PROB. IX. Two sides of a plane triangle and their included angle given to find the area. RULE. To the log. sine of the given angle (or of its supplement to 180°., if obtuse) add the logarithms of the containing sides; the sum, less radius will be the logarithm of the double area. EXAMPLES. fig. 76. B Suppose two sides, AB, AC, of a triangular lot ABC, form an angle of 30 degrees, and measure, one 64 perches, and the other 40.5, what must the content be? 2. Required the area of a triangle, two sides of which 49.2 and 40.8 perches, and their contained 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. Let AH be perpendicular to AB and equal to AC and HE, FCG, parallel to AB then making AH (=A C) 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. Q. E. D. 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. 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. 61.6 30+49=79. S Multiplied gives A. R. P. double the area 15 0.332. Note. On this 10th problem are founded most of the calculations or differences by latitude and departure, and those by off-sets, following in this treatise. 2. In the trapezoid ABCD the parallel sides are, AD, 20 perches, BC, 32, and their perpendicular 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. Let ABCD be a field in form of a trapezium, the diagonal AC, 64.4 perches, the perpendicular Bb. Proceed by the rule, and the area will be found= A. R. P. 8 0333 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, Muliply the square of the circle's diameter, of the product of the longest and shortest diameters of the ellipsis by .7854 for the area. Cr, subtract 0.10491 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. by 3.14159, produces the Circum. 2 quotes the diameter. |