20648 the difference between the quantity proposed and the quantity parted off by the guess-line, which divided by 383 (half the guess-line) gives 54 links, to be set off from n toward A. Hence DF, is the true line of division; and the irregular figure AFDE contains 2A. 3R. 20P.

Now, by the scale, Ac is found = 377 links. Measure, therefore, in the field 377 links from A to c; stake out the line DCF and the work will be completed.

NOTE 1. If the area of the irregular figure ADE be subtracted from the given quantity, and the remainder divided by half the line AD, the quotient will be the perpendicular of the triangle ADF; the side AB being nearly straight from a to F.

Now, at the distance of this perpendicular, draw a line parallel to AD, and it will intersect the side AB in F, the point to which the division-fence must be made.

2. It is not absolutely necessary to survey and plan a whole field in order to part a portion from it, as the guess-line and portion parted off may be measured in the field; but, in my opinion, the former, in general, is a more eligible method than the latter, as you have a better opportunity of proving your work.

SECTION II.

The method of dividing a piece of land among sundry claimants, in the proportion of their respective claims, or a common, &c., of variable value, among any number of proprietors, in the proportion of their respective interests.

When land becomes the property of co-heirs, co-partners, joint purchasers, &c., it is generally divided into such shares as the co-parties are entitled to; and this cannot possibly be accurately effected without the assistance of some person who is not only well acquainted with surveying, but also with the method of apportioning land.

In this process an error is evidently much more material than one committed in surveying.-When a field, &c., is to be divided into any number of parts, equal or unequal, it is necessary, first, to ascertain its dimensions, and next to enquire of the parties concerned in what part of the property in question they wish their respective shares to lie.

PROBLEM I.

To divide a square or rectangle, either equally or unequally among any number of persons, by lines parallel to one of its sides.

RULE.-If the parts into which the field is required to be divided be equal, divide the side which will be cut by the division-fences by the number of those parts, and the quotient will be the distance at which the division-fences must be placed from each other, and from the outsides to which they are parallel. But, if the parts be unequal, you must then part off each person's share as directed in Section I. Problem XI.

Examples.

1. Divide the square ABCD containing 5A. 2R. 20P. into three equal parts by fences parallel to the side AB.

Here 5A. 2R. 20P. 562500 square links; and 562500 = 750 links, the side of the square. This, divided by 3, the number of parts, gives 250 links, the distance at which the first division-fence must be placed from AB, &c. From A and B, therefore, set off 250 links to E and F; join EF, and the rectangle ABFE will be one of the parts required.

Again, from E and F set off 250 links to G and H; join GH, and the rectangles EFGH and GHCD will be the two other parts required.

2. Divide ABCD, representing a rectangular field whose length is 1500 and breadth 800 links, among three men, A, B, and C, by fences parallel to the side AD, so that A may have 3A., B 4A., and C the remainder

Here 3A. = 300000 square links, which divided by 800 gives 375 links = AE or DF; hence the rectangle AEFD contains A's share.

Again, 4A.

500 links

share.

400000 square links, which divided by 800 gives EG or FH; hence the rectangle EGHF contains B's

Now, the rectangle ABCD is found to contain 12A., consequently, the rectangle GBCH, containing 5A., is C's share.

NOTE. This and similar examples may also be performed by the following proportion: As the area of the whole rectangle is to the whole base or side cut by the division-fences, so is each person's share of the rectangle to his share of the base.

PROBLEM II.

To divide a triangular field, either equally or unequally, among any number of persons, by fences made from any of its angles to the opposite side.

RULE.-If the parts into which the field is required to be divided be equal, divide the base or side to which the division-fences are to be made by the number of those parts, and the quotient will be each person's share of the base. But, if the parts be unequal, say, as the area of the whole triangle is to the whole base, so is each person's share of the triangle to his share of the base. (See Simpson's Geom. IV. 7, and Euclid vI. 1.)

Examples.

1. Divide ABC, representing a triangular field whose sides AB, AC, and BC are 1500, 1200, and 1000 links respectively, into three equal parts, by fences made from the angle c to the side AB.

D

E

B

Here AB 1500 links, which divided by 3 (the number of parts) gives 500 links, each person's share of the base. From a, therefore, set off 500 links to D, and from D 500 links to E; draw the lines CD and CE, to represent the division-fences; and the triangles ADC, DEC, and EBC, are the three equal parts required.

2. Divide ABC, representing a triangular field whose sides AB, AC, and BC are 1450, 1150, and 960 links respectively, into three equal parts, by fences made from the angle A to the side BC.

E

B

=

Here BC 960 links, which divided by 3 (the number of parts) gives 320 links, each person's share of the side BC.

From B,

therefore, set off 320 links to D, and from D set off the same distance and the lines AD and AE will be the lines of division re

to E; quired.

3. Divide ABC, representing a triangular field whose sides AB, AC, and BC are 2200, 1700, and 1500 links respectively, among three persons A, B, and C; so that, each person partaking of a pond at C, A may have 3A., B 4A., and C the remainder.

D

B