279. To construct a scale of equal parts. · RULE I. Draw three lines A, B, and C, at convenient distances, and parallel to one another (Art. 260.); and in C, take the parts Ca, ab, bc, cd, &c. equal to one another. S II. Through C, draw DCE perpendicular to Ca (Art. 259.); and through a, c, d, &c. draw lines parallel to DCE, cutting the parallels A, B, and C; the distances Ca, ab, bc, cd, &c. are called the primary divisions of the scale. III. Divide the left hand primary divisions Ca, into 10 equal parts (Art. 265.); and draw lines through these points, parallel to DCE, across the parallels B and C ; this primary division will be divided into 10 equal parts, called subdivi sions of the scale. IV. Number the primary divisions from left to right, viz. 1, 2, 3, &c. and the scale will be complete. 280. To make a scale of which any number of its subdivisions will be equal to an inch. RULE I. Let one of the primary divisions Ca, of the scale C, be an inch; and let it be divided into 10 equal parts, as above. II. From any point D in AD, draw Da; draw DS making any angle with DA, and make DS=Ca. III. Take the number of subdivisions (which are proposed to make an inch) in the compasses from the scale C, and apply this distance from D to E. IV. Draw ES, and through C draw CG parallel to ES, and make DH=DG. 21 3 B V. Through H, draw HL parallel to Ca, cutting Da in K; then will HK be one of the primary divisions, containing 10 of the parts proposed *. VI. If lines be drawn through D to each of the subdivisions in Ca, it will divide the line HK into 10 equal parts (Art. 224.), which will be the subdivisions of the scale HL; and if the successive distances K1, 12, 23, 34, &c. be taken in KL, each equal to HK, these will form the primary divisions, and the scale HL will be constructed. EXAMPLES.-1. To construct a plane scale, having 20 of its subdivisions equal to an inch. Take the distance Cb (=2 inches =20 subdivisions of Ca) in the compasses, make DE=Cb, DS= Ca, and proceed as before. 2. To construct a scale of which 35 subdivisions make an inch. Extend the compasses from d backwards to the 5th subdivision between C and a, this extent (=35 subdivisions of the scale Cd) being applied from D in the straight line DE, proceed as before. 3. To make scales of which 15, 25, 30, and 40 respective subdivisions will equal an inch. 281. To construct scales of chords, sines, tangents, secants, &c. RULE 1. With any convenient radius CA describe the circle ABDE, draw two diameters AD, BE, perpendicular to each other (Art. 259.), produce EB indefinitely towards F, draw DT parallel to EF (Art. 260.), and join AB, BD, DE, and EA. II. Divide the quadrant BD into 9 equal parts (Art. 263.), and from the centre C, through each of the divisions, draw straight lines cutting DT in 10, 20, 30, 40, &c. this will be the scale of tangents. III. From D as a centre, through each of the divisions of the quadrant, describe arcs cutting BD in 10, 20, 30, 40, &c. this will be the scale of chords. To demonstrate the truth of this construction, let the number of subdivisions of HK contained in Ca= Ba be called n, also by construction Ca contains 10 subdivisions of itself; ·.· DE=n, DS=10; but DE: DS:: DC: (DG=) DH (4.6.) and DC: DH :: Ca: HK ; ·: DE: DS :: Ca: HK, or n :10 :: Ca: HK, ': HK= 10 Ca Ca let n=20 (as in Ex. 1.) then HK=~; 2 Ca let n=35 (as in Ex. 2.) then HK= &c. Q. E.D. IV. Through the divisions of the quadrant, draw lines parallel to BC, cutting CD in 80, 70, 60, 50, &c. this will be the scale of sines and cosines. V. If straight lines be drawn from A to the several divisions (10, 20, 30, &c.) of DT, cutting the radius in 10, 20, 30, 40, &c. CB will be a scale of semi-tangents. VI. If from the centre C, through the several divisions of DT, arcs be described, cutting BF in 10, 20, 30, &c. BF will be a scale of secants. T 70 VII. Divide the radius AC into 60 equal parts, draw straight lines through each of these divisions parallel to CB, cutting the arc AB; and from A as a centre, through the points where these parallels cut the quadrant AB, describe arcs cutting AB in 10, 20, 30, 40, &c. AB will be a scale of longitudes. VIII. Divide the quadrant AE into 8 equal parts, and through these, from E as a centre, describe arcs cutting AE in 1, 2, 3, 4, &c. AE will be a scale of rhumbs. IX. Draw straight lines from B, through the several divisions of the scale of sines (CD), these will cut the quadrant ED in as many points; from A as a centre, through each of these points, describe arcs cutting ED in 10, 20, 30, &c. ED will be a scale of latitudes. X. If the above constructions be accurately made, with a circle the radius of which is 2 inches, the several lines will exactly correspond with those on the common scales; wherefore to construct a scale, we have only to take the several lines respectively in the compasses, and apply them (with their respective divisions) to a flat ruler; and what was required will be done. 282. To find the area of a parallelogram ACDE. RULE. Let a=the altitude AB, b=the base CD; then will ab the area required ". EXAMPLES.-1. To find the area of a square whose side is 12 inches. A C B D E Here a=12, b=12, and ab=12 × 12=144 square inches = the area required. 2. To find the area of a parallelogram, the base of which is 20 inches, and its altitude 25.109. Here a 25.109, b=20, and ab=25.109 × 20=502.18 square inches the area required. 3. To find the area of a rhombus, whose base is 42, and altitude 23. 4. To find the area of a rhomboid, whose base is 10, and altitude 7. " Every parallelogram, is equal to the rectangle contained by its base and perpendicular altitude (see Euclid 35. 1; 1, 2, &c.) ; whence the rule is plain. EXAMPLES.-1. The perpendicular height of a triangle is 28 inches, and its base 16 inches; what is the area? ab 28×16 Here a=28, b=16, and -=224 square inches, the 2 area required. 2. The base of a triangle is 1.03, and its perpendicular altitude 2.11; what is the area? Ans. 1.08665. 3. The altitude 74, and the base S being given, to find the area of the triangle. 284. To find the area of a triangle, having its three sides given. RULE. Let a, b, and c, represent the three sides respectively, a+b+c and let =p; then will √p.p—a.p—b.p-c=the area of 2 the triangle". This depends on Euclid 41. 1. y Let AB=a, AC=b, BC=c, AD=x, then DC=b-x, and (Euc. 47. 1.) c2-b-a2 =BD 2 = a2 —x2, 01 c2 —b2 + 2 bx—x2 = a2 —x2, whence x= a2 + b2-c2 2 But BDAB)2 — AD) 2 = AB+AD.AB¬AD= (a+ a2+b2¬c2 26 -) × (α 26 Cor. If s=a+b, and d=bc, then will√/s2 — a2. a2 —de be the rule. Bonnycastle's Mensuration, p. 47., |