PROBLEM IX. To find the area of a trapezoid, that is, a trapezium two of whose sides are parallel but not equal. RULE. Multiply the sum of the parallel sides by their pendicular distance, and half the product will be the perarea. * EXAMPLES. 1. Required the area of a trapezoid ABCD, of which the parallel sides AD, BC measure 6.14 and 9.48 chains respectively, and their perpendicular distance BF or DE, 7.80 chains. Ch. 6.14 9.48 15,62 7.80 121960 10934 2)121.8360 60.9180Ch. == 6A. OR. 15P. + * DEMONSTRATION. The trapezoid ABCD, Fig. 74, = the tri AD X BF BC X DE angle ABD + BDC = (by prob. 2), 2 AD X BF BC X BF AD + BC X BF (because BF = DE); 2 2 2 + 2. The parallel sides of a trapezoid are 12.41 and 8.22 chains, and their perpendicular distance 5.15 chains: required the area. Ans. 5A. 1R. 10P. 3. Required the area of a trapezoid whose parallel sides are 11.34 and 18.46 chains, and their perpendicular distance 13.25 chains. Ans. 19A. 2R. 39P. PROBLEM X. To find the area of a circle, or of an ellipsis.* RULE. Multiply the square of the circle's diameter, or the product of the two diameters of the ellipsis, by .7854, for the area.t Note. 1. If the diameter of a circle be multiplied by 3.1416, the product will be the circumference; also if the circumference be divided by 3.1416, the quotient will be the diameter. 2. If the area of a circle be divided by .7854, the square root of the quotient will be the diameter. * If two pins be set upright in a plane, and a thread, the length of which is greater than twice the distance between the pins, having the ends tied together be put about the pins; and if the point of a pin or' pencil, applied to the thread, and held so as to keep it uniformly tense, be moved round, till it return to the place from which the motion began; then the point of the pin or pencil will have described on the plane, a curve line called an Ellipsis. † The demonstration of this rule is too abstruse to admit of a place in this work. The student who wishes to see a demonstration is referred to treatises EXAMPLES. 1. How many acres are in a circle a mile in diame ter? 1 mile = 80Ch. 80 Square of 80, 5026.5600 Sq. Ch. = 502A. 2R. 25P. nearly. Or by logarithms. $ 80 80 .7854 5026.56 Sq. Ch. 3.70127 2. Required the area of an ellipsis, the longer diameter of which measures 5.36ch, and the shorter 3.28ch. Ch. 5.36 3.28 4288 log. 1.90309 1.90309 -1.89509 1072 1608 17.5808 .7854 703232 879040 1406464 1230656 13.80796032 Sq. Ch. =1A. 1R. 20.9P. Or by logarithms. $ 5.36 3.28 .7854 13.808 Sq. Ch. 3. Required the area of a circular park, the diameter of which is 100 perches. Ans. 49A. OR. 14P. 5.36 x 3.28, 4. Required the area of an elliptical fish pond, the longer diameter of which is 10 perches and the shorter 5 perches. Ans. 39.37 Sq. Per. PROBLEM XI. To protract a Survey, and to find its area by dividing it into triangles and trapeziums. log. 0.72916 The method of doing this will be best understood by an example. Thus, 1.14012 Suppose the following field-notes to be given, it is required to protract the survey and find its area. Ch. 1. N. 50° E. 9.60 2. S. 32° E. 16.38 3. S. 41° W. 6.30 4. West 8.43 5. N. 79° W. 10.92 6. N. 5° E. 11.25 1 To protract the survey. Method 1st. Draw NS, Fig. 75, to represent a meridian line; then N standing for the north and S for the south, the east will be to the right hand and the west to the left. In NS take any convenient point as A for the place of beginning, and apply the straight edge of the protractor to the line, with the centre to the point A, and the arch turned toward the east, because the first bearing is easterly; then holding the protractor in this position, prick off 50° the first bearing, from the north end, because the bearing is from the north ; through this point and the point A, draw the line AB on which lay 9.60 chains, the first distance from A to B. Now apply the centre of the protractor to the point B, with the arch turned toward the east, because the second bearing is easterly, and move it till the line AB produced cuts the first bearing 50°; the straight edge of the protractor will then be parallel to the meridian NS; hold it in this position and from the south end prick off the second bearing 32° ; draw BC and on it lay the second distance 16.38 chains. Proceed in the same manner at each station, observing always, previous to pricking off the succeeding bearing, to have the arch of the protractor turned easterly or westerly according to that bearing, and to have its straight edge parallel to the meridian; this last may al. . ways be done by applying the centre, to the station point and making the preceding distance line, produced (or not as may be) if necessary, cut the degrees of the preceding bearing: It may also be done by drawing a straight line through each station, parallel to the first meridian. When the survey is correct and the protraction accurately performed, the end of the last distance will fall on the place of beginning. |