PROPOSITION X. THEOREM. The perimeters of two regular polygons, having the same number of sides, are to each oth as the radii of the circumscribed circles, and also, as the radii of the inscribed circles; and their areas are to each other as the squares of those radii. Let AB be the side of the one polygon, O the centre, and consequently a OA the radius of the circumscribed circle, and OD, perpendicular to AB, the radius of the inscribed circle; let ab, in like manner, be a side of the other polygon, o its centre, oa and od the radii of the circumscribed and the inscribed circles. The perimeters of A D B d VV the two polygons are to each other as the sides AB and ab (Book IV. Prop. XXVII.): but the angles A and a are equal, being each half of the angle of the polygon; so also are the angles B and b; hence the triangles ABO, abo are similar, as are likewise the right angled triangles ADO, ado; hence AB: ab: AO: ao :: DO: do; hence the perimeters of the polygons are to each other as the radii AO, ao of the circumscribed circles, and also, as the radii DO, do of the inscribed circles. The surfaces of these polygons are to each other as the squares of the homologous sides AB, ab; they are therefore likewise to each other as the squares of AO, ao, the radii of the circumscribed circles, or as the squares of OD, od, the radii of the inscribed circles. PROPOSITION XI. THEOREM. The circumferences of circles are to each other as their radii, and the areas are to each other as the squares of their radii. 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 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. G 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. 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. B D For, since the arcs are simi lar, 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 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. D 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 area OA=10A x circ. OA: consequently, the area of a circle is equal to the product of half the radius of the circumference. XVII. Sch. 2.), or as AMBxAC is to D D kel Cor. 2. Let the circumference of the circle whose diameter is unity, be denoted by л: then, because circumferences are to each other as their radii or diameters, we shall have the diameter 1 to its circumference π, as the diameter 2CA is to the circumference whose radius is CA, that is, 1 : : : 2CA : circ. CA, therefore circ. CA=X2CA. Multiply both terms by CA; we have CA× circ. CA =лx CA2, or area CA= =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. D M In like manner, the area of the circle, whose radius is OB, will be equal to л × OВ2; but л × CA2 : л× OB2 : : CA2 : OB2 ; hence the areas 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 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 344; hence 34 or 27 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. Let AB be a side of the given inscribed polygon; EF, parallel to E P M Q F 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':: 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'= √ A× B. |