Cor. 1. To inscribe any regular polygon in a circle, we have only to divide the circumference into as many equal parts as the polygon is to have sides, and to draw the chords of the arcs; hence, in a given circle, it is possible to inscribe regular polygons of any number of sides whatever. Having constructed any such polygon in a given circle, it is evident, that by changing the radius of the circle without changing the number of sides of the polygon, it may be made to represent any regular polygon of the same name, and it will still be inscribed in a circle. As this reasoning is applicable to regular polygons of whatever number of sides, it follows, that any regular polygon may be circumscribed by the circumference of a circle.

Cor. 2. Since ab, bc, cd, etc., are equal chords of the same circle, they are at the same distance from the center, (Th. 3, B. III); hence, if with O as a center, and Ot, the distance of one of these chords from that point, as a radius, a circumference be described, it will touch all of these chords at their middle points. It follows, therefore, that a circle may be inscribed within any regular polygon.

SCHOLIUM.-The center, O, of the circle, may be taken as the center of both the inscribed and circumscribed polygons; and the angle AOB, included between lines drawn from the center to the extremities of one of the sides AB, is called the angle at the center. The perpendicular drawn from the center to one of the sides is called the Apothem of the polygon.

Cor. 3. The angle at the center of any regular polygon is equal to four right angles divided by the number of sides of the polygon. Thus, if n be the number of sides of the polygon, the angle at the center will be expressed 360°

by

n

Cor. 4. If the arcs subtended by the sides of any regular inscribed polygon be bisected, and the chords of these semi-arcs be drawn, we shall have a regular

inscribed polygon of double the number of sides. Thus, from the square we may pass successively to regular inscribed polygons of 8, 16, 32, etc., sides. To get the corresponding circumscribed polygons, we have merely to draw tangents at the middle points of the arcs subtended by the sides of the inscribed polygons.

Cor. 5. It is plain that each inscribed polygon is but a part of one having twice the number of sides, while each circumscribed polygon is but a part of one having one half the number of sides.

BOOK V.

ON THE PROPORTIONALITIES AND MEASUREMENT OF POLYGONS AND CIRCLES.

PROPOSITION I.-THEOREM.

The area of any circle is equal to the product of its radius by one half of its circumference.

Let CA be the radius of a circle, and AB a very small portion of its circumference; then ACB will be a sector. We may conceive the whole

circle made up of a great number of such sectors; and when each sector is very small, the arcs AB, BD, etc.,

b

B

a

each one taken separately, may be regarded as right lines; and the sectors CAB, CBD, etc., will be triangles. The triangle, ACB, is measured by the product of the base, AC, multiplied into one half the altitude, AB, (Th. 33, Book I); and the triangle BCD is measured by the product of BC, or its equal, AC, into one half BD; then the area, or measure of the two triangles, or sectors, is the product of AC, multiplied by one half of AB plus one. half of BD, and so on for all the sectors that compose the circle; therefore, the area of the circle is measured by the product of the radius into one half the circumference.

PROPOSITION II.-THEOREM.

Circumferences of circles are to one another as their radu, and their areas are to one another as the squares of their radii.

Let CA be the radius of a circle, and Ca the radius of another circle. Conceive the two circles to be so placed upon each other so as to have

a common center.

Let AB be such a certain definite

portion of the circumference of the

a

D

larger circle, that m times AB will represent that circumference.

But whatever part AB is of the greater circumference, the same part ab is of the smaller; for the two circles. have the same number of degrees, and are of course susceptible of division into the same number of sectors. But by proportional triangles we have,

CA: Ca :: AB: ab

Multiply the last couplet by m, (Th. 4, B. II), and we have

CA Ca: m.AB: m.ab.

That is, the radius of one circle is to the radius of another, as the circumference of the one is to the circumference of the other.

To prove the second part of the theorem, let C represent the area of the larger circle, and c that of the smaller; now, whatever part the sector CAB is of the circle C, the sector Cab is the corresponding part of the circle c,

That is, but,

C: c :: CAB: Cab,
CAB: Cab :: (CA)2: (Ca)2,

Therefore,

C: c

:: (CA)2: (Ca)2,

(Th. 20, B. II). (Th. 6, B. II).

That is, the area of one circle is to the area of another, as

the square of the radius of the one is to the square of the

radius of the other.

Hence the theorem.

Cor. If

then,

Cc: (CA): (Ca)2,

Cc:4 (CA)2 : 4 (Ca)2.

But 4 (CA) is the square of the diameter of the larger circle, and 4 (Ca) is the square of the diameter of the smaller. Denoting these diameters respectively by D and d, we have,

C : c :: D2 : d2.

That is, the areas of any two circles are to each other, as the squares of their diameters.

SCHOLIUM. As the circumference of every circle, great or small, is assumed to be the measure of 360 degrees, if we conceive the circumference to be divided into 360 equal parts, and one such part represented by AB on one circle, or ab on the other, AB and ab will be very near straight lines, and the length of such a line as AB will be greater or less, according to the radius of the circle; but its absolute length cannot be determined until we know the absolute relation between the diameter of a circle and its circumference.

PROPOSITION III.—THEOREM.

When the radius of a circle is unity, its area and semicircumference are numerically equal.

Let R represent the radius of any circle, and the Greek letter,, the half circumference of a circle whose radius is unity. Since circumferences are to each other as their radii, when the radius is R, the semi-circumference will be expressed by «R.

Let m denote the area of the circle of which R is the radius; then, by Theorem 1, we shall have, for the area of this circle, R2 = m, which, when R = 1, reduces to

T= m.

This equation is to be interpreted as meaning that the semi-circumference contains its unit, the radius, as many