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about ABC, and inscribed in it, whose difference shall be less than any given space.

Let AB be a side of a regular inscribed and EF of a similar circumscribed polygon (VI. 4, Cor. 2) having any number of sides, F

Bisect the arc AB in H, and join AH. Then will AH be a side of a regular inscribed polygon of 2n sides. Construct (VI. 4, Cor. 2) the side KL of the similar circumscribed polygon of 2n sides. Draw (III. 15) the tangent AM, and HS parallel to it. Because the triangles LPH, MPO are similar, and (III. 15, Sch.) OP-PH..ALPH - AMPO. Similarly AKAR = ARMO; add to both ARPH; then

KAHL= MAH. Now the difference between the triangles NAB, NEF is the trapezoid ABFE. Hence the difference between the inscribed and circumscribed polygons of n sides is

nx ABFE, or 2nx AGHE. Also the difference between the inscribed and circumscribed polygons of 2n sides is

2n KAHL, or 2n MAH. Because EAM is a right angle, EM> AM. Hence (III. 15, Sch.) EM > HM. Hence (V. 1, Cor. 2) AEMA>AMAH. Also, since (I. 32) AMAH - ASAH, AAHG >AMAH. Therefore AGHE is divided into three triangles, EMA, AMH, AHG, of which MAH is the least; therefore MAH is less than one-third AGHE.

Hence 2n< MAH<j 2nx AGHE.

Therefore the difference between the polygons of 2n sides is less than one-third the difference between the polygons of n

sides. For 4n sides the difference would be less than one-ninth the original difference, and so on. By increasing the number of sides sufficiently the difference may be made 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 circumference of the circle lies between the perimeters of the polygons. Since each time the number of sides is doubled, the difference between the polygons becomes less than one-third its former amount; when the number of sides becomes infinite the difference becomes infinitely small, and the three perimeters may be regarded as coinciding. From these considerations we will hereafter speak of the circumference of a circle being the perimeter of a regular polygon of an infinite number of sides; the area of a circle being the area of such an inscribed polygon, and the radius of a circle being its apothegm.

Proposition 11. Theorem.The circumferences of circles are proportional to their radii.

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Let dBC, 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, and draw AG, BG, DH, EH.

The angles at G and H are equal, because they are equal portions of four right angles; the polygons being reg. ular, and having the same number of sides, the angles of one are equal to the angles

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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 P and 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 1.-Hence the circumference bears a constan ratio to the diameter. This ratio is universally represented by the Greek letter t.

Corollary 2.-The circumference is equal to the radius multiplied by 27.

Let C= circumference, D= diameter, and R=radius. Then

=T, C=D; and since D=2R, C2-R. D

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.

The area of the whole polygon is equal to one half the rectangle of DE and the perimeter of the polygon (VI. 9).

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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 Te.

Let S = area, C= circumference, D=diameter, and R=radius ; then from this Prop., S={R. C. But (VI. 11, Cor. 2) C=21R. Hence, S-4R.2-R = *R.

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 :: TR2 : TR':: R: R??.

Proposition 13.
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Let ABC be a circle, AC the side of a regular inscribed polygon, DE of a similar circumscribed polygon, AB, HK of regular inscribed and circumscribed polygons of double the number of sides.

Let P and p represent 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 (1 6); 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 p as 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. Now, AF is contained in P as many times as HK is contained in P'.

... (IV. 4),

p+P: 2p :: P: P'.

2pP P'

(1)

Hence,

P+P

If now we have numerical values for p and P, we may obtain the value for P by formula (1). Again, in the similar triangles ABG, HBL (V. 4),

AG : AB :: BL : BH.

But AG, AB are equal parts of p and p', and BL, HB equal parts of p' and P';

P :p :: p' : P'. Hence,

p' = VpP'. (2) If, then, we have given a numerical value for p, and have, by formula (1), found one for P', we may find a value for p by formula (2).

Now, we know that the perimeter of a circumscribed square in a circle whose diameter is 1, is 4, and the perimeter of an inscribed square in the same circle is 272; hence, substituting these values for P and p in the formulæ just proved, we have

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