two triangles, in both of which all the sides will be known. Then, find the areas of the triangles separately, and their sum will be the area of the quadrilateral.

1. Suppose that we have measured the sides and diagonal AC, of the quadrilateral ABCD, and found AB=40.05 ch, CD=29.87 ch, BC 26.27 ch, AD=37.07 ch, AC=55 ch:

and

required the area of the quadrilateral.

B

Ans. 101A 1R 15P.

REMARK. Instead of measuring the four sides of the quadrilateral, we may let fall the perpendiculars Bb, Dg, on the diagonal AC. The area of the triangle may then be determined by measuring these perpendiculars and the diagonal AC. The perpendiculars are Dg=18.95 ch, and Bb 17.92 ch.

PROBLEM V.

112. To find the content 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 content of the field ABCDE, having five sides.

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

AC=36.21 ch, EC-39.11 ch,

E

a

A

B

4.19 ch; also Ea=4.00 ch, Ed=13.60 ch, Ab=20.30 ch :

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

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

a

B C

d

At A and E, the extremities of the right line AE, erect the two perpendiculars Aa, Ee, and on each of them measure the breadth of the land. 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. Prob. VI).

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

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, 3.55 ch, 4.12 ch, 5.02 ch, and 6.10 chains required

the area.

REMARK.

Ans. 9A 2R 30P.

If it is not convenient to erect the perpendiculars 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 required area.

:

PROBLEM VII.

114. To find the area of a picce of ground in the form of a circle.

Measure the radius AC: then multiply the square of the radius by 3.1416 (Mens. Prob. a

X).

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

PROBLEM VIII.

Ans. 49.A OR 14P.

115. To find the content of a piece of ground in the form of an ellipsis.

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

by 3.1416.

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 1. 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 AB in the two points

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 content of ground, in the examples which have been given, the linear dimensions have been taken in chains and decimals of a chain. 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. 107).

2d. By dividing it by 4840 when it is in square yards (Art. 107).

3d. By dividing it by 43560 when it is in square feet (Art. 107).

OF LAYING OUT AND DIVIDING LAND.

116. 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 with 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, four equal lines at right angles to each other.

PROBLEM I.

117. 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=60 R=2400 P

First,
Add

12 P; hence,

15 A 0R 12 P=2412 P; the square root of which is 49.11.

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 1764 1R 24P.

root of which is 42.

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

PROBLEM II.

118. To lay out a given quantity of land in a rectangular form, having one of its sides 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, 240A 2400 square chains 38400 square rods.

Then,

80) 38400(480 rods; which is the required side