Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

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. × 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 = inch; Crapothem. Cx Cb=

bx=1-4.75; Cx √.75.866+, or nearly 7.

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 measures about in. so that the A

base of the rectangle (the half

1

FIG. 43.

B

perimeter) will be nearly 3 inches and the apothem about 18 in. Area about 3 × 15 sq. in. = about 218 sq. in.

10. ABX AX=3x7=2g. Ans. 2g 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.

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 × 3.14166.2832, multi

plied 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, 14, 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.

A

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" B' and AB" to locate B'". On AB' construct a B" third triangle, using A and B' as centers. and the radii previously given. Upon one side of this triangle construct a fourth.

2 in.

3 in.

2 in.

B'

B"

FIG. 44.

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. √8 inches = 2.83 inches nearly = about 218 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 π (pi) represents 3.1416.

[ocr errors]
[blocks in formation]
[merged small][ocr errors][merged small][merged small][merged small]

=

radius length of

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

height.

==

circumference × slant

[merged small][merged small][merged small][ocr errors][merged small][merged small]

11⁄2 inches. This is the circumference of the base of the cone, which makes the diameter = 11⁄2 inches ÷ = 1 inches. The

[merged small][merged small][merged small][ocr errors][merged small]

=π inches. The diameter of the base of the cone = inches ÷ π

[blocks in formation]
[ocr errors]

- 3 π.

This

12. The circumference of the base of the cone 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

= 10 π. The arc of the sector is, therefore, ence; 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 1 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.

To draw the development,

2 in.

take AY as a radius, and draw an arc.

3 in.

4 in.

FIG. 45.

of the circumfer

[merged small][ocr errors][merged small][merged small]

On this lay off succes

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

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small]

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 of 18 inches, or 9 inches, making AM, one slant height, =√144 +81 inches = 15 inches. The other slant height, drawn to the center of OY, = √144 + 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 instruction.

« ΠροηγούμενηΣυνέχεια »