Again 40.25 (BC+ CD): .75 (= BC– = BC-CD):: 1.0446136 (tang. BDC + CBD = 46° 15′): 1.0446136 x 3 = .01946485 = the tang. of 161 BDC-CBD = 1° 07'. 2 Wherefore BDC = 46o 15′ + 1° 07′ = 47° 22′. And 4 ADB = (ADC-CDB=73°-47° 22′=) 25° 38'. Their sum is 343.1479 square chains = 34A.1R. and 10.3664P. the area required. EXAMPLES FOR PRACTICE. 1. Required the area of a trapezium, whose diagonal measures 120 perches, and the perpendiculars 24 perches and 40 perches? Ans. 24 Acres. 2. Required the area of a trapezium, whose diagonals are 85 and 24 four-pole chains, and at their intersection, one of the less angles is 30°o; what is the area? Ans. 105A: 3. What is the area of a trapezium, whose south side is 27.4 chains, east side 35.75 chains, north side 37.55 chains, and west side 41.05 chains; also the diagonal from south-west to north-east 48.35 chains? Ans. 123A. 11.867P. PROB. XII. To find the area of a circle, or an ellipsis. RULE. Multiply the square of the circle's diameter, or the product of the longest and shortest diameters of the ellipsis by .7854 for the area. Or, subtract 0.104909 from the double logarithm of the circle's diameter, or from the sum of the logarithms of those elliptic diameters, and the remainder will be the logarithm of the area. Note. In any circle, the Diam. multi. Circum. div.by 3.14159, { produces the Cir. EXAMPLES. quotes the Diam. 1. How many acres are in a circle of a mile diameter? 1 Mile =320 per. log. 2.505150 2.505150 5.010300 0.104909 4|0)804215. log. 4.905391 4)2010.25 Answer, 502A. 2R. 25P. 2. A gentleman, knowing that the area of a circle is greater than that of any other figure of equal perimeter, walls in a circular deer-park of 100 perches diameter, in which he makes an elliptical fish-pond 10 perches long by 5 wide; required the length of his wall, content of his park, and area of his pond? Answer, the wall 314.16 perches, inclosing 49A. 14P. of which 39 perches, or of an acre nearly, is appropriated to the pond. PROB. XIII. The area of a circle given, to find its diameter. RULE. To the logarithm of the area add 0.104909, and half the sum will be the logarithm of the diameter. Or, divide the area by .7854, and the squareroot of the quotient will be the diameter. EXAMPLE. A horse in the midst of a meadow suppose, Made fast to a stake by a line from his nose. How long must this line be, that feeding all round, Permits him to graze just an acre of ground? Area in perches 160 log. 2.2041 20 2) 0.104909 2)2,309029 Diameter 14.2733 log. 1.154514 Answer, 7.13665 per. = PROB. XIV. Allowance for roads. It is customary to deduct 6 acres out of 106 for roads; the land before the deduction is made may be termed the gross, and that remaining af ter such deduction, the neat. 1. How much land must I enclose to have 850A. 2R. 20P. neat? 40/20 4 2.5 Acres. A. R. P. 850.625×1,06=901.6625=901. 2. 26. the ans. 2. How much neat land is there in a tract of 901A. 2R. 26P. gross? 40/26. 42/2 4 2.65 Acres. A. R. P. 1.06)901.6625(850.625=850. 2. 20. the ans. 848 &c. NOTE. These two operations prove each other. PROB. XV. To find the area of a piece of ground be it ever so irregular by divid ing it into triangles and trapezia. PL. 7. fig. 4. We here admit the survey to be taken and pro tracted; by having therefore the map, and knowing the scale by which it was laid down, the con tent may be thus obtained. Dispose the given map into triangles, by fine pecilled lines, such as are here represented in the scheme, and number the triangles with 1, 2, 3, 4, &c. Your map being thus prepared, rule a table with four columns; the first of which is for the number of the triangle, the second for the base of it, the third for the perpendicular, and the fourth for the content in perches. Then proceed to measure the base of number 1, from the scale of perches the map was laid down, and place that in the second column of the table, under the word base; and from the angle opposite to the base, open your compasses so, as when one foot is in the angular point, the other being moved backwards and forwards, may just touch the base line, and neither go the least above or beneath it; that distance in the compasses, measured from the same scale, is the length of that perpendicular, which place in the third column, under the word perpendicular. If the perpendiculars of two triangles fall on one and the same base, it is unnecessary to put down the base twice, but insert the second perpendicular opposite to the number of the trian |