Page images
PDF
EPUB

is ab. Therefore if upon the center B, with the radius 115, an arc be described ; and a line be drawn parallel to AB, touching the arc, and cutting ed in D; and if Ad be drawn, it will be the line cutting off the required quantity ADC A.

NOTE. If the first piece had been too much, then Dinust have been set below B.

In this manner the several shares of commons, to be divided, may be laid down upon the plan, and transferred from thence to the ground itself.

Also for the greater eafe and perfection in this business, the following problems may be added.

[blocks in formation]

From an Angle in a Given Triangle, to draw Lines to the Opposite Side, dividiug the Triangle into any Number of Parts, which mall be in any

afigned Proportion to each other. Divide the base into the same number of parts, and in the same proportion, by problem 1; then from the several points of division draw lines to the proposed angle, and they will divide the triangle as required *.

EX A M P L E.

Let the triangle ABC, of 20 acres, be divided into five parts, which thall be in proportion to the numbers 1, 2, 3, 5, 9; the lines of divilion to be drawn from A to CB, whose length is 1600 links.

* DEMONSTRATION. For the several parts are triangles of the fame altitude, and which therefore are as their bases, which bases are taken in the affigned proportion.

Here

A

[ocr errors]

Here 1 + 2 + 3 + 5 + 9 - 20, and 1600 = 20 = 80; which being multiplied by each of the proportional numbers, we have 80, 160, 240, 400,

and

720. Therefore I make ca = 80, ab = c abc

B 160, bc = 240, cd = 400, and db = 720 ; then by drawing the lines Aa, ab, Ac, Ad, the triangle is divided as required.

[blocks in formation]

From Any Point in one side of a Given Triangle, to draw Lines to the other Two Sides, dividing the Triangle into Any Number of Parts which

shall be in Any Aligned Ratio. From the given point

B

I D, draw DB to the angle opposite the side ac in

H which the point is taken; then divide the same side AC into as many parts

А

E F D G AE, EF, FG, GC, and in the same proportion with the required parts of the triangle, like as was done in the last problem; and from the points of division draw lines EK, F1, GH, parallel to the line BD, and meeting the other sides of the triangle in K, 1, ; lastly, draw KD, 1D, HD, so shall ADK, KDI, ID, AB, HDC be the parts required *.

The example to this will be done exactly as the last.

* DEMONSTRATION. The triangles ADK, KDI, IDE, being of the same height, are as their bates AK, KI, 16; which, by means of the parallels EK, FI, DB, are as AE, EF, FD; in like manner, the triangles CDH, HDS are to each other as cg, GD: but the two triangles IDE, BDH, having the same base ed, are to each other as the distances of i and froin , or as Fi) to do; confequently the parts DAK, DKI, DIBH, DHC are to each other as AE, EF, FG, GC.

A COL.

[blocks in formation]

I.

R

QUESTION
EQUIRED the area and plot of a figure from

the following field-book. Note, That a cipher in the place of a perpendicular, denotes that there the base line touches an angular point. So here the ciphers for the first and last perpendiculars, shew that the base line begins and ends at an angle. Also R denotes right, and i left.

Field-Book.
]
A

G

E

[ocr errors]
[blocks in formation]

II.

QUESTION Beginning at the westmost station a, of a large tract of land, and going round towards the north, suppose the lengths of the lines and the angles formed by them and che meridians, to be taken thus: AB = 1550 links, and

its direction N, 35° E, that is, 357° from the north towards the east; then proceeding to the 2d station B, I find the direction of Bc to be n E, 723°, and its length = 1870 links. In the same manner I find all the other sides and their directions as expressed in order in the 2d and 3d columns of the following table; the ift column containing only the number and mark of each station.

Required the plan and content of this pjece,

Stations Angles Sides
1. A NE 357° 1550 A B
2. B NE 72 1870 BC
3. C SE 70% 1870 CD
4. D SW 53

I 245 DE

Answer 5. E SE 83:

2410 EF 6. F SW 311

1520 FG 145ac. Ir. 39'5776p.

2260 GH 8. H NW 737

2730 HI 9. 1 NW 177 1456 1A

7. GSW 623

[ocr errors][merged small][merged small][ocr errors][ocr errors][merged small][merged small][merged small][merged small][merged small]

H a

с 1 е Note, A line is wanted to be drawn along the bottom of the figure, through the points a i bckedsf.

[ocr errors]

III.

QUESTION In a pentangular field, beginning with the south side, and measuring round towards the east first, the ift or fouth side was = 2735 links, the 2d = 3115, the 3d = 2370, the 4th = 2925, and the 5th = 2220; also the diagonal from the ist angle to the 3d was 3800, and that from the 3d to the 5th was 4010: Required the figure and area.

[merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small]

In going round a five-sided field ABCDE, the sides and angles were thus : The side AB = 1940 links, and the angle B 110° 30'; the side bc = 1555, and the angle c 117° 45'; the side cd= 2125, and the angle D 91° 20'; and the side de = 2741 : Required the figure and the content.

[ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small]
« PreviousContinue »