Scholium. If the circumference of a circle is divided into equal arcs, the chords of these arcs are sides of a regular inscribed polygon. For, the sides are equal, because they are chords of equal arcs, and the angles are equal, because they are measured by halves of equal arcs. If the vertices A, B, C, &c., of a E regular inscribed polygon be joined with the centre 0, the triangles thus formed will be equal, because their sides are equal, each to each: hence, all of the angles about the point o are equal to each other. I B DEFINITIONS. 1. The CENTRE OF A REGULAR POLYGON is the common centre of the circumscribed and inscribed circles. 2. The ANGLE AT THE CENTRE is the angle formed by drawing lines from the centre to the extremities of any side. The angle at the centre is equal to four right angles divided by the number of sides of the polygon. 3. The APOTHEM is the shortest distance from the centre to any side. The apothem is equal to the radius of the inscribed circle. Let ABCD be the given circle. Draw any two diameters AC and BD perpendicular to each other; they divide the circumference into four equal arcs (B. III., P. XVII., S.). Draw the chords AB, BC, CD, and DA: then the figure ABCD is the square required (P. II., S.). D Scholium. The radius is to the side of the inscribed square as 1 is to V2. If a regular hexagon is inscribed in a circle, any side is equal to the radius of the circle. H с Let ABD be a circle, and ABCDEH a regular inscribed hexagon: then any side, as AB, is equal to the radius of the circle. Draw the radii OA and OB. Then E D the angle AOB is equal to one sixth of four right angles, or to two thirds of one right angle, because it is an angle at the centre (P. II., D. 2). ( The sum of the two angles OAB and B OBA is, consequently, equal to four thirds of a right angle (B. I., P. XXV., C. 1); but, the angles OAB and OBA are equal, because the opposite sides OB and OA are equal : hence, each is equal to two thirds A of a right angle. The three angles of the triangle AOB are therefore equal, and consequently, the triangle is equilateral: hence, AB is equal to OA; which was to be proved. To inscribe a regular herugon in a given circle. Cor. 1. If the alternate vertices of the regular hexagon are joined by the straight lines AC, CE, and EA, the inscribed triangle ACE is equilateral (P. II., S.). Cor. 2. If we draw the radii OA and oC, the figure AOCB is a rhombus, because its sides are equal: hence (B. IV., P. XIV., C.), we have, or, taking away from the first member the quantity OA, and from the second its equal OB', and reducing, we have, or (B. II., P. XII., C. 2), AC : OA :: V3 : 1; that is, the side of an inscribed equilateral triangle is to the radius, as the square root of 3 is to 1. If the radius of a circle is divided in extreme and mean ratio, the greater segment is equal to one side of a regular inscribed decagon. Let ACG be a circle, OA its radius, and AB, equal to OM, the greater segment of OA when divided in extreme and mean ratio: then AB is equal to the side of a regular inscribed decagon. B hence, the triangles OAB and BAM have the sides about their common angle BAM, proportional; they are, therefore, similar (B. IV., P. XX.). But, the triangle OAB is isosceles; hence, BAM is also isosceles, and consequently, the side BM is equal to AB. But, AB is equal to OM, by hypothesis: hence, BM is equal to OM, and consequently, the angles MOB and MBO are equal. The angle AMB being an exterior angle of the triangle OMB, is equal to the sum of the angles MOB and MBO, or to twice the angle MOB; and because AMB is equal to OAB, and also to OBA, G the sum of the angles OAB and OBA is equal to four times the angle AOB: hence, AOB is equal to one fifth of two right angles, or to one tenth of four right angles; and consequently, the arc AB is equal to one tenth of the circumference: hence, the chord AB is equal to the side of regular inscribed decagon; which was to be proved. ************** M A C B a Cor. 1. If AB is applied ten times as a chord, the resulting polygon is a regular inscribed decagon. Cor. 2. If the vertices A, C, E, G, and I, of the alternate angles of the decagon are joined by straight lines, the resulting figure is a regular inscribed pentagon. Scholium 1. If the ares subtended by the sides of any regular inscribed polygon are bisected, and chords of the semi-arcs drawn, the resulting figure is a regular inscribed polygon of double the number of sides. Scholium 2. The area of any regular inscribed polygon is less than that of a regular inscribed polygon of double the number of sides, because a part is less than the whole. |