New plane and spherical trigonometry, surveying, and navigation (1903)
Wentworth, G. A. (George Albert), 1835-1906 · Boston : Ginn & Company · leaf 303
.20945 sq. ch. = 0.720945 a. = 115 ^ p. 5. Required the area of a field in the form of a trapezium, one ofwhose diagonals is 9 chains, and the two perpendiculars upon this diag-onal from the opposite vertices 4.50 chains and 3.25 chains. Area = 9 x |(4.5 + 3.25) sq. ch.= 34.875 sq. ch.= 3.4875 a.=… … 2 area J.F.E :2 area #DjE;JL2 area jBDC 6.4 x 9.25 = 59.2 : 13.75(4.75+ 7) = 161.5625: 10 x 4 = 40. 2 area ABCDEFarea ABCBEF130.38125 sq. ch. = 13.038125 a. = 260.7625= 130.38125= 13a. 6T\?< 7. Determine the area of the field ABCB from two interior stations,P and P, if PP = 1.50 chains,
Trigonometry · Surveying · Navigation
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