(Appendix, Prop. 18,) and OFFK-KO=71. Now OG2 = OK2 + KG2, (Appendix, Prop. 21, Cor. 2,) therefore OG 8.838834765; and OB2 OF2+ FB2, (Appendix, Prop. 21, Cor. 2,) therefore OB or OH = 12.5, and GH= 3.661165, which divided by 25 gives ⚫1464466 for the versed sine, for which the area is 07134954; and this multiplied by 252, gives 44.5934625 the area of the segment AHB, and the trapezoid ABCDEF × (AD+BC) = 306-25, which, added to twice the segment, gives the zone 395-436925. 3. Required the area of a zone, having the parallel chords 96 and 60, and their distance 26 yards. Ans. 2136-7528 square yards, 1 ro. 30 per. 19 yds. 4. Required the area of a zone, the parallels each 36, and their distance 84 feet. Ans. 6380-81726 square feet, = 23 per. 13 yds. 2 feet. 5. Required the area of a zone, the parallels 136 and 68, and their distance 248 feet. Ans. 55655.2 square feet, = 1 ac. 1 ro. 4 per. 12 yds. 8.2 ft. 6. Required the area of a zone, the parallels 157 and 216, and their distance 128 yards. Ans. 3 acres 34 perches 22 yards 73 feet. 7. Required the area of a zone, the parallels 247 and 192, and their distance 368 feet. Ans. 3 ac. 17 per. 23 yds. 63 feet. 8. Required the area of a zone, the parallels 32 and 40, and their distance 72 inches. Ans. 33 feet 138 inches. PROB. XXIV. To find the area of a ring contained by two concentric circles. RULE. Multiply the sum of the diameters by their difference, and then by 7854. NOTE. If the circumferences or similar arcs of the circles be given, multiply half their sum by the difference of the radii: the product will be the area of the ring, or of the part of it contained by the similar arcs. 1. Required the area of the ring ABC-DEF, of which the diameters are 10 and 6, or OC 5 and OF 3. E B (10+6)(10-6) ×·7854 = 50·2656 the area of the ring, 2. Required the area of the ring, of which the radii are 36 and 24 feet. Ans. 2261.952 square feet, = 8 perches 9 yards. 3. Required the area of the ring, of which the radii are 10 and 6, and similar arcs 15 and 9. 93 Ans. 12×4=48, the area contained by the arcs. 4. Required the area of the ring, of which the radii are 157 and 128 yards. Ans. 5 ac. 1 ro. 18 per. 10 yds. 7.42 feet. 5. Required the area of the ring, of which the diameters re 246 and 228 inches. Ans. 46 feet 77 inches. OF THE ELLIPSE. PROB. XXV. To find the area of an ellipse. RULE. Multiply one of the semiaxes by the other, and by 31416; or one of the axes by the other, and by 7854. Or if the circle upon either axis be given: As that axis is o the other, so is the circle to the ellipse, and so is any sector or segment of the circle to the sector or segment of the ellipse, which has the same chord perpendicular to the first-mentioned axis. See Appendix, Prop. 78, Ex. 3. NOTE. If any two straight lines be drawn perpendicular to AC, and the points be joined in which they meet the circle and the ellipse, these trapezoids are to one another as EG to EK, and their number may be multiplied, until their sum either in the circle or ellipse shall be more nearly equal to it than by any given difference. Therefore the circle and ellipse which are their limits are in that ratio; that is, the circle is to the ellipse as EG to EK, or AC: BD, or as AC2 ×·7854 : ACX BD x 7854. 1. Required the area of the ellipse ABCD, of which the semiaxes are OA 436, and OB 254 feet. 3.1416 × 436 × 254 = 347913-3504 square feet, = 7 acres 3 roods 37 perches 27 yards 7 feet. 2. Required the area of an ellipse, of which the axes are 526 and 354 inches. Ans. 112 yards 7 feet 84 inches. 3. Required the area of the sector OHAK of an ellipse, the chord HK being perpendicular to the greater axis AC; the axes AC 72 and BD 54, and the versed sine AE 18. The angle FOG is 120°. The circle = 4071-50408, and of it x = 1017.87602 the area of the sector. 4. Required the area of the segment of an ellipse, the chord being perpendicular to the less axis, the versed sine 12, and the axes 80 and 60 yards. Ans. 536 75424 square yards, 17 perches 22 yards. 5. Required the area of the segment of an ellipse, the chord being perpendicular to the greater axis, the height 25 feet, and the axes 156 and 120 feet. Ans. 5 perches 17 yards 73 feet. 6. Required the area of the segment of an ellipse, the chord being perpendicular to the less axis, the height 110, and the axes 246 and 180 yards. Ans. 4 acres 2 roods 16 perches 3 yards 81 feet. PROB. XXVI. To find the circumference of an ellipse. RULE. Add the squares of the two axes, and take the square root of half the sum, and to the half of this root add a fourth of the sum of the axes, and then multiply by 3-1416: the product will be the circumference nearly. See Appendix, Prop. 77, Ex. 3. 1. Required the circumference of the ellipse, of which the axes are 24 and 18. 241+18=21-2132, and =21, and (21-2132+21) 24+18 2 × 3·141666·3085 the circumference. 2. Required the circumference of the ellipse, of which the axes are 60 and 40 feet. Ans. 158 6351 feet,: = 9 poles 3 yards 1 foot 1.6 inches. 3. Required the circumference of the ellipse, of which the axes are 256 and 196 feet. Ans. 713-1156 feet. 4. Required the circumference of the ellipse, of which the axes are 320 and 240 yards. Ans. 884-1133 yards. 5. Required the circumference of the ellipse, of which the axes are 166 and 12.8 inches. Ans. 46.3736 inches. 6. Required the circumference of the ellipse, of which the axes are 27 and 18 poles. Ans. 1 furlong 31 poles 2 yards. OF THE PARABOLA. PROB. XXVII. To find the area of a parabola. RULE. Multiply the base by the perpendicular height, and of the product will be the area. Ex. 1. See Appendix, Prop. 78, GLB H NOTE. If EG bisect AD, the triangle AFGAFB, or it is trilineal AFBK. Also, since GK= KH, the triangle PLGALG, or trilineal AGBK; and every triangle thus formed cuts off more than the half of what was left by the preceding; therefore the trilineal AFBK is the limit of the sum of the triangles. Now the triangle AFGFD, and the triangle GPL = AGB, or of AFG, and so on; therefore A E D C 42 1 the sum of them is FD × (++, &c.), and the limit of this geometrical series is (Prop. 3, Cor. 3,) FD× = }FD = {BD × AD, and therefore AKBD = {FD. 4-1 1. Required the area of the parabola ABC, of which the base AC is 54, and the height BD 36 feet. 2. Required the area of the parabola, of which the base is 42, and the height 63 yards. Ans. 1764 yards, 1 rood 18 perches 9 yards. 3. Required the area of the parabola, of which the base is 482, and the height 320 feet. Ans. 2 acres 1 rood 17 perches 20 yards 812 feet. 4. Required the area of the parabola, the base 126, and the height 210 inches. Ans. 13 yards 5 feet 72 inches. 5. Required the area of the parabola, the base 67, and the height 98 yards. Ans. 3 roods 24 perches 213 yards. 6. Required the area of the parabola, the base 16, and the height 12 poles. Ans. 3 roods 8 perches. PROB. XXVIII. To find the area of a frustum of a parabola. A Frustum is what remains after a part has been cut off from the top by a line parallel to the base. RULE. Find a third proportional to the sum of the bases, and one of them, to which add the other base: the sum, multiplied by two-thirds of the height, gives the area. Appendix, Prop. 78, Ex. 1. 1. Required the area of the frustum of a parabola, of which the bases are 64 and 32, and the height 26 feet. B See F E G 64+32 32: 32: 103 64 742 D x26=17 Ans. Area 1294 feet, 4 perches 22.8 yards. 2. Required the area of the frustum of a parabola, of which the bases are 16 and 54, and the height 46 yards. Ans. 1768-15238 square yards, = 1 ro. 18 per. 13·65 yds. 3. Required the area of the frustum of a parabola, of which the bases are 364 and 186, and the height 280 feet. Ans. 1 acre 3 roods 12 perches 21 yards 2 feet. M 4. Required the area of the frustum of a parabola, of which the bases are 424 and 268, and the height 318 inches. Ans. 2 perches 25 yards 7 feet 75.8828 inches. 5. Required the area of the frustum of a parabola, of which the bases are 63 and 22, and the height 44 poles. Ans. 12 acres 2 roods 15 perches. 6. Required the area of the frustum of a parabola, of which the bases are 18 and 12, and the height 20 yards. Ans. 10 perches 13 yards. PROB. XXIX. To find the area of a hyperbola. RULE. Multiply half the base by the semitransverse axis, and its distance from the centre by the semiconjugate, and divide the sum of the products by the product of the two semiaxes, and take the hyperbolic logarithm of the quotient, and multiply it by the product of the semiaxes, and subtract the product from the product of half the base by its distance from the centre: the remainder will be the area. See Appendix, Prop. 78, Ex. 4. NOTE. The hyperbolic logarithm is got by multiplying the common logarithm by 2.30258509. 1. Required the area of the hyperbola ABC, of which the base AC is 24, and the altitude BD 10, and the transverse axis Bb 30, and the conjugate Pp 18 feet. 12 x 15+25 × 9 15x9 = 3, of which the logarithm 0·4771212 × 2.30258509 = 1.0986123 the hyperbolic logarithm of 3; and P this logarithm, multiplied by 15x9, gives 148.3126605, which, taken from 25 × 12, leaves 151.6873395 the area, = 16 yards 7 feet 99 inches. 2. Required the area of the hyperbola, of which the base is 208, the height 70, and the transverse semiaxis 105 yards. ((210+70) × 70): 104 :: 105: 78 the semiconjugate. Ans. 9202.365 square yards, = 1 ac. 3 ro. 24 per. 63 yds. 3. Required the area of the hyperbola, of which the base is 384, the height 250, and the axis 176 feet. Ans. 55686 feet. 4. Required the area of the hyperbola, of which the base is 156, height 196, axis 248 yards. Ans. 18449.84 yards. 5. Required the area of the hyperbola, of which the base is 48, height 22, axis 36 inches. Ans. 647-2483 inches. 6. Required the area of the hyperbola, of which the base is 96, height 110, axis 124 poles. Ans. 6324-686 poles. |