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OFF-SETS are perpendiculars measured from the amgular points of the land, to the stationpay distance on either side thereof; to do which, observe the following


Rule a table, as hereafter exemplified; take the bearing of the line from which off-sets are to be made, and from this line make off-sets at right angles to each angular point on either side; set the distance to each offset under the head of Base, and the distance of each offset, under that of Perpendicular.

Then, to find the area of these off-sets,

Set the firt base opposite to station 1, in the column headed Bases, and take the first base from the second, the second from the third, &c. and set the remainders, respectively, in said column; each of which will be the distance between the respective perpendiculars. Then add the first and second perpendiculars together, the second and third, third and fourth, &c. and set their sums respectively, in the column marked Sum of Perpendiculars. Then each of these numbers, multiplied in its respective base, will give the double area of the quadrilateral figure, and triangle, (by Problems 6 and 9, pages 107 and 110. -


Let A B C D E. F. G. H, be the boundary of a field, by which it is required to lay off a field containing 85A. 3R. 20P. in form of an oblong, by a line parallel to AI.

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A. R. P.
From 85 3 20 -
Take 21 3 30 Area of the off-sets. -

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Remains 63 3 30 To be laid off. 40)30 \ 4)3.75 - - 639.375 Square Chains.

By Problem 4, page 105, divide thus: 46)639.875(13.90 ch. nearly, from A and I to K and L.


Suppose a meadow bounded on a brook, as under specified, the area of which is required 2

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JVote. The 1st perpendicular is set opposite the 1st base, because they form a triangle.

. . . . Measured in Perches.

JVote. The perpendicular opposite the 9th station, extends to the extreme point; and that opposite the 10th, to the nearest - point. The offsets on Area of the off-sets. 91.99 the 9th station are as oblong. 355.2 above.

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