By Prob. IL, Part I., we have √59536 = 211 inches, the side of the square required. EXAM. 3. By Prob. III., Part I., we have 6432 ÷ 7854 = 8180457601; and 8189.457601 90.495 inches. the diameter required EXAM. 4. By Prob. IV., Part I., we have √782 + 112% √6084 + 12544 = √18628136.484 inches. the diagonal required. EXAM. 5. Here a perpendicular drawn from the middle of the given diagonal of the opposite angle, will be equal to half the said diagonal; consequently, we have the base and perpendicular of an isosceles right-angled triangle given, to find the hypothenuse, as in the last Example; hence 57.5% + 57.52 = 3306.25 + 3306.25 =6612.50 81.317 inches, the side required. EXAM. 6. By Prob. V., Part I., we have 1452 1162 = - 31.5, half the sum of the diameters; then 31.52 + 22.52 = √992.25 +506.25 = √1498.50 38.71 inches, the diagonal required. EXAM. 8. By Prob. VII., Part I., we have 38.7 + 29.6 68.3 2 = 34.15, half the sum of the diameters; then 2 40.32 34.152 = 21624.09 1166.2225 = 2√ 457.8675 = 21.39 × 2 = 42.78 inches, the length required. 3 41.52 inches, the 3 By Prob. II., Part I., we have 1255 x 51953125 3 x 5 = 9765625 the required vessel. 3 213.74 inches, the length of 3 3 Again, 863 x 5 = √636056 × 5√3180280 = 147.05 inches, the breadth of the required vessel. 3 3 3 Lastly, 623 x 5 = √238328 × 5 = √1191640 = 106.01 inches, the depth of the required vessel. 3 EXAM. 11. By Problem II., Part I., we have √2543÷ 3 = 3 ✓16387064 ÷ 3 = √ 5462354.666666 = 176.11 inches, the depth of the required vessel. 3 3 3 Also, 1125 ÷ 3 = √1404928 ÷ 3 = √468309.333333 = 77.64 inches, the diameter of the required vessel. EXAM. 12. By Prob. III., Part I., as 121500: 753 :: 562500 1953125, the cube of the length; and = 125 inches, the length of the required vessel. Again, 121500: 455 :: 562500 cube of the breadth; and 421875 = 75 inches, the breadth of the required vessel. Lastly, as 121500 363 3 562500 216000, 60 inches, the cube of the depth; and 216000 the depth of the required vessel. EXAM. 13. By Prob. I., Part IV., we have 62.8 x 62.8 3943.84, the area in square inches; and 13.981, the area in ale gallons. EXAM. 14. 3943.84 282 By Prob. II., Part IV., we have 115.3 x 86.4 = 9961.92, the area in square inches; 43.125, the area in wine gallons. EXAM. 15. 9961.92 231 = By Prob. III., Part IV., we have 84.6 x 63.4 = 5363.64 5363.64, the area in square inches; and 2150.42 2.494, the area in malt bushels. EXAM. 16. By Prob. IV., Part IV., we have 123.6 x 42.85 5296.260 = 5296.260, the area in square inches; and 282 18.781, the area in ale gallons. EXAM. 17. 5668 79 By Prob. V., Part IV., we have 2 203 2 = 101.5, half the sum of the sides; then 101.5 5645.5 the first remainder; 101.5 68 = 33.5 the second remainder; and 101.5 79 29.5 the third remainder; whence √101.5 × 45.5 × 33.5 × 22.5 = √3481005.9375 = 1865.745, the area in square inches; and 1865.745 = 8.076, the area in wine gallons. EXAM. 18. By Prob. VI., Part IV., we have 63.8 + 56.4 x 138.6 = 120.2 x 138.6 = 16659.72; and = $329.86, the area in square inches; then 3.873, the area in malt bushels. EXAM. 19. 16659.72 2 8329.86 2150.42 By Prob. VII., Part IV., we have 98 + 124 x 136 = 222 × 136 = 30192; and 30192 ÷ 2 = 15096, the area in square inches; then 15096 282 = 53.531, the area in ale gallons. EXAM. 20. (See Problems V. and VIII., Part IV.) Here 122104 +168 394 2 sum of the sides; then 197 remainder; 197 and 197 = = 197, half the 2 122 75, the first 104 93, the second remainder; 168 = 29, the third remainder; whence √197 × 75 × 93 × 29 = √√39848175 = 6312.541 inches, the area of the first triangle. |