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Las diagrammes suivants illustrent la mithode. 1 2 3 4 S 6 \ > 1^1' \ ]y € KEY, TO THE ^k FATIOl^AL ME:KSUEATIO]Sr; CONTAINING SOLUTIONS TO ALL TEE QTJESTIOJSTS LEFT UNSOLVED IN THAT WORK. BY 3D. MCK^Y. TORONTO: PRINTED FOR THE AUTHOR. 1860. >,. Enteved, according to the Act of the Provmca. Parha- „,en., in the year one thousand eight ^^"^^'^^ si.ty.one, by DoHAtn McKay, in the Office of the Registrar of the Province of Canada. "mmmmi--«m.,Hm^ k PKEFAOE. The following Key, being designed not only to assist those who are pursuing their studies without an in- structer, but to abridge the labour of teachers, by fur- nishing them with the means of reference, care has been taken to illustrate the solutions, by referring, when ne- cessary, to appropriate rules and observations in the Mensuration ; and so to exhibit the results of the different operations, as to make it easy to detect errors in the work of a pupil. In the solution of some of the questions, reference is made to some of the figures in the Mensuration. It will be observed, that quantities are frequently used in the form of vulgar fractions, instead of decimal, as being shorter and more accurate. Reference is also made, in the solution of some of the Miscellaneous Problems, to the Remarks in the beginning of the Key. D. McK. . mmmmwimiiM^ i\ V I REMARKS. 1st. An angle in a semicircle is a right angle. 2nd. All angles in the same segment of a circle are equal. 3rd. Triangles that have all the three angles of the one respectively equal to the three angles of the other, are callad equiangular triangles, or similar triangles. 4th. In similar triangles the like sides, or the sides oppo- site to the equal angles, are proportional. 5lh. The areas of similar triangles are to each other as the squares of their like sides. 6th. The areas of circles are to each other as the squares of their diameters, radii, or circumferences. 7th. The areas of similar figures are to each other as the squares of their like sides. 8th. Similar solids are to each other as the cubes of their like sides. ' REMARKS. 9lh. The solidities of cylinders, prisms, parallelopcpe. dons, &c., which have their altitudes equal, are to each other as the squares of their diameters or like sides. The same remark is applicable to frustrums of a cone or pyramid when the altitude is the same, and the ends proportional. 10th. In finding the content of the hoof of a cylinder, if the cutting plane passes through the centre of the base, the content of the hoof becomes f r^ fe ; where h = the height and r =-- the radius. f I MENSURATION OF SUPERFICIES. PROBLEM I. Ex. 2. 15 X 15 = 225 chains, the area. , Ex. 3'. 7| n. X 7| ft- = 60A ft., the area. Ex. 4. 4769 X 4769 = 22,743,361 links, the area. PROBLEM n. Ex. 2. 5ft. 6 inches X 7ft. 8 inches = 42ft. 2 inches. Ex. 3. 176 X 154 = 27,104 ; then 27,104 -v- 100,000 = ,27,104 of an acre. Ex. 4. 40ft. 6 inches X 29ft. 9 inches = 1,164 feet, 4 inches, 6 parts. PROBLEM IIL Ex. 2. 7.6 X 5«7 = 43.32 chains, the area. Ex. 3. 7ft. 6 inches X 3ft. 4 inches = 25 feet. Ex. 4. 3 yards = 9 feet ; then, 9 feet X 2 feet 3 inches = 20 feet 3 inches. ,,KI»i»|sW4iit;imf''- 8 MENSURATION OF SUPERFICIES. PROBLEM IV. Ex. 2. 72.7 X 36.5 ~ 2 = 1,326.775 yards, the area. Er.. 3. (1276 X 976 — 2) -^- 100,000 = 6 acres, 36.300S perches. Ex. 4. 15 feet 6 inches X 12 feet 7 inches -~ 2 ^ 97 feet 6^ inches. PROBLEM V, Ex. 2. (50 -f 40 + 30) -J- 2 = 60 ; then, 60 — 50 = 10, 60 — 40 = 20, and 60 less 30 = 30 i then, i/( 60 X 30 X 20 X 10 ) = V 260000" = 600. Ex. 3. ( 4900 + 5025 + 2569 ) -^ 2 rz^r 6247 j then, 6247 — 4900 = 1347, 6247 — 5025 = 1222, and 0247 — 2569 = 3678 ; then, t/( 3247 X 3678 X 1347 X 1222 ) = 1/37820044225844 = 6149800 -r- 100000 = 61 acres, 1 rood, 39.68 perches. MENSURATION OF SUPERFICIES. 9 Ex. 4. ( 20 -f 15 + 15 ) -- 2 == 25 ; then, 25 - 20 = 5, 25 — 15 = 10, and 25 — 15 = 10 ; i/( 25 X 5 X 10 X 10 ) = |/12500 = 111.803. Ex. 5. ( 380 + 420 + 765 ) ~ 2 = 782.5 j then, 782.5 — ^80 -— 402.5, 782.5 — 420 = 362.5, and 782.5 — 765 = 17.5 ; then, l/( 782.5 X 402.5 X 362.5 X 17.5) = V^l 998003710. 9375" = 44699.64 yards -j- 4840 = 9 acres, roods, 38 poles, nearly. Ex. 6. (13 + 14 + 15) -J- 2 == 21 ; then, 21—13=8, 21 — 14 = 7, and 21 — 15 = 6 ; then, -/( 21 X 8 X 7 X 6 ) = 1/7056 = « 84 -V- 9 =rr 9^ square yards. Ex.7. (49 + 50.25 + 25.69) -^ 2 = 62.47; then, 62.47 — 49 = 13.47, 62.47 — 50. "25 = 12.22, and 62.47 — 25.69 r= 36.78; then, |/(62.47 X 36.78 X 13.47 X 12.22) = i/37S200. 44225844"= 614.93 chains -f- 10 = 61 acres, 1 rood, 39.68 perches. 10 iMENSURATION OF SUPERFICIES. PROBLEM VI. Ex. 2. 702 ^ 4900 -f- 4 = 1225 ; then, 1225 X l/3 or 1.732 = 2121.7 ~ 160 = 13 acres, 1 rood, 1 perch. Ex. 3. The perimeter of any figure is the sum of its sides. " 27 -^ 3 = 9, the side of an equilateral triangle whose perimeter is 27. 92 = 81 -- 4 = 20.25 X l/3"==35.074. PROBLEM VII. Ex. 2. 14 acres X 4840 = 67760 yards ; then, (67760 -H 7) X 2 = 19360 yards. PROBLEM IX. Ex. 4. i/( 1202 + 2102) ^ |/58500 = 241.86 feet. Ex. 5. i/(602 — 402) = ^2000 = 44.7213; and i/(602 — 502) = ^./llOO = 33.1662; then, 44.7213 -}- 3.^.1662 = 77.8875 feet. Ans. MENSURATION OF SUPERFICIES.. H Ex. 6. ^/(102 + 142) ^ ^296 = 17.204. Ans. PROBLEM X. Ex. 2. y( 102 -f- 122)) ^ ^244 = 15.6204 feet, the breadth o- the building. 122 = 144 _i. 15.6204 = 9.2186 feet, the greater segment ; then, 15.6204 — 9.2186 = 6.4018 feet, the lesser segment. Again, i/(9.2186 X 6.4018 ) = ^59.015633 = 7.68 ; and 30 + 7.68 = 37.68 feet, the length of the prop. PROBLEM XL Ex. 2. ( 4105 -f 3755 -f 4835 ) -^- 2 = 6347.5 ; then, 6347.5 — 3755 = 2592.5, 6347.5 — 4835 = 1512.5, and 6347.5 — 4105 = 2242.5 ; then, 6347.5 X 2592.5 == 16455893.75 X 2242.5= 36902341734.375 X 1512.5 = 55814791873242.1875 j |/55814791873242.1875 = 7470929.780 links, area of one triangle. ( 4835 -I- 3575 -f 2740 ) -- 2 = 5575 ; ■-% .UN' 13 MENSURATION OF SUPERFICIES. then, 5575 — 4835 == 740, 5575 — 3575 = 2000, and 5575 — 2740 = 2835 ; then, 5575 X 2835 = 15805125 X 740 = 11695792500 X 2000 = 23391585000000 ; then, 1/23391585000000 = 4835484.777 links, the area of the other triansle. Again, (7470929.786 -f- 4836484.777) ~- 100,000 = 123 acres, roods, 11.8633 perches. Ans. E Ex. 3. ( 15 + 13 -j- 16 ) -f- 2 = 22 ; then, 22 — 15 = 7, 22 — 13 = 9, and 22 — 16 = 6 ; then, i/(22X9X7X6) = |/8316 = 91.19, area of the triangle A C D. ( 14 + 12 -f 16 ) -I- 2 = 21 ; then, 21 — 14 = 7, 21 — 'l2 = 9, and 21 ~ 16 = 5 ; then, |/( 21 X 9 X 7 X 5 ) ^6615 = 81.38, area of the triangle ABC. Then, 81.33 -f- 91.19 = 172.52. E E> MENSUBATION OF SUPERFICIES. I3 Ex. 4. j/( 2202 - 100>) = y'38400 = 196.959; and ( 265» ~ 70») = 65325 ; then, ^6532^ =255.5875; Then, (195.959 + 255.6875) x 378 == 170684.577 -- 2 = 85342.2885 yards; Then, 85342.2885 -f- 4S40 = 17 acres, 2 roods, 21 perches. Ex. 5. y( 2202 _ 195.9592) = v^JOOOO = 100 ; and |/( 2652 — 255.58752) ^ |/490o"= 70 ; then, 100 + 70 + 208 = 378, the diagonal ; Then, (195.959 + 255.5876) X 378 -i- 2 = 85342.2885. Ex.0. ^/( 262 -252)== ^51 = 7.1414; and V^(222-i92) = ^123 = 11.09; and ^(322 -252) = ^399 = 19.9754; Then, 7.1414 + 11.09 -j- 19.9754 = t 38.207, the diagonal ; and (19-f25) -j- 2 =22 X 38.207 = 840.654 square yards. PROBLEM XII. Ex. 2. ( 25 + 34 f 35 -f 16 ) -^ 2 = 66 ; 14 MENSURATION OF SUPERFICIES. Tten, 55 — 25 = 30, 55 — 34 = 21, 55 — 35 = 20, 55 — 16 = 39 ; Then, 39 X 21 X 30 X 20 = 491400 ; and v/491400 = 700.99. PROBLEM XIII. Ex. 2. ( 750 + 1225 ) -=- 2 : 1520750 -^ 100,000 33.2 perches. 9S7.5 X 1540 = 15 acres, roods, Ex. 3. ( 4 feet 6 inches + 8 feet 3 inches ) -h 2 = '6 feet ^ inches X 5 feet 8 inches = 36 feet 1^ inches. Ex. 4. ( 1476 + 2073) -^ 2 == 1774.5 X 976 = 1731912 -j- 7840 (the number of square yards in an Irish acre) = 220 acres, 3 roods, 25 perches, 7 yards, Irish. PROBLEM XIV. Ex. 2. 28 X 14 -f- 2 = 196, and ( 13 -f- 7 ) -- 2 = 10 X 35 = 350 ; Then, 350 -f 196 = 546 -^ 160 r= 3 acres. 1 rood, 26 perches. MENSURATION OF SUPERFICIES. l6 Ex. 3. 15 X 5 -T- 2 = 37.5, ( 10 -f 5 ) ~ 2 X 8 = 6p, ( 10 + 12 ) -- 2 X 14 = 154, 12 X 6 H- 2 = 36, 12 X 14 -^ 2 = 84, ( 14 + 20 ) -- 2 X 14 = 238, 17 X 20 -^ 2 = 170 ; Then, 37.5 -f 60 + 154 -f 36 + 84 + 238 -f- 170 = 779.5 -~ 160 == 4 acres, 3 roods, 19J^ perches. PROBLEM XV. Ex. 2. 14 X 6 -J- 2 X 12.1243556 = 509.22298 52. Ex. 3. 5.72 X 4.8284271 = 156.875596479. Ex. 4. 19.382 X 3.6339124 = 1364.84. Ex. 5. 102 ^ 100 X 4.8284271 = 482.84271. Ex. 6. 502 = 2500 X 6.1818242 = 15454.5605. Ex. 7. 202 X 9.3656404 = 3746.25616. Ex. 8. 402 >x 11.1961524 = 17913.84384. 16 MENSURATION OF SUPERFICIES. PROBLEM XVI. Ex. 3. As 7 : 22 ; : 10 : 31.4285. Ex. 4. As 22 : 7 : : 50 : 15.909. Ex. 5. As 1 : 3.1416 : : 7958 : 25000.8528. Ex. 6. As 3.1416 : 1 : : 25000.8528 : 7958. PROBLEM XVIL Ex. 3. 3i X 8 = 28 ; then, ( 28 — 6 ) -- 3 = Ex. 4. 40 -^ 2 = 20 '; then, y'{ 202 -f 152) = ^^625 = 25, the chord of half the arc. Then, 25 X 8 = 200 — 40 = 160 -f- 3 == 53i. Ex. 5. ( 30.5 X 8) — 48.7* = 193.26 -^ 3 = 64.42. Ex. 6. y'( 152 + 82) = 1/269 = 17, the chord of half the arc ; then, ( 17 X 8 ) — 30 = 106 ~ 3 = 35f MENSUttATIUiN OF SSUI'ERFICIES. 17 PROBLEM XVfir. Ex. 2. As 7 : 2-2 : : 7 : 22, the circu.nference ; then, 7-^2^ 3^, and, 22 -- 2! =. u . then, 11 X 3i := 38^. Ex. 3. As 7 : 22 : : IJ. ; 3^, Uie. circumference ; then, 3f -- 2 = If, and, IJ -- 2 = /^ ; tl^en, 1-g X /j -- 1/jj =-. 1.089. Ex. 4. 200- 2 .-. 100 X 16^ =^165o'feet; then, 1650 -- 3 = 550 -- 5^ yards = IQO poles ; then, 100^ X .07958 -- 160 = 4 acres, 3 roods, 35.8 perches. Ex.5. 562 X .07958^249.56298. Ex. 6. 100 X 2 = 200, diameter ; then, 200^ = 40000 -f- 4 = 10000 X .7854 = 7854. Ex. 7. 4840, the number of square yards in an acre, H- .7854 = 6162.48 ; ^^6162. 48 =^ 78^, the diameter, ~~ 2 = 39^. Ex. 8. 912 =: 8281 X .07958 = 659.00198. 18 MENSURATION OP SUPERFICIES. Ex. J>. 1.5a --:=- 225 X .7854 = 176.715. Ex. 10. 20 X '-2 = 40, the diameter ; then, 40^ X .7854 = 12o6.64, area of the wlicle circle, -5-2 = 028.34. ' PROBLEM XIX. Ex. 1. 100 X .8862269 = 88.62269. Ex.2. 200 X .8862269 = 177.24538. PROBLEM XX. Ex. 1. 100 X .2820943 — 28.20948. Ans. Ex.2. 200 X .2820948 = .^36. 41896. Ans. PROBLEM XXI. Ex. 1. 100 X .7071065 = 70.710P8. Ex.2. 200 X .7071068 = 141.42136. PROBLEM XXn. Ex. I. 100 X .0366197 = 63.';6197 ; tJ:en, V 03.66197 ^ 7.97884. ;n, 402 X cle circle, MKNSURATION OF SUPERFICIES. 10 Kx.2. 200 X .6366197 = 127.32394; then, 1/127.32394 = 11.28,37. PROBLEM XXIII. 12x. 1. 10 X 1.4142136 = 14.142136. Ex. 2. 20 X 1.4142136 = 28.284272. PROBLEM XXIV. Ex. 1. 100 X 4.4428934 == 444.28934. Ex. 2. 30 X 4.4428934 = 133.286802. PROBLEM XXV. Ex.1. 100 X 1.1283791 = 112.83791. Ex.2. 200 X 1.1283791 -225.67582. PROBLEM XXVI. Ex. 1. 100 X 3.5449076 = 354.49076. Ex.2. 300 X 3.5449076 == 1063.47228. PROBLEM XXVII. Ex. 4. 25 X 2 =. 50, the diameter ; then, 50^ X r ^ MENSURATION OP StTPERFICIRS. .7854 = 3475002.3. As 300 : 1472B :: 3475002.3 : 804.3986. Ex. 5. 3X2 = 6, thn dianiotor ; then, 6' X .7854 = 28.2744. As 360 : IS : : 28.2744 : 1.41372. PROBLEiM XXVIII. Ex. .5. 12 H- 2 = 6 ; tlu n, 6^ = 36 -^ 18 = 2 ; and, J8 -j- 2 = 20, the diameter of the cir- oie. Then, 2 -j- 20 = .100, area in table corresponding to .100 is .040875 X 20^ = 36.35, area of the lesser segment. And, 202 X .7854 = 314.16, area of the whole circle ; then, 314.16 — 16.35 = 297.81, the area of the gr-jater segment. PROBLEM XXiX. Ex. 2. (48 ^ 30) -- 2 = 9 = XB ; then, |/(DB2 + XB2) = 1/(15.81142 + 92) :. 13 .early. Then, AB — XB = 48 — 9 = 39 = i»v.. : 147JJ : : len, 62 X : IS : : I'm' + lb *early. = 39 = JIENSURATION OF SUPERFICIES. f| AX ; then, 39 X -j- 13 = 27 4- 13 = 40 -^ DF, and, y'(^W^ -\- mo^) =r- y(2^oo) = 50, the diumeter. Thon, 15.8114 X 39 -5- (13 X 2) =. 23.7171. And, .50 -- 2 = 25, rndlus; then, S.'i -- 23.7171 -. 1.2829 = YZ -f- 60 = .0250, the tabular segment answering to which is .005434. See note 2nd, page 51, in Mensuration. Then, .005434 X 502 ^ 13.535 X 2 == 27.17, areas of segments AZC and DB. Then, (30 -h 48) -- 2 X 13 = 507 ; and 507 -f- 27.17 = 534.17. Ans. Ex. 3. (20 - 15) -f- 2 =. 2.5 ; then, (20 - 2.5) X 2.5 -^ 17.5 = 2.5, and 17.5 + 2.5 = 20 = DF J then, /(202 -f 152) ^ 05^ diameter; then, 2.5 -j- 25 = .100, the tabular area corresponding to which is .040875 X 252 =:. 25.546875. Again, 20 - 15 =: 5 -4- 2.'i = .200, the tabular area corres- ponding to which is .111823 X 252 _ /■ Ex. 4. 3» MENSURATIOxN ->F SUPERFICIES. 69.889375; then, 69.869375 + 25.546875 = 95.43625 ; then, 252 >< ^^g^^ ^ 490.875; and, 490.875 - 95.43625 = 395.43875. Ans. (96 - 60) -^ 2 = IS ^ XB ; then, . V(IS^ A- 26^) -^ ^.6227 -^ DB. Agnin. (96 - 18) ::^ 78 =.. AX ; then, 78 X 18 - 26 = 54 = FX, 54 + 26 = 80 = DF: then, ^(60^ + 80^) ^ jqo, and 100 -f- 2 = 50 = GF. Then, 31.62277 X 78 -f- (26 X 2) =^ 47.434105 = GZ ; then, 50 - 47.434105 = 2.565895 = YZ, and 2.5658 -^ 100 == .025,c^Vo, the tabular area corresponding to which is .005437928 X 1002 _ 54.37928 X 2 =. 1C8. 75856 ; And (96 + 60) X 26 - 2 :^ 2028 ; then, 2028 + 108.75856 ^ 2136.75856. Answer. PROBLEM XXX. Ex. 2. 30 + 40 == 70 ; and, 40 - 30 = 10 ; MENSURATION OF SUPERFICIES. 28 25. 5468 V5 .7854 = 5.43625 = then, 70 X 10 = 700 X .7854 = 549.78. Ex. 3. 50 -f 45 = 95 ; and 50 — 45 = 5 ; then, d5 X 5 = 475 =: .7854 = ?73.065. PROBLEM XXXI. Ex. 2. (60 + 40) -^ 2 = 50 ; then, 50 X 2 = 100. 1^ Ex. S. (25 4- l^) -i- 2 ^-= 20 X 6 = 120. = 10,- PROBLEM XXXH. Ex. 2. (20'' -V- 15) + 15 = 41| ; then, 15 -- 41f = .36, the tabular area corresponding to which is .2.5455 X 41§2 :r.^ 441.9; (202 -^ 2) -f 2 = 202 ; then, 2 -i- 202 = .009, the tabular area corresponding to which is .001329 X 2022 = .53-4 ; then, 441.9 — 53.4 ^ 388. ;5. 24 . KXKRCISES I.\ PROBLEM XXXIir. Ex.2. (4.8 + 7.2) -..^e; then, (. _^ 5.2 H- 4.1 +7.3) -^4^5.65, 5.65 X 39 = 220.35. Ex. 3. (5.5 4. 8.8^ -^ o -r ,^ V ^ o.s; -^ 2 =: 7.15; then, (7.15 + 6.2 + 7.3 + 6 + 7.5 + 7) -^ G ^ 6.8583 X 50 =342.915. Ex.4. (0 + C.I)_.,^3.o,^ then, (3.05 + 4.4 + 6.5 +.7.6 + 5.4 + S + 5.2 + 6..^) H- 8.^5.83125 X 37.6 == 219.255. EXERCISES. Ex. I. 35.25 X 35.25 == 1242.5025 -,- ,0 =. 124 acres, 1 rood, 1 perch. Ex. 2. 12^ X i ---: 9|. Ex. 3. (42 _L 50 , fi2N __ «, V r o I- 0.*) = 77 . then, y^77 = 8.7749. Ex.4. 10.51 X 4.2s = 44.9828 -^10 ^4 acres, 1 rood, 39 perches. (« -h 5.2 5.65 X 1, (7.15 4- -f- G ^ (3.05 4- f 5.2 -f 219.255. =r. 124 /77 = 4 acres, MENSURATION OF SUPERFICIES. 25 Ex. 5. 12.6 X 0.4 -V- 2 = 40.32. Rx. 6. The same as question 2nd, Problem V. Ex.7. (150 -[- 200 -\- 250) ~ 2 = 300; then, 300 — 150 =:= 150, 300 — 200 = 100, and 300 — 250 = 50 ; then, ^^(300 X 150 X 50- X 100) = 1/225000000 =r 15000 -4- 4840 = 3 acres, 15 perches = 3j'j acres at 9s. 6d. = £l 9s. 5d. Ex. 8. 36 -V- 9 = 4 X 2 8. Ex. 9. 36 -- 12 ^ 3 X 2 =. 6. Ex. 10. 1/(3202 — 1032) = y'91791 --= 302.97. E\. 11. 6 inches = -^ foot : then 121 _ l :^ 12 • and y'{l2}^^ —■ 122) ^ ^'4_9 ^ feet. 1 — '\^ Ex. 12. ^"(-^2 _,_ 82) ^ ^73 ^ 8.544 = BC ; then, as 3 : 8 : : 8.544 : 22.78 ; and VT22.782 -_ 82) == ■/454.9284 = 21^. Ex. 13. (26.5 4- 30) ~ 2 = 28.35 X 199S.675. 70 . .5 ]r 26 Exercises in Ex. 14. (66 feet 3 inches + 60 feet 9 inches) -^ 2 = 58 feet 6 inches X 108 feet 6 inches = 6347 square feet 36 square inches. Ex.15. (75 + 122) -V- 2 = 98.5 X 154 = 15169. Ex. 16. (6340 -I- 4380) -v- B2 = 5360 X 121 -j- 4840 = 134 acres ; then, £207 14s. -j- 134 = £1 Us. Ex. 17. (24 + 26 + 28 + 30) -- 2 = 54 ; ^ then, 54 — 24 = 30, 54 — 26 = 28, 54 — 28 = 26, 54 — 30 = 24 ; then, V"(30 X 28 X 26 X 24) = p/524160 = 723.99. Ex. 18. Solve the same as the first question under Problem XIV, Mensuration, having double the dimensions there given. Ex. 19. 102 _ 100 X 4 = 400 ; then, 400 - |/3" = 400 -f- 1.732 = 230.99; then, 1/230.99 = 15.196. BS) -4- 2 iches = 154 = 121 — 14s. ~ = 54 ; ^ = 28, then, 160 = under ible the 00 ^ then. MENSURATION OF 8UPERFIC1ES. 27 Ex. 20. 8 X 9 -^ 2 = 36 ; then, 36 X 10.99 = 395.64. Ex. 21. 20.52 = 420.25 ; then, 420.25X7.6942088 = 3233.4912482. Ex. 22. 5280, the number of feet in a mile X 10' -j- 4400 = 12 feet -- 3.1416 = 3.819708 feet. Ex. 23. As 7 : 22 : : 9 : 2Sf. Ex. 24. 14 X 60 = 840 X .01745329 = 14.6607 feet. Ex. 25. 80 - 2 = 15 ; then, ^'(152 _|_ 32) == 17 ; then, 17 X 8 = 136 — 30 = 106 -;- 3 = 35f Ex. 26. 2002 -- 4 ^ 10000 X .7854 == 7854. Ex. 27. 15 + 10 = 25, and 15 - 10 = 5 ; then, 25 X 5 X .7854 = 98.175. Ex. 28. 628.32 -- 3.1416 ==. 200, the diameter, -^- 2 = 100, and 628.32 -^ 2 == 314.16 X 28 KXK-KCISKS 1\ 100 ---. 9I41G -:- 2 = 15708 -f- .7854 141.42, diameter, -:- 2 — . 70.71. Ex.29. 3 X .-802269 = 2.6586807. See figure, Problem XXIX. Ex. 30. (16 — 12) -T-' 2 =r 2 = XB ; tiien, v'(XD2 _}_ XB2) = y (22 _|_ 22) == 2.828 =. Dli; AB - BX = 16 - 2 == 14 :== AX, and AX X BX -:- DX = 14 X 2 -- 2 =:. 14 = YX ; then, 14 -f 2 = 16 = DF, and y^(122 -[- 1^2) = 20, the diameter. Then, 20 -^ 2 = 10 = GZ. And, DB X AX -- 2DX == 2.828 X 14 -^ 4 = 9.898 == GY ; tiien, GZ — GY = 10 — 9.893 = .102 = ZY. Then, .102 ~ 20 = .005, the tabular area segment answering to which is .00047 ; then .00047 X 202 =. .188, and .188 X 2 = MENSrjRATION OF SUPERFICIES. 29 .376. Then, (1« + 12) x 2 -v- 2 --^ 28 ; then 2S ^- ..376 :=. 28.376. . ■ See figure, Problem XXII. Ex. 81. 15 -f- 2 z= 7J = AE ; then, (AE2 -^ EC) H- EC =.-. (712 _:_ 7) ^_ 7 _ j^^^^^^ . then, 7 -- 15.035 = .465, the tabular area answering to which is .357727 X 15.0352 = 80.8626. Again, (AE2-^ ED) + ED = (7^-- 4)-f 4=18.0625; then, 4 -~ 18.0625 rz:= .221, the tabular area segment answering to which is .128942 X 18.06252 = 42.19607. Then, 80.8626 — 42.19607 = 38.7. Answer. 80 CONIC SECTIONS. CONIC SECTIONS. Ex.32. i/(P + 12) _ ^^^j,^ hypothemusej which is to be ihe diameter of the circle. Then, ^/2 squared = 2 X .7854 = 1.5708. Ex. .S3.' 100 X .6366197 = 63.66197. PROBLEM I. Ex. 1. 35 X 25 X .7854 = 687.225. Ans. Ex. 2. 70 X 50 X .7854 = 2748.9. Ex. 3. 80 X 60 X .7854 = 3769.92. Ex. 4. 50 X 45 X .7854 = 1767.5. PROBLEM n. Ex. 2. (60 X 40) - (30 X 10) = 2100 ; then, 2100 X .7854 = 1649.34. lemuse ; 5 circle. 954 =: then, I CONIC SECTIONS. 31 Ex. 3. (30 X 24) - (19 X 13) = 473 ; then, 473 X .7854 == 371.4942. PROBLEM III. Ex. 2. 10 -^ 35 = .285f, the tabular versed sine, the area segment answering to which is .185166 X 35 X 25 = 162.0202. Ex. 3. 10 -f- 30 = .333^ ; the area segment answer- ing to which is .229172 X 40 X 30 = 275.0064. Ex. 4. 10 -- 70 = .146f ; the area segment answer ing to which is .068824 X 70 X 50 = 240.884. PROBLEM IV. Ex.2. 30-1-20 = 50; then, 50 X 1.5708 = 78.54. Ex.3. 60 -f- 40 = 100; then, 100 X 1.5708 = 157.08. Ex.4. 6-}- 4 = 10; then, 10 X 1.5708 = 15.708. ^2 CONIC SECTIONS. V Ex. 5. 3 4-2^5; then, 5 X 1.5708 ^ 7.854. PROBLEM V. Ex. 2. As 152 : 102 : : (20 X 10) : (800 -- 9) = TB2 ;. then, ^/^go = 9.42278 = TB, and 1/(9.422782 + 52) = i/ii^s'=- 10.07187 = OG or OB; then, (10.67187 + 10) ~- 2 r= 10.335 = OY ; (10.67187 — 9.42278) = 1.25 nearly, the versed sine ; then, 1/(1.252 _]_ 52) ^ 5.153, half the arc. Then, by Rule II., Problem XVII. (5.153 X ^) — 10 = 31.224 -r- 3 = 10.408 = 2BG. As 10.67187 : 10.335 :: 10.408 : 10.079 = 2XY ; then, 10.079 -f- 2 = 5.039. CONIC SECTIONS. 38 » Ex.3. As il02 : 15^:: 25 X 15 : 211 ; then, ^211 = 14.5 = TB, and |/(14.52 -^ 5a) == ^/236 = 15.38 = OB or OG, nd (15.36 + 15) -4- 2 = 15-18 = OY ; then, 15.36 — 14.5 = .86, and ^/{b^ -j- .86^) = 1/25.76 = 5.075, half the arc; then, by Rule II. Problem XVII., (5.075 X 8) — 10 = 30.6 -- 3 = 10.2 = 2BG. As 15.36 : 15.18 :; 10.2 : 10.08 = 2XY = 2BC J then, 10. OS -f- 2 = 5.04. PROBLEM VI. 8 : 10.079 Ex. 2. 35 - 7 = 28 ; then, as 35 : 25 : l/(28 X 7) : 10. Ex. 3. 70 - 10 = 60 ; then, as 70 : 60 : l/(60 X 10) : 20.9956. ^' .34 CONIC SECTIONS. > PROBLEM VII. Ex. 2. 40 — 2 = 20 ; then, |/(202 — le^) = 1/144 = 12. As 40 : 120 : : 12 : 36, the distance be- tween the ordinate and the centre ; then, 120 -r- 2 = 60, and 60 — 36 =. 24, lesser abscissa, and 36 -f ^0 = 96, greater abscissa. PROBLEM VIII. Ex. 2. 25 -r- 2 = 12.5 ; then, ^/(la.S^ -- 102) = |/'56.25 = 7.5, and 12.5 — 7.5 = 5 ; then, as I02 : 25 X 28 : : 5 : 35. t PROBLEM IX. Ex. 2. 96 X 24 = 2306 ; tiien, |/2306 = 48. As 48 : 16 : : 120 : 40. PROBLEM X. Ex. 2. 24 X 4 X f = 64. I i CONIC SECTIONS. distance be- 35 Ex. 3. 12 X 2 X I = 16. ■M PROBLEM XI. Ex. 3. ao^ = 900, 25» = 025, and 26 X 30 = 750; 900 -\- 625 -{- 750 = 2275. Again, 30 -f 25 == 55 ; then, 2275 -J- 55 = 41^^ X 6 X f = 165/p PROBLEM Xn. Ex. 2. 82 = 64 J and 32 = 9 X J then, v^(64 -f 12) = y/ie = length of the single curve ; and 8.7177 X 2 = 17.4354. = 12; 8.7177, the PROBLEM XIII. Ex. 2. As 9 : 16 ; : 6^ : = 8. 64 ; then, |/64 PROBLEM XIV. Ex. 2. 82 : 92 : : 10 : 12.656. i. '■ 96 CONIC SECTIONS. PROBLEM XV Ex. 2. 42 = 16, 32 = then, as 7 : 2 : : As 7 : 2 : : : 9, then, 16 16 — 9 =: 7 ; 41 ^7' •2^ PROBLEM XVL Ex. 2. 120 -|- 40 = 160, the greater abscissa ; then, i/(160 X 40) = ^"6400 = 80. As 120 : 72 : : 80 : 48. '£x. 3. 60 + 20 = 80, the greater abscissa ; 1/(80 X 20) = 1/1600 = 40. As 60 : 36 : : 40 : 24. PROBLEM XVn. Ex.2. 72 -r- 2 = 36; and i/(362 + 482) ^ l/3600 = 60 ; then, as 72 : 120 : : 60 : 100, half the sum of the abscissa ; CONIC SECTIONS. 37 Then 120 -> 2 = 60 -^ 100 = 160, the greater abscissa ; And 100 — 60 = 40, the lesser abscissa. PROBLEM XVIII. Ex. 2. 72 -T- 2 == 36, and |/(362 -{- 432) — l/3600 == 60 ; then, 60 + 36 = 96- As 482 : 72 X 40 : : 96 : 120. Ex. 3. 36 -^ 2 = 18, and ■i//(182 -|- 242) ^^ y^GOO = 30 -f 18 — : 48 ; then, as 242 : 36 X 20 : : 48 : 60. PROBLEM XIX. Ex. 2. 60 + 20 == 80, greater abscissa ; and i//(80 X 20) = ^i6oo"= 40 ; then, as 40 : 24 : : 60 . 36. * PROBLEM XXI. Ex. 1. 14 + 25 = 39, and 15 -f- 17 + 20 -f 23 = 75, and 16 -[- 18 -f 22 = 56 ; ,/r-" // r i 38 CONIC SECTIONS. then, 75 X 4 = 300, and 56 X 2 : 112 ; then, 39 + 300 + 112 = 451 X 2 -^- = 300f. Ex. 2. 5 -{- 8 = 13, and 7X4=:-- 28 : then, 28 4- 13 = 41. AE == 10 ; then, AC == 10 ~ 2 : then, 41 X 5 -^ 3 = 68-}. 3 Ex. 3. ig + 10 = i| ; then, 4f + if + }f + 10 -_L 10 17 n^ To -239C_8.9P y 4 95P7 5C0 . 692835 A ^ — 69T03 J 10 12 4- 10 _L 10 _L. iO __ 1_375 V O 1^ 14 ~ 16 n^l8 5 04 X -i = 2J_5 . 5 0'4 f X i X t'o- = UMUmi ^vhich reduced to a decimal = .693150, PROBLEM XXII. Ex. 2. 802 = 6i00 X 19 = 121600, aad 002 3600 X 21 = 75600 ; 39 197200 X COXIC SECTIONS. then, 75600 + 121600 = 2.1637 =: 426681.04. Again, 802 = 6400 X 9 = 57600, and 57600 + 75600 = 133200 X 2.1637 = 288204.84 ; then, 602 = 3600 X 80 X 15 = 4320000, and 288204.84 -f 4320000 = 4608204.84, and 426681.64 + 4320000 =4746681.64 ; then, as 4608204 :, 4746631.64 :: 10: 10.. 3005. Ex. 2. PROBLEM XXIII. 25 X 50 = 1250, and 252 = 025 X f =- 446.42857; then, i/'(1250 + 446.42S57) .= 1/1696.42857 = 41.197S X 21 = 964.9438. Again, 1/1250 = 35.3552 X 4 == 141.4208 + 864.9438 = 1006.3640 --75 = 13.4181817 nearly ; then, 30 X 25 X 4 -- 50 = GO X 13.4181917 = 805.0909. 40 MEMSURATION OF SOLIDS. Ex. 3. 50 X 100 = 1795.7142857 5000, and 502 y^ 5 then, 1/(50000 + 1785.71428571) = V 6785.7142857 = 82.3756 nearly X 21 = 1729.8876. Again, y 5000 = 70.7104 X 4 = 282.8416 -f 1729.8876 = 2012.7292 -f- 75 = 26.8363634 nearly ; then, 60 X 50 X 4 -^ 100 = 1-20 X J6. 8303034 3220.3636. MENSURATION OF SOLIDS. PROBLEM I. Ex. 2. 5 feet 7 inches X 5 feet 7 inches X 5 feet 7 inches = 174 feet. K\ MENSURATION OF SOLIDS. 41 Ex. 3. 12 feet 3 inches X 12 feet 3 inches X 1^ feet 3 inches = 1838 feet 3 inches. PROBLEM II. Ex. 2. 10 inches f of a foot, and 8 inches f ; then, 26 X | Xf = 14|. Ex. 3. 10 inches = | ; then, 40 X 'f X ~ = 27J. Ex. 1 Ex. 5. 15 inches = 1.25 ft. ; then, 18 X 1.25 X 1.25 = 28.125, and 1.25 X 1.25 = 1.5625; then, 12 inches -f- 1.5625 = 7.68. 2 feet 9 inches = 2| feet, and 1 foot 7 inches = i| ft. ; then, 2f x if = %^ ; then, 12 -f- ^o_9 ^12 X 48 - 209 = 2.756. PROBLEM in. Ex. 2. H X H X H = 9h fl 42 MENSURATION OF SOLIDS. Ex. 3. 42 -^ 4 = 4 . then, 4 X i/3 = 6.928 X 10 = 69.28. Ex. 4. 32 = 9 X 7 =:= 68. PROBLEM IV. Ex.2. 96 --12 = 8 feet i ■". -, 8-' =: C-i X .07058 = 91.67616. Ex. 3. 1728 X 3 = 5184 ; then, 442 := 193Q X .07958 = 154.0688, and 5184 -^ 154.0688 = 33.64. « PROBLEM V. Ex. 2. 50 -^ 2 == 25, and 25 — 21.75 = 3.25 ; then, 3.25 -^ 50 = .065, the tabular versed sine ; the area segment answering to which is .021659 ; . tiien, .021659 X 502 = 54.1475 X 20 = 1082.95. MENSURATION OF SOLIDS. PROBLEM Vf. Ex. 2. 32 = 9 -f- 4 = 0.05 y^ ^/3 X 30 -f- 3 = 38.971125. 4:3 3.8971125 Ex. 3. 5 feet 10 inches =: fo. of a foot, and 4 feet • 11 inches = f| of a foot j then, {-§2 _ Y_o_o y 4.8284271 = 164.300643, and IP == 3_4j.i y 4.8V84271 == 116.720519 X 41 ~ 3 = 1595.180426; then, 164.300C43 X 45 ~ 3 := 2464.509645, and 2464.509845 — 1595.180426 = 869.32- 9219 - 27 = 32.19738 solid yards, nearly. Ex.4. 102 = 100 X 2. .5980762 = 259.80762 X 45 -- 3 = 3397.1143. PROBLEM VIL Ex.2. 122 _ 14^ y ,.954 ^ 113.0976 X lOO -f- 3 = 37'79.92. I I I 1 44 MENSURATION OP SOLIDS. Ex. 3. 37.0992 ~- 3.1416 == 12, the diameter ; then, 122 ^ 244 ^ ^7954 ^ 113.0976 X 100 ~ S =: 3769.92. Ex.4. 70 ~ 3.1416 = 22.28164, the diameter, -j- 2 — 11.1408 ; then, v(302 — • 11.14082) = |/(900 ~ 121.117) = |/775.883 = 27.855 nearly, the perpendicular. Again, 22.281642 = 496.4714810896 X .7854 = 3S9. 9283284 X 27.855 -:- 3 = 3620.4818 -^ 27 = 134.09, solid yards. Ex. '5. 202 _ (192 _j_ 82) -J- (8 X 2) == 400 — 320 -i- 16 = 5, the distance between the perpen- dicular and obtuse angle ; then, v(162 — 52) = y^231 = 15.19868, the perpendicular. Again, 82 = 64 X •'^'^54 = 50.2656 X 15.19868 ■— 3 = 254.656583. MKNSURATION OF SOLIDS. 45 50.2656 X Ex. 2. Ex. 3. PROBLEM VIII. 1 foot 6 inches =:. 1^ ft.; and 1^2 ^ i> ^ 1.7204774 = 3.8711 ; then, 6 inches = J- foot, and .i. x ^ = ^ X 1.7204774 = .43012 nearly; then, i/(3.8711 X .43012) = 1.2904, and 1.2904 + .43012 + 3.8711 = 5.5915 X 5 ~ 3 = 9.3192. , 18 inches = IJ foot, and ij x IJ = f x 2.5980762 = 5.8456714; Again, 12 inches = i foot, and V x 2.5980762 =r 2.5980762 ; Then, 1/(5.8456714 X 2.5980762) = V^15. 188499735362 = 3.89724, and 3.89724 + 5.84567 + 2.5980762 = 12.3409862 X 6 -^- 3 = 24.6819724. Ex. 4. 504 -J- 144 = 3.5 feet, and 3^2 ~ 144 = 2.583; then, 1/(2.583. X 3.5) = y^g. 0416 = 3.0069, and 3.0069 -f 2.583. -f 3.5 = 9.0902 X 31.5 -f- 3 = 95.447. J 46 MENSURATION OF SOLIDS. E\. 5. 18 inches = 1^ ft., and Ij X 1} = 2^. Again, 12 ;= i • Then, |/(2i X 1) = IJ, and 2^ + Ij + 1 = 4| X 18 -- 3 = 28.5. PROBLEM IX. Ex. 2. 18 inches = 1.6 foot, and I.52 =2.25 X .7854 = 1.76715; and 9 inches = .75 ft. ; then, .752 X .78.54 = .44178; then, 1/(1. 76715 X .44173)=|/. 780691527 = .883567, and 1.76715 + .44178 + .883567 = 3.092504 X 14.25 -f- 3 = 14.689394. Ex. 3. 2 feet 4 inches = J feet, and ^2 __ 4^9 -^ .7854 == 4.276 ; Again, 1 foot 8 inches = f ft., and f 2 __ 2_5 y^ .7Q54 ^ 2.1816- ; Tlien, |/(2.1816- X 4.276) = -/g. 32877 816 = 3.0543, and 2.181 + 4.276 + 3.054 = 9.511 ; Then, 9.511 x If -f- 3 = 5.284. Ans. MENSURATION OF SOLIDS. PROBLEiVI X. 47 E-2. (70X2) + 110^,50X24.8 = 186000 -- 6 = 31000. Ex. 3. (3, X 2) + 21 = as X 4J = 392.5 X 14 -=- 6 = 892.0. PROBLEM XI. E.V. 2. 1 foot 2 inches = 1^ ft., ,he length ; and 1 foot the breadth ; also, 6 inches = i fbot and 4 inches = ^ foot ; Then, (If X 2)4- i = 2f X 30ix-l = 14.402; Again, (1- X 2) -f J = 2J X i X 30J X i = 3.672 ; Then, 3.672 + 14.402 == 18.074. PROBLEM XII. Ex. 2. 10 -^ 2 -- .5 -L iq ,o , ■*» — & + 13 == 18 ; then 18 X 8 = 144, and 13 -^ g = 6.5 + 10 = 16.5; ^ i ill m i I i 48 MENSURATION OF SOLIDS. Then, 10.5 X 5.2 = 85.8, and 144 -}- 85.9 = 229.8 X .2618 = 60.16164 X 12 = 721.93968. Ex. 3. 12 inches = 1 foot, and 7 inches = ^'5 ft., 14 inches = J ft., 12 inches = 1 foot ; Then, 1 ^ 2 = i + | = f , and J -?- 2 = ,^, + 1 = il X T^i = m ; Then, f + Iff = H-3 X 10 X .2618 == 6.78 feet. PROBLEM XIII. Ex. 2. 243 = 13824 X .5238 = 7238.2464. Ex. 3. 25000 -^ 3.1416 = 7957.75, the diameter ; then, (7957.75 -4- 2) X (25000 -- 2) = 263858149120. Ex.4. 303 ^ 27000 X .5236 = 14137.2. PROBLEM XIV. Ex. 2. 16 -7- 2 = 8, the radius ; then (82 X 3) + 42 = 208 X 4 = 832 X .5236 = 435.6352. d 144 -- ). 16104 X = 1^ ft., 1 foot ; and J • • 3 3 . 44 ' : .2618 Ex. 3. MENSUBATION OP SOLIDS. ^g 20 X 3 == 60, and 6 X 2 = lo ; Then, 60 -- lo = 50, and 5^ == 25 X 50 = 1250 X .5236 = 654.5. PROBLEM XV. Ex.4. 3-^2=ai, and U > = 2.25 -f- 2.25 = 4.5, and 4^ x i = 5^ ; Then, Si -I- 4i = 9| X 4 = 39^ X 1.5708 = 61.7848. Ex. 5. 20 -^ 2 = 10, and 15 -^ 2 = 7.5 . Then, (10> + 7.5^) = 156.25 -f- (10^ ^ 8) X 1.5708 X 10 = 2977.92264. Ex. 6. 12 -^ 2 — fi o«^ CO • ^ - 6, and 62 =. gg . ^j^^^ ^^ ^ 2= 5, and b^ = 25. Then,(36 + 25+i^)>,2^^^^^^^^ 195.8264. Ex. 2. 48 PROBLEM XVI. -^ 2 =: 24, and 242 = 576 ^ '■ I 50 MENSURATION OF SOLIDS- (36 -^ 2) = 32, + 18 = 50, the diameter. Also, (50 — 36) -4-2 = 7, the central distance. Now, 18 -r- 50 = .36, the area segment corresponding to which is .25455 ; Then, 50= = 2500, X .25455 = 636.575, the area of the generating segment ABC, the half of which is 318.1875, X 7 = 2227.3125 ; Then, 24^ -v- 3 = 4608, — 2227.3125 = 2380.6875 X 12.5664 = 29916.6714. PROBLEM XVII. -Ex.2. First, 40 -r- 2 =r 20, and 202 == 400 ~- 10 = 40, 40 -f- 10 = 50, the diameter; Again, 25 — 18 = '7, the central distance. .. Also (36 — 16) -f- 2 = 10, and TO -f- 50 = .2, the area segment corresponding to which is .111823 X 502 =, 279.5575, area of PLQ; MEN'SURATION OP SOLIDS. 51 Then, 16 -i- 2 o v. ^^ • 2 - S X 40 = 320, and 320 -f 279.5575 = 5qQ n^-rn ,, oyy.&&7o, the area of the generating surface PDLE. Again, 50 -=- 2 = 25 ; Then, 252 — 72 __ o„^ 7 __ 625 - 49 = 576, the square of half the length of the spindle • Then, (576 ^ ^ooj >< ^^ _ ^ ' And, 599.5575 X 7 == 4196.9025 • ' Then, 8853.3. -- 4196.9025 =. 465G. 4315 X 6.2832 = 29257.2904. PROBLEMS VIII. Ex. 2. 1002 X 6 V 'soift X X .5236 = 31416, cubic feet. Ex. 3. 402 X ,0 X .5236 =. 4I888. Ex.4. 202 X 10 X .5236 = 2094.4. PROBLEM XIX. ^-3. 100X3 =.300, and 10X2 = 20; Then, 300 - 20 = 280, and 280 X 10^ X .5236 = 14660.8; Then, 1002: 602:: 14660.8: 5277.888. f«« WW 52 MENSURATION OF SOLIDS. Ex. 4. 10 X 3 = 30, and 1 X 2 = 2 ; Then, 30 - 2 = 29 X 1=* X .5236 = 14.6608 ; Then, as 10^ : 62 : : 14.6608 : 5.277888. PROBLEM XX. Ex. 2. 302 = 900 X 2 = 1800 -f 182 = 2124 X 40 X •2618 = 22242.528. PROBLEM XXI. £x. 3. 122 = 144 X .3927 X 40 = 2261.952. Ex. 4. 8' = 64 X .3927 X 30 = 753.948. PROBLEM XXn. i I I Ex. 2. (20? + 102) 2356.2. Ex. 3. (302 + 102) 19635, • Ex. 4. (152 _|_ 122) 1139.2504. = 500 X 12 X .3927 = == 1000 X 50 X .3927 = 369 X 8 X .3927 = 1 5.277888. 182 = 2124 •mensuration op solids. PROBLEM XXIII. 53 Ex. 8. 122 = 144 X .7854 = 113.0976 x 30 X tV = 1800.5616. Ex. 3. 32 = 9 X .7854 = 7.0686 X 9 x j'^ == 33.92928. Ex. 4. 62 X .7854 = 28.2744 X • 10 X i^ = 150.7968. Ex.5. 302 X .7854 == 706.86 X 60 X j% == 18849.6. PROBLEM XXIV. Ex. 2. (302 X 2 + 202) ^ 2200, and (30 - 20) = 10, and 102 = 100 x j', = 40 ; Then, 2200 - 40 == 2160 X 40 X .2618 = 22619.52 -- 282 = 80.211. Ex. 3. (402 X 2 + 302) == 41OO, and 40 - 10 ; Then, 102 ^ 100 X A = 40 ; 30 tfVff* 54 MENSURATION OF SOLIDS. Then, 4100 — 40 = 4060 X 60 X .2618 -f- 231 = 276.08, PROBLEM XXV. Ex. 2. 10 X 2 = 20, 30 X 3 = 90 ; Then, 90 + 20 = 110; And, 122 ^ 144 x 110 X 10 X .5236-v- (30 + 10) = 2073.456. PROBLEM XXVL Ex. 2. 1/310 = i/|J X 310 = i/'\^'\ squared = ^%^\*^ ; Then, ^%\*^ X 4 = 3174.4 -f 24^ + 322 ^ 4:17 4. A X 40 = 190976 X .1309 = 24998.7584. PROBLEM XXVn. Ex. 2. 102 _{_ 52 ^ 125^ and 8 X 2 = 16 ; Then, 162 ^ 256 + 125 = 381 X 20 X .1309 = 997.458. REGULAR BODIES. 55 PROBLEM XXVIII. i^^x. 2. (2 + 18) == 20 X 22 X 2.4674 = 197.392. Ex. 3. (7 -f 20) 3264.3702. Ex. 4. (2 -{- 12) 138.1744. 27 X 72 X 2.4674 = = 14 X 22 X 2.4674 = REGULAE BODIES. i PROBLEM I. Ex.2. 123 = 1728 X .1176511 == 203.6467. PROBLEM If. Ex. 1. 3X3X3= 27. PROBLEM JIL Ex. 2. 23 z= 8 X .4714045 . 3.77126949. I 66 REGULAR BODIES. PROBLEM ly. Ex. 2. 123 ^ 1728 X 7.6631189 = 13241.8675. PROBLEM V. . Ex. 2. 12^ = 1728 X 2.181695 = 3769.9689. PROBLEM VL Ex. 2. 122 _ 144 s^ 1.7320508 == 249.4153152. PROBLEM VIL Ex. 2. 42 = 16 X 6 = 96. PROBLEM VIII. Ex. 2. 122 ^ 144 ^ 3.4641016 = 498.8306304. Ex. 3. 42 = 16 X 3.4641016 = 55.4256256. PROBLEM IX. Ex. 2. 22 = 4 X 20.6457289 == 82.58292. PROBLEM X. Ex. 2. 22 = 4 X 8.660254 = 34.641. Ex. 3. 32 = 9 X 8.660254 = 77.9423. SURFACES OF SOLIDS. 5r 241.8675. 9.9689. .4153152. 1.8306304. !56256. 1292. SURFACES OF SOLIDS. PROBLEM I. Ex. 2. 3 feet 6 inches X 4 X 3 feet 6 inches = 49 feet upright surface, and (3 feet 6 inches X 3 feet 6 inches) X 2 = 24 feet 6 inches ; Then, 49 feet + 24 feet 6 inches = 73 feet 6 inches. Ex. 3. (4 feet 8 inches + 2 feet 3 inches) X 2 =. 13 feet 10 inches, the perimeter; Then, 13 feet 10 inches X 12 feet 9 inches = 176 feet 4^ inches, and 4, feet 8 inches X 2 feet 3 inches X 2 == 21 feet, the area of both ends ; Then, 176 feet 4i inches + 31 fee, =. 197 feet H inches. PROBLEM II. Ex.2. 15 inches -:- 12 = 1.25 fo„,_ ^„j j gg ^ m I 58 SURFACES OF SOLIDS, 1.0382617 (see Table, Mensuration, page 40) = 1.297827125, the perpendicular; Then, |/(1. 2978271252 -|- 13.52) ^ f>/(l. 684355246385765625 -f 182.25) = . i/183' 934355246385765625 = 13.502235, the slant height ; Then, (15 X 7 -- 2) -^ 12 = 4.375 feet, and 13.562235 X 4.375 = 59.334778075. Also, 1.297827125 X 4.375 = 5.677993; Then, 5.677993 -f- 59.334778075 = 65.012761. PROBLEM HI. Ex. 2. 9 -J- 3.1416 = 2.8648, the diameter, h- 2 = 1.4324; 1 Then, 1/(1.43242 + 10. 52)= |/112. 30176976 = 10.59725, the slant height ; Then, (10.59725 X 9) ^ 2 = 47.68762, and 92 X .07958 = 6.44598, area of the base ; Then, 47.68762 + 6.44598 = 54.1336. IIH W I SURFACES OF SOLIDS. 59 PROBLEM IV.. See Table, Mensuration, page 40. Ex.2. .1.2071068X9 -.i, = .,053301, and 1.2071068X5 -.12 ^.51029611 ft.,. Then, .9053301 -- .51029611 = .40237 nearly ; And ^(i0.5» + .40237») = ^(„o.4„9 016169) = ,0.512, nearly, .he slant heigh. • Again, 9 inches X 8 ^ 12 = e f,,,^ J 5 inches X 8 - 12 = 3J feet, And 6 + 3J = 9i, .he sum of .he peri«e- ters of .he ends, ^ 2 = 4J X 10.512 = ' 49.05; Also, 9 inches = | of a fooi, and j . = /j X 4.8281271 = 2.71, And 5 inches = /, of a fool, And i»j 2 = ^^ >< 4.g2gj2yj ^ gg _ Then. 49.05 +2.71 + .88 = 52.59.' PROBLEM r. Ex. 2. (,8 + 9)^2x 8.14Z6 = 42.4116 X '/^ I '.! '! I mm SURFACES OF SOLIDS. 00 171.0592 = 7254.894, and 182 x .7854 = 254.4696, also, 9^ x .7854 = 63.6174 ; Then, 7254.694 -f- 254.4696 + 68.6174 = 7672.981. PROBLEM VI. Ex. 2. 20 X 2 = 40, the area of the back, and (20 + 20) -f- 2 X 10 = 200, area of one face X 2 = 400, area of both faces, and i/(102 — 12) =, 9.949 x 2 = 19.898, area of both ends ; Then, 400 + 40 -f 19.898 = 459.898. PROBLEM VII. Ex. 2. 10 X 4 = 40, area of the back, 5 X 2 == 10, area of the upper section, and (4 -{- 2) -^ 2 X 20 = 60, area of one end X 2 = 120, area of both ends, and (10 -f- 5) -=- 2 X 20 = 150, area of 60 .7854 = 83.6174 ; t.6174 == SURFACES OF SOLIDS. 61 one of the faces v q — -inn A -* — 300, area of both faces. Then, (40 + 10 + ,20 + 300, ^ „,^ ickf and , area of faces, X 2 = 9.898. iction, rea of one h ), area of PROBLEM VIU. Ex. 3. 7957.75 X 25000 = 198943750. PROBLEM IX. Ex.2. 7970X3.1416== 25038.552X2143.623 5535 = 53673229.812734532. E-3. 7970 X 3.1416 x 3178.030327 =. 79573277.600166504. • Ex.4. 3 X 3.1416 X 1 =9.4248. Ex. 5. 33 X 4 = 132. Ex. 2. PROBLEM X. 21.5 -< 12 = 1.7925 feet, diameter x 3.1416 = 5.631318, circumference, x 16 = 90.101088, convex surface, i m ' n 62 SURFACES OF SOLIDS. and 1.79253 x 2 X ."7954 =-. 5.046, area of both ends j Then, 90.101 -f- 5.046 = 95.147. Ex.3. 20.75 X 3. 1416 = 65.1882 X 55 = 3585.351, convex surface, and 20.753 X 2 X .78.54 = 676.327575, arrp of both ends ; Then, (3585.351 + 676.f27.575) — 144 = 29.595 nearly. PROBLEM XI. Ex. 1. (2 4- 5) X 2 = 14 ; then, 14 X 9.8696 == 138.1744. PROBLEM Xn. Ex. 2. 5 X 4 X 2 = 40, the area of both faces ; 5 X 3 X 2 = 30, the area of both faces ; 3X4X2 = 24, the area of both ends ; Then, 40 + 30 -|- 24 = 94. TIMCER WEASURK. 68 TIMBER MEASURE. PUOnLEM I. Ex. 2. 8 feet (i inches x 1 foot '.) inches = io| ft. at /)tl. = i.3. 5->.«. (4 + 2) -. = 3 X U = 36 fee, a. CO. :^ 18. shillings. Ex. 5. (15 -|~ 17^ -^ o in ;., u \,iu -I itj ~ 4 __ iQ inches = 1^ foot x 6 = 8 feet. Ex. 6. 3 feet 3 inches X 20 = 65 feet. PROBLEM II. Ex.2. (15 X 15)-^ 144 = i,o^ foot x IS = 28^ feet. Ex. 3. 16 x 12 feet. 144 = U X 12 = 16 II! •64 ' TIMBER MEASURE. Ex. 6. 81i + 41 = 122i, and 55 + 29^ = 844 J Then, 122^ X 84^ = 10351^, and 81^ X 55 = 4482J, and 41 X 29j = 1209J ; Then, (1035U + 4482J + 1209^) -=- 6 = 2673.25 X 47.25 = 126340.59375. PROBLEM III. Ex. 1. 144 -J- 6 == 24 inches -^ 12 = 2 foet. Ex. 2. 144 -^ 8 = 18 inches -^ 12 = IJ foot X 4 — 6 feet. Ex. 3. 144 -f- 16 = 9 inches -~ 12 = | of a foot X 7 — H feet. PROBLEM IV. Ex. 2. 1728 -T- (20 X 10) =. 8.64 = Sjf inches. Ex. 3. 1728 -V- (9 X 6) == 32 inches = 2f feet X 3 = 8 feet. i = 84i ; Qd 81i X 1209J ; t) - 6 = 75. TIMBER MEASURE. PROBLEM V. 65 JI.0 fo„o«.i„, ,„e.ion. „e .oU-.S ., n.e Table in the Mensuration, page 140. ! foet. H foot X of a foot Ex. 3. 1.41 X 20 := ^x. 4. .562 X 40 = Ex. 5. .444 X 32 = Ex.6. .39 X 8.5 = Ex. 7. 5.252 X 40 = Ex. 8. 5.252 X 30 = Ex. 9. 1.129 X 25f = Ex. 10. 1.265 X 12 = Ex. 11. 1.511 X 38 =:r: 28.2. 22.48. 14.208. 3.315. 210.08. 160.186. 29.071. 15.18. 57.418. 4 inches. 2f feet / ■■■MM 66 carpenters' and joiners' work. CAEPENTEKS' AND JOINERS' WOEK. Ex.2. 51.5 X 40.75 -f- 100 = 20.98625. ^ Ex.3. 36.25 X 16.5 -^ 100 = 5.9S125 5 squares 98^ feet. Ex. 4. 86 feet 11 inches X 21 feet 2 inches 1839.73 ~ 100 18.3973. ' 1839 feet 8| inches Ex. 5. 1 foot 2 inches = li X 1 X 22 = 25f, content of the girder ; 3 inches = ^ foot, and 6 inches = ^ foot ; then, i X i X 22 X 9 = 24f, content of the bridgings ; Again, 8 inches = f foot, and 4 inches = i foot ; then, f X ^ X 10 X 9 = 20 feet, content of the binding joists ; Again, 4 inches = ^, and 3 inches = ^ ; then, ^ X I X 7 X 25 = 14j'j, content of IKS' >. IS125 = inches = ^ 100 = Ex. 6. CARPENTERS' AND JOINERs' WORK. 67 ^''^ ceiling joists ; then, 25f + 24^ + 20 -f- Uj\ = 85 feet. 30 X 20 X 3 =.1800; and 6 X 4 X 2 = 48, and * 5 feet 6 inches X 6 X 2 inches =. 66 feet, and 5 feet 6 inches X 4 X 2 =. 44 feet, and 5 feet X 4 feet = 20 feet ; Also, 10 X 8 X 8 = 240 ; Hien, 1800 - (240 + 43 + 6G + 44 + 20) =. 1382 feet -^ loo ^- 13.82 squares, at £5 = £69 2s. i t*\ 1 = 25f, = i foot, X i X dgings ; inches = : 9 = 20 s = i ; •I content of OF PAETITIOOTNG. E-^. 1. 173 feet 10 inches X 10 feet 7 inches = 1839 feet 8f inches = 1839.73 feet -f- loa == 18.3973. Ex. 2. 80 feet X 50 feet 6 inches = 4040 feet - 100 = 40| squares. Ex. 3. 10 feet 6 inches X 10 feet 9 inches = 112 ;l 111 &S!'*'"~ kii li i 68 carpenters' and joiners' work. feet 10^ inches = 112.875 -=- 1 square 12|- feet. 100 = Ex.4. 50 feet 6 inches X 12 feet 9 inches -.--=■. 643 feet 10| inches ~ 100 -^^ 6 squares 43 i'eet 10^ inches. OF ROOFING. Ex.2. 50 feet 9 inches X 30 feet = 1522 feet X 1^ = 22831 feet, at 11 shillings -f- 100 = £12 lis. 2J-Jd. Ex. 3. 40 X 18 = 720 X H = 1080 -~ 100 = 10 squares 90 feet. .Ex. 4. 14 feet 6 inches X 2 = 29 feet X <50 -f- 100 = 17 squares 40 feet. Ex. 5. 15 feet X 2 = 30 X 50 = 15 squares. Ex. 6. 13 feet X 2 = 26 X 3'3' 9 squares 62 feet. 1500 962 100 100 = CARPEMEKS' ANB JOINERs' WORK. gg E- V. 14 foe. 6 inches x 2 = 29 feet x 70 fee. 6 .nohes = 2044i foe. -.. 100 == oo ,,„„,, 44-^r feet. Ex.8. 50 X 30 =. 1.500 -- ]00 - i-, • ^^^ — l) squares. OF WAmscOTTIJVfG. E- 2. 142 ieet 6 inches x [15 feet 6 inches =. 22081 feet -. 9 :=. 245,^ yards. Ex. 3. 60 feet 6 inches v f\ f . . - , licnes X 6 feet 4 inches = 333 ^eet 2 inches X H = 574| -- 9 ^ ^m yards. Ex. 4. 129 fee. 6 inches x 16 fee. 3 inche. = 2104 fee. 4^ inches, .he superficial con.ent of the room ; / And 3 fee. 9 inches x 7 feet = 26 feet 3 inches -^ 2 = ,3 fee. 1 J inches, half woric of door. And 7 fee. 3 inches x 4 fee. 6 inches = 32 fee. 7^ inches x 2 = 85 fee. 3 inches, f, li i 'Li ST'l ■I i ■ & 70 CARPENTERS AND JOINERS WORK. the superficial content of the shutters -v- 2 == 32 feet 7^ inches, for half work of shut- ters ; And 15 iacbes = 1 foot 3 inches X 2 = 2 feet 6 inches -f- 7 feet 3 inches = 9 feet 9 inches X 1 foot 2 inches = 11 feet 4J inches X 4 = 45 feet 6 inches, superfi- cial content of the cheek-boards of the win- dows ; And 4 feet 6 inches X 1 foot 2 inches X 2 = 10 feet 6 inches, superficial content of the boards over the top of the windows ; And 16 inches = 1 foot 4 inches X 7 X 2 = 18 feet 8 inches ; And 3 feet 9 inches X 1 foot 4 inches X 2 = 10 feet ; Then, 10 feet + 18 feet 8 inches = 28 feet 8 inches, superficial content of the lining- boards round the door ; Then, 3 feet 9 inches X 3 feet = 11 feet 3 inches, the chimney ; And 2104 feet 4-J inches + 13 feet Ij in. M^ carpenters' and joiners' work. 71 + 32 feel 7^ inches + 28 feet 8 inches + 45 feet 6 inches + lo feet 6 inches = 2234 feet 9^ inches — H feet 3 inches = 2223 feet G-J- inches -- 9 =. 247JJ feet, at 3s. 6d., = £43 4s. 8d. Ex. .5. 7 feet 8 inches X 12 feet 6 inches r:rr 1045 ieet 10 inches, superficial content of the room ; And 7 feet 8 inches X 3 feet 9 inches X 3 = 80 feet 6 inches, the superficial content of the shutters, -- 2 = 40 feet 3 inches, for half work ; And 3 feet 6 inches x 7 == 24 feet 6 h, -f- 2 = 12 feet 3 indies, half work of door ; Then, 1045 feet 10 inches + 40 feet 3 in. + 12 feet 3 inches = 1098^ feet ~- Q ~ 122J^ feet, at 6 shillings per foot, = ^36 12s. 2|d. (Hi fer^ — ?■ 72 „? J5KICKLAYEBS WORK. BRICKLAYERS' WORK. OF TILING OR SLATING. E\. 2. 40 feet 9 inches X 47 feet « inches = 1035 feet 7^ inches -- 100 — ly squares 35f feet. Ex. 3. 43 feet 9 inches X 2*^ ft'et 5 inches X ij ~ J802 feet 7^ inches, and 43 feet 10 inches X 1 foot 4 inches X 2 = 116 feet lOf inches ; Then, 1802 feet 7f inches -j- 116 feet IDf inches = 1919 feet 6^% inches = 1919j-Y^ feet ; As luO : lOlOiV^ : : ^£1 .^s. 6d. : £24 9s. 5|d. Ex. 4. 45 feet inches X -^4 t^^et 3 inches = 1566 feet \\\ inches -^ 9 = 174.104 yards. BUiCKLAVERs' WORK. 73 OF WALLING. 1 r. nches = \) squares : X ij - 10 inches i feet lOf 1 feet lOf 1919//^ £24 9s. s = 1566 ards. Ex. 2. 57 feet 3 inclies X 24 feet inches X 5 half bricks ■- 3 = 2337.7083 — 272.25 = 8.5866 rods. , Ex.3. 28 feet 10 inches X 20 feet X 5 half- bricks -f- 3 =: 9611 ; And 28 feet 10 inches X 20 feet X 4 half- bricks 3 = 768| ; And 42 ~- 4 === lOJ feet, the height of the gable-end ; ' Tiien, 28 feet 10 inches X lOJ -v- 2 = Then, (96 1 A -L 768f -j- lOOjJ) - 9 — 253.62. 25312 7 7 m 48*] ..• M kJ^ ^"// '11 74 masons' work. MASONS' WOKK, Ex. 2. 120 feet 4 inches X 30. feet 8 inches 3390 feet 2f inches ^ 3390J feet. Ex.3. 112 feet 3 inches X 16 feet 6 inches = 1852 feet Ij inches -^ 63 = 29 rods, 25 » feet. Ex. 4. 5 feet 7 inches X 1 foot 10 inches 101 J feet, at 8 shillings per foot, £4 Is. lO^d. plasterers' work. PLASTEEEKS' WOEK. 3 inches t. inches = 29 rods, 25 inches ' foot, Ex. 2. 14 feet 5 inches + 13 feet 2 inches X 2 = 55 feet 2 inches x 9 feet 3 inches = 5io feet 3} inches -f- 9 = 53 yards 5 feet, of rendering ; 5 indies x 2 == 10 inches ; Then, 14 feet 5 inches - 10 inches =. 13 feet 7 inohes - 10 inches = 12 feet 4 in. X 13 feet 7 inches = 167 feet -j- 9 :~ 28 yards 5 feet of ceiling. Ex. 3. (275 feet 5 inciies X 105 feet 6 inches) -^ 9 = 3228-/,3^ yards, at Is., = £m 8s. 5fd. Ex. 4, (18 feet 6 inches H- 12 feet 3 inches) x 2 = 61 feet 6 inches X 10 feet 6 inches = 645 feet 9 inches ; And .3 feet S inches X 7 = :>5 feet 8 inches ; And 5 feet v 5 ^ 25 feet ; Then, 25 ,eet 8 inches -f- 25 feet .== 50 feet 8 inches ; ii • )M I 76 J'LlJMLEnS WORK. And 645 feet 9 inches 50 feet 8 inches — 505 feet 1 inch -^- 9 = QG^fy yards, at 3 pence, r^ 16s. 4j|d ; Agiun, (18 feet inches X 12 feet 3 inches) -:- ™ ^oij yards, at S pence, ~ 16s. 9jd ; ■Hien, 16s. 4jfd. -|- 16s. 9jd. = £1 ms. ad. PLLMBEKS' WOKK. Ex. 2. 15 feet 6 inches X 10 feet 3 inches — 15SI feet X lbs. :.-: 953f lbs. = 8 cwt. 2 qrs. 1^ lbs. Ex. 3. 43 feet X 32 feet :r^ 1376 feet, and 57 X 2 = 114 ; Then, 1376 + 114 = 1490 feet, at 8f lbs. per square foot = 13037 J lbs. ; Then, as 112 : 13037J : : 18 shillings : jei04 15s. 3^d. Ex. 4. 130 yards X 18 ibs. == 2340, at 4d. = c£39. I'AINTERs' WORK, t 8 inches J- yards, at it 3 inches) 16s. 9jd ; I. =^ .£1 77 inches r— — 8 cwt. nd 57 X at 8f lbs. gs : jei04 pai:nters' wokk. Es. 2. (24 feet G inches -f 18 feet 3 inches) X 2 == 81 feet 6 inches X 12 feet 9 inchrs = 1039 feot IJ inches, 3 feet 6 inches X 7 .-= 24 feet inches, and (7 feet 9 inches X 3 feet 6 inches) x 2 = 54 feet 3 inciies, and (8 feet 6 inches X 1 foot 3 inches) X 4 = 42 feet 6 inches ; and 3 feet 6 inches X 1 foot 3 inches X 4 = 17 feet 6 inches ; Also, 5 feet 6 inches X 5 feet = 27 feet 1 inch ; Then, 1039 feet l-i inch + 24 feet 6 inches 4- 54 feet 3 inches + 42 foet 6 inches + 17 feet 6 inches - 27 feet 1 inch = 1150 feet ~- 9 = 127f yards ; Then, 127? x 3 coats, at 26. .= £s 3s, lOJd. I*- 78 glaziers' work. Ex. 3. (20 feet -f- 14 feet 6 inches) X 2 X 10 feet 4 inches = 713 feet ; And 4 feet 4 inches X i feet = 17 feet 4 inches ; Also, £ feet 2 inches X 6 feet X 2 = 38 feet + 17 feet 4 inches = 55 feet 4 inches ; Then, 713 — 55^ feet = 657f feet -=- 9 = 7322^- yards. GLAZIERS' WORK. Ex. 2. 4 feet 8 inches 9 parts X 1 foot 4 inches 3 parts = 6 feet 4 inches lO^^g parts X 10 = G4 feet inches 5-J parts : 6'1.0407. fid „ 8 1.6 _ 8 ■ Ex. 3. 3 feet 6 inches ,9 parts X 1 foot 3 inches 3 parts X 20 = 90 feet, 6 inches, 6| parts = 90.546875. Ex. 4. 7 feet 6 inches X 3 feet 4 inches — 25 feet. 5 X 10 feet 17 feet 4 X 2 = ^ 55 feet I feet -=- 9 PAVERS'" WORK. ^g Ex. 5. 14 feet 6 inches X 4 feet 9 inches = 68 feet, 10 inches. Ex. 6. (12 fee. 6 inches X 6 feet 9 inches) ^ 2 = 42 feet 2 inches 3 parts = 42/, feet, a. Is. 8d. = £3 103. 3jd_ PAYEES' WOEK. ■Mi 4 inches 3 .rts X 10 '^20736 3 inches 3 6| parts = — 25 feet. E.X. 2. 62 feet 7 inches X 44 feet 5 inches = 2T79 feet 8ii inches, and 62 feet 7 incites X 5 feet 6 inches ^ 344 feet 2i inches -=_ o -_ oo 53 • ^ — ^^ii\ yards, at 3 shillings = £5 14s. 8|. ^d. ; Then, 2779 feet 8- inches 1 344 feet 2 inches 6 parts = 2435^^, feet .^ 9 ^ '^^^m, yards, at 2s. Cd. = £33 16s. 6^ i|d. + ^5l4s. 8|H = je39lls. 3id. :r I ' U 4i 80 .,; PAVERS WORK. Ex. 3. 27 feet 10 inches X 14 feet 9 inches =: 410 feet 6 J inches -^ 9 = 45^ ?3 yjj,,^g^ ^^ 3s. i>d. = £7 4s. 5^c]. Ex. 4. 5 X '-i === 10 + 40 :--= 50, grealer diameter ; and 40 -f- .50 = 90 ; also, 50 — 40 = 10 ; Then, 90 X 10 = 900 X .'i'854 -:- 9 = 78.54, at 2s. 4d. = 183.26 shillings =.- £9 3s. ^2%^' Ex. 5. 4 X 2 = 8 -1- 60 = 68, and 50 + 8 = 58 ; . Then, 68 X 58 = 3944, and 60 X 50 = 3000 ; Then, 3944 — 8000 = 944 X .7854 -=- = 82.3797 yards. iches = 410 'i yards, at VAULTED AND ARCHED ROOFS. 81 VAULTED AND ARCHED ROOFS. Di- diameter ; 54 -:- y = ihillings =z:r 50 H- 8 60 X 50 .7854 PROBLEM I. ^-- 3. 15 X 30 X .7854 = 353.43 x 90 - 31808.7. PROBLEM II. E-"- 2. 40 feet 6 inches x 100 = 4050 feet. Ex.3. 40., 5 X 60 = ^430 feet. PROBLEM III. E- 2. 20» = 400 ; then, (,y Table, Mensuration. Pag. 40.) 4.8284371X400=27039.19176. E-3. 4X2 = 8 + 40 = 48, and 30 + 8 = 33, .he diameters fro,« the outer e.vtremi.y of the wall ; Also, 17.32 4-4 = 9] .39 ,1 , . , Ill 82 VAULTED AND ARCHED ROOFS. Then, 48 X 38 = 1824 X -7854 X 21-32 X I = 20361.599248 ; Again, 40 X 30 X .7554 = 942.48 X 17.32 X f = 10882.5024; Then, 20361.58924 — 10882.5024 = 9479.086848. PROBLEM IV. Ex.2. 40 X 30 X 1-5708 = 1884.96. Ex. 3. 102 == 100 X 2 X 2.5980762 (see Table, Mensuration, page 40) = 519.61524. PROBLEM V. Ex. 2. DE ==3.2 feet, DF=: 4.5 feet, 9 inches = .75 of a foot ; Then, 4.5 -^- 2 = 2.25, and i/(2.252 _|_ ^^52) == -j/5.635 = 2.37, half the arc ; Again, -1/(4.52 __ 3.22) _ -^lO.Ol = 3.1638 = FE or CD ; Again, 3.1638 X 3.2 X i = 5.06208; >< 21.32 12.48 X )024 = SPECIFIC GRAVITY. And 2.37 X | =3.16 + 4.5 = 7.66 X I X .75 = 2.298; Then, 5.06208 - 2.298 = 2.76408 X 50 = 138.204. PROBLEM VI. Ex. 2. 98 X 12 = 1176 feet. ee Table, 4. 9 inches (2.252 _j_ le arc ; 10.01 = 06208 ; SPECIFIC GEAVITT. PROBLEM I. Ex. 2. 10 — 6| == 3^ . Then, as 3^ : 10 : : lOOO : 3077 nearly. Ex.4. 120-80 = 40; and 20 +120 = 140 - m = 123i - 40 = 83^ ; Then, as 83^ ; 20 : : 1000 : 240. Ex.6. 166f — 42| = 123^; Then, as 16^J 123| : : 1333 : 991. us ft ft tlVVv 84 SPECIFIC GRAVITY PROBLEM II. Ex. 2. 112 lbs. = 1792 ounces ; one cubic foot = 1728 inches ; Then, a? 2520 : 1792 : : 1728 : 1228^%. Ex. 3. 1 lb. = 16 ounces ; Then, as 1745 i 16 : : 1728 : 15{f}|. Ex. 4. One ton = 35840 ounces ; Then, as 925 : 35840 : : 1728 : 38}f|. PROBLEM m. Ex. 2. 10 X 3 X 2^ = 75 ; Then, as 1 : 75 : : 925 : 69375 ounces -f- 16 = 4335JI lbs. Ex. 3. 12 X 12 X 63 = 9072, content ; by the table, the specific gravity of marble is 2742, and one ton = 35840 ounces ; Then, as 1 : 9072 : : 2742 : 24875424 ounces -=- 35840 == 6941^0 tons. PROBLEM IV. Ex. 2. 6 — 5 = 1, and 9 — 6 = 3 ; Then, 3 + 1=4; foot — iQ 8 4 .3.8 [85* iices by the is 2742, 4875424 SPECIFIC GRAVITY. '^« 4 : 3 : : lOO : 75, ounces of gold. 86 PROBLEM V. Ex- 2. As 1 : 1 . . 970 : 970, ounces ; Ex. 2. Then, as 1000 : 1728 :: 970 : 1676.16, cubic inches immersed -f- 144 = 11.64. PROBLEM VL solid foot ; and 1000 — 070 = go ; Then, 30 x i'/^ = 2|j omoes. Ex. a. An anker contains 10 gallons ; then the cask contained 5 eallons • •in^ „ n feai.ons, and a gallon contains 231 cubic inches X 5 = 1155 + 216 = 1371 cubic inches contained iu the cask of brandy ; As 1728:1371: ..,030:817.20486 ounces, the weight of an equal bulk of sea-water ; m 86 ) Ex. 2. SPECIFIC GRAVITY. As 1728 : 216 : : 932 : 116.5 ounces, weight of the cask ; As 1728 : 1155 :: 927 : 619.609375 ounces, the weight of the brandy ; 619.609375 + 116.5 = 736.19375 ounces, the weight of both, and 817.20486 — 736.19375 = 81.01111, difference ; The specific gravity of lead ts 11325, and of sea- water 1030 ; Then, 11325 — 1030 = 10295 ; As 10295 : 81.01111 :: 11325 : 89.09495 ounces -^- 16 = 5 lbs. 9 ounces. PROBLEM VII. 49| lbs. =790 onnces ; As 11325 : 790 : : 1728 : 120 cubic inches of lead ; As 1728 : 120 : : 1030 ^ 71Jf ounces ; 1030 — 240 = 790, and 790 ounces — 71|| = 718JJ ; As 790 : 718JJ : : 1728 : 1571.54. >s, weight : : 927 : J brandy ; b ounces, 0486 — TONNAGE OF SHIPS. 87 TONNAGE OF SHIPS. 25, and PROBLEM VIII. 89.09495 uc inches ces ; unces Ex. 2. 20 -f- 2 = 10 ; Then, 50.5 X 20 X 10 = 10100 -J- 94 = 107.4. Ex. 3. 30 -^. 2 = 15 ; Then 100 X 30 X 15 -f- 94 = 478. I I 88 WEIKHT AND DIMENSIONS OF BALLS AND SHELLS. WEIGHT AND DIMENSIONS OF BALLS AND SHELLS. I I PROBLEM I. E^. 2. 33 = 27 X 9 -^ 64 = 3.8 lbs. Kx. 3. 5.543 = 170.030464 X 9 -^ 64 = 24 ibs. PROBLEM II. Ex. 2. 6.63 = 287.490 X 2 -- 9 = 63.888 lbs. Ex. 3. 3.53 =._= 42.875 x 2 -- 9 = 9.53 lbs. Ex.4. 6r= 216 X 2 ~ 9 = 48 lbs. PROBLEM III. Ex. 2. 24 X 7^ = 170.030464 ; Then, ^'170.030464 = 5.54. Ex. I ELLS. OF 64 38 lbs. bs. WEIGHT AND DIMENSIONS OF i3AtLS Ex. 3. 3.8 AND SHELLS. 89 X 7J := 27 Then, ^21 .^ 3. PROBLEM IV. E^- '^- 'in X 9 - 2 = 125 ; Then, ^'125 = 5. fi-3. 63.888X9-^2 = 287.496; Then, ^287.496 = 6.6. PROBLEM V. diameter ; '^' = 729, cube of e.xternal diameter, and 63 === 216, cube of Then, 729 — 216 — 72.14 lbs. internal diameter ; 5L3 X 9 -- 64 == Ex. 3. 9.83 = 941.192, and 7^ ^ 343. Then, 941.192 -343 ==598. 192' X 9 64 =r 84.12 lbs. f PROBLEM VI. Ex. a. 133 = 2197 ^ 59_3.^ ^ g^ 1^^^ IMAGE EVALUATION TEST TARGET (MT-3) // >./> /[^'^.4is '*" ^ ^% z 1.0 ^«- « 1.1 1.25 50 1^ 1^ 2.5 2.2 1.8 i-4 IIIIII.6 6" V] '^y ^ >^ Photographic Sciences Corporation \ iV 4^ \ \ X;' 2.) WEST MAIN STREET WEfSTER.N.Y. 14580 (716) 972-430G .: N '> fp f/j ^ 90 WEIGHTS AND DIMENSIONS OF BALLS AND SHELLS. Ex. 3. 73 = 343 -J- 57.5 = 6 lbs. PROBLEM VII. Ex.2. 10 X 5 X 2 = 100 X .0322 = 3.22 lbs. Ex. 8. . 5 X 2 X 10 = 100 X .0322 = 3.22 lbs. PROBLEM VIIL Ex. 2. 52 = 25 X 40 -f- 40 = 25 lbs. Ex. 3. 5a = 25 X 12 -5- 40 = 7.5 lbs. PROBLEM IX. Ex. 2. 14' = 196 ; And, 10 X 40 = 400 ; Then, 400 -5- 196 = 2.05. Ex. 3. 12» = 144 X 40 -^^ 144 = 40. 5LLS. miNO OP BALLS AND SHELLS. 91 PILIJ^G OF BALLS AKD SHELLS. .22 lbs. .22 lbs. PROBLEM I. Ex.2. 15 X 16 X 17 = 4080-^-6 = E^- 3. 50 X 31 X 32 = 29760 ^ balls. 680 balls. 6 = 4960 PHOBLEM li. Ex.2. Ex, 3. (19 H- 1) = 20, and (19 x 2) -f. j ^ 39; / -r 1 ~ Then, 19 X 20 X 39 = 14820 ^ 6 ^ 2470. \ Th"^ Ir ''' "■" (^' X =') + ' = «; The». 21 X 82 X 43 = ,9866 ^ 8 - 3311. ■ " — PROBLEM III. Ex.8. («X8) + , = ,39_,. 16 = 184, 92 PILING OF BALLS AND SHELLS. and 15 + 1 = 16 ; Then, 124 X 15 X 16 --- 6 = 4960. E PROBLEM IV. X. 2- (24 X 3) + 1 = 73 - 24 = 49, and 24 -f 1 = 25 X 49 = 1225 X 24 ~ 6 pil< 4900, the number of balls in a complete Again, 8 — 1=7. Then (7 X 3) + l = = 15X8X7-=-6 Then, 4900 — 140 = Ex. 3. (40 X 3) If- 1 = 121 and 20 -f 1 = 21 : 22, and 22 — 7 140. 4760. 20 = 101 Then, 101 X 21 x 20 -j- 6 = 7070 Again, 40 — 12 28, and .20 — 12 iS," and 8 -f- 1 = 9 . Then, (28 X 3) -f 1 = 85 — 8 = 9X9 — 6 = 924; Then, 7070 — 924 = 6146. 77 X 960. 9, and 24 — 6 complete GAVGISG. ^^TEmmm distances bv sound. 2. llAn V. - 93 Ex.2. 1142 X e ^ gjQfl ^^ ~~ 9136 -^ 5280, the number of ^«et in a mile == i^r^ ^,^ 22 — 7 ro; - 12 =£ = 77 X Ex. 2. 40 X 4Q gallons. ^^•'^- 37 X 40 5.33762. ^•^•4. 144 X 144 74.785. GAVGING. PROBLEM JJ. = 1600 ^ 277.274 5.77 ^4^0 X .0036065 ^^^6 X .0036065 Ex.5. 64X6-^2^^9^ ^ ^. ^ ^ 192X50.42-. 10640.64 • '''-^^^^ 3^.38 gallons pandas there "" ' ^^"^"^ ^" - bushel, 38.38 -, s ^ 4.8 bushels nearly. 94 GAUGING. Ex. 2. 103 Ex. 3. 303 Ex. 4. 243 PROBLEM III. = 100 X .002832 = .2832. = 900 X .002832 = 2.548. = 656 X .002832 = 1.631. PROBLEM IV. Ex. 2. 10 X 20,= 200 X .002832 = .566, of a gallon. Ex. 3. 70 X 50 X .002832 = 9.912 gallons -- 8 = 1.24 bushels, nearly. Ex.4. 24 X 18 = 432 X .002832 = i.2234 ^ gallons. PROBLEM V. Ex. 2. 20 X 20 X 10 gallons. Ex. 3. 15 X 20 X 10 gallons. 4000 -:- 262 = 14.2 3000 -^ 282 = 10.638 »66, of a ns -i- 8 i.2234 = 14.2 GAUGING. E*. 4. 20 X 8 V ,0 ^^ -^ « X 10 = 1600 -^ 231 — « n^ • '^''1 = 6.92 gnllons. PROBLEM Vf. Ex. 2. 100 -f 80 - iftn ^ - 190, sum of the lengths • and 70 + 56 — lofi ^, ^^^^' ^^'^ of the breadths; Then, 180 X 126 = 22680 .h . " ^^OHO, the product, "--ox 70 = .000, ..a of. He ^,.„„, A-, 56 X 80 = 4480, «.ea of ,he .op • Then. 2.080 + .000 + 44S0 = 34I6o'x *^-« = 3.-.9mH- 277.274 = 862.3. PROBLEM VJI. Ex. 3. 30,2 X 38 = 767 6 V , '"^•6 X 3 = 2302.8, and .38 _ 20.2 = 17.8 . . and 17.82 = 3i6.g4. Then, 2302.8 + 316.64 = 2619 44 v. -^ d = I83;)6.08 • f*« 96 ' GAUGING. the other divisors by Problem I., Gauging ; Then, 18336.08 -?- 359.05 = 51.07 gallons. , . PROBLEM IX. Ex. 2. 20 X 20 X 20 = 8000 X .5236 = 4188.8 -7- 282 = 14.85, old ale gallons. And 4188.8 -r- 231 = 18.133. Ex.3. 100' X .3236 = .523600 -f- 277.274 = 1888^ gallons. PROBLEM XIV. Ex. 2. Here (122 ^ 152 y 3) = 656 X 20 = 13120 J A.id, 13120 X .0009i = 12.136, old ale gallons. Also, 13120 X .OOIH = 14.869, wine gallons. LAND SURVEYING. 07 LAND SURYEYING. Ex. 2. PROBLEM HI. O^M- 91) -^ 2 = 74, and 785 - 634 == 151 ; Then, 151 X 74 = 11,74, area of BNMG (57 + 88) -^ 2 == 72.5, and 634 ^ 510 Then, 124 X 72.5 ::= poon ,h« c»yu, the area of GMLF ; (88 + 70) -.- 2 = 79, and 510 -340=. no ; Then 79 x 170 = 13430, area of EPLK • (84 + 70) -^ 2 = „, ^„, g^^ _ ' = 120 ; Ihen, 120 X 77 = 9240, area of DEKI • (94 + 62) ^ 2 = 73, and 320 - 40I 175 ; 88 LAND SURVEYING. Then, 175 X 73 = 1277r3, area of CDIM ; 62 -f- 2 X 45 = 1395, area of ACH ; Then, 11174 -f 8990 -j- 13430 + 9240 -f- 12775 +• 1395 == 57004 links ~ 100,000 = .57004 of an aero =: acre, 2 roods, 12 perches, nearly. PROBLEM IV. Ex. 2. (12.25 X 8.5) -f- 2 = 52.0625 -:- 10 chains = 5,20625 == 5 acres, rood, 33 perches. problem" V. Ex. 2. DF = 342, and CE = 625 ; Then, (625 -f- 342) — 2 = 483.5 X 1360 = 65756 links -^- 100,000 = 6.5756.=: .6 acres, 2 roods, 12 perches. Ex. 3. (2.25 + 3.6) -- 2 = 2.925 X 4.75 == ' 13.89375 -J- 10 = 1.389375 acres = I acre, 1 rood, 22.5 perches. l-AND SVRVEYINQ, PROBLEM VI. 9$ 2- EG .. 900, and FS =. 268 ; Then, 268 X 900 -^ 2 = isoflon ^, ~ X-^0600, area of tne triangle GFE • 280, and GW 140, and 10 im =. 1100, DV = 410; Then, (410 + 280) -^ 2 ^ .^r ^ - S70^nn ^ - 345 X 1100 - 379500, area of HGED ; ^^ = 1180, DN , 280; Then, (280 -f 140) ^ 2 - o,n = 247900 2 = 210 X 1180 ^47900, area of CDHI ; ;^^ = ;oo, B.M = ,00, a„au_-=,ao. — 292500, area of AICB • -- - ...... .* ; ,::: : 1U.404 acres = m o -10 acres, i rood, 24.64 perches. * f i A'. 100 MISCELLANKOUS PROBLEMS. MISCELLANEOUS PKOBLEMS. Ex. 1. (12 4- 20 + 28) -:- 2 = 30 ; Then, 30 — 12 --=: 18, 30 — 20 = 10, and 30 — 28 = 2; Then, ,/(30 X 18 X 10 X 2) =:V 10800 == l/(3flOO X 3) = 60 y^JT Ex. 2. (3 -f 4 -f 5) ^ 2 = 6 ; Then. 6-3 = 3. 6-4 = 2, and 6 — 5 = 1, and i/(6 X 3 X 2 X I) — |/36 = 6, area of the triangle ; Again, (3 +4+5) -r- 8 = 4, the side of the equilateral triangle of equal perimeter ; Then, 42 ==: 16 -I- 4 = 4 X V^^ 0.928, area of the equilateral triangle ; Then, 6.928 - 6 = . 928 of a square foot. 10, MISCKLLANKOUS rSOBLEMS. IQJ Kx. y. 24 -^ 2 --== 12 ,. tl.on, 12^ =:. 144 X 3 =r- 432, and 10> =. mo -^ 4:J2 -= C32 X 10 X .5230 = 2785.552. Kx. 4. I8i =.: 18.5, and 18.r,2 ._^, 342.25 ; Then, as 7^ : 8 : : 342.25 : .M65.00 ; Then, I "385.06 =. 19. 107. ^ce Relark - 33») = ^5-ir= 22.603, and v/(40» = 2P) = ^1159 = 34.044 . Thon, 31.044 + 22.603 = 56.640 J 56 (let 7J inches. Ex. 0. The eircnnforenco a,„| .lian.olfr of ,|,e circle desorihcd by ,he culcr whcd, is ,0 ,ha. de,. cibcd by >he inn.r, „., a is ,„ , ; „„d .ho radius of the circle doscibed by .he eu.er wheel is 5 feet greater .!,«„ ,hat described by the inner ; therefore the diameter is 10 (feet greater, and consequently the diameter of the circle described by the outer wheel is 20 feet ; r Then, as 7 : 20 • • on . n.-m ' — . . 20 ; 62f, or 63 feet nearly. , ' m I ^^2 MISCELLANEOUS PROBLEMS. Ex.7. Here 9^x .07958 x 3 == 19.33794 ^ tJie co/itent of the smaller rope ; and 122 X .07958 X 6 = 68.75712, con- tent of the larger rope ; Then, as 19.33794 : 68.75712 : : 22 lbs. ; 78| lbs. See Remark 9. Ex. 8. 83 = 512, and 4=* .::=r 04 ; Then, 512 -^ 64 =r 8. Ex. 9. The perimeter of a ^gnre is the sum of its sides ; Then, 40 - 3 =: 13.333, side of the equila- teral triangle, and 13.3332 ~- i X \/S = 76.980035 = area of the iriangle ; Again, 40 -:- 4 = 10 ; Then, 102 ,^ j(^o ^^^^ ^^ ^^^ ^^^^^^ ^ Also, 40 -- 6 = 0.666 X .8660254 (see Table, iV/onsu ration, page 40) == 5.772953, and 40 -•- 2 = 20 X 5.772953 =. 115.47 = area of the hexagon. Ex. 10. See Remark' 9th. Here 3^ = 3.5, and r.62 :r= 12.25; Ex of its and MISCELLANKOU,.^ PROBLEMS. 103 Then, as U3 : 5 : : 12.25 : 40.8333; (1:2 ' Then, v/40.8333 .^ 6.39 feet. Ex- 11. See Remark 8th. By similar cones we ha.e 3 : 1 : : 203 : 2666.066 ; Then, tK2G66.fi66 =. 13.867, altitude of the upper part. Also, 3 ; 2 ;: 203 : 5333.333; Then, ^5333.333 =. 17.472, the altitude of the middle and upper part ; Then, 17.472 - 13.867 = 3.604, and ^0 - 17.472 == 2.528, altitude of the lower part, Ex, 12. 21 i„ci,es -~ 12 =r 1.75 feet ; Then, 1.752 == 3.0G25 X 4.8284271 (see Tahle, Mensuration, page 40) = 14.788 nearly ; ' Again, 9 inches -^- 12 -^- 7^ /• , ^^2 . 1^ — .75 foot, and •7.^ - .5625 X 4.8284271 = 2.716 nearly. * AUo 1/(14.788 X 2.716) ^ ,/,o.i64208 - 0.,?3^ the mean proportional between the areas of the ends ; Then, 6.337 + ,4.788 + 2.716 = .r,.8i X 15 ~ 3 = 119.2. ■'! 1 ^^i ii ^^4 MISCELLANEOUS PROBLEMS. Ex.13 As 4.1 : oo :: 5 : 79.26. Ex 14, Since there^are 160 square poles in an acre, X 2 -^ 40 = 8, perpendicular of tlio triangle ; ' , * And i/(202 — 82\ ___ ..,„^ y ^ ^ ; — ]/.336 =r 18.3303, the distance between the perpendicular and obtuse angle ; And 40 — 18.3303 == 21.07; Then, ^(21.672 + 8^) ^ ^'533.5880 - 23.099 ; Again, 40 -f 18.3303 = ..8.3303, and the 1/(58.33032+ 82)^ 1/^3466.388990 - 58.876. Ex. 15. Here half.an-acre = 2420 square yards, and the area of a circle whose diameter is 1 ::= • . 7854 ; Then, as .7854 : 2420 : ; p . 2420 _ ■ .7854 ~' . square of the diameter = 3081.2325 • Then, ^.3081.2325 =. 55.5 diameter -^ 2 — 27.75, radius. Ex.16. ^(100= + 80-) = 128.062-18, and 100- 80 = 20 ; then, 128.06248 — 20 = 108.06248-^2=54.0312, and SO _ WBHiwwuiUlJtl.WlW an acre, '• of the 8.3303, liar and and the J99 = Js, and s 1 = = the -4- 2 100— } = ) — MISCELLANEOUS PROBLEMS. 105 54.0312 = 25.9688 = twice the breadth of the walk -- 2 = 12.9844 feet. Ex. "17. 4 -f- 2 :.-= 2, the height of the base of the hoof, and 2 -- 4 =. .5, tho area segment corresponding to which is .392699 X 4^ =.. 25.132736, first content; Again, 4 1, and 2 — I :r- ] • Then, 1 ~ 3 = .333}, the area correspond, ing to which is .229172, which is found thus:— Area segment answering to .333 = .228858. and to .334 is .229801 • Then, .229801 - .228858 = .000943 X } = .000314 -f .228858 == .229172, as above : Then, 33 == and 4 — 3 27 X .229172 r=n 6.187644, Then, 2 -- 1 =. 2, and i/2 = 1.4142; Then, 6.187644 X 2 x 1.4142 = 17.5011323, and 25.132730-17.5011323 = 7.63 -^ (4 - 3) =. 7.63 X 6 -^ 3 =: 15.2 cubic inches, the quantity of liquor left in the cup ; Again, 3^ = 9 X .7854 =: 7.O686. .ud^ ^' -- 16 X .7854 ==: 12.5664; !■ (I ^^^ MISCELLAiNEOUS PROBLEMS. Then,|/(12 = 5664 X 7 . 0686)==y^88 . 82686504 = 9.425 nearly, the mean proportional be- tween the areas of the ends ; Then, (12.5664 -f- 7.0686 -f- 9.425) = 29.05 X 6 -- 3 = 58.1 cubic inches, the content of the frustrum ; Then, 58.1 - 15.2 = 42.9 cubic inches, the quantity of liquor the person drank, ~- 282 = .152127 ale gallons. Ex. 18. 52 X .7853 = 19.635 X 8 -- 3 = 52.36, the solidity of a cone whose base is 5 and altitude 8 ; Again, £^ 13s. 7d. -- 10s. = ^^^9> 120 And by Remark 8th, As 52.36 : ij3^6_3 .. 33 , ni^067, the 1^111.067 = 4.8068, the altitude of the cone ; As 8 : 5 : : 4.8068 : 3.005, nearly, the base ; the tbe MISCELLANEOUS PROBLEMS. iQj Again, 3.005 :^ 2 ^ 1.5025, and 1/(1.50252 4. 4.80flfla\ /^^ -r *.W008a) == 1/25.36232624 - 5.0361, the slant height of the whole cone ,' Also, 8 + 5 = 13 . Then, by Remark 9th, As 13 : 8 :: 25.36232624 : ,5.6077, nearly ; Then, |/15.6077 = 3 q-.nfi , 35.75. = ,278.0635 x .07958 = 10..7083,375. area of .he base „f .^^ fi'ustrum. And 23.75= x .07958 = 44.88809375, area of the top. Then, v^()01.70S21375 X 44.88809375) = v/4565.4878339550359625 = 67.5683, .he mean proportional be.ween .he areas of ,he ends ; ■•t< 112 WISCKLLANKOUS PRODLEMS. And (101.7082 -J- 44.8-!8 -f- 67.508:>) 214.1045 X 9.6 ~ li = 685.3264. Ex. 23. 40 -~ 2 ==^ 20, radius ; Tlien, as 20 : 5 : : 20 2 : 1 00, and |/100 ==r 10, the part of ilie radius tiie third man must grind down ; C -j- 5 = 11 shillings ; Then, as 20s. : 11 : : 202 . 220, and |/220 — 14.832397; Then, 14.832397 — 10 = 4.832397, the part of the radius the second must grind ; And 20 - 14.832.^97 .- 5.167603, the part the first must grind. Ex. 24, 52 ... 25 X .7654 = 19.635, area of the base, and 22 = 4 X .7854 = 3.1416, area of tne top j ^ MISCELLANEOUS PKOHLKiMS. H^ Then, K(19.635X3.141C) = ^ 61.685316 = 7.8539, the mean proportional ; And 7.8539 + ij.uio -f- 19.635 = 30.6305; then, 30.6305 X 12 -f- 3 = 122.522, the solidity. * _ Then, by Remark 9th, As 122.522 : 20 : : 5^ : 4.0809 ; Then, 1/4.0809 = 2.02012, greater diame- ter. Then, as5 : 2:: 2.02012: .80804 feet. Ex. 25. 63 = 216 cubic feet ; and 4^ =: 64 x 2 = 128 ; Then, 216-128 = 88 cubic feet still due. Ex.26. Let ^ — the depth of the bowl, a = half the transverse diameter. * = half the conjugate diameter, y — half the diameter of the top of the bowl ; b^ Then, yJ == >< ^^^^ _ ^,^ ^ 114 MISCELLANEOUS PROBLEMS. 8a» Hence, y' = , but y' = 100 ; 4 Then, = 100, and *»=. 133.3- ; 4 Then, y'lii^.d = 11.547, Imlf the conju- gate, X 2 = 23.094, the conjugate diameter. And, as 3 : 4 :: 23.094 : 30.702, the trans, verse diameter ; Then, 30.792 ~ i = 7.698, the depth of the bowl ; Again, .?0.792 X 3 = 92.370, and 7.698 X 2 = 15.i396 ; Then, 92.376 — 15.396 = 76.98 ; Also, 32 -T- 42 = .5625 ; Then, 76.98 X .5625 = 43.30125 ; And 7.6892 = 59.25929 X 43.30125 = 2566.00012 X ..''1236 = 1343.65822832, cubic inches, content of the bowl ; Again, I.52 = 2.25 X .7854 = 1.76715 ,1 MISCELLANEOUS PROBLEMS. HO X 2 -5- 3 = 1.1781, cubic inches, content of the gloss ; Then, 1343.55822S32 -\- 1.1781 = 1140.4 4497 -f- 10 = 114.044497. Ex. 27. Her(>, 1 cubic foot :- 1728 cubic inches ; .7854 ^ And i I \ 40 / X .7854 .3927 800 1600 the nrea of one end of the wire ; .3927 1382400 Hence, 1728 -^ = __-«« _ 800 .3927 , 3520244.4614 inches, = 97784.5684 yards, = 55.5 = 55| miles. . Ex. 28. 48 X 3 = 144 + 1 = 145 — 30 = 115, and 30 + 1 = 31 ; Then, 115 X »1 = 3565 X 30 = 106950 -f- 6 = 17825 balls, the number in a com- plele pile whose base is 48 by 30 ; 116 6 == MISCELLANEOUS PROBLEMS. Again, 24 X 3 = 72 + 1 :=. 73 67, and 6 -|- 1 = 7 . Then, 67 X 7 X 6 V 6 = 469 balls in . pile whose base is 24 by e . And 17825 - 469 == 17356 balls, the answer in the Mensuration, w iich is incorrect. * The correct answer is found thus :— 24 — 1 = 23, and 6 — l — 5 . Then, (:13 X 3) + l = 70 _ 5 ^, gg ^ 6 X 5 -^ 6 = 325 balls in a pile whose base is 23 by 5 ; 'nen, 17823 _ 325 = 1750O balls ; the correct answer. Ex.29. 12-1^1,, and40+n=5, balls in the length of the base; And 10 + u = 21 balls in the breadth of the base ; Then, (51 X 3) -f i == 154 _ ^j ^ 133, and 21 -|- i ~ 22 • . Vhen, (133 X 22 X 21) -^ 6 .. 10241 6 == MISCELLANEOUS PROBLEMS. 117 balls, the number in a pile whose base is 51 by 21. Again, 40 - 1 = 39, ..^^ 10 _ 1 ^ 9 . Then, (39 X 3) + 1 = 118 _ 9 _ iqO X 10 X 9 -f. 6 = 1635 balls in a pile whose base is 39 by 9 ; Then, 10241 — 1635 = 8606 balls. Ex. 30. See question 2ml, Problem III., in Weight and Dimensions of Balls and Shells. Since ^^th of an inch ==.1. then 5.54 -f- .1 =5.64. Ex.31. Here, (9.43 + 7.92 ^ 6.15 -|- 3.16) 31,4 -f- 5 = 6.28, mean girt; Then, 6.28 ^ 4 = I..57, and 1.572 2.4649 X 17i = 42.5195. 118 Ex. 32. MISCELLANEOUS PEOBLEMS. Let ABC be the equilateral cone circumscribing the sphere EHG ; Then, since EH = i of the cir- cumference, the angle EDH = 120 degrees ; Then, the angle EBH = 60 degrees, and the angle EBD = 30 degrees, and DB or DA = 2DE ; Hence, AH = 3DE ; Let r = radius of the sphere, and Z = 3.1416 ; Tljen, 2Zr3= content of the cylinder circum. scribing the sphere EHG ; And 2Zr3 X f = tent of the sphere ; 4Zr< 3 = the con. 1 MISCELLANEOUS PROBLEMS. HQ Also, BH2 = BD2 — DH7 = 3r^ ; Then, Z X BH2 x AH = Z3r' — content of the circumscribing equilateral cone ; Then the sphere, cylinder, and circumscribing cone are as *, 2, 3, or as the numbers 4, 6, and 9. Ex, 33. 102 = 100 -. 4 = 25 x Vr= 43.301 27, area of the equilateral triangle ; . Then, 43.30127 -^ 3 = 14.43376, area of each of the triangles into which the equilateral triangle is divided by bisecting the angles and producing the lines until they meet ; Again, 10 -=- 2 = 5 ; Then, 14.43376 - 5 = 2.8868, nearly, the perpendicular of each of the triangles into which the equilateral triangle is divided, or the radius of the inscribed circle ; 120 . MISCELLANEOUS PflOBLEMS. But the radius of the circumscribing circle is double the radius of the inscribed ; Then, 2.8868 X 2 == 5.7736. Ex. 34. Let the perpenaicular = ^, the sum of the Sides = s, and the difference of the seg- ments of the base = d ; Also, let y -^ ^d = greater segment, and y — id = lesser segment ; Then, ^ j (y + -|rf)2 -f p2 | == greater side ; And 1^ I (y - ^d)2 ^ j,2 ? ^ lesser side ; And, by the question, V i (y^^d)^_^pil Squaring both sides, and transposing, = *2 _ 2y2 _ , ^^2 __ 2p2 ; tiie the I circle is m of the the seg- nt, and the the '+P'\ f P'l MISCELLANEOUS PROBLEMS. ^^l ^'•' ^*n order to simplify the expression, Let, .2 __ .j^, .__ 2^3 ^ ^^ ^^ Then, 1/ j (y 4. .^)2 ^ p, ) Squaring again both sides, and actually per- forming the multiplication of the first two factors, we have And, by involving and collecting the terms, tV^^ - Wy^ + 2p^,2 ^ ^^2^2 _^ ^, _ Whence, ^ =. V rlZ^^J^'- p'\ V 2m + 2;?2 __'J^ ) '" "'""^'^ ^*^'''^"'«' -^^'ituting the values of P^ d, and. m, vve have y ==. 4731. ^2*^ MISCELLANEOUS PROBLEMS. TJien, y -jr id =: 720, greater segment, and y • -Jd == "225, lesser segment ; And T^/(7202 + 3002) .. 780, greater side ; Also, y/(225=^ -I- 300^) = 375, lesser side ; And 720 -f 225 :^: 915, .the base. Ex. 35. Here, (12-^ + 10^') =.. 244, the square of the distance from the top of the higher wall, to that of the lower ; and 40 —- 30 = 10 • Then, lO-^ == lOO ; and y (244 ~ 100) = 12, breadth of the building ; Again, v^244 = 15.6204, and 12= = 144 -J- 15.6204 = 9. 2186 feet, greater segment; Then, 15.6204 — 9.2186 = 6.4018, lesser segment; and 9.2186 — 6.4018 = 2.8168; Then, by Remark 4th, 16.6204 : 2.8168 : : 12 : 2.163; and i/(12^ — 2.163^) = 1/139.321431 == 11.803; then, 11.803 -f 30 = 41.803, the length of the upright. 144 E BRAT A. The attention of the Teacher is respectfully directed to the following typographical errors in the Key : — Page 12, Ex. 2, 4835484.777 should be 4836484.777 18, " 10, 628.34 should be 628.32 67, « 6, 6X2 inches, should be 6 X 2 72, " 3, 43 ft. 9 in. sliould be 43 ft 10 inches. 74, " 2, 3390 feet should be 3690 feet. 75, « 2, after 13 ft. 7 in. should be and 13 ft. 2 in. ~ 10 inches. 80, " 5, -i- should be ~ 9 -~~ 83, " 6, 166J should be 166f 95, " 2, 7854 should be .7854. •* 101, '' 5, I (402=212) should be ^(40->— 212) '' 106, " 18, .7853 should be .7854 " 108. " 19, 5w — 325007 should be on ----= 325007 (( (( (t « (( a ecled to r: — 184.777 2 I inches. id 13 ft. (—212) |ttMis|tr of % iatimial ^t|oaI ^ooh, (By Auihoriiy of the Council of Public Instractlon.) 66 KING STBEET EAST, TORONTO, Would call the attention of School Teachers and the Trade to the reduced Price at which he is now supplying these popular School Books. • The National Mensuration (of which this Work is the Key), is now offered to Teachers and the Trade, at the low price of 82 • 00 per dozen, nett cash. Cap and Post Copy Books, and every School Requisite furnished at the lowest prices. ;i25007 Printed by Thomag Cutlell & Sod, C5 King Street East, Toronto.