of the base multiplied by one-third of the altitude; and 118784359.05 330.828, the content of the given vessel, in ale gallons. Then by Proportion, as 3; 2: 330.828: 220.552, the content of the frustum; and as 3:1 :: 330.828 110.276, the content of the segment, in ale gallons. EXAM. 94. By Rule II., Prob VI., Part V., we have 60 x 60 =3600, the square of the side of the base; 40 × 40 = 1600, the square of the side of the top; and 60 × 40 2400, the product of the sides; then (3600 + 84 1600 + 2400) x = 7600 × 28 = 212800, the 3 content in cubic inches; hence 212800 x .004329 = 921.2112, the content in wine gallons. Then by Proportion, as 9: 5 :: 921.2112: 511.784, the content of the greater vessel; and as 9 4921.2112: 409.4272, the content of the less vessel, in wine gallons. EXAM. 95. By the question, AB 60, and CF72; then by Note 1, Prob. VII., Part V., we have 60 × 60 × .7854 = 3600 x .7854 2827.44, the area of the base in square inches; and 2827.44 X 24- 67858.56, the content of the cone ABC, in cubic inches; also, 67858.56 ÷ 2 = 33929.28 = the content of the segment or cone DEC. Now, by similar solids, Theo. 20, Part III., as 67858.56 603 :: 33929.28 : 108000 = DE3; and 47.622 inches = DE, the diameter of 3 108000 the ton of the Sustum, and also that of the base of the segment. Ain, as 6938.36:79: 33929.28: 186624 = C: and 136624 = 57.146 inches = CG, the altude of the segment; and 72 — 57-146 = 1485+ FC, the altitude of the frustum. inches Exam. 96. Let 43CD represent the given vessei; and AFB, the pyramid when completed: then by the question AB = $. CD = 4, DH = 50, AG = 20 and AH = 1) inches. Now, by similar triangles, Theo. 11, Part III., s AH: HD :: AG: GF; that is, as 10: 50 :: 30: 1X = GF; and 150 50 100 inches = EF. By Prob. V., Part V, we have 40 x 40 x 100 = the content of the pyramid Also, by Rule II, Prob. VI, Part V., we have (60 x 30 380000 60 + 40 × 40 + 60 × 40) × = 3 = the content of the given vessel ABCD; and this being divided by 2, we obtain 190000 the content of the frustum KLCD: the Ene KL being the dividing 160000 190000 plane; then + tent of the pyramid KFL Now, by similar solids, Theo. 20, Part III., as 160000 : 403 :: 3 350000 : 140000 = KL; and 3 140000 51.924 inches KL, the top diameter of the vessel ABLK, and the bottom diameter of the vessel KLCD. FM; then FM FE 2187500 = 129.812 129.812 100 29.812 inches EM, the altitude of the vessel KLCD; and 29.812 20.188 inches MG, Now, by Theo. 9, Part III., the angle BAD, in the semi-circle, is a right angle; and by Theo 12, BE × ED = AE2; that is, 49 x 25 = 1225 the square of the radius AE. By Rule II., Prob. X., Part V., X74) 538.58 = 5476 × 74 we have (74 × 74 538.58 = 405224 538.58752.393 ale gallons, the content of the given vessel ABCD. By Rule I., Prob. XI., Part V., we have 1225 × 3 3675, three times the square of the radius AE; and 25 × 25 = 625, the square of the depth BE; then (3675 625) × 25 × .5236 4300 × 25 × .5236 107500 × .5236 = 56287, the content in cubic inches; hence 56287282 = 199.599 ale gallons, the content of the less segment ABC; con. sequently, 752.393 199.599 = 552.794 ale gallons, the content of the greater segment ADC. EXAM. 98. Let ABC represent the given semi-spheroidal vessel; then by the question, AB = 50, FC D 60, and FG 30; the line DE being the dividing plane. By the Note at the end of Prob. XIII., Part V., we have 60 + 30 = 90, the altitude of the given vessel ABC, added to that of the frustum ABED; 603030, the difference between their altitudes; and 290 x 30 = 2√2700 = 103.923, twice the square root of the sum and difference. 51.9615 x 2 = product of the Then, as 120 50: 103.923: 43.30125 inches, the diameter DE. Now, by the Remark at the end of Prob. XIII, Part V., we have 50 x 50 x 60 = 2500 × 60 = 150000, the square of the diameter AB, multiplied by the perpendicular FC; then 150000 ÷ 441.18 = 339.997 wine gallons, the content of the given vessel ABC. Again, by the Remark at the end of Prob. XIV., Part V., we have 2500 x 2 + 43.301252 = 5000 + 1874.9982515625 = 6874.9982515625, twice the square of the diameter AB, added to the square of the diameter DE; then 6874.99825 × 30882.36 = 206249.9475882.36 = 233.748 wine gallons, the content of the frustum ABED; and 339.997 233.748 = 106.249 wine gallons, the content of the segment DEC. EXAM. 99. - Let ABC represent the given parabolic vessel; then by the question, AB = 50, FC = 60, D and FG 30; the line DE being the dividing plane. By Note 4, Prob. XVII., Part V., we have 60 3030, the reserved remainder; then, as 60: 50% :: 30: 1250, the square of the diameter DE. By Prob XVII., Part V., we have 50 x 50 x 30 = 2500 × 30 75000, the square of the diameter AB, multiplied by half the altitude FC; and 75000 ÷ 294.12 254.997 wine gallons, the content of the given vessel ABC. Again, by Prob. XVIII., Part V., we have (2500 + 1250) × 30 = 3750 × 30 : 112500, the sum of the squares of AB and DE multiplied by FG; and 112500588.24 191.248 wine gallons, the content of the frustum ABED; and 254.997 191.248 |