Neither can AB be to CD, as the circumference of OR to one greater than that of FK. For if it could, we should have, by inversion, CD to AB, as a circumference greater than that of FK to the circumference of OR, or, which is equivalent, as the circumference of FK to one less than that of OR. But this has been proved to be impossible: wherefore AB is to CD, as the circumference of OR to the circumference of FK.

Cor. Hence the arcs of sectors which have equal angles, are as the radii of the sectors. For the arcs are evidently proportional to the entire circumferences, the angles at the centres being equal.

PROF. XXXIX. THEOR.-The area of a circle is equal to the rectangle under its radius, and a straight line equal to half its circumference.

Let AB be the radius of the circle BC: the area of BC is equal to the rectangle under AB, and a straight line D equal to half the circumference.

E

H

B

A

K

For, if the rectangle AB.D be not equal to the circle BC, it is equal to a circle either greater or less than BC. First, suppose, if possible, the rectangle AB.D to be the area of a circle EF, of which the radius AE is greater than AB, and let GHK be a regular polygon described about the circle BC, such that its sides do not meet the circumference of EF. Then, by dividing this polygon into triangles by radii drawn to G, H, K, &c. it would be seen that its area is equal to the rectangle under AB and half its perimeter. But (APP. I. 37) the perimeter of the polygon is greater than the circumference of BC: and therefore the area of the polygon is greater than the rectangle AB.D, that is, by hypothesis, than the area of the circle EF, which is absurd, since the polygon is only a part of that circle, being contained within it." The rectangle AB.D, therefore, is not equal to the area of any circle greater than BC.

M

D

In the second place, this rectangle cannot be less than BC. For suppose, if possible, that it is equal to the area of a circle described from A as centre, within BC. Then, by describing about that circle a regular polygon, the sides of which would not meet BC, it would be shown as before, that the area of the polygon would be equal to the rectangle under half its perimeter and the radius of the interior circle, and would therefore be less than the rectangle AB.D, because AB is greater than the radius, and (APP. I. 37) D greater than half the perimeter. Therefore the polygon being less than AB.D, would be less than the circle about which it is described, which is absurd.

Hence the rectangle under the radius AB and half the circumference, is not equal to a circle greater than BC, nor to one less: it is therefore equal to BC.

Cor. 1. Hence the surface of a sector ALM is equal to the For the circle BC is rectangle under its radius and half its arc. to ALM, as the circumference of BC to the arc LM, sectors and arcs being proportional (VI. 33) to their angles at the centre. Hence, by halving the third and fourth terms, BC is to ALM, as D to LM, or as D.AB to LM.AB: therefore (V. 14) μLM.AB is equal to the sector ALM, because BC is equal to D.AB.

Cor. 2. Let the circumference of the circle whose diameter is unity, and consequently half the circumference of the circle whose radius is unity, be denoted by T. Then (APP. I. 38) 1 is to π, as 2 AB to the circumference of BC. Hence we find the circumference to be 2X AB; and multiplying half of this by the radius AB, we find the area to be equal to TX AB. Hence the area of any circle is found by multiplying the square of its radius by the constant number T, which, as will be shown in the scholium to the next proposition, is 3.141593, nearly.

PROP. XL. PROB.-The area of a regular inscribed polygon, and that of a regular circumscribed one of the same number of sides being given; to find the areas of the regular inscribed and circumscribed polygons having double the number of sides.

Let A be the centre of the circle, BC a side of the inscribed

D

G

K

C

polygon, and DE parallel to BC, a side of the circumscribed one, Draw the perpendicular AFG, and the tangents BH, CK. and join BG: then BG will be a side of the inscribed polygon of double the number of sides; and (IV. B) HK is a side of the similar circumscribed one. Now, as a like construction would be made at each of the remaining angles of the polygon, it will be sufficient to consider the part here represented, as the triangles connected

with it are evidently to each other, as the polygons of which they are parts. For the sake of brevity, then, let P denote the polygon whose side is BC, and P' that whose side is DE; and, in like manner, let Q and Q', represent those whose sides are BG and HK: P and P', therefore, are given; Q and Q', required.

:

Now (VI. 1) the triangles ABF, ABG are proportional to their bases AF, AG, as they are also to the polygons P, Q; therefore AF AG PQ. The triangles ABG, ADG are likewise as their bases AB, AD, or (VI. 4) as AF, AG; and they are also as the polygons Q and P: therefore AF: AG: Q: P'; wherefore (V. 11.)PQ:Q: P'; so that Q is a mean proportional between the given polygons P, P'; and, representing them by numbers, we have Q1=PP', so that the area of Q will be computed by multiplying P by P', and extracting the square root of the product.

:

D

H

G

K

E

Again, because AH bisects the angle GAD, and because the triangles AHD, AHG are as their bases, we have GH: HD :: AG: AD, or AF: AB :: AHG AHD. But we have already seen that AF AG :: P: Q; and therefore (V. 11) AHG : AHĎ :: P: Q. Hence (V. E, cor.) AHG: ADG:: ::P:P+Q; whence, by doubling the antecedents, 2AHG: ADG :: 2P P+Q. But it is evident, that whatever part the triangle ADG is of P', the same part of the polygon Q' is the triangle AHK, which is double of the triangle AHG. Hence the last analogy becomes Q': P' :: 2P:P+ Q. Now (V. Sup. 2, cor. 1) the product of the extremes is equal to the

B

F

C

product of the means; and therefore Q' will be computed by divid ing twice the product of P and P' by P+Q; and the mode of finding Q has been pointed out already.

Schol. This proposition enables us to find the approximate ratio of the diameter and circumference of a circle. Thus, let the radius be 1, and it is plain that the area of the circumscribed square is 4, and that of the inscribed one 2, being (1.41) half of the circumscribed one. Put therefore P-2 and P'=4; and by the methods found above, we have the area of the inscribed octagon Q = √ 8, 16* and the area of the circumscribed octagon Q'=;

2+8

: or by actual

computation, Q=2·8284271, and Q'=3.3137085. By using these numbers instead of P and P', we should find the areas of the inscribed and circumscribed regular polygons of sixteen sides. From them, in like manner, we should derive the areas of regular polygons of thirty-two sides: and thus, we may proceed as far as we please, each succeeding operation giving the areas of the inscribed and circumscribed polygons of double the number of sides of those employed in finding them. By this means the following table has been calculated, the decimals having been carried out to eight places, and the last rejected, so as to secure their accuracy for seven places.

✦ By an easy and well-known algebraic operation (see the Author's Algebra, page 86), this may be reduced to the preferable form, 4(8—2).

Now, as the area of the circle is intermediate between the areas of the inscribed and circumscribed polygons, and as the areas of the last two in the table are the same in all the figures to which they are carried out, it follows that the area of the circle must also be 3.1415926: and as, by the last proposition, this is equal to the product of the radius 1, and half the circumference, the half circumference to the radius 1, or the whole circumference to the diameter 1, is the same number 3.1415926. This number is usually denoted by .*

PROP. XLI. THEOR.-If the diameter of a circle be divided into any two parts, AB, BC, and if semicircles ADB, BEC be described on opposite sides of these, the circle is divided by their arcs into two figures GDE, FED, the boundary of each of which is equal to the circumference of FG; and which are such that AC: BC :: FG: FED, and AC: AB :: FG: GDE.

F

For (APP. I. 38) the circumferences of circles, and consequently the halves of their circumferences, are to one another as their diameters: therefore AB is to AC, as the arc ADB to AFC, and BC is to AC, as the arc BEC to AFC. Hence (V. 24) AC is to AC, as the compound arc ADEC to AFC therefore ADEC is equal to half the circumference; and the entire boundaries of the figures GDE, FED are each equal to the circumference of FG.

Again, (XII. 2) circles, and consequently semicircles, are to one another as the squares

A

C

B

of their diameters: therefore AC2 is to AB2 as the semicircle AFC to the semicircle ADB. Hence, since (II. 4) AC2= AB2+

* This number, when carried out to more places, is found to be 3.14159265358979, &c.; and by the method now pointed out, or by others still easier, especially by series' furnished by the modern analysis, it may be continued to any extent whatIt cannot be determined, however, with complete precision; and hence the celebrated problem of the quadrature of the circle can be solved only by approximation.

ever.

The ratio of 1 to is nearly that of 7 to 22, which gives by division 3·1428. This agrees with the value of 7, so far as to produce an error in excess in the circumference, which is only a little more than the eight hundredth part of the radius: and on account of the smallness of the numbers, it is much used, when great accuracy is not required. The approximate ratio of 113 to 355, discovered by Metius, is also very convenient, and is so near the truth, as to give an error in excess in the circumference, not much exceeding a four millionth part of the radius.

2AB.BC+BC2, we find by conversion, that AC2 is to 2AB.BC+ BC2, as the semicircle AFC to the remaining space BDAFC; whence, by inversion, 2AB.BC+BC2 is to AC, as BDAFC to the semicircle AFC. But BC2 is to AC2, as the semicircle BEC to the semicircle AFC; and therefore (V. 24) 2AB.BC+2BC2 is to AC2, as the compound figure FDE to the semicircle AFC. But (II. 3) 2AB.BC+2BC2 = 2AC.BC, and (VI. 1) 2AC.BC: AC: BC AC. Hence the preceding analogy becomes BC to AC, as FED to the semicircle, or by doubling the consequents, and by inversion, AC to BC, as FG to FED: and it would be proved, in the same manner, that AC: AB:: FG GDE.

Cor. Hence we can solve the curious problem, in which it is required to divide a circle into any proposed number of parts, equal in area and boun

K

G

E

B

D

H

L

F

dary; as it is only necessary to divide the diameter into the proposed number of equal parts, and to describe the semicircles as in the annexed diagram. Then, whatever A part AB is of AD the same part is LEF of the circle KL; and AC being double of AB, LGH is two of these parts: GEFH is therefore one of them; and so on, whatever is their number. Their boundaries are also equal, the boundary of each being equal to the circumference of the circle.*

PROP. XLII. PROB.-To divide a given circle ABC into any proposed number of equal parts, by means of concentric circles.

G

H

B

Divide the radius AD into the proposed number of equal parts, suppose three, in the points E, F, and through these points draw perpendiculars to AD, meeting a semicircle described on it as diameter in G, H; from D as centre, at the distances DG, DH, describe the circles GI, HK: their circumferences divide the circle into equal parts.

E

F D

L

M

K

Join AH, DH. Then (VI. 20, cor. 2) AD, DH, DF being continual proportionals, AD is to DF, as a square described on AD is to one described on DH. But (XII. 2) circles are proportional to the squares of their diameters, and consequently to the squares of their radii. Hence (V. 11) AD is to FD as the circle

* Another solution would be obtained, if the circumference were divided into the proposed number of equal parts, and radii drawn to the points of division. This division, however, can be effected only in some particular cases by means of elementary geometry.