29. To find the area of a five-sided field ABCDE, measurements were made as follows:

measurements, and also by measuring on the plat the base and altitude of the equivalent triangle.

Area, 6 A. 3 R. 20 P.

REMARK.-A much less laborious plan of measuring the above field would be to measure the diagonals AD and DB, noting on them the distances to the points m and n of intersection with the diagonal EC, and then measure Em, mn and nC, also a side, as AB, for use as a proof-line.

30. Plan and execute a survey like that of the last Example, pursuing either method, or each, at pleasure, keeping the notes in regular form, constructing the plat and computing area.

ulars from the points upon the axis YY'.

The distances Oa1, Oa2, Oa, are called Abscissas of the points P1, P2, P3; and the distances Ob1, Ob2, Ob, are called Ordinates of the points.

The point O is called the Origin.

The abscissa and ordinate of a point are together called Coordinates of the points.

Coordinates at right angles with each other are called Rectangular Coordinates.

It is customary to denote abscissas by a and ordinates by y, coordinates of different points in connection with each other being distinguished by use of subscripts.

Thus, of the point P1, the coordinates Oa1 and Ob, or "P, may be denoted by x, and y1; of the point P2, the codinates Oa, and Ob, or a,P, may be denoted by x, and y2; and so on.

It will be seen that the coordinates of a point afford the means of locating it with respect to the axes.

The use of longitude and latitude in Geography is an illustration. By use of the signs + and -, the coordinates of any point in the plane of the axes are readily expressed.

EXERCISES.

56. 1. Construct the point of which x = 4 and y = 7. 2. Given x = — 5 and y = 3, to construct the point. 3. Given x = - 6, to construct the

point.

3 and y =

4. Given x = 6 and y

=

-4, to construct the point.

5. Given x = 0, y = 2; x = -5, y = 0; x = 0, y = 0. Required the points.

57. Application to Area.-Let it be required to find

the area of a series of trapezoids included between perpendiculars from the points of a broken line upon a

straight line. Suppose the straight line, as OX', to be an axis of abscissas, and the first perpendicular at the left, as OA, to be an axis of ordinates.

Let x1, x2, x3, etc., be the abscissas of the points A, B, C, etc., and y1, y2, y3, etc., the corresponding ordinates. Accordingly, the area of the several trapezoids is [X2 (Y1 + Y2) + (X3 — X2) (Y2+Y3)

+ (X4 — X3) (Y3 ·+ Y4) +

-(xn− xn−1) (Yn-1+Yn)],

in which n is the number of trapezoids plus one. The above formula may be changed to the form [X (Y1-Y3) + X3 (Y2 — Y1) + X1 (Y3-Y5)

+-----Xn-1 (Yn-2- Yn)+xn (yn-1+Yn)]. (a). Whence, for the area included between a straight line, as a base, and a broken line whose points are given by their coordinates upon the base, we have the following

RULE. From each ordinate subtract the second succeeding one and multiply the remainder by the abscissa corresponding to the intervening ordinate.

Also, multiply the sum of the last two ordinates by the last abscissa.

Divide the algebraic sum of the products by 2.

The above formula and Rule have been deduced independently of any supposition as to the relative directions of the parts of the broken ine. They are therefore true whatever may be the form of the broken ine. That is, whether any part should be perpendicular to the base, either toward or from it, or whether any part should be turned backward respecting the preceding one.

to the base, CD as perpendicular toward it, and FG as being turned backward from EF.

Find how it would be, if one or more of the ordinates were zero; if one or more were negative.

EXERCISES.

58. 1. Given y1=12, y2=12, y=6, y=8, and y=6, also, x=10, x=18, x=24, x=30, and x=20, to find area.

Given the following, to find area:

59. As a second example of the application of coordinates in finding area, let there be taken an ordinary polygon, as ABCDEF.

Let x1, x2, xs, etc., be the abscissas of the points A, B, C, etc., and y1. Y2, ys, etc., the corresponding ordinates.

Now since formula (a) is true for any broken line, it holds for the case in which the broken line beginning, as at A, returns to the same point, forming thus a polygon as ABCDEFA.

In this case, the last term of (a) vanishes, and we have, as the area a polygon of n sides,

[x1(yn―y1)+X12 (Y1Y3) + X3 (у2—Y1) + xX1 (Ys — Y5) + etc., to n terms].

or, factoring with respect to y, we have the form

(b)

-1 (xnx) + Y2 (X1—X3) + Y3 (X2-X1) + Y1 (2z-Xz) + etc., to n terms].

(c)

Whence, for the area of a polygon whose vertices are given by their coordinates, we have the following

RULE.-From the ordinate of each vertex subtract the second succeeding one, and multiply the remainder by the abscissa of the intervening vertex; or, from the abscissa of each vertex subtract the second succeeding one, and multiply the remainder by the ordinate of the intervening vertex.

Divide the sum of the products by 2.

SCH.-Formulas (b) and (c) will be seen to be in accordance with any situation of the coordinate axes, agreeably with convenience of field work. In particular cases, one or more terms will be found to disappear. Due attention to algebraic signs is important.

The formulas are easy to remember, and simple of application. With an instrument adapted to laying off right angles, they afford the most practical means of computing the content of irregular tracts.

EXERCISES.

60. 1. Required the area and a plat of a field the coordinates of whose corners are

xn = x=0, x1= 7 ch., x,= 12} ch., x,= 18 ch., x= 15 ch., x=10 ch.; and

Yn =y=6 ch., Y1=12 ch., y,=20 ch., y=15 ch., y=84 ch., y=0 ch. Area, 16 A. 0 R. 28 P.

Find the area, supposing a different situation of the