A complete treatise on practical land-surveying, in seven parts; . square Area 3a. 3r. 35p. 7. Required the plan and area of a field from the followingequidistant ordinates. AB 217 1096 187 1000 150 900 125 800 107 700 98 600 95 500 100 400 114 300 130 200 167 100 190 000 Begin «.t A, and 202 150 112 84 66 58 57 65 80 110 148 200 go X. 106 land-surveying. (Part III, Answer. 727 The first and last Four times the sum, & Twice the sum, &c. 6651 Sum total. 100 The common distance. 3)665100 22170036288 Trapezoid at the end. Area in square link
A complete treatise on practical land-surveying, in seven parts; . square Area 3a. 3r. 35p. 7. Required the plan and area of a field from the followingequidistant ordinates. AB 217 1096 187 1000 150 900 125 800 107 700 98 600 95 500 100 400 114 300 130 200 167 100 190 000 Begin «.t A, and 202 150 112 84 66 58 57 65 80 110 148 200 go X. 106 land-surveying. (Part III, Answer. 727 The first and last Four times the sum, & Twice the sum, &c. 6651 Sum total. 100 The common distance. 3)665100 22170036288 Trapezoid at the end. Area in square 2^3195240 Area 2a. 2r. 12|p. Note.—Whenever the rule given in this Problem can be applied, it willbe found more easy, expeditious, and accurate, in finding the areas of offsets,and of narrow pieces of land, than the rules for triangles and trapezoids.(See my Mensuration, page 274.) Part IILj LAND-SURVEYING. 107 PROBLEM FIND THE BREADTH OF A RIVER. EXAMPLE. Let the following figure represent a river, the breadth ofwhich is required. B. Fix upon any object B, close by the edge of the river, on theside opposite to which you stand. By the help of your cross,make A D perpendicular to A B; also make A C = C D, anderect the perpendicular D E; and when you have arrived at thepoint E, in a direct line with C B, the distance D E will be =A B, the breadth of the river; for by Theo. 1, Part I, the angleA C B = D C E, and as A C = C D, and the angles A and Dright angles, it is evident that the triangles A B C, C D E arenot only similar but equal. Note 1.—The distance between A and the edge of the river, must be de-duced from D E, when it is not convenient to fix A close by the rivers edge. 2.—This Problem may also be well applied in measuring the distance ofany inaccessible object; for let A C equal 8, C D equal 2, and D E equal 10chains ; then, by similar triangles, as CD :DE :: A C : A B ; that is, as2 : 10 :: 8 : 40 chains = A B. (Sec Thco. 11, Part I.) 108 LAND-
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