2. Required the area of a triangle, two sides of which are 49.2 and 40.8 perches, and their con tained angle 144 degrees? Answer, 3A. 2R. 22P. 3. What quantity of ground is inclosed in an equilateral triangle, each side of which is 100 perches, either angle being 60 degrees? Answer, 27A. 10P. Demonstration of this problem. PL. 11. fig. 3. Let AH be perpendicular to AB and equal to AC, and HE, FCG, parallel to AB; then making AH (= AC) radius, AF (= CD) will be the sine of CAD, and the parallelograms ABEH (the product of the given sides) and ABGF the double area of the triangle) having the same base AB, are in proportion as their heights AH, AF; that is, as radius to the sine of the given angle; which proportion gives the operation as in the rule above. PROB. X. To find the area of a trapezoid, viz. a figure bounded by four right lines, two of which are parallel, but unequal. RULE. Multiply the sum of the parallel sides by their perpendicular distance, and take half the product for the area. NOTE. On this 10th problem are founded most of the calculations of differences by latitude and departure, and those by offsets, following in this treatise. Ff EXAMPLES. 1. Required the area of a trapezoid, of which the parallel sides are, respectively, 30 and 49 perches, and their perpendicular distance 61.6? 2. In the trapezoid ABCD the parallel sides are, AD, 20 perches, BC, 32, and their perpendicu lar distance, AB, 26; required the content? Answer, 4A. 36P. PROB. XI. To find the Content of a trapezium. RULE. Multiply the diagonal, or line joining the remotest opposite angles, by the sum of the two perpendiculars falling from the other angles to that diagonal, and half the product will be the area. EXAMPLE. PL. 7. fig. 3. Let ABCD be a field in form of a trapezium, the diagonal AC 64.4 perches, the perpendicular Bb 13.6 and Dd 27.2, required the content? NOTE. The method of multiplying together the half sums of the opposite sides of a trapezium for the content is erroneous, and the more so the more oblique its angles are. To draw the map set off Ab 28 perches and Ad 34.4, and there make the perpendiculars to their proper lengths, and join their extremities to those of the diagonal. 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. quotes the diam. EXAMPLES. 1. How many acres are in a circle of a mile diameter ? 1 Mile =320 per. log. 2.505150 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 494. 14P. of which 391 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 square-root of the quotient will be the diameter. EXAMPLES. A horse in the midst of a meadow suppose, Permits him to graze just an acre of ground? Area in perches 160 log. 2.204120 2) 1.154514 Diameter 14.2733 log. Answer, 7.13665 per. 117F. 9 In. = 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 after such deduction, the neat. 1. How much land must I inclose to have 850A. 2R. 20P. neat? |
