Triangle is in Proportion to the Radius of Circumscribing Circle,

0,57735027 &c, its Inscrib'd Circle, As I : To 30,28867513&c. Perpendicular Height,

0,80602540 &c. AB:CD::1:0,57735027 i, e. AB:CG::1:0,28867513

А
A B:AG::1:0,86602540

Demontration. Let AB = BD=1, then will BG =GD=0,5; but o AB - BG = 0 AG by Theorem 11. That is, I -0,25 = 0,75 = O AG, consequently, v 0,75=0,86602540=AG: Then ÁG: AB :: AB: AH, by Theorem 13, that is, 0,8660254: 1::1:1,15470054 &c. = AH, then ÄH= 0,57735027 = AC. Again, ÄG: DG :: DG: CG, that is, 0,8660254 : 0,5 :: 0,5: 0,28867513=CG. Q. E. D.

Now, by the Help of the First of these Proportions, it will be easy to resolve the following Problem.

PRO

H

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