By making n = 3, a = R √3, we find

A = 3R2√3,

for the area of the inscribed equilateral triangle.

In a similar manner we proceed for the other cases.

THEOREM X.

The side of the inscribed square is to the radius in the ratio of √2 to 1.

Draw the two diameters AB, CD perpendicular to each other, and draw the chords AC, CB, BD, DA. The figure ADBC is evidently a square; and we have

AC2=AO2 + OC2 = 2R2;

consequently AC=R √2, and

AC: R:: √2:1.

P

A

B

M

Scholium. In drawing through the points A, D, B, C, tangents, we form the circumscribed square; and we have

MN=AB=2R.

That is-The side of the circumscribed square is equal to the diameter of the circle.

The areas of these two polygons are expressed respectively by 2R2 and 4R2.

THEOREM XI.

The side of a regular inscribed decagon is equal to the greater segment of the radius, when divided into mean and extreme ratio.

Draw the radii OA, OB to the extremi

ties of a side AB. The angle O of the triangle OAB is equal to of a right angle; there remains, then, 2-3=} of a right angle, for the sum of the other two angles; and as OA = OB, it follows that the angle OAB=OBA = of a right angle.

L

B

Thus, each of the angles at the base of the triangle OAB is double the angle at the vertex.

This being premised, draw BL bisecting the angle OBA, and we shall have this proportion,

AL: LO:: AB: OB (B. III., T. XII.).

But since the angle LBO = LBA = LOB = 3, and consequently ALB=2LOB, the two triangles OLB, ALB are also isosceles, and give OL = LB = AB, and also having OB=OA, the above proportion becomes

AL: OL:: OL: OA.

From which we see that the point L divides the radius OA into mean and extreme ratio (T. III., S. I.), and that the greater segment OL is equal to AB a side of the regular decagon.

Scholium I. We will deduce the numerical value of this side algebraically. Denoting it by x, we have (T. III., S. I.),

That is, the numerical value of the side of a regular inscribed

Scholium II. If for a in formula (2) of Scholium to T. IV. we substitute

R

(√5 1), which we have just found for the value of

R

a side of a regular inscribed decagon, we shall find (10 — 2√/5)* for the side of a regular inscribed pentagon.

R

Now, since [ (10 − 2 √5)'] = [}(√5—1)]*+ R2, it fol

lows, that the square of the side of a regular inscribed pentagon is equal to the square of the side of the regular inscribed decaagon increased by the square of the radius.

THEOREM XII.

The side of a regular inscribed pentadecagon is the chord of the difference between the arcs subtended respectively by the sides of a hexagon and a decagon.

3

=

= 30 30 30

=

For, we have 10%-%-%-%; which shows that -The difference between the arcs subtended by the sides of a hexagon and of the decagon is equal to one fifteenth of the entire circumference.

If we wish the numerical value of the side of the pentadecagon, we may employ formula (4) of T. IV., in which, for c, we must substitute the value of the side of a hexagon, and for 6 we must use the side of a decagon.

General Scholium. It results from what has been said in the last four Theorems, and of the principles before established, that we have geometrical methods for constructing regular inscribed and circumscribed polygons of 3, 6, 12, 24, etc., of 4, 8, 16, 32, etc., of 5, 10, 20, 40, etc., in short, of 15, 30, 60, etc., number of sides.

That we also have arithmetical methods for calculating the sides, and afterwards the areas, of all these polygons, if not exactly-since nearly all the numbers which enter into our calculations are incommensurable-yet to as close a degree of approximation as may be desired.

MEASURE OF THE CIRCLE AND OF ITS CIRCUMFERENCE.

THEOREM XIII.

The area of a circle is equal to half the product of the circumference into its radius.

In effect, let us consider a series of regular circumscribed polygons, each having double the number of sides of the preceding one. The superficies of these polygons have for their respective measures half the product of each perimeter into its apothem—that is, into the radius of the circle. Hence, the area of the circle, which is the limit (T. VIII., S.) of these polygons,

has for its measure half the product of the circumference, which is the limit of the perimeters, into its radius.

Or, considering the circle as a regular polygon of an infinite number of sides, it immediately follows that its area is equal to half the product of its perimeter, that is, of its circumference, into its radius.

Scholium I. Let R, C, and S denote respectively the radius, circumference, and surface of any circle. We shall have

S=C×R=C × R.

We must keep in mind that S is an abstract number, expressing the ratio of the surface of the circle to the unit of surface; C and R are ratios of the circumference and of the radius to the linear unit.

Scholium II. We may also say that the area of a circle is equal to that of a triangle having for its base the rectified circumference (D. IV.), and for its altitude the radius of the circle (B. III., T. XXIII.).

THEOREM XIV.

In two circles, the circumferences are proportional to their radii or to their diameters, and their areas are proportional to the squares of their radii.

First. Let us conceive a series of regular polygons, each having double the number of sides of the preceding one, to circumscribe the circumference C; also another series of similar polygons to circumscribe the circumference C'; and let us designate by R, R' the constant apothems of these two series of polygons; by p, p' the perimeters of two similar polygons, taken, the one in the first series, the other in the second.

This being supposed, we shall have this proportion,

p: p'::R: R',

which is applicable to any two similar polygons whatever; it must therefore hold good at the limit of the perimeters, that is, when 22, p' become C, C'; hence,

C: C':: R: R', or, :: 2R: 2R'.

Secondly. Multiplying this last proportion by the obvious proportion,

R:R'::R: R', or, : : 2R: 2R',

it becomes, C × }R : C' × }R' :: R2: R2, or 4R2: 4R22.

Or, since (T. XIII., S. I.) C × }R = S, C' × R' = S', this becomes S: S':: R2: R2, or 4R2: 4R22.

Or, considering the circle as a regular polygon of an infinite number of sides, the above follows immediately from T. XXVIII., B. III.

Cor. The proportion C: C':: 2R: 2R', may be written

C: 2R:: C': 2R',

and being applicable to any number whatever of circumferences, we have

C: 2R:: C': 2R':: C": 2R" :: C": 2R""::, etc.,

from which we infer that

The ratio of the circumference to the diameter is a constant number.

We usually denote by this constant ratio.

Scholium. If R, C, and S denote respectively the radius, the circumference, and the surface of any circle whatever, we shall have this proportion

:1::C:2R;

consequently, C=2R, from which, by multiplying each term by R, we find CxRS

«R2.

From this, we see that the circumference of any circle may be found by multiplying its diameter (2R) by ; that its surface may be obtained by multiplying the square of its radius by .

THEOREM XV.

The area of a circular sector is equal to half the product of its arc into its radius.

In effect, from the definition of a circular sector, it follows that in the same circle two sectors are proportional to their angles, and consequently to their corresponding