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FIG. 1.

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, 1 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

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 Fig. 2. . they will obtain by memorizing many pages of definitions or listening to numerous "explanations."

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

8. If x represents one side of a cube, x2 = the surface of one face, and 6.x = the entire surface = 216 sq. in. Ans.

9. The convex surface = 4x2 = 144 sq. in. Ans. 11. The perimeter = (600 -- 15) ft.

Or, let x= one side of base ; the perimeter=4x; the convex surface = 4x x 15 = 60 x = 600 ; x = 10; etc.

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12. Let x = the altitude. (15+15+15+15) x2 + 15+ 152 = 1650; 60 x + 225 + 225 = 1650; 60 x = 1650 – 225 — 225 = 1200; x = 20; the convex surface = 60x = 1200. 20 in.; 1200 sq. in. Ans.

13. Let x = one side of base; 4x = the perimeter; 4x x 15 = 60 x = the convex surface = 540; x=9. The entire surface == 540 sq. in. +81 sq. in +81 sq. in.

14. Circumference of base = 3.1416 ft. Convex surface = (3.1416 x 1) sq. ft. = 3.1416 sq. ft. Radius of base = { ft.; area =(1 x x 3.1416) sq. ft. = .7854 sq. ft. Entire surface = 3.1416 sq. ft. +.7854 sq. ft. +.7854 sq. ft. = 4.7124 sq. ft. Ans. 15. See Arithmetic, Art. 1290. While pupils should be per

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 con

struct a hollow pyramid. Fig. 3.

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

18. One side of base = V144 in.

19. Either calculate the slant height, which is V22 – 1?=V3; or employ the method given in 1, of Measurements, Arithmetic, Art. 1107.

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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.

The circumference of the base of the given cone =(4 times 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, f of the circle, and its arc measures of 360°, or 120°.

Any sector of 120° will form a hollow Bitte 1200 mm cone of the proper proportions.

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

Fig. 4. determining the number of degrees it should contain. The compasses or the protractor may be employed to construct an angle of 120° at C.

The convex surface of a cone is equal to the circumference of the base x i 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 x altitude); that of a sector = } (base X 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 V144 + 25 in., or 13 in. The convex surface =į (10 x 3.1416 x 13).

23. The entire surface = [1 of (6 x 3.1416 x 10)] + (32 X 3.1416) = (30 x 3.1416) + (9 3.1416) = 39 x 3.1416.

Using a (pi) instead of 3.1416, the circumference of the base = 67 inches. The radius of the sector representing the development, is 10 in., and the circumference of the whole circle = 20 inches. As the arc of the sector must be 67 inches, it measures in degrees of 360°, or 108°,

24. The slant height will be of 6 in. The circumference of the base will equal the arc of the semicircle, 31 inches; its diameter will, therefore, be 3 in.

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

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 1 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. With grace, $4+ zbo of $4, or $4.20. Ans.

12. of cost = $2; etc. 13. The principal is unimportant. Time=(100 = 8) yr. 17. (100 = 6) 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) x 10] + [twice (4 x 3)]. 28. $ 1.121 + 1 of $1.121 =$1.124 +$.281. 30. Selling price = of cost = 1¢; cost =1$= = 14. Ans.

1144. 4. 14% profit=7&; cost = 76= .14=50$; selling price = 576. Ans.

7. x+60+200 = 7828 ; etc

17. Cost of 350 tons @ $3.50 = $1225. Selling price = $4.25 X 350 x 2446

NotE.— The scholars should not use pencils to obtain answers to problems that can be solved at sight. 1145. See Art. 1290. 3. Area of base = 6 sq. in.; etc. 5. See Art. 1107, Measurements, 7.

8. Changing given dimensions to inches, the number of gallons will be 36" X 3.1416 x 66 = 231.

9. 1 cu. ft. = 1000 lb. Cubical contents = (32 x 3.1416 x 51) cu. ft.

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

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