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half the product of its base and perpendicular; and their sum is the area of the whole figure. (Mens. 13.)

Example I.

Let the plan Fig. 32 be the same as Fig. 31, the sides of which with their bearings, are given in Art. 111.

Then the triangle ABC-BCAP -3.84 sq. chains.
ACE=AC-EP' =1.53

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The contents of the whole=8.26

114. This method cannot be relied on, where great accuracy is required, if the lines are measured by a scale and compasses only. But the parts of the several triangles may be found by trigonometrical calculation, independently of the projection; and then the area of each may be computed, either from two sides and the included angle, or from the three sides. (Mens. 9, 10.)

The sides of the field and their bearings being given by the survey, the angles of the original figure may all be known. (Art. 112.) Then in the triangle ABC (Fig. 32.) we have the side AB and BC, with the angle ABC, to find the other parts. (Trig. 153.) And in the triangle CDE, we have the sides DC and DE, with the angle CDE. Subtracting the angle BAC from BAE, we shall have CAE; and subtracting DEC from DEA, we shall have CEA. There will then be given in the triangle ACE, the side EA and the angles. (Trig. 150.)

The sides and bearings, as given in Art. 111, are

1. AB N. 78° E. 2.46 chains.

2. BC S. 16 W. 3.54

3. CD N. 83 W. 2.72

4. DE N. 12 E. 2.13

5. EA N. 60 E 0.95

Then by Mensuration, Art. 9,

R: Sin. ABC :: AB×BC: 2 area ABC=7.69 sq. chains. R: Sin. AEC:: AEXEC: 2 area AEC=3.06

R: Sin. CDE::CD×DE: 2 area CDE=5.77

Contents of the whole field,


Or the areas of the several triangles may be found by the rule in Mensuration, Art. 10; viz. If a, b, and c, be the sides of any triangle, and h=half their sum;

The area=vh×(h—a)×(h—b)×(h—c.)

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Contents of the field, 69.735 acres.

The method which has been explained, of ascertaining the contents of a piece of land by dividing it into triangles, is of use in cases which do not require a greater degree of accuracy, than can be obtained by the scale and compasses. But if the areas of the triangles are to be found by trigonometrical calculation, the process becomes too laborious for common practice. The following method is often to be preferred.


115. Let ABCDE (Fig. 33.) be the boundary of a field. At a given distance from A, draw the meridian line NS. Parallel to this draw L'R', AG, BH, and DK. These may be considered as portions of meridians passing through the points A, B, D, and E. For all the meridians which cross a field of moderate dimensions, may be supposed to be parallel, without sensible error. At right angles to NS draw the parallels AL, BM, CO, EP, and DR. These will divide the figure LABCDR into the three trapezoids ABML, BCOM, and CDRO; and the figure LAEDR, into the two trapezoids DEPR and EALP. The area of the field is evidently equal to the difference between these two figures.

The sum of the parallel sides of a trapezoid, multiplied into their distance, is equal to twice the area. (Mens. 12.)


(AL+BM)×AG=2 area ABML.

Now AL is a given distance, and BM=AL+BG. But BG is the departure, and AG the difference of latitude, cor

responding to AB one of the sides of the field. (Arts 39. 40.) And by Art. 44,

Rad.: Dist. AB:: Cos. BAG: Diff. Lat. AG.
(Sin. BAG: Depart. BG


Or the departure and difference of latitude may be taken from the Traverse table, as in Navigation. (Art. 50.)

In the same manner, from the sides BC, CD, DE, and EA, may be found the departures CH, CK, DR', AL', and the differences of latitude BH, DK, ER', and EL'. We shall then have the parallel sides of each of the trapezoids, or the distances of the several corners of the field from the meridian NS. For






If the field be measured in the direction ABCDE, the differences of latitude AG, BH, and DK, will be Southings, while R'E and EL' will be Northings. The former are the breadths of the three trapezoids which form the figure LABCDR; and the latter are the breadths of the two trapezoids which form the figure LAEDR. The difference, then, between the sum of the products of the northings into the corresponding meridian distances, and the sum of the products of the southings into the corresponding meridian distances, is twice the area of the field.

It will very much facilitate the calculation, to place in a table the several courses, distances, northings, southings, &c. We have, then, the following


116. Find the northing or southing, and the easting or westing, for each side of the field, and place them in distinct columns in a table. To these add a column of Meridian Distances, for the distance of one end of each side of the field from a given meridian; a column of Multipliers, to contain the pairs of meridian distances for the two ends of each of the sides; and columns for the north and south Areas. See Fig. 33, and the table for example I.

Suppose a meridian line to be drawn without the field, at any given distance from the first station; and place the assumed distance at the head of the column of Meridian Distances. To this add the first departure, if both be east or both west; but subtract, if one be east and the other west;

and place the sum or difference in the column of Meridian Distances, against the first course. To or from the last number, add or subtract the second departure, &c. &c.

For the column of Multipliers, add together the first and second numbers in the column of Meridian Distances; the second and third, the third and fourth, &c. placing the sums opposite the several courses.

Multiply each number in the column of Multipliers into its corresponding northing or southing, and place the product in the column of north or south areas. The difference between the sum of the north areas, and the sum of the south areas, will be twice the area of the field.

This method of finding the contents of a field, as it depends on departure and difference of latitude, which are calculated by right-angled trigonometry, is sometimes called Rectangu lar Surveying.

117. If the assumed meridian pass through the eastern or western extremity of the field, as L'ER' (Fig. 33.) the distance EP will be reduced to nothing, and the figures AEL' and EDR' will be triangles instead of trapezoids. If the survey be made to begin at the point E, cipher is to be placed at the head of the column of meridian distances, and the first number in the column of multipliers will be the same, as the first in the column of meridian distances. See example II.

118. When there is a re-entering angle in a field, situated with respect to the meridian as CDE; (Fig. 34.) the area EDM, being included in the figure BCRA, will be repeated in the column of south areas. But, as it is also included in the figure DCRM, it will be contained in the column of north Therefore the difference between the north areas and the south areas, will be twice the area of the field, in this case, as well as in others.


119. If any side is directly east or west, there will be no difference of latitude, and consequently no number to be placed against this course, in the columns of north and south areas. See example II, Course 1. AB. (Fig. 34.)

The number in the columns of areas will be wanting also, when any side of the field coincides with the assumed meridian. See example II, Course 5. EA. (Fig. 34.)

120. In finding the departure and difference of latitude from the traverse table, the numbers for the links may be looked out separately; care being taken to remove the decimal point two places to the left, because a link is the 100th part of a chain.

Thus if the course be 29°, and the distance 23.46 chains; The diff. of lat. and depart. for 23 chains are 20.12 and 11.15

for 46 links

for 23.46

Example I. See Fig. 33.

.40 20.52

.22 11.37

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The contents of the field 1050.8461 square chains, or

105.0846 acres. (Art. 109.)


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