RULE. Multiply the chains by 2, and the links by 4, which will give hundredths of a rod; thus, 17 two rod chains and 21 links, make 34 rods and 84 hundredths; expressed thus, 34.84 rods. If the links exceed 25, add one to the number of rods, and multiply the excess by 4: thus, 15 two rod chains and 38 links make 31.52 rods. PROBLEM III. decimal parts. To reduce four rod chains to rods and RULE. Multiply the chains, or chains and links, by 4; the product will be rods and hundredths: thus, 8 chains and 64 links, make 34.56 rods. NOTE. The reverse of this rule, that is, dividing by 4, will reduce rods and decimals to chains and links: thus, 105.12 rods, make 26 chains and 28 links. PROBLEM IV. To reduce square rods to acres. RULE. Divide the rods by 160, and the remainder by 40, if it exceeds that number, for roods or quarters of an acre: thus 746 square rods make 4 acres, 2 roods, and 26 rods. PROBLEM V. To reduce square chains to acres. RULE. Divide by 10; or, which is the same thing, cut off the right hand figure: thus, 1460 square chains make 146 acres; and 846 square chains make 84 acres and 6 tenths. PROBLEM VI. To reduce square links to acres. RULE. Divide by 100000; or, which is the same thing, cut off the 5 right hand figures: thus, 3845120 square links make 38 acres and 45120 decimals. NOTE. When the area of a field, by which is meant its superficial contents, is expressed in square chains and links, the whole may be considered as square links, and the number of acres contained in the field, found as above. Then multiply the figures cut off by 4, and again cut off 5 figures, and you have the roods; multiply the figures last cut off by 40, and again cut off 5 figures, and you have the rods. EXAMPLE. How many acres, roods, and rods, are there in 156 square chains and 3274 square links? 15)63274 square links. 4 2)53096 40 21)23840 Answer. 15 acres 2 roods and 21 rods. PROBLEMS for finding the area of right lined figures, and PROBLEM VII. To find the area of a square or rectan gle. RULE. Multiply the length into the breadth; the product will be the area. PROBLEM VIII. To find the area of a rhombus or rhomboid. RULE. Drop a perpendicular from one of the angles to its opposite side, and multiply that side into the perpendicular; the product will be the area. PROBLEM IX. To find the area of a triangle. RULE 1. Drop a perpendicular from one of the angles to its opposite side, which may be called the base; then multiply the base by half the perpendicular, or the perpendicular by half the base; the product will be the area. Or, multiply the whole base by the whole perpendicular, and half the product will be the area. RULE 2. If it be a right angled triangle, multiply one of the legs into half the other; the product will be the area. Or, multiply the two legs into each other, and half the product will be the area. RULE 3. When the three sides of a triangle are known, the area may be found arithmetically, as follows: Add together the three sides; from half their sum substract each side, noting down the remainders; multiply the half sum by one of those remainders, and that product by another remainder, and that product by the other remainder; the square root of the last product will be the area.* EXAMPLE. Suppose a triangle whose three sides are 24, 20, and 18 chains. Demanded the area. 24+20+18=62, the sum of the three sides, the half of which is 31. From 31 substract 24, 20, and 18; the three remainders will be 7, 11, and 13. 31X7-217; 217×11-2387; 2387X13-31031, the square root of which is 176.1, or 17 acres, 2 roods, and 17 rods. BY LOGARITHMS. As the addition of logarithms is the same as the multiplication of their corresponding numbers; and as the number answering to one half of a logarithm will be the square root of the number corresponding to that logarithm: it follows, that if the logarithm of the half sum of the three sides, and * Better expressed thus. From half the sum of the sides substract each side separately. Multiply the half sum and the several remainders together, and the square root of the product will be the RULE. Multiply the chains by 2, and the links by 4, which will give hundredths of a rod; thus, 17 two rod chains and 21 links, make 34 rods and 84 hundredths; expressed thus, 34.84 rods. If the links exceed 25, add one to the number of rods, and multiply the excess by 4: thus, 15 two rod chains and 38 links make 31.52 rods. PROBLEM III. decimal parts. To reduce four rod chains to rods and RULE. Multiply the chains, or chains and links, by 4; the product will be rods and hundredths: thus, 8 chains and 64 links, make 34.56 rods. NOTE. The reverse of this rule, that is, dividing by 4, will reduce rods and decimals to chains and links: thus, 105.12 rods, make 26 chains and 28 links. PROBLEM IV. To reduce square rods to acres. RULE. Divide the rods by 160, and the remainder by 40, if it exceeds that number, for roods or quarters of an acre: thus 746 square rods make 4 acres, 2 roods, and 26 rods. PROBLEM V. To reduce square chains to acres. RULE. Divide by 10; or, which is the same thing, cut off the right hand figure: thus, 1460 square chains make 146 acres; and 846 square chains make 84 acres and 6 tenths. PROBLEM VI. To reduce square links to acres. RULE. Divide by 100000; or, which is the same thing, cut off the 5 right hand figures: thus, 3845120 square make 38 acres and 45120 decimals. links NOTE. When the area of a field, by which is meant its superficial contents, is expressed in square chains and links, the whole may be considered as square links, and the number of acres contained in the field, found as above. Then multiply the figures cut off by 4, and again cut off 5 figures, and you have the roods; multiply the figures last cut off by 40, and again cut off 5 figures, and you have the rods. EXAMPLE. How many acres, roods, and rods, are there in 156 square chains and 3274 square links? 15)63274 square links. PROBLEMS for finding the area of right lined figures, and PROBLEM VII. To find the area of a square or rectan gle. RULE. Multiply the length into the breadth; the product will be the area. PROBLEM VIII. To find the area of a rhombus or rhomboid. RULE. Drop a perpendicular from one of the angles to its opposite side, and multiply that side into the perpendicular; the product will be the area. PROBLEM IX. To find the area of a triangle. RULE 1. Drop a perpendicular from one of the angles to its opposite side, which may be called the base; then multiply the base by half the perpendicular, or the perpendicular by half the base; the product will be the area. Ör, multiply the whole base by the whole perpendicular, and half the product will be the area. RULE 2. If it be a right angled triangle, multiply one of the legs into half the other; the product will be the area. Or, multiply the two legs into each other, and half the product will be the area. RULE 3. When the three sides of a triangle are known, the area may be found arithmetically, as follows: Add together the three sides; from half their sum substract each side, noting down the remainders; multiply the half sum by one of those remainders, and that product by another remainder, and that product by the other remainder; the square root of the last product will be the area.* EXAMPLE. Suppose a triangle whose three sides are 24, 20, and 18 chains. Demanded the area. 24+20+18=62, the sum of the three sides, the half of which is 31. From 31 substract 24, 20, and 18; the three remainders will be 7, 11, and 13. 31x7=217; 217×11-2387; 2387×13-31031, the square root of which is 176.1, or 17 acres, 2 roods, and 17 rods. BY LOGARITHMS. As the addition of logarithms is the same as the multiplication of their corresponding numbers; and as the number answering to one half of a logarithm will be the square root of the number corresponding to that logarithm: it follows, that if the logarithm of the half sum of the three sides, and * Better expressed thus. From half the sum of the sides substract each side separately. Multiply the half sum and the several remainders together, and the square root of the product will be the the logarithms of the three remainders, be added together, the number corresponding to one half the sum of those logarithms will be the area of the triangle. The half sum, 31 The first remainder, 7 The third remainder, 13 1.491362 0.845098 1.041393 1.113943 The square of the area, 31030.083 nearly 4.491796 Area, 176 square chains 150 square links 2.245898 RULE 4. When two sides of a triangle, and their contained angle, that is, the angle made by those sides, are given, the area may be found as follows: Add together the logarithms of the two sides, and the logarithmic sine of the angle; from their sum substract the logarithm of radius, the remainder will be the logarithm of double the area. EXAMPLE. Suppose a triangle, one of whose sides is 105 rods, and another 85, and the angle contained between them 28° 5'. Demanded the area. One side, 105 2.021189 1.929419 9.672795 13.623403 10.000000 3.623403 Double area, 4200 rods nearly Answer, 2100 rods. NOTE. Radius may be substracted by cancelling the left hand figure of the index, or substracting 10, without the trouble of setting down the ciphers. BY NATURAL SINES. Multiply the two given sides into each other, and that product by the natural sine of the given angle; the last product will be double the area of the triangle. Nat. sine of the angle 28° 5', 0.47076. 105X85-8925, and 8925X0.47076-4201, the double area of the triangle. |