A very small pair of dividers going with a screw, or, simply, a fork cut in metal by a file, or even in hard wood with a knife, will be found convenient.

PROBLEM XIII

To reduce or enlarge a plot.

From any point conveniently situated, draw lines passing through the several stations-then draw, in succession and terminating in these lines, parallels to the sides of the plot.

The student will exercise himself in plotting all the following fields, measuring the last sides and their bearings.

D

Fig. 99.

P

In order to determine a line rendered inaccessible by intervening declivities, I trace the line ABCDEFG through a neighboring ravine, and find AB, N 23° E, 25 chs.; BC, N, 31 chs.; CD, N 50 W, 10 chs.; DE, N 12° W, 15 chs.; EF, N 10° E, 35 chs., to the top of the ravine, thence to the second extremity, G, of the required line, S 45° W, 51 87 chs. Required, AG.

VI.

Find where the meridian, passing through the middle point of the side a of IV., will intersect the opposite part of the perimeter. Ans. At a point in e distant 55 links from the extremity of d.

SECTION THIRD.

Computation of Areas.

PROPOSITION I.

If the sides of any polygon be projected on the same (418) line, the sum of these projections, taken in order with their proper signs, will obviously be equal to zero.

Thus if the sides AB, BC, CD, DE, EA, of the polygon ABCDEA, be projected on the line LL, in ab, bc, cd, de, ea, we evidently have

ab+be+cd+(-de) + ( − ea) = 0.

And, generally, if we regard the perimeter as described by a point revolving about the polygon

B

A

E

Fig. 100.

in a constant direction, as ABCDEA, then will the projection of this point upon any line given in position, as LL, describe, by its motion, the projection of the perimeter, which projection, increasing and diminishing, will obviously become nothing when the revolving point returns to its first position, as A.

Scholium. The principle enunciated in (180), applied to this problem, serves to distinguish the plus and minus projections, which will be found to be measured to the right or left, corresponding with the motion of the revolving point. Thus, as the point revolves through ABC, the projection, measured towards d, decreases, becoming = 0, when the point arrives at D, and then reappears, measured in the contrary direction, as the point returns through DEA.

1,

It follows that if a, b, c, j, k, denote the sides of a polygon taken in order, a', b', c', ..., j', k', l', their projections north and south, a", b", c", ..., j", k", l', the corresponding projections east and west, and m a meridian line; then will the sum of the meridional projections, taken with their proper signs, be = 0, as also the sum of the projections at right angles to the meridian; and we shall find

or

since

and

or

since

C

a' + b + c' + ... + j' + k' + l' = 0, acos(a,m)+bcos(b,m) + ccos(c,m)+...

+jcos(j,m)+kcos(k,m)+lcos(l,m)=0, ♪ (419)

a' = acos(a,a') = acos(a,m), b' = cos(b,m), ...,

l' = lcos(l,m);

a"+b" +c"+ ... + j" + k" + l′′ = 0, asin(a,m)+bsin(b,m)+csin(c,m)+.

...

+jsin(j,m)+ksin(k, m) + lsin(l,m)=0, (420)

a" = acos(a,a") = asin(a,a') = asin(a,m), ...,

H

l" = lsin.(l,m);

and the last side, l, is completely determined—that is, its bearing and length are drawn from the bearings and lengths of the other sides. Enunciate the above forms. How is the denominator of (421) formed from the numerator?

Cor. 1. If the sides of a polygon, except one, vary in such (423) way as to preserve their mutual ratios and inclinations constant, then will this excepted side bear to any one of the others a ratio and inclination also constant.*

For, denoting the constant ratios which b, c,..., k, bear to a by Tb, rcs ....., T, or putting

which is in accordance with the condition that any one of the sides, a, b, c, ... 9 k, shall have a constant ratio to any other, since the ratio of b to k, for instance, is b: k = r2a : r1a=r: rk, we have

tan(l,m) =

asin(a,m)+ar,sin(b,m)+ar,sin(c,m) +...+arsin(k,m)

acos(a,m)+arcos(b,m)+arcos(c,m)+...+arcos(k, m)' sin(a,m)+sin(b,m)+r.sin(c,m)+...+ sin(k, m)

or, tan(l,m) =

... 9

cos(a,m)+cos(b,m)+r, cos(c,m)+...+cos(k,m)' which fraction is obviously a constant quantity, being independent of a, or of the absolute length of any of the sides, a, b, c, k, and depending only upon the constant quantities, rb, Tc, ..., Tki sin(a,m), sin(b,m), sin(k,m), cos(a,m), cos(b,m), ..., cos(k,m). The tan(l,m) being constant, the (l,m) is constant;

and

▲s, (l,a), (l,b), (l,c), (l,k), are constant;

1=

asin(a,m)+ar, sin(b,m) + ... + ar sin(k, m)
- sin(l,m)

= (constant). a. Q. E. D.

Polygons, which, like the above, have the same number of sides proportional in the same order and their homologous angles equal, are said to be similar.

Cor. 2. In similar polygons, like diagonals are to each (424) other as the homologous sides; for the diagonal is obviously in the same condition as a last side, l.

Cor. 3. The perimeters, or their like portions, in similar (425) polygons, are to each other as homologous sides or diagonals.

* See Variation, Part 1, Book 1.