all the angles of the field are visible. The bearing and distance of the stations, and the bearings of the angles, from each station, are as follow. What is the area of the field?

The station G bears from the station F, N. 43° W. 20 ch.

PROBLEM III.

To lay out a given quantity of land in a rectangular form, having the length to the breadth in a given ratio.

RULE.

As the less number of the given ratio,

Is to the greater;

So is the given area,

To a fourth term.*

The square root of this fourth term will be the length required. Having the length, the breadth may be found by the preceding problem. Or it may be found in the same manner as the length. Thus,

As the greater number of the given ratio,

Is to the less;

So is the given area,

To a fourth term.

The square root of this fourth term will be the breadth required.

EXAMPLES.

1. It is required to lay out 864 acres in a rectangular form, having the length to the breadth in the ratio of 5 to 3.

* DEMONSTRATION. Let ABCD, Fig. 85, be a rectangle, and let ABFE and AHGD be squares on the greater and less sides respectively: then (1.6) AD : AE (AB) :: the rectangle AC : square AF. Also AB : AH (AD) :: the rectan

As 3:5: 138240: 230400

✔230400-480 Perches, the length required.

✔82944

Sq. P. Sq. P.

As 5:3: 138240: 82944

288 Perches, the breadth required.

2. It is required to lay out 27 A. 3 R. 20 P. in a rectangular form, having the length to the breadth in the ratio of 9 to 7.

58.897 P.

Ans. Length 75.725 P.

Breadth

PROBLEM IV.

To lay out a given quantity of land in a rectangular form, having the length to exceed the breadth by a given difference.

RULE.

To the given area, add the square of half the given difference of the sides, and extract the square root of the sum; to this root add half the given difference for the greater side, and subtract it therefrom for the less.*

* DEMONSTRATION. Let ABCD, Fig. 86, be a rectangle; in DC let DE be taken equal DA or BC, and let EC be bisected in F; then (6.2) DF2=DC × DE +FC2=DC× AD+FC2=the rectangle AC+the square of half the difference of the sides DC, DA; also DF+FC=DC, the greater side, and DF-FC=DE or DA, the less side.

This problem may be neatly constructed thus: take EC equal the given difference of the sides and bisect it in F; make EG perpendicular to EC and equal to the square root of the given area, and with the centre F and radius FG describe the arc DG meeting CE produced in D; make DA perpendicular to DC and equal to DE, and complete the rectangle ABCD, which will be the one required. Since (47.1.) FG2=EG2 +EF2 the given area+the square of half the given difference of the sides, the truth of the construction is plain from the

1

EXAMPLES.

1. It is required to lay out 47 A. 2 R. 16 P. in a rectangle, of which the length is to exceed the breadth by 80 perches.

2. It is required to lay out 114 A. 2 R. 33,4 P. in a rectangular form, having the length to exceed the breadth by 15.10 chains.

27.15 Ch.

Ans. Length 42.25 Ch. Breadth

PROBLEM V.

To lay out a given quantity of land in the form of a triangle or parallelogram, one side and an adjacent angle being given.

RULE.

For a triangle.

As the rectangle of the given side and sine of the given angle,

Is to twice the given area;

So is radius,

To the other side adjacent to the given angle.

Then having two sides and the included angle given, the other angles and side, if required, may be found by

For a parallelogram..

As the rectangle of the given side and sine of the given angle,

Is to the given area;

So is radius,

To the other side adjacent to the given angle.*

EXAMPLES.

1. Let AB, BC, Fig. 87, be two sides of a tract of land; the bearing of AB is S. 87° W. dist. 16.25 ch. and the bearing of BC, N. 27° E.; it is required to lay off 10 acres by a straight line AD, running from the point A to the side BC.

* DEMONSTRATION. It is demonstrated, prob. 3, sect. 1, Measuring Land, that rad. sin. B: ABX BD: 2.ABD (see Fig. 87.); therefore (1.6 cor.) rad. XAB: sin. BX AB :. AB × BD : 2.ABD, or (16.5.) sin. B× AB : 2.ABD :: rad. × AB : AB× BD :: rad. : BD. Since ABDF is equal to 2 ABD, the truth of the rule for the parallelogram is evident.

This problem may be constructed as follows; take AB equal the given side and draw BC making the angle B equal the given angle; make BE perpendicular to AB and equal twice the given area of the triangle divided by the given side, or equal the given area of the parallelogram divided by the given side; and parallel to AB, draw EF cutting BC in D; join DA, then will ABD be the triangle required, or complete the parallelogram ABDF, for the one required. The reason of the construction is plain.

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