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PROBLEM V.

11. To find the contents of a field having any number of sides.

Measure the sides of the field and also the diagonals: the three sides of each of the triangles into which the field will be thus divided will then be known, and the areas of the triangles may then be calculated by the preceding rules. Or, measure the diagonals, and from the angular points of the field draw perpendiculars to the diagonals and measure their lengths: the base and perpendicular of each of the triangles will then be known.

1. Let it be required to determine the contents of the field ABCDE, having five sides.

Let us suppose that we have measured the diagonals and perpendiculars, and found,

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a E

d

Aa4.19 ch.; required the area of the field.

Area of triangle ABC= 73.8684 square chains,

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12. To find the contents of a long and irregular figure, bounded on one side by a straight line.

Suppose the ground, of which the contents are required, to be of the form ABEeda, bounded on one side by the right line AE, and on the other by the curve edca.

At A and E, the extremities of the right line AE, erect the two per- a pendiculars Aa, Ee, and on each of

d

b

Then divide the base into any convenient number of equal parts, and measure the breadth of the land at each point of division.

Add together the intermediate breadths and half the sum of the two extreme ones: then multiply this sum by one of the equal parts of the base line, and the product will be the required area very nearly (Mens. Art. 11).

1. The breadths of an irregular figure, at five equidistant places, being 8.20 ch., 7.40 ch., 9.20 ch., 10.20 ch., and 8.60 chains, and the whole length 40 chains, required the

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2. The length of an irregular piece of land being 21 ch., and the breadths, at six equidistant points, being 4.35 ch., 5.15 ch., 2.55 ch., 4.12 ch., 5.02 ch., and 6.10 chains: required the area.

Ans. 9A. 2R. 30P.

3. The length of an irregular piece of land is 80 ch., and the breaths at nine equidistant points are 5.75 ch., 6.12 ch., 4.80 ch., 5.09 ch., 3.87 ch., 5.17 ch., 6.00 ch., 3.94 ch., and 5.95 ch.: what is the area?

Ans. 40A. 3R. 14P.

4. The length of an irregular field is 39 rods, and its breadths at five equidistant places are 4.8, 5.2, 4.1, 7.3, and 7.2 rods: what is its area? Ans. 220.35 sq. rods.

REMARK. If it is not convenient to erect the perpen. diculars at equal distances from each other, the areas of the trapezoids, into which the whole figure is divided, must be computed separately: their sum will be the re

PROBLEM VII.

13. To find the area of a piece of ground in the form of a circle.

Measure the radius AC: then multiply the square of the radius by 3.1416 (Mens., Art. 15.).

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1. To find the area of a circular piece of land, of which the diameter is 25 ch.

PROBLEM VIII.

Ans. 49A. OR. 14P.

14. To find the contents of a piece of ground in the form of an ellipse.

Measure the semi-axes AE, CE. Then multiply them together, and their product by 3.1416.

A F EG B

1. To find the area of an elliptical piece of ground, of which the transverse axis is 16.08 ch., and the conjugate axis 9.72 ch.

Ans. 12A. 1R. 4P.

REMARK I. The following is the manner of tracing an ellipse on the ground, when the two axes are known.

From C, one of the extremities of the conjugate axis as a centre, and AE half the transverse axis as a radius, describe the arc of a circle cutting AE in the two points F and G: these points are called the foci of the ellipse.

Then, take a tape, the length of which is equal to AB, and fasten the two ends, one at the focus F, the other at the focus G. Place a pin against the tape and move it around, keeping the tape tightly stretched: the extremity of the pin will trace the curve of the ellipse.

REMARK II. In determining the contents of ground, in the examples which have been given, the linear dimensions

If the linear dimensions were taken in terms of any other unit, they may be readily reduced to chains. For, a chain is equal to 4 rods, equal to 22 yards, equal to 66 feet. Hence,

1st. Rods may be reduced to chains and the decimal of a chain, by dividing by 4.

2d. Yards may be reduced to chains and the decimal of a chain, by dividing by 22.

3d. Feet may be reduced to chains and the decimal of a chain, by dividing by 66.

REMARK III. If it is thought best to calculate the area, without reducing the linear dimensions to chains, the result can be reduced to acres:

1st. By dividing it by 160 when it is in square rods (Art. 5).

2d. By dividing it by 4840 when it is in square yards (Art. 6).

3d. By dividing it by 43560 when it is in square feet (Art. 6).

OF LAYING OUT LAND.

15. The surveyor is often required to lay off a given quantity of land, in such a way that its bounding lines shall form a particular figure, viz., a square, a rectangle, a triangle, &c. He is also often called upon to divide given. pieces of land into parts containing given areas, or bearing certain relations to each other.

The manner of making such divisions must always depend on a judicious application of the principles of geometry to the particular case.

If, for example, it were required to lay out an acre of ground in a square form, it would first be necessary to find, by calculation, the side of such a square, and then to trace, on the ground, a figure bounded by four equal line:

PROBLEM I.

16. To lay out a given quantity of land in a square

form.

Reduce the given area to square chains, or square rods: then extract the square root, and the result will be the side of the required square. This square being described on the ground, will be the figure required.

1. To trace a square which shall contain 15A. OR. 12P. 15A=60R=2400P

First,
Add,

of which is 49.11.

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Therefore, if a square be traced on the ground, of which the side is 49.11 rods, it will be the required figure.

2. To trace a square which shall contain 176A. 1R. 24P.

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176A 1R 24P=1764 square chains: the square root of which is 42. Hence, if a square be traced on the ground, of which the side is 42 ch., it will be the required. figure.

PROBLEM II.

17. To lay out a given quantity of land in a rectangular form, having one of the sides of the rectangle given. Divide the given area, reduced to square chains or square rods, by the given side of the required rectangle, and the quotient will be the other side. Then, trace the rectangle on the ground.

1. To lay off 240

acres in a rectangular form, one of

the sides being given, and equal to 80 rods.

First, 24042400 square chains 38400 square rods. Then, 80)38400(480 rods; which is the required side

of the rectangle.

18. A great number of similar problems might be proposed. The solution of them does not, however, properly belong to surveying. The laying out of the ground, and

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