gon (VI. 6, Cor.). If the whole circumference were to be divided into 15 equal parts, AB would subtend 5, and AC, 3. Therefore BC must contain 2; bisect (III. 23) BC in D, and lay off BD around the circle; we would have a regular pentedecagon.

Scholium.-If we bisect (III. 23) the arcs of a circle which are subtended by the sides of a square, we may form a regular polygon of 8 sides; from this, one of 16, and so on. Similarly from a decagon, we can construct polygons of 20, 40, etc., sides; from a hexagon, polygons of 12, 24, etc.; from a pentedecagon, figures of 30, 60, etc.

Proposition 9.

Theorem. The area of a regular polygon is equal to onehalf the product of its perimeter and apothem.

Theorem. About a given circle a polygon may be described, which shall differ from a similar inscribed polygon, by a space less than any given space.

Let ABC be a circle; similar polygons may be described about ABC, and inscribed in it, whose difference shall be less than any given space.

Let AB be a side of an inscribed, and EF of a similar circumscribed polygon (VI. 4, Cor.

ence of the polygons that is, the sum of all trapezoids, AF, etc. -is (P+p)HG, and is therefore less than P× HG.

Now because DG+GA>AD, subtracting DG from both, GA > AD – DG, or GA>HG. By successively doubling the number of sides of the polygons, AG, and consequently GH, which will always be less than AG, may be made as small as desired, and the space PxHG may be made less than any given space; but the difference between the polygons is always less than PxHG. Hence this difference may be less than any given space.

Corollary. A circle may be regarded as a regular polygon of an infinite number of sides, and whatever is true of one is true of the other.

For the area of the circumscribed polygon will always be greater than the area of the circle, and of the inscribed polygon less; as we increase the number of their sides, they approach more nearly to each other, and consequently to the circle. The circles is therefore their limit, as the number of their sides increases indefinitely. For any finite number of sides the difference between a polygon and the circle is finite; the number of sides may increase so that the difference will become less than any given quantity; hence, as this number increases above any finite quantity-that is, to infinity-the difference becomes less than any finite quantity; that is, zero. The area of the polygon coincides with the area of the circle, the perimeter with the circumference, and the apothem with the radius. Hence, the truth of the corollary.

Proposition 11.

Theorem. The circumferences of circles are proportional to their radii.

Let ABC, DEF be two circles, and AG, DH their radii; their circumferences are proportional to AG, DH.

In ABC, DEF describe the polygons ABC, DEF of the

same number of sides, the angles of one are equal to the angles of the other; hence the angles GAB, GBA are equal to the angles HDE, HED; therefore the triangles GAB, HDE are similar,

and (V. 4) alternately, AB: DE :: AG: DH.

If P represent the perimeter of the polygon ABC, and P' of DEF, then Pand P' being equimultiples of AB and DE, (IV. 3) P:P':: AG: DH.

And this is true whatever the number of sides; it is therefore true when that number becomes infinite, in which case the perimeters of the polygons become the circumferences of the

circles (VI. 10, Cor.). Hence the circumferences of circles are proportional to their radii.

Corollary. Hence the circumference bears a constant ratio to the diameter. This ratio is universally represented by the Greek letter π.

If C represent the circumference, D the diameter, and R the radius, then

C
D

= π, C=D; and since D = 2R, C=2R.

Proposition 12.

Theorem. The area of a circle is equal to one half the rectangle contained by its radius, and a straight line equal to its circumference.

Let ABC be a circle and AD its radius; the area ABC

is equal to the rectangle contained

by AD, and a straight line equal to one half its circumference.

Inscribe in ABC the regular polygon ABC; join DF, DB, and draw the perpendiculars DE, DG.

A

C

E

F

G

D

B

The area of the whole polygon is equal to one half the rectangle of DE and the perimeter of the polygon (VI. 9). Now this being true, whatever the number of sides of the polygon, it is true if that number be infinitely increased. But in this case the polygon becomes a circle (VI. 10, Cor.), its perimeter agrees with the circumference, and its apothem with the radius. Hence the area of the circle is equal to one half the rectangle of the radius, and a straight line equal to the circumference, or the product of half the diameter and half the circumference.

Corollary 1.- The area of a circle is equal to the square of the radius multiplied by π.

Let S = area, C = circumference, D = diameter, and R = radius; then from this Prop., S=4R.C. But (VI. 11, Cor.) C= 2R. Hence, SR. 2R = πR2.

Corollary 2.- Circles are proportional to the squares of their radii.

For, let S, S' be areas of two circles, and R, R' their radii. Then (last corollary) S : S' : : πR2 : πR'2 :: R2 : R'2.

Proposition 13.

Problem. To find the approximate value of π.

Let P and prepresent the perimeters of the given circumscribed and inscribed polygons, and P' and p' the perimeters of the circumscribed and inscribed polygons of double the number of sides.

Now, because HB = HA (V. 27, Cor. 2), the angles HAB, HBA are equal (I. 13); also the angles HBA, BAG are equal (I. 27); therefore the angle FAG is bisected by AB; ... (V.3), AG: AF or DB :: BG : BF.

But AG is contained in pas many times as DB is contained in P;

... (IV. 4),

p:P::GB: BF :: (V.2), AH : HF;

by composition (IV. 9), p+P:p ::AF: AH;

hence (IV. 5),

p+P: 2p :: AF : 2AH or HK.