EXERCISES.

I. To construct a triangle equal to a given quadrangle. 2. To construct a triangle equal to a given polygon.

3. To bisect a triangle by a line drawn through a given point in one of the sides.

4. To construct a rhombus equal to a given parallelogram, and with one of the sides of the parallelogram as its side.

5. The three connectors of the middle points of the sides of a triangle divide the triangle into four equal triangles.

6. Any line concurrent with the diagonals of a parallelogram bisects the parallelogram.

7. The triangle having one of the non-parallel sides of a trapezoid as base and the middle point of the opposite side as vertex is one-half the trapezoid.

8. The connector of the middle points of the diagonals of a quadrangle is concurrent with the connectors of the middle points of opposite sides.

9. ABCD is a parallelogram and O is a point within. Then ▲AOB+COD={O.

What does this become when O is without?

10. ABCD is a parallelogram and O is a point within. Then AAOC AAOD-AAOB.

What does this become when O is without? (This theorem is important in the theory of Statics.)

11. Bisect a trapezoid by a line through the middle point of one of the parallel sides. By a line through the middle point of one of the non-parallel sides.

12. The triangle having the three medians of another triangle as its sides has three-fourths the area of the other.

POLYGON AND CIRCLE.

146°. Def.—The sum of all the sides of a polygon is called its perimeter, and when the polygon is regular every side is at the same distance from the centre. This distance is the

apothem of the polygon.

Thus if ABCD...LA be a regular polygon and O the centre (132°, Def. 2), the triangles OAB, овс, are all congruent, and L/ OP=OQ=etc.

...

B

AB+BC+CD+...+ LA is the perimeter and OP, perpendicular upon AB, is the apothem.

147°. Theorem.—A regular polygon is equal to one-half the rectangle on its apothem and perimeter.

...

on

Proof. The triangles AOB, BOC, LOA have equal altitudes, the apothem OP,.. their sum is one-half the OP and the sum of their bases AB + BC+...LA. (144°, Cor. 3) But the sum of the triangles is the polygon, and the sum of their bases is the perimeter.

.. a regular polygon on its apothem and perimeter.

148°. Of a limit.—-A limit or limiting value of a variable is the value to which the variable by its variation can be made to approach indefinitely near, but which it can never be made to pass.

Let ABCD be a square in its circumcircle. If we bisect the arcs AB, E BC, CD, and DA in E, F, G, and H, we have the vertices of a regular octagon AEBFCGDHA. Now, the area of the octagon approaches nearer

F

K

B

H

to that of the circle than the area of the square does; and the

perimeter of the octagon approaches nearer to the length of the circle than the perimeter of the square does; and the apothem of the octagon approaches nearer to the radius of the circle than the apothem of the square does.

Again, bisecting the arcs AE, EB, BF, etc., in I, J, K, etc., we obtain the regular polygon of 16 sides. And all the foregoing parts of the polygon of 16 sides approach nearer to the corresponding parts of the circle than those of the octagon do.

It is evident that by continually bisecting the arcs, we may obtain a series of regular polygons, of which the last one may be made to approach the circle as near as we please, but that however far this process is carried the final polygon can never become greater than the circle, nor can the final apothem become greater than the radius.

Hence the circle is the limit of the perimeter of the regular polygon when the number of its sides is endlessly increased, and the area of the circle is the limit of the area of the polygon, and the radius of the circle is the limit of the apothem of the polygon under the same circumstances.

149°. Theorem.-A circle is equal to one-half the rectangle on its radius and a line-segment equal in length to the circle. Proof. The is the limit of a regular polygon when the number of its sides is endlessly increased, and the radius of the is the limit of the apothem of the polygon.

But, whatever be the number of its sides, a regular polygon is equal to one-half the on its apothem and perimeter. (147°) .. a is equal to one-half the □ on its radius and a linesegment equal to its circumference.

EXERCISES.

1. Show that a regular polygon may be described about a circle, and that the limit of its perimeter when the number of its sides is increased indefinitely is the circumference of the circle.

2. The difference between the areas of two regular polygons,

one inscribed in a circle and the other circumscribed

about it, vanishes at the limit when the number of sides of the polygons increases indefinitely.

3. What is the limit of the internal angle of a regular polygon as the number of its sides is endlessly increased?

SECTION II.

MEASUREMENT OF LENGTHS AND AREAS.

150°. Def.-1. That part of Geometry which deals with the measures and measuring of magnitudes is Metrical Geometry.

2. To measure a magnitude is to determine how many unit magnitudes of the same kind must be taken together to form the given magnitude. And the number thus determined is called the measure of the given magnitude with reference to the unit employed. This number may be a whole or a fractional number, or a numerical quantity which is not arithmetically expressible. The word "number" will mean any of these.

3. In measuring length, such as that of a line-segment, the unit is a segment of arbitrary length called the unit-length. In practical work we have several such units as an inch, a foot, a mile, a metre, etc., but in the Science of Geometry the unit-length is quite arbitrary, and results obtained through it are so expressed as to be independent of the length of the particular unit employed.

4. In measuring areas the unit magnitude is the area of the square having the unit-length as its side. This area is the unit-area. Hence the unit-length and unit-area are not both

arbitrary, for if either is fixed the other is fixed also, and determinable.

This relation between the unit-length and the unit-area is conventional, for we might assume the unit-area to be the area of any figure which is wholly determined by a single segment taken as the unit-length: as, for example, an equilateral triangle with the unit-length as side, a circle with the unit-length as diameter, etc. The square is chosen because it offers decided advantages over every other figure.

For the sake of conciseness we shall symbolize the term unit-length by u.l. and unit-area by u.a.

5. When two magnitudes are such that they are both capable of being expressed arithmetically in terms of some common unit they are commensurable, and when this is not the case they are incommensurable.

E

B

F

A

C

P

H

D

G

Illus. Let ABCD be a square, and let EF and HG be drawn to BD, and EH and FGL to AC. Then EFGH is a square (82°, Cor. 5), and the triangles AEB, APB, BFC, BPC, etc., are all equal to one another.

If AB be taken as u.l., the area of the square AC is the u.a.; and if EF be taken as u.l., the area of the square EG is the u.a.

In the first case the measure of the square AC is 1, and that of EG is 2; and in the latter case the measure of the square EG is 1, and that of AC is . So that in both cases the measure of the square EG is double that of the square AC. .. the squares EG and AC are commensurable.

Now, if AB be taken as u.l., EF is not expressible arithmetically, as will be shown hereafter.

.. AB and EF are incommensurable.

151°. Let AB be a segment trisected at E and F (127°), and let AC be the square on AB. Then AD=AB. And