Examples. 1. Required the area of a heptagon, whose side is 400. 4002=160000, and 3.633912×160000=581425.92 Ans. 2. Required the area of a pentagon whose side is 12. .688191×128.258292 = perpendicular of the polygon. 12×5=60, and 60÷2 half sum of the sides. 8.258292×30= 247.74876 Answer. = 3. What is the area of a hexagon whose side is 10? Ans. 259.8076. OF THE CIRCLE. A circle is a plane superficies, bounded by one line called the circumference, which is every where equidistant from a point within it, which is called the centre. The circumference given, to find the diameter; or the diameter given, to find the circumference. RULE. 1. As 7 is to the diameter, so is 22 to the circumference. Or, As 113 is to the diameter, so is 355 to the circumference. Or, As 1 is to the diameter, so is 3.1416 to the circumference. 2. As 22 is to the circumference, so is 7 to the diameter. Or, As 355 is to the circumference, so is 113 to the diameter. Or, As 3.1416 is to the circumference, so is 1 to the diameter. Examples. 1. If the diameter of a circle be 31.8309, what is the circumference? As 1 : 31.8309 :: 3.1416 100 Ans. 2. If the circumference of a circle be 29, what is the diameter ? As 3.1416 29 :: 1 : 9.23 Answer. 3. If the diameter of the earth be 7958 miles, (as it is very nearly,) what is the circumference, supposing it to be exactly round? Ans. 25000.8528 miles. The diameter or circumference of a circle given, to find the area; or the area given, to find the diameter or circumference. RULE. 1. Multiply half the circumference by half the diameter, and the product will be the area. Or multiply the square of the diameter by .7854, and the product will be the area. Or multiply the square of the circumference by .07958, and the product will be the area. 2. Divide the area by .7854, and the square root of the quotient will be the diameter. Or, divide the area by .07958, and the square root of the quotient will be the circumference. Examples. 1. If the diameter of a circle be 12, what is the area? 12×12-144, and 144x.7854-113.0976 Answer. 2. If the circumference of a circle be 12, what is the area? 12×12=144, and 144x.07958-11.45952 Answer. 3. If I drive a stake in my meadow, and fasten my horse to it by a rope of such a length that he may graze exactly half an acre, how long must the rope be? Ans. 27.75 yards + To find the area of an oval or ellipsis. RULE. Multiply the product of the two diameters by .7854, and this last product will be the area. Examples. 1. Required the area of an elliptical fish pond whose diameters are 12 and 10.6 yards. 12×10.6=127.2, and 127.2x.7854-100 sq. yds. Ans. 2. What is the area of an ellipsis whose diameters are 12 and 9? Ans. 84.8232. To find the side of a square piece of timber, that may be sawn or hewn from a round piece. RULE. Extract the square root of half the square of the diameter. Or, multiply the girth or circumference by 2, and divide the product by 9, the quotient will be the side of the square, near enough for common purposes. Examples. 1. The girth of a tree is 5 ft. 7 in., required the side of a square post that may be hewn from it. 2. What must be the girth of a piece of round timber that will hew 2 feet square? Ans. 9 feet. 3. I wish to cut a tree that will hew 15 inches squareWhat must be its circumference? Ans. 5 ft. 7 in. To measure the length of standing timber. Provide a slight pole, about your height; measure off on your pole a length exactly equal to the height of your eye, and at this place cut a notch. Fix your pole opposite that side of the tree which affords the best view of the summit of its main stem, that is, up to that part which can be squared, or made into merchantable timber. Proceed to such a distance from the tree as to you may seem equal to its height; set the pole in the ground perpendicularly, up to the notch, and the station fixed upon must be level with the surface of the ground at the foot of the tree; lie down on your back, with the soles of your feet against the pole and your head directly from the tree; look over the top of the pole, and see where the sight strikes the tree; the pole may be removed nearer to or further from the tree, until the sight taken strikes that part of the tree up to which you calculate timber. Measure the distance from the tree to the foot of the pole, that added to the height of the pole from the notch, is the length of the timber. An allowance must, however, be made for the stump. Example. 1. The distance from the foot of the tree to the pole is 31 feet; the height of the pole or eye 5 feet 6 inches, and an al S lowance of 2 feet 6 inches is to be made for the height of the stump and waste in cutting-Required the length of merchantable timber. 34 0 = length of the timber. 2. From the foot of a tree to the pole measures 47 feet; height of the pole 5 feet, allowance for stump 2 feet-Required the length of the stick. Ans. 50 feet. BOARD MEASURE. CASE I. To find the superficial contents of boards. RULE. Multiply the length in feet by the width in inches, and divide the product by 12, the quotient will be the superficial content. Note. It frequently happens in measuring boards, that they are found to be wider at one end than the other; in which case take the width in the middle, if it be a straight edged board; or which results the same thing, add the width of the two ends together, and take half their sum for the mean breadth; but if it be not a straight edged board, and wider in some places than others, take the width in several places, and divide their sum by the number of breadths for the mean breadth; then proceed as before. Examples. 1. What number of feet is there in a board 17 feet long and 11 inches broad? 17×11-187, and 187÷÷12-15 ft. 7 in. Answer. 2. Required the number of feet in a board 13 feet long, 16 inches wide at one end and 20 inches at the other. 16+20=36, and 36÷÷2-18 mean breadth. 13×18=234, and 234÷÷12=19 ft. 6 in. Ans. 3. An irregular board of 14 feet long measures at the ends 15 and 9 inches, the intermediate widths are 16, 13, 8, 6 and 10 inches-Required the number of feet in said board. 16+13+8+6+10+15+9=77, and 77-7-11 mean breadth. 14x11=154, and 154÷12-12 ft. 10 in. Ans. CASE II. To find the superficial contents of any square piece of timber, scantling, plank, &c. having the length, breadth and thickness given. RULE. Multiply the length in feet by the breadth in inches, and divide by 12; the quotient will be the content in feet, at one inch thick, which multiply by the thickness in inches, the product will be the whole content. If the piece of scantling, plank, &c. be wider at one end than the other, or be irregular, find the mean breadth as taught in Case I. Examples. 1. Required the number of feet in a piece of timber 20 feet 6 inches long, 7 inches wide, and 5 inches thick. 20 ft. 6 in. x7=1431, and 143÷12-11 ft. 11 in. |