X117. A Regular Polygon is a polygon whose sides are equal each to each, and whose angles are equal each to each.

118. The Perimeter of a polygon is the distance around it, or the sum of the bounding lines.

119. Theorem.-Any polygon may be divided by diagonals drawn from any angle, into as many triangles as the polygon has sides, less two sides.

A

FIG. 97.

ILL. In the figure the polygon has 7 sides. By drawing the diagonals from C to the other angles, we divide the polygon into 5 (7—2) triangles.

120. Theorem.-The sum of the angles of any polygon is twice as many right angles as the polygon has angles (or sides), less four right angles.

ILL.-Draw a polygon, as ABCDEFG, and the arcs a, b, c, d, e, f, g, measuring

its angles. With the same radius draw a circle. Beginning at some point, as O, lay off OA = a, AB = b, BC = c, CD = d, DE = e, EF = ƒ, and FG = g. It is found in this case that the sum of these measures is two circumferences and a half. Now, one circumference is the measure of 4 right angles. Hence, 2 circumferences measure 2 × 4 = 10 right angles. Thus it appears that the sum of all the angles of the polygon is 10 right angles. This agrees with the theorem; for, by that, the sum should be 2 right angles × 7 — 4 right angles, which is 10 right angles.

FIG. 98.

121. Prob.-To draw a regular polygon.

SOLUTION.-Draw a circle, and divide the circumference into as many equal arcs

as the polygon has sides. The chords of these arcs will constitute the perimeter

of the polygon.

The practical difficulty lies in dividing the circumference as required. The circumference can be divided into 6 equal arcs by (55). Drawing radii to these points of division, and bisecting the included angle, a division into 12 equal parts is effected. These can be again bisected, and the division into 24 equal parts effected, etc. Again, the circumference can be divided into 4 equal parts by drawing two diameters at right angles to each other (see Fig. 95). These arcs can be bisected as indicated above, and the division into 8 equal parts effected. Bisecting the latter arcs, we have 16 equal parts, etc. There is also a way to divide the circumference into 10 equal parts, but it is too difficult to be given here. For all regular polygons except those of 3, 6, 12, 24, etc., and 4, 8, 16, etc., sides, the pupil, at this stage of his progress, is expected to effect the division by trial.

EXERCISES.

1. By drawing diagonals from any one angle, into how many triangles can a pentagon* be divided? Show it with a figure. Into how many an octagon ? A dodecagon? A nonagon? A hexagon? 2. What is the sum of the angles of a hexagon? Determine the number mentally, and then measure the angles geometrically, as in the solution of (120), observing that the latter result verifies the former. In like manner determine the sum of the angles of a pentagon. Of an octagon. Of a decagon. Of a nonagon. Of a triangle. Of a quadrilateral.

3. If the angles of a hexagon are equal each to each-that is, if the hexagon is equiangular-what is the value of any one angle? Ans. 1 right angles.

[NOTE-A regular polygon is equiangular.]

4. What is the value of any angle of a regular octagon? Of a regular pentagon? Of a regular dodecagon?

Answer to the last, 1 right angles.

5. Construct a regular dodecagon.

6. Construct a regular heptagon.

SUG's-Observing that as the chord for the hexagon is the radius, and hence the chord for the heptagon is a little less, we can readily find by trial just how wide to open the dividers so that they shall step around the circumference at 7 steps. This is not a very scientific way of constructing a figure, it is true, but it is the only way we can get the chord in this case.

FIG. 99.

* Polygons are not to be assumed regular unless they are so designated.

7. Construct a regular octagon.

H

SUG.-See the general solution (121).

FIG. 100.

8. Construct a regular nonagon.

SOLUTION.-First get a quarter of the cir cumference by marking the points where two diameters at right angles to each other would cut the circumference. AX is an arc of 90°. Then from A take AY = 60° by using radius as a chord. YX is therefore an arc of 30°. Divide this into three equal parts by trial. Measure YB equal to two-thirds of YX, and AB and BC are arcs of 40°, and the chords AB and BC are chords of the regular nonagon.

9. To draw a five-point star.

SOLUTION.-Draw a circle, and dividing the circumference into five equal parts, join the alternate points of division, as in the figure.

FIG. 101.

10. To circumscribe a square about a circle (56). Also an equilateral triangle, and a regular hexagon.

These are inserted simply to give completeness. Of course, the student is not expected

to know more than their names.

PART II.

THE FUNDAMENTAL PROPOSITIONS OF ELEMENT-
ARY GEOMETRY, DEMONSTRATED, ILLUS-
TRATED, AND APPLIED.

CHAPTER I.

PLANE GEOMETRY.

SECTION I

OF PERPENDICULAR STRAIGHT LINES.

PROPOSITION I.

122. Theorem.-At any point in a straight line, one perpendicular can be erected to the line, and only one, which shall lie on the same side of the line.

DEM.-Let AB* represent any line, and P be any point' therein; then, on the same side of AB there can be one and only one perpendicular erected at P. For from P draw any oblique line, as PC, forming with AB the two angles CPB and CPA. Now, while the extremity P, of PC, remains at P, conceive the line PC to revolve so as to increase the less of the two angles, as CPB, and decrease the greater, as CPA. It is evident that for a certain position of CP, as C'P, these

FIG. 102.

B

* In class recitation the pupil should go to the blackboard, after having had his proposition assigned him, and first draw the figure required for the demonstration. This should be dor neatly, accurately, with dispatch, and without any aids. The figure being complete stands at the board, pointer in hand, enunciates the proposition, and then gives the do stration as it is in the text, pointing to the several parts of the figure as they are referrec