the object D, and let the point, at which the line AD cuts the staff BE, be marked. Measure the distance BE on the staff; then say, As . AB : BE :: AC : CD, then, CD becomes known. If the line AC cannot be measured, on account of intervening objects, it may be determined by calculation, as in the last problem, and then, having found the horizontal distance, the vertical line is readily determined, as before. CHAPTER III. Of the area or content of ground. Of laying out and dividing land. 103. We come next to the determination of the area or content of ground. The surface of the ground being, in general, broken and uneven, it is impossible, without great trouble and expense, to ascertain its exact area or content. To avoid this inconvenience, it has been agreed to refer every surface to a horizontal plane : that is, to regard all its bounding lines as horizontal, and its area as measured by that portion of the horizontal plane which the boundary lines enclose. For example, if ABCD were a D piece of ground having an uneven surface, we should refer the whole to a horizontal plane, and take for the measure of the area that part of the plane which is included between the bounding lines AB, BC, CD, DA. A B In estimating land in this manner, the sum of the areas of all the parts into which a tract may be divided, is equal to the area estimating it as an entire piece : but this would not be the case if the areas of the parts had reference to the actual surface, and the area of the whole were calcu. 104. The unit of a quantity is one of the equal parts of which the quantity is composed (Arith. In. VI). Thus, a line of three feet in length is made up of three single feet, and of this line, 1 foot is the unit. The unit of a line may be 1 foot, 1 yard, 1 rod, 1 chain, or any other known distance. If, on the unit of length, a square be described, it will form the unit for computing areas. 1 foot. Thus, is 1 square foot, 1 yard=3 feet. 1 square yard, or 9 square feet, 1 chain=4 rods. 1 square chain, or 16 square rods. Thus it is seen that there are two kinds of quantity to be considered, viz. lines, and areas or surfaces; and each kind has its own unit of measure. When, therefore, the linear measures of ground are feet, yards, rods, or chains, the superficial measures will be square feet, square yards, square rods, or square chains; and the number expressing the area will be nothing else than the number of times which the unit of superficial measure is contained in the land measured. It has been already observed (Art. 83), that Gunter's chain of four rods or 66 feet in length, and which is divided into 100 links, is the chain in general use among surveyors. We shall, therefore, take the length of this chain for the unit 105. An acre is a surface equal in extent to 10 square chains; that is, equal to a rectangle of which one side is ten chains, and the other side one chain. One-quarter of an acre, is called a rood. Since the chain is 4 rods in length, 1 square chain contains 16 square rods; and therefore, an acre, which is 10 square chains, contains 160 square rods, and a rood contains 40 square rods. The square rods are called perches. 106. Land is generally computed in acres, roods, and perches, which are respectively designated by the letters A. R. P. When the linear dimensions of a survey are chains or links, the area will be expressed in square chains or square links, and it is necessary to form a rule for reducing this area to acres, roods, and perches. For this purpose, let us form the following TABLE. 1 square chain=10000 square links. links. 1 acre=4 roods=160 perches. 1 square mile=6400 square chains=640 acres. Now, when the linear dimensions are links, the area will be expressed in square links, and may be reduced to acres by dividing by 100000, the number of square links in an acre : that is, by pointing off five decimal places from the right hand. If the decimal part be then multiplied by 4, and five places of decimals pointed off from the right hand, the figures to the left will express the roods. If the decimal part of this result be now multiplied by 40, and five places for decimals pointed off, as before, the figures to the left will express the perches. If one of the dimensions be in links, and the other in chains, the chains may be reduced to links by annexing two ciphers : or, the multiplication may be made without annexing the ciphers, and the product reduced to acres and decimals of an acre, by pointing off three decimal places at the right hand. When both the dimensions are in chains, the product is reduced to acres by dividing by 10, or pointing off one deci From which we conclude ; that, 1st. If links be multiplied by links, the product is reduced to acres by pointing off five decimal places from the right hand. 2d. If chains be multiplied by links, the product is reduced to acres by pointing off three decimal places from the right hand. 3d. If chains be multiplied by chains, the product is reduced to acres by pointing off one decimal place from the right hand. 107. Since there are 16.5 feet in a rod, a square rod is equal to 16.5 x 16.5=272.25 square feet. If the last number be multiplied by 160, we shall have 272.25 X 160=43560=the square feet in an acre. Since there are 9 square feet in a square yard, if the last number be divided by 9, we obtain 4840=the number of square yards in an acre. PROBLEM I. 108. To find the area of a square or rectangular piece of ground. Multiply the two sides together, and the product will express the area (Geom. Bk. IV, Prop. IV.). 1. To find the area of the rectangular D field ABCD. Measure the two sides AB, BC: let us suppose that we have found AB=14 chains 27 links, and BC=9 chains 75 links. Then, А. B BC= 975 links, =13.91325 acres. 4 3.65300 rods, 40 26.12000 perches. Ans. 13 A 3R 26P. 2. What is the area of a square field, of which the sides are each 33 ch 81} 3. What is the content of a rectangular field, of which the longest side is 49 ch 271, and the shorter 38 ch 71? Ans. 187A 2R 11P. PROBLEM II. 109. To find the content of a piece of land in the form of a triangle. FIRST METHOD. А Measure either side of the triangle as BC, and from the opposite angle A let fall a perpendicular AD, and measure this perpendicular ; then, multiply the base and perpendicular to B C gether, and divide the product by 2, the result will express the area of the triangle. Or, the area is equal to the base multiplied by half the perpendicular, or to the perpendicular multiplied by half the base (Geom. Bk. IV, Prop. II). 1. What is the content of a triangle whose base is 25 ch il, and perpendicular 18 ch 141? Ans. 22A 2R 29P. 2. What is the content of a triangle whose base is 15.48 chains, and altitude 9.67 chains ? Ans. 7A IR 38 P SECOND METHOD. Measure two sides and their included angle. Then, add logether the logarithms of the two sides and the logarithmic sine of their included angle ; from this sum subtract the logarithm of the radius, which is 10, and the remainder will be the logarithm of double the area of the triangle. Find, from the table, the number answering to this logarithm, and divide it by 2 ; the quotient will be the required area (Geom. Mens. Prob. II). 1. In a triangle ABC, suppose that we have found AB= 57.65 ch, AC=125.81 ch, and the included angle CAB= |