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a

C

Area ABCD

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Given a, b, c, d and A&C to find the

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℗ = { ad. sin A + 1⁄2 Bc. sin C

O2 - 4 (a2d" sin2 A + 2 a b c d sin A. sinc + b3c3èm *C)

sin 2

= |

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ABCD.

(note to 65).

and sin A. sinα = cos A. cosС - cos (A+C) .

®2- ¿[aˆd3- a`dˆ cos3 A + 2 abcd co: A. cos C-2 abed cos (A+C)+b*o2-b'e2cos^C].

C

Draw the diagonal

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x2- a2 + λ =- 2 ad. cos A = b2 + c2 - 2 bc. cos. C··2(ad.com A-bc.com. C) = a2+ d2-b-c2

···4 (a`dˆcœ1A - 2 abed Cas A. cos C + b^c? Cos 'C) = a* 2a'dˆsd" qa'b1- 20°c 2= ib°d=nc`d>2]}%°•%•«c° Dividing each member by

16 and subtracting each from 4 (a`d2 - zabed. Cos (A +C) +b2c2), and adding and subtracting

the second member 1⁄2 abed, we have

1 [a2 d2 - a2d", cos^ A + 2 abed. cos.com C-2abcd.cos (A + C) + b2 c2 - b*c2 cos* C]

to

- ®ˆ = 1⁄2 (2 ad ̃-d-a ̃+2c3d2 + ¿aˆc2¬e + 2b ̃d ̃+ 2 a ̃b2+ 2bc? bTM, 8 abcd)- ÷ abcd.cxf+C) - { abcał

= — ¿ (a + b + c − α) ( a + b + d − c )( a + c + d − b )( b + c + d − a) - { abcd[cos (A + C) + 1]

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Area 4 10B = 1⁄2 A0x 08 x sinß; Area 4 BOC = ± BO × OC + ain B: COD = {CO2OD xain ß; DOA - {DO x OA x iin ß. , we have, Area ABCD = 1⁄2 (A0»OB+BO×OC+C OxOD+ODxOA) ainß - 1⁄2 (40+0CX (80+00) ein B = &ACx BDain B.

Adding,

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to find the

oraw DE parallel to BC.

Area AADE - ); S ( S − a X ¦ S - c )( ¦ § - Z + μ)

1⁄2 perpendicular

Area ABCD

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put 15.q

S

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7 q (q- a) q - c) (q -b +

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{ (b + d) x perpendicular

b-a

= 1 =
d) x 2 ) q (q. a) q - ?) ( q - b + d)

=

1⁄2 (b +

b+a

ъ-à

A)

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a

NOTE. Given b.d, a and angle A, to find the

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Perpendicular

(68) NOTE 1.

Given two opposite side and all

the

D

2 A

angles, to find

the area of a trapezium.

Given a and c and angles A,B,C and D.

Produce & and a

to meet in E.

By (65) note 8, area EDC - c"..

area EAB = a2.

ain C. an o

2 i'm (C + D)

ein (180°-A). ain (180°-B)
2 ain [(180° - A) + (180°-8)]

Area ABCD = c2 ein Cain D
2 im(C + D)

Б

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ein A. ein B

2 cm (A + B)

Given a b. c. I and angle AOB = 3, to find area of

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÷ v x.em ß = ¿(v2 + x2-d") tau (B3, and 1⁄2 xy sin B = (a2 - x - y 1 tauß.

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(69) Surfaces bounded by irregularly curved lines. The rules for these will be more appropriately given in connection with the surveys which measure the necessary lines; as explained in Part II, Chap. III.

(70) SECOND METHOD.-GEOMETRICALLY. surements of the necessary lines upon the plat.

From mea

(71) Division into Triangles. The plat of a piece of ground having been drawn from the measurements made by any of the methods which will be hereafter explained, lines may be drawn upon the plat so as to divide it into a number of triangles. Four Fig. 31.

Fig. 32.

Fig. 33.

Fig. 34.

0 900

ways of doing this are shown in the figures: viz. by drawing lines. from one corner to the other corners; from a point in one of the sides to the corners; from a point inside of the figure to the corners; and from various corners to other corners. The last method is usually the best. The lines ought to be drawn so as to make the triangles as nearly equilateral as possible, for the reasons given in Part V.

One side of each of these triangles, and the length of the perpendicular let fall upon it, being then measured, as directed in Art. (43,) the content of these triangles can be at once obtained by multiplying their base by their altitude, and dividing by two.

The easiest method of getting the length of the perpendicular, without actually drawing it, is, to set one point of the dividers at the angle from which a perpendicular is to be let fall, and to

the cosine. Subtract this product from the reserved product, and take the square root of the remainder. It will be the area of the quadrilateral.

When the four sides, and the angle of intersection of the diagonals of the quadrio teral are given; square each side; add together the squares of the opposite sides; take the difference of the two sums; multiply it by the tangent of the angle of intersection, and divide by four. The quotient will be the area.

When the diagonals of the quadrilateral, and their included angle are given, mul tiply together the two diagonals and the sine of their included angle, and divide by two. The quotient will be the area

open and shut their legs till an arc described by the other point will just touch the opposite side.

Otherwise; a platting scale, (described in Art. (49) may be placed so that the zero point of its edge coincides with the angle, and one of its cross lines coincides with the side to which a perpendicular is to be drawn. The length of the perpendicular can then

at once be read off.

The method of dividing the plat into triangles is the one most commonly employed by surveyors for obtaining the content of a survey, because of the simplicity of the calculations required. Its correctness, however, is dependant on the accuracy of the plat, and on its scale, which should be as large as possible. Three chains to an inch is the smallest scale allowed by the English Tithe Commissioners for plats from which the content is to be determined.

In calculating in this way the content of a farm, and also of its separate fields, the sum of the latter ought to equal the former. A difference of one three-hundredth (3) is considered allowable.

Some surveyors measure the perpendiculars of the triangles by a scale half of that to which the plat is made. Thus, if the scale of the plat be 2 chains to the inch, the perpendiculars are measured with a scale of one chain to the inch. The product of the base by the perpendicular thus measured, gives the area of the triangle at once, without its requiring to be divided by two.

Another way of attaining the same end, with less danger of mistakes, is, to construct a new scale of equal parts, longer than those by which the plat was made in the ratio 2:1; or 1.414:1. When the base and perpendicular of a triangle are measured by this new scale and then multiplied together, the product will be the content of the triangle, without any division by two. In this method there is the additional advantage of the greater size and consequent greater distinctness of the scale.

When the measurement of a plat is made some time after it has been drawn, the paper will very probably have contracted or expanded so that the scale used will not exactly apply. In that case a correction is necessary. Measure very precisely the present length of some line in the plat, of known length originally. Then

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