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FIG. 3

FIG. 4

MENSURATION MENSURATION OF SURFACES

DEFINITIONS 1. Mensuration treats of the measurement of lines,

angles, surfaces, and solids.

2. A line expresses length or distance without breadth or thickness.

3. A straight line, Fig. 1, is one that does not change its direction throughout its FIG. I

whole length.

4. A curved line, Fig. 2 , changes direction at every point. FIG. 2

5. Parallel lines, Fig. 3, are those that are equally distant from each other at all points.

6. A line is perpendicular to another, Fig. 4, when it meets that line so as not to incline toward it on either side.

7. A vertical line, Fig. 5, is one that points toward the center of the earth; it is also known as a plumb - line.

8. A horizontal line, Fig. 5, is one that makes a right angle with a vertical line.

*ontat FIG. 5

9. A surface is that which has length and breadth without thickness.

For notice of copyright, see page immediately following the title page

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10. A plane surface is one in which if two points be taken, a straight line connecting them will be wholly in the surface.

11. A curved surface is one no part of which is plane.

12. An angle is the inclination of two lines one to the other and is measured in degrees 4 ° ) .

13. A right angle, Figs. 4 and 5, is formed by two lines that are perpendicular to each other, and is 90° in magnitude.

14. An acute angle, Fig. 6, is an angle of less than 90°.

FIG. 6 FIG. 7

15. . An obtuse angle, Fig. 7, is an angle of more than 90°.

16. A plane figure is any part of a plane surface bounded by straight or curved lines.

17. The area of a plane figure is its surface contents.

TRIANGLES

18. A triangle is a plane figure bounded by three straight lines and having three angles.

19. The altitude of a triangle is the distance from its apex to base measured perpendicularly to the base. In the tri-angle a b c, Fig. 8, the dotted line b d repre-sents the altitude of the triangle, while

c the base of the triangle is represented by the line a c.

20. An equilateral triangle, Fig. 9, is one that has all of its sides equal and each of its angles of 60° magnitude,

FIG. 9 FIG. 10

FIG. 11 FIG-. 12

21. An isosceles triangle, Fig. 10, is one that has two equal sides and two equal angles.

22. A scalene triangle, Fig. 11, is one that has all of its sides and all of its angles unequal.

23. A right tri-angle, Fig. 12, is one that has one angle a right angle.

24. To find the area of a triangle:

Rule.—Afultiftly the base by the altitude and divide the product by 2.

EXAMPLE.—The base of a triangle is 14 inches in length and the altitude is 12 inches; what is the area?

SOLUTION.— 2

14 in. X 12 in. 84 sq. in Ans.

NOTE.--In the above example it will be noticed that by multiplying inches by inches the product obtained is square inches; similarly. feet multiplied by feet or rods by rods equals square feet or square rods, etc. It must be remembered that only like numbers can be multiplied together and that feet can never be multiplied by inches, nor rods by feet; consequently, in all problems dealing with mensuration, all dimensions must be reduced to like terms before multiplying.

EXAMPLES FOR PRACTICE

Find the areas of the following triangles, the length of the base and altitude being respectively:

(a) 10 inches and 8 inches. (a) 40 sq. in. (b) 34 feet and 42 feet. Ans.

(bi 714 sq. ft. (c) 114 inches and 212 inches. (c) 12,084 sq. in. (d) 34 miles and 18 miles. (d) 306 sq. mi.

25. To find the area of a triangle when the alti-

tude is unknown but the length of each side is given:

Rule.—From one-half the sum of the Three sides, subtract each of the sides seParately and multifily the remainders together

and by one-half the sum of the sides; the square root of the

4 MEN SURATION 6

EXAMPLE.—What is the area of a triangle the sides of which are, respectively, 16, 16, and 12 feet in length?

SOLUTION. — 16 + 16 + 12 ----- 44; 44 ÷ 2 . 22; 22 --16 = 6; 22 — 16 = 6; 22 — 12 = 10;

6 X 6 X 10 X 22 = 7,920; li7,920 = 88.99 sq. ft. Ans.


1. Find the area of a triangle the sides of which are, respectively, 32, 32, and 24 inches in length. Ans. 355.977 sq. in.

2. Find the area of a triangle the sides of which are, respectively, 18, 24, and 22 feet in length. Ans. 189.314 sq. ft.

QUADRILATERALS

26. A quadrilateral is a plane figure bounded by four straight lines.

27. A parallelogram is a quadrilateral the opposite sides of which are parallel.

FIG. 13

FIG. 14

ing all of its angles right angles length.

30. A rhomboid, Fig. 15, none of its angles right angles.

28. A rectangle, Fig. 13, is a parallelogram having all of its angles right angles.

29. A square, Fig. 14, is a parallelogram hay-

and all of its sides of equal

is a parallelogram having

FIG. 15 FIG. 16

31. A rhombus, Fig. 16, is a parallelogram having all th but none of its angles right

FIG, 17

e.2 -AL

32. The altitude of a parallelogram is the distance between two opposite sides measured perpendicularly, as indicated by the dotted lines in Figs. 15 and 16.

33. To find the area of a parallelogram:

Rule. —Multiply the altitude by the base and the product will be the area.

EXAMPLE.—Find the area of a parallelogram the base of which is 345 inches and the altitude 423 inches.

SOLUTION.— 423 in. X 345 in. = 145,935 sq. in. Ans.

EXAMPLES FOR PRACTICE Find the areas of the following parallelograms, the lengths of the

bases and altitudes being respectively:

(a) 145 inches and 136 inches. (a) 19,720 sq. in. (b) 2,034 feet and 23 feet.

Ans . (b) 46,782 sq. ft.

(c) 135 rods and 4+ rods. (c) 567 sq. rd. (d) 39 feet and 141 feet. (d) 559 sq. ft.

34. A trapezoid, Fig. 17, is a quadrilateral having only two of its sides parallel.

35. The altitude of a trapezoid is always measured perpendicularly between the parallel sides as shown by the dotted line in Fig. 17.

36. To find the area of a trapezoid:

Rule. — Multifily one-half the sum of the fiarallel sides by the altitude.

EXAMPLE.—The parallel sides of a trapezoid are, respectively, 12 and 28 feet in length, and the altitude is 30 feet; what is the area of

the figure?

SOLUTION.— 12 ft. + 28 ft. = 40 ft.; 40 ft. 4- 2 == 20 ft. 20 ft. X 30 ft. = 600 sq. ft. Ans.


1. What is the area of a trapezoid the parallel sides of which are, respectively, 54 and 78 feet in length, and the altitude of which is

4 224 s • ft.

b MtdiN 6

2. What is the area of a trapezoid the parallel sides of which are, respectively, 8 and 18 inches in length, and the altitude of which is 23 inches? Ans. 299 sq. in.

37. A trapezium, Fig. 18, is a quadrilateral that has no two sides parallel.

38. A line joining two oppo-site corners of a quadrilateral, as for instance the line a b, Fig. 18, is known as a diagonal.

39. To find the area of a trapezium:

FIG. 18 two triangles by means of a diagonal;

the sum of the areas of these triangles equals the area of the trapeziunz.

EXAMPLE.—What is the area of a trapezium whose diagonal is 43 inches long, the length of the perpendicular lines dropped on. the diagonal from the opposite corners being 22 and 26 inches, respectively?

NOTE. — The perpendicular lines drawn from opposite corners of a quadrilateral to its diagonal constitute the altitudes of the two triangle into which the diagonal divides the quadrilateral. Thus, in Fig. 18, the line fd represents the altitude of the triangle a db, and the line e c the altitude of the triangle a cb.

SOLUTION.— 43 in. X 22 in. = 946 sq. in.; 946 sq. in. 2 = 473 sq. in., area of one triangle; 43 in. X 26 in. = 1,118 sq. in.; 1,118 sq. in.

2 = 559 sq. in., area of other triangle. 473 sq. in. + 559 sq. in. = 1,032 sq. in., area of trf--peziurn. Ans.


1. The diagonal of a trapezium is 26 feet in length and the per-pendiculars from the opposite corners to the diagonal are 8 and 14 feet, respectively, in length; what is the area? Ans. 286 sq. ft.

2. The diagonal of a trapezium is 78 inches and the perpendicu-ctivel in length; what is the area of the

the figure into

POLYGONS

40. A polygon is a plane figure bounded by straight lines. The term is usually applied to a figure having more than four sides. The bounding lines are called the sides, and the sum of the lengths of all the sides is called the per-imeter of the polygon.

41. A regular polygon is one in which all the sides and all the angles are equal.

42. A polygon of five sides is called a pentagon; one of six sides, a hexagon; one of seven sides, a heptagon, etc. Regular polygons having from five to twelve sides are shown in Fig. 19.

00000 Pentagon Hexagon Heptagon Octagon Decagon Dodecagon

FIG. 10

43. To find the area of a regular polygon:

nule.—lifultifily the perimeter by one-half the length of The PerPendicular from its center to one a iis sides.

EXAMPLE.—The perimeter of a regular polygon is 28 inches in length and the perpendicular distance from its center to one side is 8 inches; what is its area?

SOLUTION.— 8 in. ÷ 2 = 4 in.; 28 in. X 4 in. = 112 sq. in. Ans.


1. If the perimeter of a regular polygon is 78 feet in length and the distance from its center to one side measured perpendicularly is 21 feet, what is its area? Ans. 819 sq. ft.

2. The perimeter of a regular polygon is 112 inches in length and the perpendicular distance from the center to one side is 32 inches;

ea? Ans. 1,792 sq. in.

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THE CIRCLE 44. A circle, Fig. 20, is a plane figure bounded by a

curved line, called the circumference, every portion of which is equally distant from a point within called the center.

FIG. 20 FIG.21 FIG. 22

45. The diameter of a circle is any straight line drawn through its center and terminating at each end in the circumference. Thus the line a b, Fig. 21, is a diameter of the circle.

46. If a circle is divided into halves, each half is called a semi-circle, and each half of the circumference is called a semi -circumference.

47. Any straight line terminating at each end in the circumference but not passing through the center is called

a chord, as for instance the line ' a e, Fig. 22.

48. A straight line drawn from the center to the circumference of a circle (as a c, Fig. 23) is

Fxo. 24 called a radius.

49. An arc of a circle (see a d e, Fig. 24) is any part of its circumference.

50. To find the circumference of a circle:

Rule.--11fullifily the diameter by 3.1416.

NOTE.— 3.1416 is the approximate length of the circumference of a circle whose diameter is 1.

ExAMPLE. --What is the circumference of a circle the diameter of

which is 48 inches?

-- 48 in. X 3.1416 150.7968 in. Ans.

a

FIG. 23

4-.4 4. .1 N. 1 .1 .1 1 AL I

51. To find the diameter of a circle with a given

length of circumference:

Rule.—Divide the circumference by 3 ;1416 .

EXAMPLE.—What is the diameter of a circle the length of circum-ference of which is 8 feet?

SOLUTION.— 8 ft. ÷ 3.1416 = 2.5465 ft. Ans.


Find the circumferences of circles having the following diameters: (a) 24 inches. (a) 75.3984 in. (b) 35 feet.

Ans (b) 109.956 ft.

. (c) 27 rods. (c) 84.8232 rd. (d) 79 yards. (d) 248.1864 yd.

52. To find the area of a circle:

Rule.—MultiPly the square of the diameter by .7854.

NOTE.— .7854 is the area of a circle whose diameter is 1.

EXAMPLE. —What is the area of a circle the diameter of which is 75 inches?

SOLUTION.— 75 in. X 75 in. X .7854 = 4,417.875 sq. in. Ans.


Find the areas of circles of the following diameters: (a) 22 inches. (a) 380.1336 sq. in. (b) 47 yards. (b) 1,734.9486 sq. yd. (c) 768 rods.

Ans. (c) 463,247.7696 sq. rd.

(d) 176 inches. (d) 24,328.5504 sq. in.

53. To find the. length of one side of a square

Inscribed in a given circle:

Rule.—Multiply the diameter of the circle by .707107.

NOTE.—A square is said to he inscribed in a circle when the vertices of all its angles lie in the circumference of the circle. .707107 is the length of the side of a square inscribed in a circle whose diameter is 1.

EXAMPLE.—How thick is the largest square stick that can be inserted

in a pipe 5 inches in diameter on the inside?

SOLUTION.— 5 in. X .707107 = 3.5355 in Ans.

'FIG. 25

111 1.Y.1.12i IN


1. What is the thickness of the largest square iron rod that will just fit in a round hole 12 inches in diameter? Ans. 1.06 in.

2. What is the thickness of a square stick of timber that may be cut from a log 28 inches in diameter? Ans.. 19.7989 in.

54. To find the length of one side of a square

equal in area to a given circle:

Rule.--illultiply the diameter of the circle by .886227.

NOTE.— .886227 is the length of the side of a square equal in area to a circle whose diameter is 1.

EXAMPLE.—What is the length of one side of a square that is equal in area to a circle 15 inches in diameter?

SQLUTION.— 15 in. X .886227 = 13.293 in. Ans.


1. What is the length of one side of a square that will have an area equal to that of a piston 20 inches in diameter? Ans. 17.7245 in.

2. What is the length of one side of a square field that will contain the same number of acres as a circular field 700 feet in diameter?

Ans. 620.3589 ft.

MENSURATION OF SOLIDS

THE PRISM

55. A solid, or solid body, is one that has three dimensions; viz., length. breadth, and thickness.

56. A prism is a solid body the ends of which are formed by two similar plane figures that are equal and parallel to each other, and whose sides are parallelograms. Prisms are triangular, rectangular, square, etc. according to the char- acter of the figure forming the ends.

57. A paralielopipedon, Fig. 25, is a • rism whose bases (ends) are parallelograms.

FIG. 26

' MEIN SURATION

11

58. A cube, Fig. 26, is a prism whose faces and ends are squares. All the faces of a cube are equal.

59. In the case of plane figures, perimeters and areas must be considered. In the case of solids, the areas of their outside surfaces and their contents or volumes must be considered.

60. The base of a prism is either end, and of solids in general, the ends on which they are supposed to rest.

61. To find the surface area of a prism:

Rule.—Mulliftly the length of the perimeter of the base by the altitude, and to the product add the area of both ends.

EXAMPLE.—What is the surface area of a square prism the base of which is 14 inches square and the altitude 95 inches in length?

SOLUTION.— 14 in. X 4 = 56 in., perimeter of base 56 in. X 25 in. = 1.400 sq. in., area of sides 14 in. X 14 in. ,----- 196 sq. in., al-ea of one base 196 sq. in. X 2 = 392 sq. in., a -i.ea of both bases

1,400 sq. in. + 392 sq. in. 1,792 sq. in., total surface area. Ans.


1. Find the surface area of a rectangular prism whose base is 10 inches long and 12 inches wide and which is 13 inches high.

Ans. 812 sq. in.

2. What is the surface area of a square prism the base of which is 175 inches square and whose altitude is 342 inches?

Ans. 300,650 sq. in.

62. To find the contents or volume of a prism

or rectangular box:

Rule.—Multiply the width by the dePth and by the length; or find the area of the base according to the rule Previously given, which when multiplied by the height equals the contents or solidity of the Prism.

ExAmPLE.—What is the capacity of a box 36 inches long, the ends being 14 inches by 28 inches?

TION.— 28 in. X 14 in. x 36 in. = 14,112 cu. in. Ans.

NOTE.—It has been stated that inches multiplied by inches equals square inches or, similarly, yards multiplied by yards equals square yards. Continuing still further, as is necessary in finding the contents, volume, solidity, or capacity of solids; square inches or square yards multiplied by inches or yards equals cubic inches or cubic yards, etc.

From this it will be seen that by multiplying together the two dimensions of a surface, such as a rectangle, the area of the figure will be expressed in square units, while if the three dimensions of a solid, as for instance, a parallelopipedon, are multi-plied together the contents, or solidity, of the solid is expressed in cubical units.


1. What is the capacity of a box the ends of which are 24 inches square and the length of which is 44 inches? Ans. 25,344 cu. in.

2. What is the capacity, in cubic feet, of a freight car 34 feet long, 8 feet wide, and 7 feet high? Ans. 1,904 cu. ft.

3. How many cubic inches are there in a stick of timber 82 feet long, 7 inches wide, and 41- inches thick? Ans. 3,213 cu. in.

SUGGESTION.—Reduce the 8 feet to inches.

4. What is the contents of a tank 13 feet square and 12 feet high? Ans. 2,028 cu. ft.

5. How many cubic feet are there in a cube whose sides are 12 feet in length? Ans. 1,728 cu. ft.

63. THE CYLINDER

A cylinder, Fig. 27, is a body of uniform diameter the ends, or bases, of which are equal parallel circles.

64. To find the surface area of a cylinder:

Rule. —Afulti/ly the circumference of the base by the height of the cylinder and to this firoduct add the FIG, 27

area of the ends.

EXAMPLE.—What is the surface area of a cylinder 6 inches in

diameter and 13 inches high?

S OLUTION.—

62 X .7854 = 28.2744 sq. in., area of one end

28.2744 sq. in. X 2 = 56.5488 sq. in., area of both ends

6 in. X 3.1416 = 18.8496 in., length .of circumference

18.8496 X 13 = 245.0448 sq . in., area of convex surface 245.0448 + 56.5488 = 301.5936 sq . in., total surface area. Ans.

NOTE.—The convex surface of a solid is the curved surface; thus, the area of the of a c Under is its total surface area less the area of the ends.

1.)1(1-1 I 1 VA 1


1. What is the surface area of a cylinder 7 feet long and 21 inches in diameter? Ans. 6,234.5052 sq. in.

2. What is the surface area of a cylinder 21 feet long and 27 inches in diameter? Ans. 22,520.5596 sq. in.

3. What is the surface area of a 21-foot boiler, 6 feet in diameter? Neglect the curvature of the heads. Ans. 152.3904 sq. ft.

65. To find the contents or volume of a cylinder:

Rule.--First find the area of the base according to the rule in Art. 52, and then multifily the area of the base by the altitude.

EXAMPLE.-HOW many cubic feet of water will a cylindrical tank 12 feet in diameter and 14 feet high hold?

SOLUTION.- 122 X .7854 = 113.0976 sq. ft., area of base; 113.0976 sq. ft. X 14 ft. ------ 1,583.3664 cu. ft. Ans.


1. What is the contents of a cylinder 35 inches in diameter and ti feet high? Ans. 51,954.21 cu. in.

2. What is the capacity of a cylindrical tank 8 feet in diameter and 9 feet deep? Ans. 452.3904 cu. ft.

3. If the water in a circular tank 44 inches in diameter is 27 inches deep, how many cubic feet of water are there in the tank?

Ans. 23.7583 cu. ft. 4. How many cubic inches in a cylinder 19 inches long and 11 inches

in diameter? Ans. 1,805.6346 cu. in.

THE PYRAMID AND CONE

66. A pyramid, Fig. 28, is a solid the base of which

is a polygon and the sides of which taper uniformly to a point called the apex, or

vertex.

67. A cone, Fig. 29, is a solid having a circle as a

base and a convex surface uniformly to a point called the apex, or vertex.

FIG. 28 PIG. 29

14

MENSURATION

68. The altitude of a pyramid or cone is the perpe dicular distance from the vertex to the base.

69. To find the contents or volume of a cone pyramid:

Rule.—Multifily the area of the base by one-third the altitua

EXAMPLE.—What is the solid contents of a cone 30 feet high ai 5 feet in diameter at the base?

SOLUTION.— 5 2 X .7854 = 19.635 sq. ft. area of base; I of 30 ft. = 10 ft.

19.635 sq. ft. X 10 ft. = 196.35 cu. ft. Ans.


1. Find the contents of a cone 39 inches high and 12 inches diameter at the base. Ans. 1,470.2688 cu. i

2. Find the contents of a square pyramid 300 feet high and 325 fe square at the base. Ans. 10,562,500 Cu.:

THE FRUSTUM OF A P YltAMID OR CONE

70. If a pyramid be cut by a plane parallel to the bas so as to form two parts, as Fig. 30, the lower part is call( the frustum of the pyramid.

If a cone be cut in a similar ma ner, as in Fig. 31, the lower pa is called the frustum of the con

71. TO find the contents (

volume of the frustum of FIG. 30 pyramid or cone:

Rule.—Find the areas of the two ends of the frustum; multi' them together and extract the square root of the product. the result thus obtained add the two areas and multi:Ply the su by one-third of the altitude.

EXAMPLE. —What is the capacity of a tank shaped like the frusta of a cone, the inside diameter of the top being 10 feet and of t

14 feet, and the depth of the tank being 12 feet?

FIG. 31

FIG. 32

V., ■") t(2-1 li.J.N 16

SOLUTION.— 10 ft. X 10 ft. X .7854 = 78.54 sq. ft., area of small end; 14 ft. X 14 ft. X .7854 = 153.9384 sq. ft., area of large end; 153.9384 X 78.54 = 12,090.321936; Ni12,090.321936 = 109.956 sq. ft.; 109.956 ± 153.9384 + 78.54 = 342.4344 sq. ft.; 12 ft. ÷ 3 = 4 ft.

342.4344 sq. ft. X 4 ft. = 1,369. 7376 cu. ft. Ans.


1. What is the volume of the frustum of a square pyramid the length of which is 30 feet, the top being 10 feet square and the bottom 20 feet square? Ans. 7,000 cu. ft.

2. What is the contents of a round stick of timber 20 feet long, 1 foot in diameter at the larger end, and 2 foot at the small end?

Ans. 9.162 cu. ft.

THE SPHERE

72. A sphere, Fig. 32, is a solid bounded by a continuous convex surface, every part of which is equally distant from a point within called the center.

73. The diameter, or axis, of a sphere is a line passing through its center and ter-minating- at each end at the surface.

74. To find the surface area of a sphere:

Rule.—Square the diameter and multiply the result by 3.1416.

EXAMPLE.—What is the surface area of a sphere 14 inches in diameter?

SOLUTION. — 14 2 X 3.1416 = 14 X 14 X 3.1416 = 615.75 sq. in. Ans.


1. What is the surface area of a sphere 10 inches in diameter? Ans. 314.16 sq. in.

2. The diameter of the earth is about 8,000 miles: what is its approximate surface area? Ans. 201,062,400 sq. mi.

75. To find the contents or volume of a sphere:

the cube of the diameter by .5236..

10 DI P., IN

S

EXAMPLE.--How many cubic inches of ivory in a billiard ball 2 inches in diameter?

SOLUTION.-- 23 X .5236 = 4.1888 cu. in. Ans.

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1. How many cubic inches are there in a sphere 20 inches in diameter? Ans. 4,188.8 cu. in.

2. What is the contents of a spherical float 9 inches in diameter? Ans. 381.7044 cu. in.

MENSURATION OF LUMBER 76. Lumber is measured by board measure, which is

an adaptation of square measure.

77. A board foot is considered as 1 square foot of board 1 inch thick; therefore 1,000 feet of lumber is equal to 1,000 square feet of boards 1 inch thick.

78. To find the number of feet of lumber in 1-inch boards:

Rule. —lifultifily the length of the board, in feet, by the width, in inches, and divide the firoduct by 12.

EXAMPLE. —HOW many feet of lumber are there in a 1-inch board 18 feet long and 8 inches wide?

18 X 8 SOLUTION.— = 12 ft. Ans.

12


1. How many board feet in a 1-inch board 21 feet long and 18 inches wide? Ans. 312 ft.

2. How many board feet in a 1-inch board 12 feet long and 24 inches wide? Ans. 24 ft.

79. To find the number of feet of lumber in

joists, beams, etc.:

Rule.---AfultiPy the width, in inches, by the thickness, in

inches, and by the length, in feel. Divide this firoduct by 12 and the quotient is the number of feet of lumber in the stick.

MENSURATION 17

EXAMPLE.-HOW many feet of lumber in a joist 4 inches wide, inches thick, and 12 feet long?

4 X 3 X 12 SOLUTION.- = 12 ft. Ans.

12


1. How many feet of lumber in a beam 8 inches wide, 3 inches tick, and 16 feet long? Ans. 32 ft.

2. How many feet of lumber in a timber 12 inches wide, 16 inches .ep, and 24 feet in length? Ans. 384 ft.

MENS IRATION 1,010.0/011.70■110....111mr•ROMOSOMMIim

EXAMINATION QUESTIONS (1) How many samples, each containing 720 square

inches, can be cut from a piece of cloth 45 yards long and 40 inches wide? Ans. 90 samples

(2) The respective lengths of the sides of a triangular piece of sail cloth are . 5 yards, 7 yards, and 8 yards; how many square yards does it contain? Ans. 17.32 sq. yd,

(3) What is the diameter of a roll that is 142 inches in circumference? Ans. 44615 in.

(4) A size box is 54 inches long, 20 inches deep, and 2 feet 3 inches wide; if there is 231 cubic inches in 1 gallon, how many gallons will it contain? Ans. 126.234 gal.

(5) A certain roll is 2 inches in diameter and makes 137 revolutions per minute; how many inches of yarn will it deliver per minute? Ans. 860.798 in.

(6) What is the difference in the lengths of the circumferences of two pulleys, one having a radius of 7 inches and the other having a diameter of 7 inches? Ans. 21.991 in.

(7) How many tons of cotton may be stored on the floor of a room 200 feet long and 100 feet wide, if the floor will hold 180 pounds to the square foot? Ans. 1,800 T.

(8) The altitude of a certain triangle is 10 feet and itE base is 12 feet; how many square inches does it contain?

Ans. 8,640 sq. in,

6

2 MENSURATION

(9) If a square yard of cloth weighs 20 ounces (avoirdu- pois), how many pounds will a cut of 50 yards weigh, the cloth being 54 inches wide? Ans. 93.75 lb.

(10) How many cubic feet will a cylindrical tank 4 feet in diameter and 5 feet deep contain? Ans. 62.832 cu. ft.

(11) How many bags of wool weighing 400 pounds each can be stored in a room 200 feet long and 100 feet wide, without loading the floor more than 250 pounds to the square foot? Ans. 12,500 bags

(12) How many gallons of oil are there in a cylindrical tank 22 inches in diameter if the depth of the oil in the tank is 151 inches? Ans. 25.5068 gal.

(13) If a cubic inch of lead weighs .41 pound, what is the weight of a ball of lead 7 inches in diameter?

Ans. 73.634 lb.

(14) A sphere is 16 inches in diameter; what is its surface area? Ans. 804.250 sq. in.

(15) What is the average weight per square foot on a floor carrying 150 cards, each weighing 5,000 pounds, if the room is 125 feet wide and 300 feet long? Ans. 20 lb.

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