29. How many times can 3 yd. 1 ft. 7 in. be subtracted in succession from 115 yd. 2 ft. 11 in., and what I will be the last remainder? 30. A bar of metal weighing 100 oz. 16 dwt. is made into coins, each weighing 1 oz. 8 dwt.; how many coins. are made from the bar? 31. Simplify 1 of 6-23 31-18 {2}-# of 4층 32. A surveyor measured some ground and found it to be 10 ch. long and 4 ch. broad; how many A. were there? 33. What is the smallest number of exact acres that can have the form of a square? 34. What decimal of 1 mi. is 119 yd. 2 ft. 4 in.? 35. Find the value of 2 lb. 6 oz. 10 dwt. 12 gr. of gold at $216 per lb. 36. Find 1052; 48 x 331; 850 ÷ 163. 37. Express lb. 1 as the decimal of 1 lb. Avoir. 38. Having given that a meter is 39.37 in., prove that the difference between 5 mi. and 8Km is nearly 51 yd. 41. Express .88125 cwt. in lb. and oz. 42. Express 15 yd. 2 ft. 8 in. as the decimal of a mi. 43. Reduce 11.2765625 lb. to lb., oz., pwt., and gr. 44. Find to the nearest cent $48.96 × 72.8967. 45. Reduce 1000 sq. yd. to qm. 46. Reduce 10001 to pt. 47. Express .136 x 7.3.43 as a decimal. 48. Find the value of 43 sq. rd. 244 sq. yd. of building land at $1815 per acre. 49. Find the greatest length of which both 1 mi. 4 fur. 16 rd. 2 yd. and 1 mi. 1 fur. 10 rd. 2 yd. are multiples. 54. Find the annual cost of repairing a road 9 mi. 120 rd. 177 yd. long at $88 per mi. 55. A vessel steams 18 knots an hour; to how many statute miles is this equivalent ? 56. If a ccm of iron weighs 7.7888, what will be the weight of a cu. ft.? 57. How many pieces each .17 in. long can be cut from a wire 21.09 in. long; and how long will be the piece left over? 58. Add .5125 of a yd., .62734 of a rd., and .018325 of a fur.; subtract the result from .0049 of a mi., and express the answer in yd., also in dm. 59. Find 4900546043.21156004. 60. What is the least number which when divided by 15 leaves a remainder 3, when 186. A plane figure [Art. 149] bounded by four straight lines, and whose four angles are equal, is called a Rectangle. An equilateral rectangle is a Square. The amount of surface included within the bounding lines of a figure is called its Area, and the area is measured by some square unit, one sq. in., one sq. yd., or one qm, etc. 187. To find the Area of a Rectangle. Let ABCD be the rectangle whose area is required. Suppose, for example, that AB is 4 in. and that AD is 3 in. Divide AB into four equal parts and AD into three equal parts, and draw lines parallel to the sides as in the figure on the left. Then the rectangle is divided into squares each of which is a sq. in.; and the number of these squares is clearly the product of the number of in. in AB by the number of in. in AD. The above reasoning applies to all cases, both the length and the breadth of the rectangle being an integral number of in. Now suppose, for example, that in the figure on the right AB is in., and that AD is in. Let AEFG be one sq. in. Divide AE into two equal parts, and AG into five equal parts, and GD into two equal parts. Then the subdivisions of AB will be all equal, as also those of AD. Hence, if lines be drawn as in the figure, ABCD will be divided into 3 × 7 equal rectangles, such that the square inch AEFG will 3 x 7 contain 2 x 5 of these rectangles. Hence AB will contain square inches; that is, (×) square inches. 2 x 5 From the above it follows that the number of square inches (or square feet, etc.) in a rectangle is equal to the product of the number of inches (or feet, etc.) in the length by the number of inches (or feet, etc.) in the breadth. It should be noticed that the length and breadth must both be expressed in terms of the same unit. For example, the area of a rectangle whose length is 2 ft. and breadth 6 in. is (2×) sq. ft, or (24 × 6) sq. in. The above rule for finding the area of a rectangle is often expressed shortly by the statement that area = length x breadth. 188. Now that we find the area of a rectangle, we can see that the relations between the different units given in the Table for Square Measure, on page 149, follow at once from the relations between the corresponding units in linear measure. For, since 12 in. make 1 ft., (12 × 12) sq. in. make 1 sq. ft. Since 5 yd. make 1 rd., (5 × 5) sq. yd. make 1 sq. rd. Again, 22 yd. make 1 ch., therefore (22 × 22) sq. yd. yd. make 1 sq. ch. Thus, 4840 sq. yd. = 10 sq. ch. = 1 A. Also, 1 sq. mi. = = 640 A. = 484 sq. (1760 × 1760) sq. yd. 1760 × 1760 ÷ 4840 A. Ex. Find the acreage of a rectangular field whose length is 132 yd. and whose breadth is 38 yd. 189. If the area of a rectangle be known, and also the length, the breadth can be at once found. For example, to find the breadth of a rectangle whose length is 15 ft. and whose area is 200 sq. ft. Since the product of the number of ft. in the breadth by the number of ft. in the length is equal to the number of sq. ft. in the area, we have |