PROPOSITION IX. THEOREM. to each The perimeters of similar regular polygons are other as the radii of their circumscribed or inscribed circles; and their areas are to each other as the squares of those radii. 1o. Let ABC and KLM be similar regular polygons. Let OA and QK be the radii of their circumscribed, OD and QR be the radii of their inscribed circles: then will the perimeters of the polygons be to each other as to QK, or as OD is to QR. For, the lines OA and QK are homologous lines of the polygons to which they belong, as are also the lines OD and QR hence, the perimeter of ABC K M OA is is to the perimeter of KLM, as OA is to QK, or as OD is to QR (B. IV., P. XXVII., C. 1); which was to be proved. 2o. The areas of the polygons will be to each other as OA is to QK', or as OD2 is to QR2. For, OA being homologous with QK, and OD with QR, we have, the area of ABC is to the area of KLM, as ŌA is to QK', or as OD is to QR2 (B. IV., P which was to be proved. XXVII., C. 1); : Two regular polygons of the same number of sides can be constructed, the one circumscribed about a circle and the other inscribed in it, which shall differ from each other by less than any given surface. Let ABCE be a circle, O its centre, and the side of a square which is less than the given surface ; then can two similar regular polygons be constructed, the one circumscribed about, and the other inscribed within the given circle, which shall differ from each other by less than the square of 2, and consequently, by less than the given surface. Inscribe a square in the given circle (P. III.), and by sion, regular polygons of 8, 16, Construct a similar circum scribed polygon abcde: then b D will these polygons differ from each other by less than the square of Q. For, from a and b, draw the lines a and b0; they will pass through the points A and B. Draw also OK to the point of contact K; it will bisect AB at I and Prolong AO to E. be perpendicular to it. Let P denote the circumscribed, and p the inscribed polygon; then, because they are regular and similar, we shall have (P. IX.), hence, by division (B, II., P. VI.), we have, b D KI aA E But P is less than the square of AE (P. VII., C. 4); hence, P p is less than the square of AB, and conse quently, less than the square of Q, or than the given surface; which was to be proved. Cor. 1. If the number of sides of the polygons be made. greater than any assignable number; that is, infinite, the difference between their areas will be less than any assignable surface; that is, it will be zero*. Cor. 2. When the number of sides of the polygons is infinite, either polygon differs from the circle by less than any assignable quantity; for, the circumference of the circle lies between the perimeters of the polygons: hence, the eircle differs from either polygon by less than they differ from each other. * Univ. Algebra, Arts. 72, 73. Bourdon, Art. 71. Scholium 1. The circle may be regarded as the limit of the inscribed polygons; that is, it is a figure towards which a polygon may be made to approach as near as desirable, but beyond which it cannot be made to pass. Scholium 2. The circle may be regarded as a regular polygon of an infinite number of sides, and because of the principle, that whatever is true of a whole class, is true of every individual of that class, we may affirm that whatever is true of regular polygons, is also true of circles. Scholium 3. When the circle is regarded as a regular polygon, the circumference is to be regarded as its perimeter, and the radius as its apothem. PROPOSITION XI. PROBLEM. The area of a regular inscribed polygon, and that of a similar circumscribed polygon being given, to find the areas of the regular inscribed and circumscribed polygons having double the number of sides. 1 Let AB be the side of the given inscribed, and EF that of the given circumscribed polygon. Let be their common centre, AMB a portion of the circumference of the circle, and M the middle point of the arc AMB. Draw the chord AM, and at A and B draw the tangents AP and BQ; then, will AM be the side of the inscribed polygon, and PQ the side of and CF. E P M F B A Denote the area of the given inscribed polygon by p, the area of the given circumscribed polygon by P, and the areas of the inscribed and circumscribed polygons having double the number of sides, respectively by p' and P'. 2o. Because the triangles CPM and CPE have the common altitude CM, they are to each other as their bases: hence, СРМ : CPE :: PM and because CP bisects the angle ACM, we have (B. IV., P. XVII.), PM : PE :: CM: CE :: CD : CA; hence (B. II., P. II.), СРМ : CPE :: CD : CA or CM. But, the triangles CAD and CAM have the cominon altitude AD; they are therefore, to each other as their bases: hence, CAD CAM :: CD CM; |