= whole circle 1386 sq. ft. The area of the base of the monument is 346 sq. ft. The difference, or the area of the walk, is 1039 sq. ft., or 115 sq. yd. 3. Answer: $27.10. The area of the floor is 616 sq. ft. Multiply by 2 for the thickness, and you have 1232 feet of lumber. 4. Answer: 560 yd. One mile by 34 gives 560. 5. Answer: (a) (b) = 1760 yd. This divided 308 sq. in., or of circle. 224 sq. in. The area of a quarter of the 2. Answer: 17,760 lath. There are 4320 sq. ft. in the walls and ends; 4800 sq. ft. in the ceilings; and 1536 sq. ft. in the gables. 3. Answer: $61.18. The surface to be plastered is practically 437 yd. 4. Answer: 6555 lath. 5. Answer: $1523.52. Each hall contains 2070 square feet is 57,132. Take 437 yd. Each office contains 828 sq. ft. sq. ft. The entire number of Lesson No. 17 1. Answer: 6380 yards of wire. 10 acres = 48,400 sq. yd. Length of field = 48,400 ÷ 88, or 550 yd. 2. Answer: 21,121 ties. 12 miles = 21,120 yd. Add one tie for end. 3. Answer: 4.5094 miles. Find the number of square yards of walk ($ 25,000 ÷ $ 2.10), and then divide by the width of the walk in yards, and you have the length in yards. 4. Answer: $31,600. The average length is 869 yd., and the perpendicular width is 22 yd. 5. Answer: $1530.90. Lesson No. 18 1. Answer: 174,240 stones. 2. Answer: 14,000 ft. There are 28 ft. of lumber in each scantling. 3. Answer: $7040, considering 23,466 loads. An estimater would call this 23,467 loads, or perhaps 23,500 loads. 4. Answer: $6400. 5. Answer: 132,000 loads. Find the area of the crosssection in yards, and multiply by the length (440) in yards. Lesson No. 19 1. Answer: $396. The fence is 2640 ft. in circumference, that is, in length, and 10 ft. high. 2. Answer: $326.70. 3. Answer: 1850 ft. 4. Answer: 14 in. by 71⁄2 in. by 9 in. It is possible to make a box of a different, but not so convenient a shape, as, for example, 10 in. by 10 in. by 9 in. 5. Answer: $79.20. Each of the smaller circles is onequarter of the larger circle in area, and one-half in circumference. Lesson No. 20 1. Answer: 79 perches. 2. Answer: $62.92. Find the circumference in yards, multiply by the depth in yards, and this by 2 and 24, and you will have the number of square yards of surface in the sides. Add to this the number of square yards inside and outside in the bottoms, and multiply by the price. 3. Answer: 146,455 tons (nearly). Find the area of the circle in feet, multiply by 14, and this by 57, and you will have the number of pounds. Divide by 2000. 4. Answer: $19.53. 5. Answer: 1232 ft. 6. Answer: 3168 pickets. 7. Answer: $4,428,000. 8. Answer: $45. 9. Answer: 4400 loads. 10. Answer: $27.50. The area of the painted section of a post equals the circumference (44 in.) multiplied by the height, or (3 × 71) ft. Lesson No. 21 1. Answer: 20,736 gal. 2. Answer: 144,375 lb. 3. Answer: (a) 115,500 lb.; (b) 13,824 gal. 5. Answer: 1176 gal. in each tank, or 3528 gal. This problem is a difficult one, involving the finding of the volume of what is known in mathematics as the frustum of a cone. The vats are cone shaped, and half their volume is filled. The rule is as follows: To the sum of the areas of the ends add the square root of their product, and multiply the sum thus found by one-third the height. Lesson No. 22 1. Answer: 1240 in. multiplying by 144; by 11. Reduce 2170 ft. to inches by then divide by (14 × 12), and 2. Answer: $118.27. The total length of the outside walk if in a straight line would be 624 ft., and of the cross walks the length would be 296 ft. 3. Answer: 640 ft. Reduce to cubic inches, divide by 144, because there are 144 cu. in. in a foot of lumber, and deduct one-sixth. 4. Answer: $43.56. 5. Answer: $3.56. 6. Answer: 4992 ft. 7. Answer: 15 ft. 8. Answer: 111 ft. 9. Answer: $112. 10. Answer: $38.88. The veranda is 18 ft. wide and 54 ft. long. Lesson No. 23 1. Answer: 17,920 acres. 2. Answer: 31,360 acres. 3. Answer: 63 ft. by 45 ft. 4. Answer: 3884+ yd. Note the diagram. The lower right-hand quarter represents a square mile, or a piece of land 80 ch. on each side. To find the distance from the center to B, square the two sides, add, and take onehalf of the square root. There are 80 ch. in a mile. 802 + 802 = 12,800. The square root of 12,800 equals about 113 (113.12). One-half of this, or 56 ch., equals the distance from the center to B. The whole distance is 1761⁄2 ch. There are 22 yd. in a chain. The answer, therefore, is 3883 yd., or, allowing for the omitted decimal, 3884 yd. NOTE. For assistance in the solution of this question the student is referred to Lesson I. in Part II. of this work, "Mensuration for Beginners." 5. Answer: 20 mi. to the inch. A map 4 in. by 2 in. would on this scale represent a section of country 90 mi. by 40 mi. Lesson No. 25 1. Answer: $33.84. The walls and partition equal 256 ft. for each floor, or 6144 in. for the two floors. Divide by 16, and we get 384 studding. Deduct 4 for each floor for the four points shown in the figure. If partitions cross at B, for instance, the same studding at the crossing point will answer for each partition. 384-8=376. 376 × 14 ÷ 700 and multiplied by the wages = $33.84. 2. Answer: $17.76. There will be 148 joists needed, each 16 ft. in length. 3. Answer: $16.80. 5. Answer: $159.84. 4. Answer: 85 days. Lesson No. 26 1. Answer: 11,232 bricks. Find the number of cubic inches in the pile and divide this by the number of cubic inches in a brick. 2. Answer: 16,212 bricks. There are 2316 sq. ft. in the wall, and each square foot required 7 bricks. 3. Answer: $447.55. Bricks, $288; lime, $3.75; sand, $1.80; labor, 40 d. at $3.85, $154. 4. Answer: $ 4698.04. There will be 519,120 bricks required. These will cost $2803.25. The lime will cost $259.56. The sand will cost $181.69. The labor will cost |