Hence, too, we conclude it to be equally impossible that CA can be to OB as circ. CA is to a circumference greater than circ. OB for if this were the case, by reversing the ratios, we should have OB to CA as a circumference greater than circ. OB is to circ. CA; or, what amounts to the same thing, as circ. OB is to a circumference less than circ. CA; and therefore one radius would be to another as the circumference described with the former radius is to a circumference less than the one described with the latter radius,—a conclusion just shewn to be

erroneous.

And since the fourth term of this proportion CA: OB:: circ. CA: x can neither be greater nor less than circ. OB, it must be equal to circ. OB: consequently the circumferences of circles are to each other as their radii.

By the same construction, a similar train of reasoning would shew, that the surfaces of circles are to each other as the squares of their radii. We need not enter upon any farther details respecting this proposition, particularly as it forms a corollary of the next theorem.

Cor. The similar arcs AB, DE are to each other as their radii A AC, DO; and the similar sectors ACB, DOE are to each other as the squares of those radii.

For, since the arcs are similar,

the angle C (Def. 3. III.) is equal to the angle O; but C is to four right angles (17. II.), as the arc AB is to the whole circumference described with the radius AC; and O is to four right angles, as the arc DE is to the circumference described with the radius OD: hence the arcs AB, DE are to each other as the circumferences of which they form part; but these circumferences are to each other as their radii AC, DO; hence arc. AB: arc. DE:: AC: DO.

For a like reason, the sectors ACB, DOE are to each other as the whole circles; which again are as the squares of their radii; therefore sect. ACB; sect. DOE: AC2 : DO2.

PROPOSITION XII. THEOREM.

The area of a circle is equal to the product of its circumference

by half the radius

Let us designate the surface of the circle whose radius is CA by surf. CA; we shall have surf. CA CA X circ. CA.

For if CAxcirc. CA is not the area of the circle whose radius is CA, it must be the area of a circle either greater or less. Let us first suppose it to be the area of a greater circle; and, if possible, that CAxcirc. CA= surf. CB.

About the circle whose radius is CA describe a regular polygon

B

DEFG &c., such (10. IV.) that its sides shall not meet the circumference whose radius is CB. The surface of this polygon will be equal (7. IV.) to its perimeter DE+EF+FG+ &c. multiplied by AC: but the perimeter of the polygon is greater than the inscribed circumference enveloped by it on all sides; hence the surface of the polygon DEFG &c. is greater than ACX circ. AC, which by the supposition is the measure of the circle whose radius is CB; thus the polygon must be greater than that circle. But in reality it is less, being contained wholly within the circumference: hence it is impossible that CAX circ. AC can be greater than surf. CA; in other words, it is impossible that the circumference of a circle multiplied by half its radius can be the measure of a greater circle.

In the second place, we assert it to be equally impossible that this product can be the measure of a smaller circle. To avoid the trouble of changing our figure, let us suppose that the circle in question is the one whose radius is CB: we are to shew that CBX circ. CB cannot be the measure of a smaller circle, of the circle, for instance, whose radius is CA. Grant it to be so; and that, if possible, CBxcirc. CB=surf. CA.

Having made the same construction as before, the surface of the polygon DEFG, &c. will be measured by (DE+EF+FG+ &c.) ×CA; but the perimeter DE+EF+FG+ &c. is less than circ. CB, being enveloped by it on all sides; hence the area of the polygon is less than CA× circ. CB, and still more than CBX circ. CB. Now by the supposition, this last quantity is the measure of the circle whose radius is CA: hence the

polygon must be less than the inscribed circle, which is absurd; hence it is impossible that the circumference of a circle multiplied by half its radius, can be the measure of a smaller circle. Hence, finally, the circumference of a circle multiplied by half its radius is the measure of that circle itself.

Cor. 1. The surface of a sector is equal to the arc of that sector multiplied by half its radius.

For, the sector ACB (17. II.) is to the whole circle as the arc AMB is to the whole circumference ABD, or as AMBX AC is to ABDX AC. But the whole circle is equal to ABDX AC; hence the sector ACB is measured by AMB × AC.

D

B

Cor. 2. Let the circumference of the circle whose diameter is unity be denoted by : then, because circumferences are to each other as their radii or diameters, we shall have the diameter 1 to its circumference as the diameter 2CA is to the circumference whose radius is CA, that is, 1: : : 2CA: circ. CA, therefore circ. CA=2 × CA. Multiply both terms by CA; we have ¿CA× circ. CA== × CA2, or surf. CA=XCA: hence the surface of a circle is equal to the product of the square of its radius by the constant number which represents the circumference whose diameter is 1, or the ratio of the circumference to the diameter.

In like manner, the surface of the circle, whose radius is OB, will be equal to XOB2; but × CA2: «×OB2 : : CA2 : OB2; hence the surfaces of circles are to each other as the squares of their radii, which agrees with the preceding theorem.

Scholium. We have already observed, that the problem of the quadrature of the circle consists in finding a square equal in surface to a circle, the radius of which is known. Now it has just been proved, that a circle is equivalent to the rectangle contained by its circumference and half its radius; and this rectangle may be changed into a square, by finding (6. III.) a mean proportional between its length and its breadth. To square the circle, therefore, is to find the circumference when the radius is given; and for effecting this, it is enough to know the ratio of the circumference to its radius or its diameter.

Hitherto, the ratio in question has never been determined except approximately; but the approximation has been carried so far, that a knowledge of the exact ratio would afford no real advantage whatever beyond that of the approximate ratio. Ac cordingly, this problem, which engaged geometers so deeply, when their methods of approximation were less perfect, is now degraded to the rank of those idle questions, with which no one

G

possessing the slightest tincture of geometrical science will occupy any portion of his time.

Archimedes shewed that the ratio of the circumference to the diameter is included between 348 and 39; hence 34 or 2 affords at once a pretty accurate approximation to the number above designated by ; and the simplicity of this first approximation has brought it into very general use. Metius, for the same number, found the much more accurate value. At last the value of x, developed to a certain order of decimals, was found by other calculators to be 3.1415926535897932, &c.; and some have had patience enough to continue these decimals to the hundred and twenty-seventh, or even to the hundred and fortieth place. Such an approximation is evidently equivalent to perfect correctness: the root of an imperfect power is in no case more accurately known.

The following problems will exhibit two of the simplest elementary methods of obtaining those approximations.

PROPOSITION XIII. PROBLEM.

The surface of a regular inscribed polygon, and that of a similar polygon circumscribed, being given; to find the surfaces of the regular inscribed and circumscribed polygons having double the number of sides.

struction will take place at each of the angles equal to ACM, it will be sufficient to consider ACM by itself, the triangles connected with it being evidently to each other as the whole polygons of which they form part. Let A, then, be the surface of the inscribed polygon whose side is AB, B that of the similar circumscribed polygon; A' the surface of the polygon whose side is AM, B that of the similar circumscribed polygon: A and B are given; we have to find A' and B'.

First. The triangles ACD, ACM, having the common vertex A, are to each other as their bases CD, CM; they are likewise to each other as the polygons A and A', of which they form

. part: hence A: A :: CD: CM. Again, the triangles CAM, CME, having the common vertex M, are to each other as their bases CA, CE; they are likewise to each other as the polygons A and B of which they form part: hence A': B:: CA: CE. But since AD and ME are parallel, we have CD:CM:: CA: CE; hence A: A':: A': B; hence the polygon A', one of those required, is a mean proportional between the two given polygons A and B, and consequently A=AXB.

Secondly. The altitude CM being common, the triangle CPM is to the triangle CPE as PM is to PE; but (17. III.) since CP bisects the angle MCE, we have PM: PE::CM:CE :: CD:CA:: A: A': hence CPM: CPE:: A: A'; and consequently CPM: CPM + CPE or CME :: A: A+A'. But CMPA or 2CMP and CME are to each other as the polygons B' and B, of which they form part: hence B: B::2A: A+A'. Now A' has already been determined; this new proportion will serve for determining B', and give us B'=

2A.B

A+A; and thus by means of the polygons A and B, it is easy to find the polygons A and B', which have double the number of sides.

PROPOSITION XIV. PROBLEM.

To find the approximate ratio of the circumference to the dia

meter.

2+√8

Let the radius of the circle be 1; the side of the inscribed square will (3. IV.) be 2, that of the circumscribed square will be equal to the diameter 2; hence the surface of the inscribed square is 2, and that of the circumscribed square is 4. Let us therefore put A=2, and B=4; by the last proposition, we shall find the inscribed octagon A'=/8=2.8284271, and the 16 circumscribed octagon B/= =3.3137085. The inscribed and the circumscribed octagon being thus determined, we shall easily, by means of them, determine the polygons having twice the number of sides. We have only in this case to put A=2.8284271, B-3.3137085; we shall find A'√A.В= 2A.B 3.0614674, and B= -3.1825979. These polygons of A+A 16 sides will in their turn enable us to find the polygons of 32; and the process may be continued, till there remains no longer any difference between the inscribed and the circumscribed polygon, at least so far as that place of decimals where the computation stops,—so far as the seventh place, in this example. Being