1. The side of a regular pentagon is 12 feet; what is the area? = Thus, 12 x 12 144; 1·720477 x 144 247-748688 feet, × = the area. 2. The side of a regular hexagon is 24 feet; what is its area? Ans. 1496-4917 feet. 3. What is the area of a regular nonagon whose side is 36 inches? Ans. 8011-6439 inches. 4. The side of a pentagon is 25 feet; required the area. 1075-298. Ans. 5. What is the area of a heptagon whose side is 16 feet? 930-28. Ans. 6. The side of a hexagon is 24 feet; required the area. 1496 49. Ans. 7. How many square yards in an octagon whose side is 12 feet 6 inches? Ans. 83-8268+ sq. yards. 8. How many pieces each 4 inches square, may be cut from a decagon whose side is 12 inches? Ans. 69-248 pieces. 9. Each side of a duodecagon is 9 inches; how many square feet does it contain? Ans. 62.98. 10. Required the area of a pentagon whose side is 15? Ans. 387-107325, area. 11. Required the area of an octagon whose side is 16. 1236-0773. Ans. 8011-6439. Ans. 12. Required the area of a nonagon whose side is 36.. PROBLEM 26. When the area of any regular polygon is given, to find the side. RULE. Divide the area by the number in the table corresponding with the figure, and the square root of the quotient will be the length of the side. 1. The area of a regular pentagon is 4 acres; how perches are contained in the side? many Thus, 160 × 4 acres = 640 perches; then 6401-720477 371.989819-2870 perches, the length of the side. = 2. Required the length of the side of a hexagon containing 1 acre. Ans. 7-8475 perches. 3. The area of a regular heptagon is 1356-6 yards; what is the side? Ans. 19-3214 yards. 4. The area of a regular octagon is 1642.7 perches; required the length of the side. PROBLEM 27.-IRREGULAR FIGURES. To find the area of a long and irregular figure, bounded on one side by a straight line. RULE. Divide the right line or base into any number of equal parts, and measure the breadth of the figure at the points of division, and also at the extremities of the base. Add together the intermediate breadths, and half the sum of he extreme ones. Multiply this sum by the base line, and divide the product by the number of equal parts of the base. = a A B d e 1. The breadths of an irregular figure at five equidistant places, A B C D and E, being 8-20 chains, 7.40 chains, 9-20 chains, 10-20 chains, and 8.60 chains, and the whole length 40 chains; required the area. Thus, 8.208.60 16·80; 16.80 ÷ 2 = 8.40, mean of the extremes; then 8.40 +7.40 +9.40 +10.2035.20, sum; then 35.20 x 40 chains length 14084352, square chains. 2. The length of an irregular field is 39 rods, and its breadths, at 5 equidistant places, are 2-4, 2·6, 2·05, 3.65, 3.6 rods respectively; what is the area? Ans. 111.54 rods. 3. The length of an irregular field is 50 yards, and its breadths, at 7 equidistant points, are 5.5, 6·2, 7·3, 6, 7.5, 7, and 8.8 yards; what is the area? Ans. 342.916 sq. yards. PROBLEM 28.-THE CIRCLE. To find the circumference of a circle when the diameter is given, or the diameter when the circumference is given. RULE.-Multiply the diameter by 3.1416, and the product will be the circumference; or divide the circumference by 3.1416, or multiply the circumference by 31831, and the result will be the diameter. NOTE.-The numbers 3.14159, or 3-1416, 7854, 5236, &c., should be made perfectly familiar, in consequence of their frequent use in the solution of superficies and solids in men suration. The first expresses the ratio of the circumference of a circle to the diameter. The second, the ratio of the area of a circle to the square of the diameter: and the third, the ratio of the solidity of a sphere to the cube of the diameter. 1. What is the circumference of the circle A B C D, whose diameter, A B, is 7 feet? Thus, 3-1416 × 7 = 21.9912 ft. the circumference. A 2. What is the diameter of a circle whose circumference is 100 yards? Thus, 1003141631-831 yards, the diameter. Thus, 3·1416 ÷ 4 = Or, 100 × 31831 = 31.831 yards, as before. D B 3. If the circumference of a circle be 354, what is the diameter? Ans. 112.681. 4. If the diameter of a circle be 17, what is the circumference? Ans. 53 4072. 5. If the circumference of the earth be 25000 miles, what is the diameter ? 7958 miles, nearly. Ans. 6. What is the circumference of a wheel whose diameter is 5 feet 2 inches? Ans. 16.2316. 7. What is the diameter of a circle whose circumference is 11652-1944 feet? Ans. 3709 feet. PROBLEM 29. To find the area of a circle. RULE.-Multiply the square of the diameter by 7854; or, the square of the circumference by 07958; and the product in either case will be the area. 1. How many square feet are there in a circle whose diameter is 6.5 feet? = Thus, 6.52 42.25 × 7854 = 33·18315 square feet. Ans. 2. The circumference of a circle is 11 yards; required the area. Thus, 112 = 132 × 0795810-50456 square yards. Ans. 3. How many square feet are contained in a circle whose diameter is 4 feet 3 inches? Ans. 14.1862875 sq. feet. 4. What is the value of a circular garden whose diameter is 6 rods, at the rate of 8 cents per square foot? Ans. D. 615-81-6432. 5. The diameter of a circle is 16 chains; how many acres does it contain? Ans. 20 a. 0 r. 16-9984 po. PROBLEM 30. The area of a circle being given, to find the diameter or circumference. RULE. Divide the area by 7854, and the square root of the quotient will be the diameter. Or divide the area by 07958, and the square root of the quotient will be the circumference. 1. The area of a circle is 5 acres, 3 roods, 26 poles; what is the diameter ? = Thus, 5 a. 3 r. 26 po. 946 po.7854: =1204:4879271 = 34-7056 poles, the diameter. = 2. The area of a circle being 2 acres, 3 roods, 12 poles; required the circumference. Thus, 2 a. 3 r. 12 po. 452 poles. = Then, 452079585679.6975-3637 poles, the circumference. 3. The area of a circular garden being 1 acre, what is the length of a wall which will enclose it? Ans. 44-839 poles. 4. The area of a circle being 2 acres, 3 roods, 12 poles, what is the circumference? Ans. 75-3657 poles. 5. The area of a circle being 9 acres, 3 roods, 22 poles, what is the diameter ? PROBLEM 31. To find the area of a circular ring, or the two concentric circles. space included between RULE. Find the areas of the two circles separately. Then the difference of these areas will be the area of the ring. Or multiply the sum of the diameters by their difference, and this product again by 7854, and it will give the area required. And, 314-16 113.0976 = 201-0624 yards, area of the ring. 2. The diameters of two concentric circles are 8 and 12 yards; what is the area of the ring contained between their circumferences? Ans. 62.832 yards. 3. If the diameters are 20 and 15, what will be the area included between the circumferences? 4. Two diameters are 21.75 and 9.5; circular ring. Ans. 137-445. required the area of the Ans. 300-6609. 5. If the two diameters are 4 and 6, what is the area of the Ans. 15.708. ring? PROBLEM 32. The diameter or circumference of a circle being given, to find the side of an equivalent square. RULE.-Multiply the diameter by 8862, or the circumference by 2821; the product in either case will be the side of an equivalent square. 1. The diameter of a circle is 300 yards; what is the side of a square of equal area? Thus, 300 x 8862 = 265.86. Ans. 2. The circumference of a circle is 316 yards; what is the side of a square of equal area? Ans. 316 x 282189.1436 yards. 3. A man has a circular meadow, of which the diameter is 875 yards, and wishes to exchange it for a square one of equal size; what must be the side of the square? Ans. 775-425. 4. The diameter of a circle is 100; what is the side of a square of an equal area? Ans. 88.62. 5. The circumference of a circular walk is 64 rods; what is the side of a square containing the same area? Ans. 18.0544 rods. PROBLEM 33. The diameter or circumference of a circle being given, to find the side of the inscribed square. RULE.-Multiply the diameter by 7071, or the circumference by 2251, and the product in either case will be the side of the inscribed square. |