A treatise on surveying ; in which the theory and practice are fully explained (1860)
Alsop, Samuel · Philadelphia, E. C. & J. Biddle & co. · leaf 259
to the quotient. DrawPE parallel to AB. Make DFperpendicular to AD and equal toAE. Lay off FC = DE. ThenCPB will be the division line. Demonstration.—Complete the parallelogram ADHI. By construction, APD is half the required area; and, therefore, AIHD con-tains the required area. Now, because the… … Sec. L] LAYING OUT LAND. 257 these equals add AIPKD, and we have AIHD = ABC; whence ABC containsthe required area. If the directions of AB and AC and the position of the point P be given bybearings, AC maybe calculated as follows:—In API find PI; also find theperpendicular PL. Then AD = area -r-…
Surveying
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