Also 4400.16-231-19.048 wine gallons; and 4400.16 ÷2150.42 2.046 malt bashels. By the Sliding Rule. As the square divisor on A, is to half the sum of the sides on B; so is the perpendicular on A, to the area on B. On A. On B. On A. Ón B. 282 15.60 ale gallons. 2150.42 As 231 123.6 :: 35.6:19.05 wine gallons. 2. Admit the foregoing figure to represent the base of a guil-tun; what number of ale gallons does the vessel contain, when the depth of the liquor is 38.9 inches ? Ans. 606.9567 ale gallons. 3. What is the area of a regular pentagon, in ale gal lons; the side measuring 61.8, and the perpendicular 44.6 inches? Ans. 25.621 ale gallons. 4. The side of the base of a heptagonal wine vatmeasures 46.3, and the perpendicular 48.1 inches; required the area and content in wine gallons, when the depth of the liquor is 48.6 inches? Ans. The area is 33.742, and the content 1639.8612 wine gallons. PROBLEM X. To find the area of a regular polygon, when the side only is given. RULE. By the Pen. Multiply the square of the given side by the number or area standing opposite the name of the polygon, in the following Table, and the product will be the area in square inches, ale gallons, wine gallons, or malt bushels respectively. Sq. Inches. Ale Gall. WineGall. Malt Bush. 1. The side of a regular pentagon is 38 inches, what is the area in ale and wine gallons, and malt bushels? Inches. 304 114 1444 square of the side. .006101 multiplier. 1444 1444 8.809844 ale gallons. Also, 1444×.007448=10.754912 wine gallons; and 1444x.0008=1.1552 malt bushels. By the Sliding Rule. As 1 on D, is to the multiplier on C; so is the side of the polygon on D, to the area on C. 2. Required the area and content of a hexagonal guiletun, in ale gallons; the side of the vessel measuring 21.7 and its depth 41.6 inches. Ans. The area is 4.3383, and the content 180.47328 ale gallons. 3. What is the area, in wine gallons, of a heptagonal wine-vat whose side measures 24.3 inches? Ans. 9.2889 wine gallons. 4. If the side of an octagonal cistern measures 31.2 inches, what is the area in malt bushels ? Ans. 2.1853 malt bushels. PROBLEM XI. Given the diameter of a circle to find the circumference ; or, the circumference to find the diameter. RULE I. As 7 is to 22, so is the diameter to the circumference: or, as 22 is to 7, so is the circumference to the diameter. RULE II. Multiply the diameter by 3.1416, and the product will be the circumference: or, divide the circumference by 3.1416, and the quotient will be the diameter. Note. The second Rule is more correct than the first, and is therefore generally preferred.. EXAMPLES. 1. If the diameter AB of a circle be 84.5 inches, what is the circumference? As 7: 22 84.5; 265.571 inches, the circumference required. By Rule II. Here 3.1416x 84.5=265.4652 inches, the circumference required. By the Sliding Rule. RULE I. As 7 on A, is to 22 on B; so is the diameter on A, to the circumference on B: and as 22 on B, is to 7 on A; so is the circumference on B, to the diameter on A. On A. As 7 required. On B. On A. On B. : 22 :: 84.5: 265.57 inches, the circumference RULE II. As 1 on A, is to 3.1416 on B; so is the diameter on A, to the circumference on B; and vice versa. cumference required. 2. The circumference of a circle is 265.4652 inches, what is the diameter? 3. The diameter of a cylindrical what is its circumference? Ans. 84.5 inches. vessel is 36.9 inches, Ans. 115.925 inches. 4. The circumference of a globe or sphere is 86.7 inches, what is its diameter ? Ans. 27.5974 inches. Note. There is no figure that affords a greater variety of useful properties than the circle; nor is there any that contains so large an area within the same perimeter. The ratio of the diameter of a circle to its circumference has never yet been exactly determined; although this celebrated Problem, called the squaring of the circle, has engaged the attention and exercised the abilities of the ablest Mathematicians, both ancient and modern. But though the relation between the diameter and circumference cannot be exactly defined in known numbers; yet approximating ratios have been determined sufficiently correct for practical purposes. Archimedes, a native of Syracuse, who flourished about 200 years before the Christian æra, after attempting in vain to determine the true ratio of the diameter to the circumference, found it to be nearly as 7 to 22. The proportion given by Vieta, a Frenchman, and Metius, a Dutchman, about the end of the 16th century, is as 113 to 355, which is rather more accurate than the former; and is a very commodious ratio, for being reduced into decimals, it agrees with the truth to the sixth figure inclusively. The first, however, who ascertained this ratio to any great degree of exactness, was Ludolph Van Ceulen, a Dutchman. He found that if the diameter of a circle be 1, the circumference will be 3.141592653589793238462643383279502884 nearly, which is true to 36 places of decimals. This was thought so extraordinary a performance, that the numbers were cut on his tomb-stone, in St. Peter's Church-yard, at Leyden. Since the invention of fluxions, by the illustrious Sir Isaac Newton, the squaring of the circle has become more easy; and the late ingenious Mr. Abraham Sharp, of Little Horton, near Bradford, in York. shire, has not only confirmed Ceulen's ratio, but extended it to 72` places of decimals. Mr. John Machin, Professor of Astronomy in Gresham College, London, has also given us a quadrature of the circle, which is true to 100 places of figures; and even this has been extended, by the French Mathematicians, to 128. PROBLEM XII. To find the length of the arc of a circle. RULE I. By the Pen. From 8 times the chord of half the arc subtract the |