1141. The "developed " entire surface of a square prism is shown in Arithmetic, Art. 818, 20. In drawing the development of the convex surface, the upper and lower squares, denoting the bases, will be omitted. The drawing should be done with reasonable care, to a scale of, say, inch to the inch.

In making a model of a solid, narrow strips for pasting should be added, as shown in Fig. 1. In the development of a triangular prism, the bases are usually drawn above

and below the middle rectangle, but the pupils should learn that they may be placed in other positions, one of which is here shown. It will be noticed that the pasting flaps do not form rectangles, the sides being inclined at an acute angle to make neater work in the completed model.

The shape and the arrangement of the gumming flaps for the bases of a

FIG. 1.

cylinder are shown in Fig. 2. Interested pupils may be safely left to themselves to ascertain the length of the rectangle that is needed for the model of a given cylinder.

The scholars will learn more geometrical facts while constructing these models than they will obtain by memorizing many pages of definitions or listening to numerous "explanations.'

FIG. 2.

5. The entire surface includes the convex surface and the surface of the bases.

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8. If x represents one side of a cube, x2 the surface of one face, and 6x2 the entire surface

=

=

216 sq. in. Ans.

144 sq. in. Ans.

11. The perimeter = (600 ÷ 15) ft.

Or, let x = one side of base; the perimeter = 4x; the convex

=

=

225

225

60x1200. 20 in.;

13. Let x = one side of base; 4x the perimeter; 4x × 15

540; x=9.

sq. in.

The entire surface

14. Circumference of base 3.1416 ft. Convex surface = (3.1416 × 1) sq. ft. 3.1416 sq. ft. 3.1416 sq. ft. Radius of base ft.; (×× 3.1416) sq. ft.=.7854 sq. ft. Entire surface: sq. ft.+.7854 sq. ft.+.7854 sq. ft. = 4.7124 sq. ft. Ans.

area

3.1416

=

15. See Arithmetic, Art. 1290. While pupils should be per

FIG. 3.

mitted to "develop" these solids in their own way, provided it be a correct one, they should be advised in making drawings for models to use a pattern that will require a minimum of pasting. While Fig. 3 would serve for the development of the entire surface, it would not answer as a pattern from which to construct a hollow pyramid.

16. The convex surface of a pyramid is equal to the perimeter of the base × the slant height. The slant height of a regular pyramid is the altitude of one of the equal triangles that constitute its convex surface.

19. Either calculate the slant height, which is √22 - 12= √3; or employ the method given in 1, of Measurements, Arithmetic, Art. 1107.

20. The developed convex surface of a cone is a sector, whose radius is the slant height of the cone and whose arc is equal in length to the circumference of the base of the cone.

(4 times

The circumference of the base of the given cone = 3.1416) in. The circumference of the circle of which the sector forms a part, is (2 x 6 times 3.1416) in., or (12 times 3.1416) in.; the sector is, therefore, of the circle, and its arc measures 360°, or 120°.

Any sector of 120° will form a hollow cone of the proper proportions.

A

6 in.

C 1200

6 in.

120

in.

12.5664

4 in.

inches

of

B

The base shown in Fig. 4 is not required by the terms of this problem; it is merely introduced here to show the development of the entire surface. As it is difficult to lay off the required number of inches for the arc AB, the pupil will appreciate the foregoing method of determining the number of degrees it should contain. The compasses or the protractor may be employed to construct an angle of 120° at C.

2.8664

FIG. 4.

The convex surface of a cone is equal to the circumference of the base X slant height.

An examination of Fig. 4 will show the resemblance between the methods of calculating the surface of a sector and of a triangle. The area of a triangle = 1 (base × altitude); that of a sector = } (base × radius).

22. The altitude, one-half the base, and the slant height, form a right-angled triangle; and the lengths of the two first being 12 in. and 5 in., respectively, the length of the latter is √144+25 in., or 13 in. The convex surface = (10 × 3.1416 × 13).

23. The entire surface

of

(6 x 3.1416 x 10)] + (3 × 3.1416) = (30 × 3.1416) + (9 × 3.1416) = 39 × 3.1416.

Using (pi) instead of 3.1416, the circumference of the base

20π

6 inches. The radius of the sector representing the development, is 10 in., and the circumference of the whole circle inches. As the arc of the sector must be 6 inches, it measures in degrees of 360°, or 108°.

π

inches; its

24. The slant height will be of 6 in. The circumference of the base will equal the arc of the semicircle, 3 diameter will, therefore, be 3 in.

21

1142. 3. Due Sept. (21) 24. Term of discount from July = 65 (62) da.

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4. $600 yearly interest represents a principal of $10000.

5. Length of one fence, (20+20 +20 +20) rd.; of the other, (40+ 10+40 + 10) rd.

6. The distance between the center of the first and of the last post = 10 ft. x (11-1) = 100 ft. Adding of the diameter of each post, gives 100 ft. 6 in.; and an additional 3 in. at each end to fasten the wire, makes a total of 101 ft. of wire required for each length, or 303 ft. in all. Ans.

1143. 7. Troy weight.

8. 43% mo. @25¢ per mo. 43 quarter dollars = $10.75; of 25¢10%. Total $10.85. Ans.

11. Without grace, 1% of $400, or $4. Ans.

$4+

of $4, or $4.20. Ans.

12. of cost $2; etc.

With grace,

Time = (100÷8) yr.

13. The principal is unimportant. Time =

17. (1006) yr. 23. 1% of $1234.

24. 2% of $1234.

25. The number of rings 60 pwt. ÷ 24 pwt.

26. 2° 3' x 15. 27. [Twice (4+3) × 10] + [twice (4 × 3)]. 28. $1.12+ of $1.12 $1.121+ $.28.

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30. Selling price of cost; cost = 1¢ ÷ 3 = }¢. Ans.

1144. 4. 14% profit = 7; cost = 7¢÷.14 = 50%; selling price 57. Ans.

=

17. Cost of 350 tons @ $3.50 $1225. Selling price $4.25 x 350 x 3348.

00

=

NOTE. The scholars should not use pencils to obtain answers to problems that can be solved at sight.

5. See Art. 1107, Measurements, 7.

8. Changing given dimensions to inches, the number of gallons will be 362 × 3.1416 × 66 ÷ 231.

10, 11. Careful pupils will be much interested in ascertaining how closely their calculations as to the contents of the measure, agree with the number of cubic inches it is supposed to contain. There should be 231 cu. in. ÷ 8, in a quart. The cup used must be cylindrical. Tapering measures should be left until the frustum of a cone has been studied, Art. 1295. The paper box used for ice cream, a frustum of a pyramid, can also be employed at that time. Some of these measurements should be made out of school, and comparisons made as to the results obtained by different pupils and the methods employed by them to secure accuracy. A random measurement will not obtain the correct diameter of a quart measure.

After calculating the altitude of an equilateral triangle or the diagonal of a square, the pupil should draw the figure to a scale, measure the altitude or the diagonal, and compare the measured length with the length obtained by calculation.

Pupils should ascertain the weight of a cubic foot of water by weighing a quart of water, for instance, etc.

12. Measure the height to which the water rises in the box, etc.

If a solid heavier than water, is placed in a rectangular or a cylindrical vessel containing sufficient water to cover it, and the difference in the depth of the water before and after immersion is noted, the volume of the solid