EXAMPLES. 6. Required the area of a square whose side is 5 feet 9 Inches? Ans. 35ft. I in.9". 7. Required the area of a rectangle, whose sides are 56 feet, and 18 feet 6 inches ? Ans. 1036 feet. 8. Required the area of the rhombus or rhomboides, whose length is 12 feet 6 inches, and its height 9 feet 3 inches ? Ans. 115 ft. 7 in. 61. II. To find the area of a triangle. ROLE-Multiply the base by the perpendicular height, and half the product will be the area. EXAMPLES. 9. Required the area of the triangle whose base is 10 feet 9 inches, and height 7 feet 3 inches. Ans. 38 ft. 11 in. 7". 10. What is the area of a triangle, whose base-is 18 feet 4 inches, and height 11 feet 10 inches? Ans. 108 ft. 5 in. 8'. III. To find the area of a trapezium. . RULE-Multiply tħe diagonal by the sum of the two perpendiculars falling upon it from the opposite angles, and half the product will be the area. EXAMPLES. 11. What is the area of a trapezium whose diagonal is 108 feet 6 inches, and the perpendiculars 56 feet 3 inches, and 60 feet 9 inches ? Ans. 6347 feet, 3 inches. 12. Required the area of a trapesium whose diagonal is 60 feet, and the perpendiculars 50 fect and 40 feet ? Ans. 2700 feet. IV. To find the area of a trapezoid, or quadrangle, two of whose opposite sides are parallel. RULE-Multiply the sum of the parallel sides by the perpendicular distance between them, and half the product will be the area. EXAMPLES. 13. Required the area of a trapezoid whose parallel sides are 25 feet 6 inches, and 18 feet 9 inches, and the perpendicular distance 10 feet, 5 inches. Ans. 230 ft 5 in. 7'en 14. How many square feet are in a plank 13 part of the one end, and 15 at the other, the lingth ben. Wked AW. chests mark ches? Ane, 19 feet, I inch, SB. No. 1 OF THE CIRCLE. 1. The diameter of a circle being given to find the circumference ; or, the circumference being given to find the diameter. RULE.. 7:22 : : so is the diameter : to the circumference ; or, 113 : 355 : : so is the diameter : to the circumference. 22 : 7 :: so is the circumference : to the diameter. or, 355 : 113 : : so is the circumference : to the diameter. EXAMPLES. 1. The diameter of a circle is 9 feet : What is the circumference ? Ans. 28.27 feet. 2. If the diameter of a circle be 10 feet: What is the circumference ? Ans. 31.41 feet. 3. If the circumference of a circle be 354, what is the di. ameter? Ans. 112.681. 4. The circumference of the globe is known to be 25000 miles : Required its diameter ? Ans. 7958 nearly. II. To find the area of a circle. 'RULE-Multiply half the circumference by half the diam. eter, and the product will be the area. EXAMPLES 5. What is the area of a circle whose diameter is 42, and circumference 131.946 ?, Ans. 1385.433. 6. What is the area of a circle whose diameter is 10 feet 6 inches, and circumference 31 feet 6 inches? Ans. 82 ft. 8 in. SOLIDS. 1. To find the solidity of a cube. RULE-Multiply the side of the cube by itself, and that product again by the side, and it will give the solidity. EXAMPLES. 1. Required the solidity of á cube whose šides are 15 inches Ans. 1 ft. 11 in. 5'1. 2. The side of a cube is 25 feet 6 inches : required th lidity ? Ans. o 16581 ft. 4. Te Iches II. To find the solidity of a parallelopipedon. RULE-Multiply the length by the breadth, and th 6d. duct again by the depth, or heighth, and it will give high, lidity.. feet EXAMPLES. hoy 3. Requed the solidity of the parallelopipedon, whose 1 length is 8 let, breadth 43 feet, and depth 64 feet? Ans. 243 ft. 20 4. What is the solid content of a block of marble, whose length is 10 feet, its breadth 54 feet, and the depth 3 feet? Ans. 2014 ft. III. To find the solidity of a triangular prism. RULE-Multiply the area of the base into the perpendicular height, and the product will be the solidity. EXAMPLES. 5. Required the solidity of the triangular prism whose length is 10 feet, and either of the equal sides of one of its equilateral ends 2} feet ? Ans. 27.063 ft. 6. Required the solidity of a prism whose length is 18 feet, and one side of the equilateral end 1{ feet? Ans. 17.50859 ft. IV. To find the solidity of a cylinder. Rule-Multiply the area of the base by the perpendicular height, and the product will be the solidity. EXAMPLES. 7. What is the solidity of the cylinder, the diameter of whose base is 30 inches, and the height 50 inches ? Ans. 20.4531 ft. 8. What is the solidity of the cylinder whose height is 20 feet, and the circumference of its base 20 feet also ? Ans. 636.64 V. To find the solidity of a pyramid or cone. Rule-Multiply the area of the base by }, the perpendicular altitude, and the product will be the solidity. EXAMPLES. 9. Required the solidity of a square pyramid, each side of whose base is 30, and the perpendicular height 20 ? Ans. 6000. 10. What is the solidity of a cone, the diameter of whose base is 18 inches, and its altitude 15 feet ? Ans. 8.83575 ft. If the circumference of the base of a cone be 40 feet, he height 50 feet : What is the solidity? Ans, 2120 ft. per to find the solidity of the frustum of a pyrarnid or cone. "E-Add into one sum the area of both ends, and the proportional between them ; multiply-the sum by the 13. licular height, and the product will be the solidity Are 25 EXAMPLES. licular hat is the solidity of the frustum of a quare pyrane side of the greater end being 18 inches that of the ek , sser end 15 inches, and the height 60 inches ? V 1680 izche Nil Anss 13. What is the solidity of the frustum of the cone, the diameter of the greater end being feet, that of the lesser end 2 feet, and the altitude 9 feet? Ans. 65.9736. 14. What is the solidity of the frustum of a cone, the cir. cumference of the greater end being 40, that of the lesser end 20, and the length or height 50 ? Ans. 3713.64. VII. To find the solidity of a sphere or globe. RULE--Multiply the cube of the diameter by .5236, and the product will be the solidity. EXAMPLES. 15. Required the solidity of a sphere of 10 inches diame. ter. Ans. 523.6. 16. How many solid miles are in the terraqueous globe, its 7958 ? 303 ARTIFICERS WORK. EXAMPLES. Ans. 2. There is a house with 3 tier of windows, 4 in a tier, the height of the first tier is 6 feet 6 inches, the second 5 feet 3 inches, and the third 4 feet 9 inches, the breadth of each window is 3 feet 9 inches : what will the glazing come to at 2s. 8d. per foot ? Ans, 331. 3. What is the price of a marble slab whose length is 65 feet, and breadth 3 feet, at 81 per foot ? Ans. $21.12 11. Painting, Plaistering, Paving, &c. is ineasured by the yard square. Rule-Divide the square feet by 9 the quotient will be the number of square yards. EXAMPLES. 4. What will the paving of a street come to at 12 cents per yard; the length of the street being 1764_feet, and the breadth 56 Get? Ans. $139.116. 5. What is the content of a piece of wainscotting in square yards, that is 9 feet 6 inches in height, and 8 feet 3 inches broad, and what will it come to at 128. per yard. Ans. 8 ft. 6', 41!. 6111 and 51. 4s. 6d. 6. There is a room 84 feet round, and 9 feet 6 inches high, in which are three windows, each 6 feet high and 3 feet 5 inches wide, and the fire.place 4 feet by 4 feet: I demand how many yards of paper, half yard wide will hang it? Ans. 160 jde. 6 in. 7. What will the plaistering of a criling, at 21d. per yard come to, supposing the length 343 feet, and the breadth 20 feet? Ans. 61. 145. 2d. III. Flooring, Partitioning, Roofing, Tyling, &c. is measured by the square of 100 feet. RULE-Divide the area in square feet by 100, the quotient will be the number of squares.“. EXAMPLES. 8. In 120 feet 6 inches in length, and 12 feet 9 inches in height of partitioning, how many squares ? Ans. 15 sq. 36 ft. 4 in. 611. 9. What difference is there between a floor 28 feet long by 20 broad, and two others that measure 14 feet a piece by 10; and what do all three come to at $114 per square ? Ans. 280 ft. diff. and 894.50. 10. How many planks with floor a room 604 feet long, and 33 wide, supposing the plank 15 feet long, and 15 inches wide ? Ans. 100 early. 11. Suppose a house measures, within the walls, 64 feet in length, and 36 feet in breadth, and to be a true pitch, what will it come to roofing at 25s. the square ? Ans. 431. 45. IV. Bricklayers Work is measured by the rod, of 272. square feet. This work is always computed at the rate of a brick and a of thick, and if the thickness of the wall is more or less, it must be reduced to that thickness, by the following RULEL-Multiply the area of the wall in feet, by the number of half bricks in the thickness; divide the producto by 816), and the quotient will be the rods : or, 2. Multiply the area of the wall by the half bricks in the thickness of the wall; the product divided by 3, gives the area in feet, which divide by 272, the quotient will be the rods required. Note The fraction : in rule 1, or } in rule 2, commonly Tejected in practise. EXAMPLES. 12. How many square rods are there in a wall 523 feet long, 12 feet 9 inches high, 21 bricks thick. Ans. 4r. 27ft. 7in. 13. A gentleman built a wall round his garden, which is 840 feet, and 9 feet high, and 2) brick thick; how many rods doth it contain and what will it come to at $30 per rod ? Ans. 46 r. 88 ft. and $1389.7649. 14. If each side wall of a building be 45 feet long on the outside, each end wall 15 feet broad on the inside, the height of the building 20 feet and the gable at each end of the wall 6 feet high, the whole being 2 bricks thick : required the content in standard rods? Ans. 12.1761. 15. The end wall of a house is 24) feet in breadth, and 40 fret to the roof; of which is 2 bricks thick, 1 more 13 brick thick, and the rest 1 brick thick: now the gable rises 38 |