into 3; a hexagon, into 4; a heptagon, into 5; an octagon, into 6, being 2 triangles less in each case than the number of sides in the polygon.

96. The number of degrees in each angle of a regular octagon = [180° × (8 − 2)] ÷ 8.

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97. At each end of the 2-inch line, draw a 2-inch line at an angle of 108°. At the farther extremity of each of those lines draw a line at an angle of 108°. These last lines meet at an angle of 108° if the work is correctly done, and are each two inches long to the point of their intersection..

98. A line drawn to each extremity of the base at an angle of 60° will form an equilateral triangle. Use angles of 90° for the square, 108° for the pentagon, 120° for the hexagon, nearly 129° for the heptagon, 135° for the octagon, 140° for the nonagon, etc.

The first line should be placed near the center of the bottom of the paper to give room for the successive polygons. See Fig. 17. Bright pupils will calculate the necessary angles and continue to construct polygons as far as the space will permit.

FIG. 17.

99. By drawing the diameters AB and XY (Fig. 18), the inscribed square will be divided into four 1

triangles, while the circumscribed square contains eight, being double the area of the inscribed square.

1270. These problems are given to enable the pupils to learn how to bisect lines, erect perpendiculars, construct angles, etc., by means of the ruler and the compasses,

B

FIG. 18.

and incidentally to learn a number of geometrical facts. The use of other instruments is unnecessary, and should, therefore, not be tolerated.

1. The distance between the centers

11⁄2 in. + 1 in.

2. 11⁄2 in. - 1 in.

4. The line XY joining the two points of intersection of the equal circles (Fig. 19), bisects the line AB connecting the centers. The radii AX, BX, AY, BY are each 2 inches long.

A

B

FIG. 19.

5. The previous exercise should lead the pupils to see the steps necessary to the construction of the required triangle. The 3-inch base AB is first drawn. The next requirement is to find a point X (Fig. 19) 2 inches from A and from B. A circle of 2-inches radius with B as a center, will contain every point that is 2 inches from B. A similar circle with A as a center, contains every point 2 inches from A. The intersections, X and Y, are each 2 inches from both points A and B. Using either one as a vertex, draw lines to A and B, forming the required triangle.

Authorities differ somewhat as to the advisability of requiring pupils to employ circles rather than arcs in geometrical problems. While it may be better at first to use circles, the point to be finally reached is the employment of the shortest lines possible. This should not be inconsistent with the acquirement of geometrical knowledge. A scholar should know after a very few exercises that each point of an arc is 2 inches from the center, as well as he does when he draws the entire circle.

In his later construction of an isosceles triangle, the pupil should know that the vertex is above (or below) the center of the base. For this reason the first arc need not be a very long one. The second should be still shorter.

6. Fig. 19 will suggest the necessary steps. Using each end of the 3-inch base as a center, and with a radius of 4 inches, draw two circles. Draw lines corresponding to XA and XB for the required triangle. Placing the ruler on X and Y will give the perpendicular, which should not extend below the base.

If arcs are used, the upper intersection determines the position of the vertex. A lower intersection is used to determine the

direction of the perpendicular. The pupil will gradually learn that while a definite radius, 4 inches, is required to locate the vertex of the triangle, intersecting arcs, each of 3 inches or 5 inches or any other radius, will serve to locate the second point, used with the vertex to determine the direction of the perpendicular. The point of their intersection may be above the base or below it, according to convenience. A point below secures greater accuracy, by being probably at a greater distance from the vertex than one above is likely to be.

7. This is a variation of 6, but without directions as to length of sides. If the perpendicular is correctly drawn, it will bisect the base. See Exercise 48, Art. 1263. The compasses should be employed to determine the equality or inequality of the segments of the base.

8. The same procedure is required as in 7, except that the sides of the triangle are not drawn. The bisecting line should be extremely short.

9. With a 2-inch radius, draw intersecting arcs, using as centers the two extremities of a 2-inch line.

11. Either side may be used as the base; and different pupils should use a different base, although the greatest number will probably take the longest side.

Using the 2-inch side as a base, draw from one end, as a center, an arc with a radius of 1 inch; and from the other, an arc with a radius of 11⁄2 inches. The intersection of the arcs will be the vertex of the required triangle.

With the 3-inch line as a base, the radius of the arcs will be 2 inches and 2 inches, respectively.

Besides employing different bases, the pupils should use oblique lines and vertical lines as bases, and the vertex in some instances should be below the base; etc.

12. A scholar should be permitted to discover for himself that the intersection of the 2-inch arcs will be at the center of the 4-inch base. After endeavoring, also, to make a triangle whose

sides shall measure 1, 2, and 3 inches, respectively, he will understand that the third side of a triangle must be shorter than the combined lengths of the other two.

13. If the pupil, in drawing arcs to locate the bisecting line, employs the radius used in drawing the circle, he will discover that one intersection will take place at the center of the circle. This will lead him to see that only one set of intersecting arcs is necessary the one beyond the circle, the center of the circle serving as a second point.

The teacher should be in no hurry to inform the pupil using two sets of arcs, that a single set will be sufficient; knowledge that comes in some other way than merely by direct telling, is apt to last longer.

14. Before dividing a sector (Figs. 8 and 9) into two equal parts, some scholars may consider it necessary to draw the chord. Permit them to do so at first.

15. The pupils have already learned that a line drawn from the vertex of an isosceles triangle to the center of the base, is perpendicular to the base. The method employed in bisecting a line is practically to consider it the common base of two isosceles triangles having their vertices on opposite sides of the base, although the equal sides of the triangles are not drawn. This bisecting line is perpendicular to the assumed base. The pupils have probably discovered that the perpendicular line bisecting the chord in 13, would, if produced, pass through the center of the circle. From this previous knowledge, they will doubtless be able to answer the last question of 15.

The scholars that have drawn the chord in dividing the sector in 14 into two equal parts, will be likely to see that the radius CX (Fig. 20) is perpendicular to the bisected chord AB.

C

YI

A

B

X

FIG. 20.

16 shows that a perpendicular that bisects a chord, AB (Fig. 20), also bisects the arc AXB.

It will readily be seen by the pupils that an arc AXB can be bisected without drawing the chord AB.

17. This problem is the same as 8. The drawing, however, should show a longer line, and one that does not cut the given line, the requirement being that the perpendicular be drawn to the latter. A perpendicular drawn to a horizontal or to an oblique line from below it, would be correct; although it might not be accepted as satisfactory if the more common wording of the problem were employed: Erect a perpendicular at the middle point of a line.

18 requires no explanation.

19. Bisecting one of the four divisions of a circumference gives the dividing point between two one-eighths. A ruler placed upon this point and on the center of the circle will indicate the location of another. The distance between two points can be ascertained by the compasses, which can then be used to locate the remaining two dividing points.

The lines used to divide a circumference should not be too long. The thoughtless pupil sometimes fails to see, when he employs his compasses to measure the distance between two points on a circumference, that he is measuring the chord of an arc, and not the arc itself. It may be necessary for the teacher to explain this in connection with 25.

22. Unless the pupils have learned in their drawing work the method of erecting a perpendicular at the end of a line, some of them may experience a little difficulty in solving this problem. One way of drawing the circumscribed square is to construct on AY (Fig. 18) an isosceles triangle A2Y equal to ACY. Produce 24 to 1, making Al equal to 24. Produce 2Y to 3 in the same manner. Lines from 1 through X, and 3 through B, will intersect at 4, which completes the square.

24. The object of 23 and 24 is to lead the pupils to see again that the side of an inscribed hexagon is equal in length to the radius of the circle.