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Diagrams should be employed, unless the pupils can do satisfactory work without them. Since the arc of 180° is three times as long as the arc of 60°, the more careless members of the class may jump to the conclusion that the chord of 180° is three times as long as the chord of 60°. This mistake cannot be made if the circle is constructed, and the chords are drawn and measured.

5. See Exercise 98, Art. 1269.

6. Each side of the hexagon measures 1 inch; the perimeter =1 inch x 6=6 inches. Ans. Circumference=2 in. x 3.1416 = 6.2832 inches. Ans.

7. The pupil should use the ruler to ascertain the length of the apothem, which is about } in. It can be calculated as follows:

Cb (Fig. 38) is the radius=1 inch; bx=one-half the side of the inscribed hexagon=1 inch; Csc = apothem. Các* = CO bxo =1-1=.75; Cx= V.75=.866+, or nearly 3.

8. The base of each triangle measures one inch, so that AB (Fig. 43) = 3 in. AX (apothem) x ={ in. nearly.

9. The base of each triangle v 1 measures about & in. so that the A base of the rectangle (the half

Fig. 43. perimeter) will be nearly 3 inches and the apothem about 15 in. Area about 3 x 16 sq. in. = about 21. sq. in.

10. ABX AX=3 xś=24. Ans. 25 sq. in. 11. See 9.

12. The perimeter of a 16-sided polygon will be greater than that of an octagon. The 16-sided polygon will have the greater apothem.

13. As the number of sides increases, the perimeter approaches more and more closely the circumference, 6.2832 inches; and the apothem approaches the radius, 1 inch.

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14. The base of the rectangle will be 3.1416 inches, one-half the perimeter (circumference); the apothem will be 1 inch (radius). Area = 3.1416 sq. in. Ans.

15. One-half the circumference=2 x 3.1416=6.2832, multiplied by the radius, 2, gives answer in square inches, 12.5664 sq. in.

16. 78.54 sq. in. Ans.
17. 3.1416 sq. in. Ans.
18. 314.16 sq. in. + 6 = Ans.

19. Subtract from the area of the outer circle, 113.0976 sq. in., the area of the inner circle, 28.2744 sq. in. Ans. 84.8232 sq. in.

1282. Right prisms, cylinders, etc., are meant when the word oblique is not used.

1. See Arithmetic, Art. 818, Problem 20. The upper squares need not be drawn, as only the convex surface is required.

2. Three rectangles and two triangles. See 1.

3. Three rectangles, each 3 inches high, bases 1, 17, and 2 inches, respectively.

5. A hollow paper cylinder, without bases, can be opened out into a rectangle whose base is the circumference of the base of the cylinder, and whose height is the altitude of the cylinder.

6. The slant height of a square pyramid is the distance from the apex to the center of one side of the base. This problem requires the pupil to draw, side by side, four isosceles triangles, the base of each being 2 inches and the altitude 3 inches. After constructing the first (Fig. 44), by erecting a 3-inch perpendicular at the center of a .2-inch base, he should use B" and A as centers, and radii equal to B" Bland AB" to locate B". On AB' construct a B''N third triangle, using A and B' as centers and the radii previously given. Upon one side of this triangle construct a fourth.

Fig. 44.

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The pupils can discover for themselves the method of drawing geometrically the required convex surface.

The altitude should be carefully measured. It will be equal in length to the perpendicular of a right-angled triangle whose hypotenuse is the slant height of a pyramid, 3 inches; and whose base is the distance from the center of one edge of the base of the pyramid (the foot of the slant height) to the foot of the altitude, 1 inch. Ans. V8 inches =2.83 inches nearly = about 21% inches.

7. The area of each convex face of a regular pyramid is found by multiplying one side of the base by one-half the slant height; therefore, the area of all the faces forming the convex surface, is obtained by multiplying the sum of all the sides of the base, that is, the perimeter of the base, by one-half the slant height.

8. The pupil requires the slant height in order to proceed as in 6; and he should obtain it by drawing it rather than by calculating it. He should be able to see that the altitude is a line drawn from the vertex to the center of the base, and that its foot is 1 inch distant from the foot of the altitude. Constructing a right-angled triangle whose base is 1 inch, and whose perpendicular is 3 inches, will give a hypotenuse equal to the required slant height.

Some scholars will bring in a prism whose slant height is 3 inches. The mistake should be pointed out, but not the mode of correcting it; and a pyramid of the required dimensions should be insisted upon.

1283. When the diameter of the base of a cone is 2 inches, the arc BDC, which forms the circumference when folded, measures 2 inches, or 6.2832 inches.

NOTE. — The Greek letter a (pi) represents 3.1416.

9. The semi-circumference of paper = 37 inches, which is the circumference of base of cone. The diameter of base of cone - 3 7 inches = 7=3 inches. The radius of base = 17 inches. The slant height = radius of paper = 3 inches = diameter of base of cone=twice radius of base of cone.

10. The area of any sector of a circle = radius x } length of arc. The arc when folded becomes the circumference of base, and the radius becomes the slant height; so that convex surface = } circumference X slant height = circumference x slant height.

11. Length of arc of 90° = circumference=1 of 6 7 inches =117 inches. This is the circumference of the base of the cone, which makes the diameter = 11 a inches = 1 = 1; inches. The slant height is 3 inches.

Length of arc of 60° = i circumference = 1 of 6 7 inches =a inches. The diameter of the base of the cone =a inches = 7 = 1 inch; slant height, 3 inches.

12. The circumference of the base of the cone = 37. This equals the length of the arc of the required sector. If the slant height is 5 inches, the circumference of which the sector is a part = 107. The arc of the sector is, therefore, of the circumference; and its length is % of 360° = 108°

13. XY represents the base of the pyramid; and AC, its altitude. The slant height of two faces, AM (Fig. 45), is the hypotenuse of a right-angled triangle, base 11 in., perpendicular 4 in. Using this as the perpendicular of a new triangle, with a base MY, gives as the hypotenuse AY (Figs. 45 and 46), one of the edges.

Fig. 45. To draw the development, take AY as a radius, and draw an arc. On this lay off succes

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sive chords of 3 in., 2 in., 3 in., and 2 in., connecting the extremity of each with the center A. These four triangles

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constitute the convex faces of the pyramid. On one of them, construct a rectangle 3 inches by 2 inches for the base.

14. In this pyramid, AC (Fig. 45) measures 12 inches, CM measures 1 of 18 inches, or 9 inches, making AM, one slant height, = V144 +81 inches = 15 inches. The other slant height, drawn to the center of OY,=V144 + 25 inches = 13 inches.

The pupils should be encouraged to construct stout paper pyramids and cones of required dimensions, making the necessary calculations themselves. The previous fourteen problems will present no difficulty whatever to pupils that are interested. As examples in dry calculation, they may prove somewhat tiresome. Many scholars will, of themselves, construct models of solids much more complicated than the foregoing.

1284. If there are no solids at hand, the pupils should construct a supply of paper ones, or make them of modeling clay, turnips, etc. Drawings are not sufficient for effective instruc

tion.

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