RULE. Multiply one of the sides by one half of the perpen. dicular let fall from the opposite angle. 10. How many acres in a triangular meadow, one side measuring 1273 rods, and the perpendicular 82.41 rods? 11. The base of a triangular lot is 96f rods, and the perpendicular distance from the opposite angle is 35 rods. What is the area ? 12. What is the area of a triangle whose base is 11% inches, and perpendicular, 28.49 inches? Any surface bounded by straight lines, may be divided into triangles, and the area of each triangle obtained. The sum of the several areas is the area of the whole surface. PROBLEM IV. To find the area of a circle. RULE. Multiply half the diameter by half the circumference, or multiply the square of the diameter by .785398. 13. The diameter of a circle is 363 feet. What is the area? 14. What is the area of a circle whose diameter is 97 miles ? PROBLEM V. To find the area of an ellipse, the two diameters being given. RULE. Multiply the longer by the shorter diameter, and the product by .785398. 15. A house lot in the form of an ellipse has one diameter 110 feet, and the other 45 feet. What is the area ? 16. What is the area of an ellipse, whose diameters are 25 and 17.5 feet? PROBLEM VI. To find the surface of a sphere. RULE. Multiply the square of the diameter by 3.1415926; or, multiply the diameter by the circumference. 17. What is the area of the earth's surface? 18. The circumference of a globe is 252 inches. What is the area? 19. The diameter of a globe is 11.5 inches. What is the area ? To find the area PROBLEM VII. RULE. Multiply the circumference of the base by the height of the cylinder. 20. The diameter of a cylindrical column is 6 feet, and the height 60 feet. What is the area of the convex surface? 21. What is the area of the whole surface of a cylinder, whose diameter is 7.5 feet, and height 49 feet ? PROBLEM VIII. RULE. Multiply the area of the base by the height. 22. The diameter of a cylinder is 13 inches, and the height 69 inches. What are the solid contents ? PROBLEM IX. To find the solid contents of any cylindrical body, whose sides taper uniformly,* as the trunk of a tree. * Such a body is called the frustrum of a cone. RULE. Multiply together the diameters of the two extremities, and to the product add one-third of the square of the difference of the diameters. Multiply this sum by .785398, and the product will be the mean area between the two extremities. The mean area multiplied by the length, will give the solid contents. 23. What are the solid contents of a stick of timber, whose length is 50 feet, the diameter of the larger end 36 inches, and the diameter of the smaller end 30 inches? 24. What are the solid contents of a ship's mast, whose length is 35 feet, the diameter at the base 24 inches, and the smaller diameter 18 inches? PROBLEM X. To find the solid contents of a sphere. RULE. Multiply the cube of the diameter by .5236. 25. The diameter of a globe is 4 feet. What are tho solid contents ? 26. What are the solid contents of the earth, and what does it weigh, supposing the mean weight to be twice that of water ? PROBLEM XI. To gauge, or find the dimensions of a cask. RULE. Find the diameter at the bung, the diameter at the head, and the length of the cask, all in inches. Subtract the head diameter from the bung diameter, and note the dif. ference. If the staves of the cask be much curved, multiply the difference by .7; if little curved, by .6; if of a medium curve, by .65; and if nearly or quite straight, by .55, and add the product to the head diameter. The sum will be a mean diameter, by which the cask is reduced to a cylinder. Multiply the square of the mean diameter by the length, and divide the product by 359 for the contents in beer gallons, or by 294 for the contents in wine gallons. 27. How many wine gallons will fill a cask whose bung diameter is 40 inches, the head diameter 30 inches, and the length 50 inches ? 28. How many beer gallons will a cask contain, which measures 31 inches at the head, 33 inches at the bung, and 47 inches in length ? 29. What are the contents in wine measure of a tub, whose inner diameter at the bottom is 29 inches, at the top 36 inches, and the height 30 inches ? (The tub is a frustrum of a cone, and the solid contents are found by Problem IX.) 30. How many gallons will fill a churn, that is 18 inches in diameter at the bottom, 12 inches at the top, and 3 feet in height? PROBLEM XII. To find the carpenters' tonnage of a vessel. RULE. Multiply the breadth at the main beam, half the breadth, and the length, together. Divide the product by 95 and the quotient is the tonnage. This is probably the best general rule for forming estimates ; but no rule can be given that will produce a perfectly accurate result. The rule employed by government, in the collection of revenue, gives about of the tonnage thus obtained. 31. What is the tonnage of a vessel, whose length is 70 feet, and breadth 25 feet ? 32. The length of a vessel is 163 feet, and the breadth 31 feet. Required the tonnage. 33. Find the tonnage of a vessel that is 1134 ft. long, and 244 ft. wide. 34. What is the tonnage of a ship that is 116 feet long, and 31 feet wide ? NATURAL PHILOSOPHY. PROBLEM I. To find the specific gravity of a body. RULE, If the body is heavier than water, weigh it both in water and out of water, and the difference will be the weight lost in the water. Then, the weight lost in the water : the whole weight : : the specific grarity of water* : the specific gradity of the body. But if the body is lighter than water, attach to it another body heavier than water, so that the two may sink together. Weigh the two together, and the heavier by itself, both in water and in the air, and find the loss of each in the water. Subtract the less loss from the greater, and say, the last remainder : the weight of the body in air : : the specific gravity of water : the specific gravity of the body. 1. A piece of gold weighed 364 dwt. in water, and 384 dwt, in the air. What was the specific gravity ? 2. What is the specific gravity of a body that weighs 13) pounds in the air, and 98 pounds in water? 3. What is the weight of a block of oak, that contains 13} cubic feet, the specific gravity being .925 ? PROBLEM II. To find the distance at which bodies may be seen at sea, or on level ground, the height being known. * The specific gravity of water is 1. A cubic foot of water weighs about 1000 oz., Av. Therefore the specific gravity of any body in thousandths, will represent the weight of a cubic foot in ounces. |