EXAMPLES. Ex. 1. To find the area of a triangle, whose base is 625, and perpendicular height 520 links? Here 625 × 260 = 162500 square links, or equal 1 acre, 2 roods, 20 perches, the answer. Ex. 2. How many square yards contains the triangle, whose base is 40, and perpendicular 30 feet? Ans. 66 square yards. Ex. 3. To find the number of square yards in a triangle, whose base is 49 feet, and height 25 feet? Ans. 68, or 68.7361. Ex. 4. To find the area of a triangle, whose base is 18 feet. 4 inches, and height 11 feet 10 inches? Ans. 108 feet, 5 inches. RULE II. When two sides and their contained angle are given: Multiply the two given sides together, and take half their product: Then say, as radius is to the sine of the given angle, so is that half product, to the area of the triangle. Or, multiply that half product by the natural sine of the said angle, for the area*. Ex. 1. What is the area of a triangle, whose two sides are 30 and 40, and their contained angle 28° 57'? By Natural Numbers. First,× 40 × 30 = 600, then, 1: 600 :: 484046 sin. 28° 57′ Answer 600 By Logarithms. log. 9.684887 290-4276 the area answering 2-463038 *For, let AB, AC, be the two given sides, including the given angle A. NOW AB X CP is the area, by the first rule, CP being the perpendicular. But, by trigonometry, as sin. P, or radius: AC:: sin. 4 A: CP, which is therefore AC X sin. A, taking radius 1. Therefore the area AB X CP is AR X AC, X sin. LA, to radius 1; or, as radius sin. 4 A :: AB X AC: the area. Ex. 2. How many square yards contains the triangle, of which one angle is 45°, and its containing sides 25 and 214 feet? Ans. 20.86947. RULE III. When the three sides are given: Add all the three sides together, and take half that sum. Next, subtract each side severally from the said half sum, obtaining three remainders. Then multiply the said half sum and those three remainders all together, and extract the square root of the last product, for the area of the triangle. * For, let ABC be the given triangle. Draw the parallels AE, BD, meeting the two sides AC, CB, produced, in D and E, and making CD CB, and CE CA. Also draw CFG bisecting DB and AE perpendicularly in F and G; and FHI рarallel to the side AB, meeting AC in H, and AE produced in 1. 1 D B H Lastly, with centre н, and radius HF, describe a circle meeting AC produced in K; which will pass through &, because G is a right angle, and through 1, because, by means of the parallels, AIFB DF, therefore HD HA, and HF HIAB. Hence HA or HD is half the difference of the sides AC, CB, and HC half their sum or AC +CB; also HK HIIF or AB; conseq. CK = AC + CB + AB half the sum of all the three sides of the triangle ABC, or CKs, calling s the sum of those three sides. Again HK = Hг = { IF theref. CL CK — KL = s AC, and AL = DK = CK — CD AB, Or KL AB; AB, and AKCK—CA = S - CB. Now, by the first rule, AG. CG the A ACE, and AG. FG = the A ABE, theref. AG. CFA ACB. Also by the parallels, AG: CG: DF or IA: CF, theref. AG. CF (AACB =) CG. IA = CG. DF, conseq. AG. CF. CG. DFA ACB. But CG. CFCK. CL = s. sAB, and AG. DF AK. AL AC.S BC; theref. AG. CF.CG. DFA ACB = AC. S EC is the square of the area of the A. E. D. Otherwise. Because the rectangle AG CF the AABC, and since CG: AG :: CF : DF, drawing the first and second terms into CF, and the third and fourth into AG, the propor. becomes CG.CF; AG. CF :: AG, CF: AG, DF, or CG.CF: A ABC :: A ABC: BG. DF, that is, the A ABC is a mean proportional between CG. CF and AG. DF, or between 1s. 1s AB and is AC. S BC., Q. E. D. EXAMPLES. Ex. 1. To find the area of a triangle, whose base is 625, and perpendicular height 520 links? Here 625 x 260 = 162500 square links, or equal 1 acre, 2 roods, 20 perches, the answer. Ex. 2. How many square yards contains the triangle, whose base is 40, and perpendicular 30 feet? Ans. 663 square yards. Ex. 3. To find the number of square yards in a triangle, whose base is 49 feet, and height 254 feet? Ans. 685, or 68.7361. Ex. 4. To find the area of a triangle, whose base is 18 feet. 4 inches, and height 11 feet 10 inches? Ans. 108 feet, 5 inches. RULE II. When two sides and their contained angle are given: Multiply the two given sides together, and take half their product: Then say, as radius is to the sine of the given angle, so is that half product, to the area of the triangle. Or, multiply that half product by the natural sine of the said angle, for the area*. Ex. 1. What is the area of a triangle, whose two sides are 30 and 40, and their contained angle 28° 57'? * For, let AB, AC, be the two given sides, including the given angle A. NOW AB X CP is the area, by the first rule, cr being the perpendicular. But, by trigonometry, as sin. P, or radius: AC :: sin. 4 A: CP, which is therefore AC X sin. A, taking radius 1. Therefore the area AB X CP is ARX AC, X sin. A, to radius 1; or, as radius: sin. 4 A :: AB X AC the area. Ex. 2. How many square yards contains the triangle, of which one angle is 45°, and its containing sides 25 and 214 feet? Ans. 20-86947. RULE III. When the three sides are given: Add all the three sides together, and take half that sum. Next, subtract each side severally from the said half sum, obtaining three remainders. Then multiply the said half sum and those three remainders all together, and extract the square root of the last product, for the area of the triangle. * For, let ABC be the given triangle. Draw the parallels AE, BD, meeting the two sides AC, CB, produced, in D and E, and making CD CB, and CE CA. Also draw CFG bisecting DB and AE perpendicularly in F and G; and FHI parallel to the side AB, meeting AC in H, and AE produced in 1. Lastly, with centre H, and radius HF, describe a circle meeting AC produced in K; which will pass through G, because G is a right angle, and through 1, because, by means of the parallels, AI FB DF, therefore HD HA, and HF = HI AB. Hence нA or HD is half the difference of the sides AC, CB, and HC half their sum or= ACCB; also HK HI = {IF or AB; conseq. CK = AC + CB + AB half the sum of all the three sides of the triangle ABC, or CKs, calling s the sum of those three sides. Again нK = Hr=&ff = 1⁄2 AB, Or KL = AB; theref. CL CK KL = s. AB, and AKCK-CA = {s $ CB. . AC, and AL DK = CK — CD Now, by the first rule, AG. CG the A ABE, theref. AG. CFA AG: CG: DF or IA: CF, theref. AG CG. DF, conseq. AG. CF. CG. DF But CG. CFCK. CL = s. s = S AC. S the ▲ ACE, and AG. FG = ACB. Also by the parallels, CF (AACB) CG. IA A2ACB. AB, and AG. DFAK. AL BC; theref. AG. CF.CG. DE A2 ACB = AC. S EC is the square of the area of the A. E. D. Otherwise. AG, the Because the rectangle AG. CF the ▲ ABC, and since CG: AG:: CF: DF, drawing the first and second terms into CF, and the third and fourth into propor. becomes CG.CF; AG. CF: AG. CF: AG. DF, or cc. CF: A ABC:: A ABC: BG. DF, that is, the A ABC is a mean proportional between co. CF and AG. DF, or between is. s AB and s AC. BC., Q. E. D. Ex. 1. To find the area of the triangle whose three sides are 20, 30, 40. Then 45 x 25 x 15 x 5 = 84375, The root of which is 290-4737, the area. Ex. 2. How many square yards of plastering are in a triangle, whose sides are 30, 40, 50 feet? Ans. 663. Ex. 3. How many acres, &c. contains the triangle, whose sides are 2569, 4900, 5025 links? Ans. 61 acres, 1 rood, 39 perches. PROBLEM III. To find the Area of a Trapezoid. ADD together the two parallel sides; then multiply their sum by the perpendicular breadth, or the distance between them; and take half the product for the area. By Geom. theor. 29. Ex. 1. In a trapezoid, the parallel sides are 750 and 1225, and the perpendicular distance between them 1540 links: to find the area. 1225 750 1975 x 770 = 152075 square links = 15 acr. 33 perc. Ex. 2. How many square feet are contained in the plank, whose length is 12 feet 6 inches, the breadth at the greater end 15 inches, and at the less end 11 inches? Ans. 131 feet. Ex. 3. In measuring along one side AB of a quadrangular field, that side, and the two perpendiculars let fall on it from the two opposite corners, measured as below: required the content. AP |