PART II. OF SUPERFICIES, an acre. The smallest superficial measure with us is the square, each Part II. side of which is an inch in length ; 144 square inches make a square foot; 9 square feet make a square yard ; 304 square yards a pole ; 40 poles a rood; and 4 roods an acre. In Scotland, they meaļure land by the square ell, which contains 1369 of the Scots square inches, or 1383.84 English square inches : 36 square ells make a fall; 40 falls a rood, and 4 roods But surveyers in Scotland often make the Scots ell to consist of 37 English inches ; and thus make the acre too little, by 593.6 English square feet; for the Scots acrc is to the English, as ï0,000 to 78,694. 6 acres, arable land, make a husband-land ; 13 acres an oxgate; 4 oxgate a pound-land, and 8 oxgate a fortyfhilling land. In measuring land by the chain, the dimensions are expressed in links, and the contents in square links. Now, the chain being 100 links, a square chain is 10,000 square links, and 10 square chains, or 100,000 square links make an acre; therefore, to reduce square links to acres, cut off 5 decimal places, and the rest are acres : and the decimals may be reduced to roods, by multiplying by 4, and cutting off the same number of decimals from the product; and so on. The French arpent contains 32,400 square feet of Paris, and is equal to 3.37 English roods. PRO B. I. To find the area of a rectangle ABCD. Pl. III. Fig. 1. Let the side AB, for example, be 5 feet, and let it be divided into s equal parts, one of which is AG; and let AE be equal to AG, and complete the parallelograms GE, EB. Then the rectangle EG: EB :: AG : AB”, that is, as I to 5; therefore EB a 1. 6. E. is s times EG. Let AD be three feet; then, because BE: BD :: AE : AD, that is, as I to 3, BD is 3 times BE, and BE is 5 times EG ; therefore BD is 15 times EG. Therefore the area is found, by multiplying the length by the breadth. Cor. If the parallelogram be not a rectangle, it is equal to a rectangle, on the same base, and between the fame parallels ; b 35. 1. E, Tt therefore, PART II. therefore, multiply its base AB into its perpendicular height ĄD, (not into its fide AH), and the product is the area. pl. 111. To find the area of a triangle. Fig. 2. a 41. 1. E. A triangle being the half a of a rectangle upon the same base with it, and between the same parallels : Therefore, from the angle A, draw AD, perpendicular to the oppofite fide BC, and measure BC and AD. Then multiply one of them by the half of the other, or multiply the two together, and take the half of the product. Thus, if BC be 9 feet long, and AD 6 feet, the area of the triangle will be 27 square feet. When the three sides are given, the perpendicular may be found, by protracting the triangle, and measuring the perpendicular upon the same scale, or it may be found by the 12th and 13th 2. Eucl. or by Trigonometry. Pl. III. THE fides of a triangle being given to find the Fig. 3. area, without finding the perpendicular. Let ABC be the triangle, and make AD equal to AC, and a 10. 1. E. AE to AB, and join DC, BE, and bisect a DC in F, and join b31. 1. E. AF, meeting BE in G, and draw b FL parallel to BC, meeting h 34. 3. E. AB, BE in K, L; and LB is equal h to FC or DF: and because C 2. 6. E. AD is equal to AC, and DB to CE, BE is parallel to DC o: d D. 5. E. and because DC is double of DF, DB is double of DK ", and BC or LF dauble of FK, and BE of BG: and because BA, AG are equal to EA, AG, and the base BG is equal to GE; the © 8. 1. E. angle BAG is equal to GAE, and the angle BGA to f 4. 1. E. EĞA; therefore they are right angles, and the circle descri bed from the centre K, at the distance KF, shall pass through 7 2. Cor. G%; let this be the circle FGL, meeting AB in H, M; thereA. 3. E. fore HM is equal to FL, or BC the base : and because AD is equal to AC, and DK to KB, AK is half the sum of AB, AC; and KH is the half of BC; therefore AH is half the perimeter, and AM is its excess above HM or BC: also, because HK is equal to KM, and KD to KB, BM is equal tc HD the excess of AH above AD or AC, and HB is its excess above AB: But the rectangle e 3. E. rectangle HB, BM is equal k to the rectangle GB, BL, or GB, Párt ]Í. DF; and the rectangle HA, AM is equal to the rectangle GA, AF1: and because BG is equal to GE, the triangle ABE is k 35. 3. E. equal ► to the rectangle BG, GA, and the triangle ČBE to the Cor: 36. recta:gle BG, GF; therefore the triangle ABC is equal to the m41.1. E. rectangle BG, AF: and because BG is to GA, as DF to FA ?, n 4.6. E. the rectangle BG, AF is equal o to the rectangle AG, DF: But 0 16.6. E. as the rectangle GA, AF to BG, AF, so is AG to GB P, and so p 1.6, E: is AG, DF or BG, AF to BG, DFP: and the rectangle GA, AF is equal to HA, AM; and BG, AF to the triangle ABC; and BG, DF to HB, BM; that is, the triangle is a mean proportional between the rectangle contained by half the perimeter, and its excess above the base, and the rectangle contained by the excesses of half the perimeter above the other two sides. Wherefore, add the three sides together, and from half the sum, subtract the sides separately; then multiply the half sum, and the three differences into one another, and the square-root of the product is the area of the triangle. For example, let the sides be 10, 17, and 21; the half of their sum is 24, and its excesses above the sides are 14, 7, and 3; and 24 multiplied by 3, and the product by 7, and this last product by 14, gives 7056; the Iquare-root of which 84 is the area. To find the area of any redilineal figure. any Pliit: Fig. 4. Divide the figure into triangles, by joining its angles ; and find the area of each of the triangles, by the ad or 3d Prob. and the sum of these areas will be the area of the figure. If the figure be quadrilateral, and have two fides patallel Fig. 5: to one another, multiply half the sum of the parallel fides by the perpendicular drawn from one of them to the other, and the product is the area. For example, let the parallel fides be 14 and 12, and the perpendicular 15, then 15 multiplied by 13, which is half the sumn of 14 and 12, gives 195 for the area. Note, If a field is to be measured, let it first be laid down on paper, and divided into triangles; then any line, or the distance of two points of it, may be measured, by applying it to the scale from which the figure is drawn. But the scale gives the horizontal line, or the nearest distance between two points, and not their distance in the field, which is commonly uneven ; if this be wanted, it must be measured in the field. But the horizontal plane, on which an uneven or hilly field flands, is Tt2 PART II. that upon which buildings are erected, and is that alone which is occupied by trees or corn. PROB. V. Mig!!: To find the area of a regular polygon ABEFGH. Let C be the centre of the circle described about the polygon, and it is evident, from 12. 4. E, that all the perpendiculars drawn from C to the sides are equal; therefore, if lines be drawn from C to the angles, the polygon will be resolved into as many equal triangles as it has fides; and since the product of the perpendicular into the half of one of the fides is the area of one of the triangles, the product of the perpendicular into half the sum of the fides, will be the sum of the triangles; that is, the polygon. Wherefore, multiply half the perimeter, by the perpendicular from the centre to one of the sides ; the product is the area. Cor. Since 90 degrees, multiplied by the number of fides, a Cor. 32. wanting two, is half the sum of the angles of the figure a, if this product be divided by the number of sides, the quotient is one of the angles at the base of the triangles, as CAB: and, by Trigonometry, R : tan. CAB : : { AB : perpendicular CD. By this corollary, and the problem, the following table was constructed, when the fide is unit: and by it the area may be found, by multiplying the tabular number by the square of the fide. 1. E. Numb. 01 Angles at Angle: Perpendiculars. Areas. 1200 30° 0.2886752 0.4330127 Square. 4 go 45 0.5000000 1.0000000 Pentagon. 5 72 54 0.6881910 | 1.7204774 Hexagon. 6 60 60 0.8660254 2.5980762 Heptagon. 7 513 | 643 1.0382601 3.6339124 Octagon. 8 45 674 1.2071069 4.8284271 Nonagon. 9 40 70 1.3737385 6.1818242 Decagon. 36 72 1.5388418 7.6942088 Undecagon. II 32117317 1.7028439 9.3656399 Dodecagon. 30 75 1.8660252 11.1961524 IO 12 Let the side of a regular hexagon be 14 ; the square of the fide is 196, which, multiplied by 2.5980762, gives 509.2229352 for the area of the hexagon, PROB. To find the area of a circle. PART IT. Let ADCE be the circle, of which the radius AB is 7, and the circumference 44. Multiply 22, the half of the circumfe. Pl. III. Fig. 7. rence, by 7, and the product 154 is the area of the circle a. a 2. 12. E. Note, When the diameter is 1, the circumference has been Hence, it appears, that the area of any sector of a P1. ITA It is proved, by the writers on Conic Sections, that an ellipse is a mean proportional between the circles, which have its axes for their diameters. Therefore, multiply the product of the diameters by-7854, and the product is the area of the ellipse. For example, let the greater axis BD be 10, and the leffer AC 7, their product 70, multiplied by .7854, gives 54.978, which is the area of the ellipse. Cor. Hence, as one of the axes is to the other, so is the circle upon that axis to the ellipse; and so is any segment of the circle cut off by a perpendicular to that axis, to the segmert of the ela lipse cut off by the same line. PRO3. -1 |