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Let us designate the circumference of the circle whose radius is CA by circ. CA; and its area, by area CA: it is then to be shown that

T

circ. CA circ. OB

area CA area OB :: CA2 : OB2

B

E

MAN

P

DO

S

CA: OB, and that

Inscribe within the circles two regular polygons of the same number of sides. Then, whatever be the number of sides, their perimeters will be to each other as the radii CA and OB (Prop. X.). Now, if the arcs subtending the sides of the polygons be continually bisected, until the number of sides of the polygons shall be indefinitely increased, the perimeters of the polygons will become equal to the circumferences of the circumscribed circles (Prop. VIII. Cor. 2.), and we shall have circ. CA circ. OB :: CA : OB.

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

Again, the areas of the inscribed polygons are to each other as CA2 to OB2 (Prop. X.). But when the number of sides of the polygons is indefinitely increased, the areas of the polygons become equal to the areas of the circles, each to each, (Prop. VIII. Cor. 1.); hence we shall have

area CA area OB :: CA2 : OB2.

A

G

[blocks in formation]

C

For, since the arcs are similar, the angle C is equal to the angle O (Book IV. Def. 3.) ; but C is to four right angles, as the arc AB is to the whole circumference described with the radius AC (Book III. Prop. XVII.); and O is to the 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

C

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 ACDE be a circle whose centre is O and radius OA: then will

F

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

area OA=40A x circ. OA.

For, inscribe in the circle any E regular polygon, and draw OF perpendicular to one of its sides. Then the area of the polygon will be equal to OF, multiplied by the perimeter (Prop. IX.). Now, let the number of sides of the polygon be indefinitely increased by continually bisecting the arcs which subtend the sides: the perimeter will then become equal to the circumference of the circle, the perpendicular OF will become equal to OA, and the area of the polygon to the area of the circle (Prop. VIII. Cor. 1. & 3.). But the expression for the area will then become

For, the sector ACE is to the whole circle as the arc AMB is to the whole circumference ABD (Book III. Prop. XVII. Sch. 2.), or as AMB × AC is to ABDAC. But the whole circle is equal to ABD × AC; hence the sector ACB is measured by AMB × AC

A

area OA=!0A × circ. OA : consequently, the area of a circle is equal to the product of half the radius into the circumference.

M

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=TX 2CA. Multiply both terms by CA; we have CA x circ. CA =л× CA2, or area СA=л× CA2: hence the area 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 arca of the circle, whose radius is OB, will be equal to × OB2; but л× CA2: × OB2 :: CA2: OB2; hence the areas of circles are to each other as the squares of their radii, which agrees with the preceding theorem.

D

M

B

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 a mean proportional between its length and its breadth (Book IV. Prob. III.). 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 approximatively; 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. Accordingly, 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 possessing the slightest tincture of geometrical science will occupy any portion of his time.

Archimedes showed that the ratio of the circumference to the diameter is included between 34% and 341; lence 34 or 22 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 355 At last the value of π, 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 problem will exhibit one 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.

E P M Q

Let AB be a side of the given inscribed polygon; El', parallel to AB, a side of the circumscribed polygon; C the centre of the circle. If the chord AM and the tangents AP, BQ, be drawn, AM will be a side of the inscribed polygon, having twice the number of sides; and AP+PM=2PM or PQ, will be a side of the similar circumscribed polygon (Prop. VI. Cor. 3.). Now, as the same construction 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: ČD 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'= √ ̈Ã × B.

D

F

B

Secondly. The altitude CM being common, the triangle CPM is to the triangle CPE as PM is to PE; but since CP bisects the angle MCE, we have PM: PE : : CM CE (Book IV. Prop. XVII.): 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 been already determined; this new proportion will and thus by

2A.B

serve for determining B', and give us B'=

EP M

A

PROPOSITION XIV. PROBLEM.

D

A+A'
,;

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

VA.B 3.0614674, and B'

=

Q

=

B

To find the approximate ratio of the circumference to the diameter.

Let the radius of the circle be 1: the side of the inscribed square will be ✓2 (Prop. III. Sch.), 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,

16

2+8=3.3137085. The

and the circumscribed octagon B'=; inscribed and the circumscribed octagons 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'2 A.B

=

3.1825979. These polyA+A' gons of 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, and so far as the seventh place, In this example. Being arrived at this point, we shall infer

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