West, 12.70 ch.; it is required to divide the trapezoid into two parts by a line EF parallel to AB or DC, so that the part AEFB

may

be to the part EDCF as 3 to 2.

2. The boundaries of a trapezoidal field ABCD are given as follow: viz. AB, N. 80° W. 60 per. BC, N. 3901 W. 45.5 per. CD, S. 80° E. 89.4 per., and DA, °

, South, 30 per., and it is required to divide it into two equal parts by a line EF parallel to AB or CD; what will be the distance AE? Ans. 16.46 per.

PROBLEM XVII.

The bearings and distances of the sides AB, BC, CD,

DA, Fig. 105, of any quadrilateral tract of land being given to divide it into two parts having a given ratio, by a line ĘF running parallel to one of the sides as AB.

RULE.

Suppose the sides DA, CB procluced to meet in G, and as in the preceding problem, find GD and GA.

Let HI parallel to AB be a line making the trapezoid AHIB equal to the given trapezium ABCD. Then,

1

As the rectangle of the sines of the angles H and C,
Is to the rectangle of the sines of the angles D

and I;
So is the square of GD,
To the
square

of GH.

The square root of which gives GH; and GH less GA gives AH; then as in the preceding problem it will be,

As the sum of the numbers expressing the ratio of

the parts,

Is to the greater or less of them, according as the

greater or less part is to be adjacent to CD; So is the rectangle of AH and sum of GH and GA, To a fourth term.

square root of the remainder will give GE, and GE less GA will be the distance AE*.

Note. If the division line is to run parallel to CD, then HI must also be supposed to be drawn parallel to CD, and the manner of working varied accordingly,

EXAMPLES.

1. Let the bearing of AB be North, dist. 12 ch. BC, N: 56°; E. 20.78 ch. CD, S. 33° E. 22.21 ch. and DA, S. 80°į W. 30 ch.; it is required to divide the tract into two parts by a line EF parallel to AB, so that the part AEFB may be to the part EDCF as 3 to 5.

The truth of this rule is evident from the demonstrations to the 11th and preceding problems.

Construction. Draw CM, Fig. 106, parallel to AB, and on GM describe the semicircle GLM; then proceed in every other respect as in the construction to the preceding problem.

Note. When the division line is to run parallel to CD, the construction is exactly as in the preceding problem.

2. The boundaries of a field ABCD are given as follow ; viz. AB, S. 10°1 W. 7.20 ch. BC, S. 67° W. 12.47

; ch. CD, N. 23° W. 13.33 ch. and DA, S. 890 E. 18 ch. and it is required to divide it into two parts by a line EF parallel to the side AB, so that the part AEFB may be to the part EDCF as 3 to 4; what will be the distance

PROBLEM XVIII.

The boundaries of a tract of land ABCDEFGHIA, Fig.

107, being given, to divide it into two equal parts by a line IN running from the corner I and

falling on the opposite side CD.

重

RULE.

Suppose lines drawn from I to C and D, and calculate the area of the whole tract.

Take the latitudes and departures of IA, AB, and BC, and by balancing find the latitude and departure of CI; also calculate the area of the part IABCI; from half the area of the whole tract, subtract the area of the part IABCI, the remainder will be the area of the triangle ICNI.

Take the latitudes and departures of IC and CD, and by balancing, find the latitude and departure of DI, and calculate the area of the triangle ICDI. Then,

As the area of the triangle ICDI,
Is to the area of the triangle ICNI;
So is the latitude of CD,
To the latitude of CN.

Also,

As the area of the triangle ICDI,
Is to the area of the triangle ICNI;
So is the departure of CD,
To the departure of CN.