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A Field so situated, that all the angles ean be seen at one view, may be accurately measured by the Chain only. As for instance, suppose a field, or piece of land, bounded as in the following figure.—Then,

Measure from B to G, 6 chains, and perpendicular thereto measure aa 2 chains, 20 links, and perpendicular to BG, measure co 2 chains, 27 links, and Be 1 chain, 14 links; then CE 4 chains, 21 links, and perpendicular thereto dB, 3 chains, 19 links. Again; perpendicular to BG, measure hE4 chains, 90 links, and perpendicular thereto gR, 89 links, and hC 1 chain, 53 links; then, consequently, he will be 3 chains, 33 links; hg being measured, will be 3 chains, 20 links, leaving gF 1 chain, 70 links. And as each line is measured, draw it on paper with a pencil, and the several angles by the eye, marking each line with its own distance. This rough draught will greatly assist the memory in calculating.

By the preceding Problems, the area of the several triangles, &c. will be found to be —

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By Problem 9 110. CEhcC do y Problem 9, page 110

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By measuring Ba, and Cd, with those already made, a

true Map of the field may be constructed. For BG, being. laid down from a scale of equal parts, and a 4, from the

same scale, it’s proper distance from B, will determine the

point A, and the perpendieular cc, will six the point C ;

then with cF in the compasses, and one foot in C, de

scribe an arch, as at E ; and with he in the compasses,

and one foot in h, intersect the arch at E, which fixes the

point E, and the perpendicular dB, it’s proper distance from C, determines the point D; then from g lay off the

perpendicular gF, and draw the lines AB, BC, CD, &c.

and a Map of the Field is completed.

Admit a field bounded as in the following figure; the dimensions of which are, AB 27 chains, 35 links; BC 22 chains, 20 links ; CD 29 chains, 25 links ; DE 23 chains, 70 links; and EA 31 chains, 15 links: The diagonal BE 38 chains, and EC 40 chains, 10 links; and from this measure it is required to make the Map, and find the area thereof. f

TO MAKE THE MAP.

Draw AB it’s proper length ; then, with the diagonal BE in the compasses, and one foot in the point B, describe an arch at E.; then take AE in the compasses, and with one foot in A, cross the former arch, and the place of intersection fixes the point E. Again ; take EC, and with one foot in E, make an arch at C ; then, with BC, and one foot in B, intersect the arch at C, which fixes that point. Again; with CD in the compasses, and one foot in C, make an arch at D ; then, with DE, and one foot

F f

in E, intersect the arch at D, for the point D. Lastly; draw lines from point to point, and the Map is made. Draw the diagonals, and the Map is divided into three oblique angled triangles. Then divide each of these into two right angled triangles, by Case 6, Oblique Angled Trigonometry, thus

f As the sum of Ba, and alo 38 1.57978 Is to the diff. AB and AE 3.8 0.57978 So is their sum 58.5 1.76716

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From half the sum of Ba, and ae 19
Take half their diff. 2.92

Remains the least base Ba 16.08

(See the following Map.)

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From the square of AB, take the square of Ba, leaves 489.4561, the square of Aa: the square root of which, is 22 chains, 12 links = Aa. Thus proceed with the others, and they will be found as in the figure.

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Area of ABEA, is 420.28 by Problem 6, page 107. BCDEB, is 756.687 by Problem 8, page 109.

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