SURVEYING. 
either directly, or circuitously, be com- 
muted into a triangle, of corresponding 
area : but it may be necessary, at the same 
time, to observe, that the squaring of the 
circle has not hitherto been perfected; 
though we have arrived so nearly to the 
completion of that object as to leave no 
room for regret at the want of absolute 
precision. 
These points being completely under- 
stood, the learner may proceed to the rudi- 
ments of surveying ; supposing him to be 
grounded in the few preliminary problems 
which enable him to describe the ordinary 
figures : should he not have obtained any 
previous information on that subject, we 
recommend that he turn back to the heads 
of Geometry and Mathematical instru- 
ments; under which he will find various 
items indispensable towards his progress. 
We shall submit a few propositions which 
the student may work with his compasses, 
plain scale, and protractor : when able to 
do all that may be needful on paper, be 
may then try his hand with one or other of 
the various instruments in use among sur- 
veyors. 
Propusition 1. “To ascertain the contents 
of the square field A B CD, fig. 1. Plate XV. 
Miscel.” Here little is to be done ; one of 
the sides being measured, say 70 yards, and 
multiplied by itself, will give 4,900 square 
yards for the area ; or one acre (i. e. 4,800 
square yards) and 100 square yards. 
Proposition II. “To survey the field 
A BCD, fig. 2.” This figure having the 
sides AB and CD parallel, and at right an- 
gles to A D, add the lengths of those parallels, 
say 70 and 90 yards, together; divide half 
their sum (i.e. 80) and multiply that half by 
the depth of AD, say 70 ; v/hich being mul- 
tiplied by the medium length, GF, gives an 
area of 5,600 square yards. The parallelo- 
gram, ABED, might have been computed 
by simply imdtiplying its length by its 
breadth ; and the triangle, BCE, might be 
taken separately, thus: the depth, (or alti- 
tude) BE, 70 yards, to be multiplied by 
half CE, (i.e. 10 yards) this would give 
700 ; and the produce of A B, which is 70, 
by B E which is also 70, would be 4,900 ; 
niaking in all 5,600, as above shown. 
Proposition III. “ To survey the inclined 
parallelogram A B C D, fig. 3.” It is to be 
observed that, in all inclined figures, the 
altitude is ascertained by a perpendicular 
from the base, as at C, to the parallel of 
that base, as at E on the line A B. Now 
the triangle B EC being equal to the tiian- 
gle DFA, and likewise similar thereto, it 
is evident that by transposing the former 
from the right to the left of the figure, it 
would make it rectangular ; as shown by the 
dotted line: therefore multiply the base 
D C, say 100 yards, by the altitude C E, 
say 80 yards, and the area will be found to 
contain 8,000 square yards. 
Proposition IV. “ To survey the irregulars, 
*fig. 4.” Here AC and BD are parallel, 
butn either C D nor A B are perpendicular 
thereto, nor parallel between themselves. 
We must, therefore, cut off the triangles 
A E B and C F D ; whose areas will be 
found by multiplying half their respective 
breadths, by tlieir whole depths; or their 
whole breadths, by half their depths : the 
centre part, ECBF, is treated as a paral- 
lelogram, already described; The whole of 
the calculations being added together give 
the area of the entire figure AC B D. ” 
Proposition V. “To ascertain the area of 
the trapezium, A BCD.” This figure is no 
where parallel, and has all its sides of un- 
equal lengths. The easiest mode of survey- 
ing it, is by drawing a diagonal between the 
two most distant points, C and B; and 
making olf-sets, rectangular to that diago- 
nal, from E to A, and from F to D. These 
ofiF-sets give the altitudes ; E A being the 
altitude of the triangle CAB, say 40 yards ; 
and F D be.ing the altitude of the triangle 
B D C, which we will take at 80 yards, 
Now the diagonal, CB, becomes a base 
common to both triangles; theiefore add 
the two altitudes together; namely, 40 to 
80, which make ISO ; take their half = 60, 
and multiply by the base, which we will 
call 140: the area will contain 8,400 square 
yards. It will be seen that this proposition 
IS, in a great measure, the foundation of all 
horizontal computation; and the student 
should remark that all figures, having many 
sides, may be divided into trapezia, and 
those again into triangles : each figure will 
have two sides more than the numbers of 
triangles it contains : thus, fig. 5, has four 
sides, and contains two triangles; fig. 6, has 
five sides, and contains three triangles. 
Proposition VI. “To survey the penta- 
gon, or five-sided field, A B C D E, fig. 6 .'’. 
Divide it into the three triangles, ABC, 
D A C, E A D, and having found their re- 
spective altitudes, as already shown, by per- 
pendiculars drawn to their summits from 
their respective bases, multiply half those 
altitudes by those bases, and the three pro- 
ducts will amount to the whole area of thq 
pentagon. 
