(Why ?) (Why ?) Proof. A ABO is equilateral. .. arc BA= 1 of the circumference, and ABCDEF is a regular hexagon. Q.E.F. 398. Cor. 1. By joining the alternate vertices of an inscribed regular hexagon, an inscribed equilateral triangle is formed. 399. CoR. 2. Polygons of 3, 6, 12, 24, etc., sides may be inscribed in and circumscribed about a given circle. Ex. 926. The side of an equilateral triangle is equal to its radius multiplied by V3. Ex. 927. If the radius of a circle is r, the apothem of the inscribed hexagon is A equal to B V3. Ex. 928. To circumscribe a regular hexagon about a given circle. Ex. 929. To construct a regular polygon of twelve sides, having given a side. Ex. 930. In Ex. 929 how many degrees are in the angle at the center? Ex. 931. Find the area of a regular hexagon if its radius is equal to r. Ex. 932. The apothem of an equilateral triangle is equal to one-half its radius. Ex. 933. The area of an inscribed equilateral triangle is equal to onehalf the area of the inscribed regular hexagon. Ex. 934. The areas of triangles inscribed in equal circles are to each other as the products of their three sides. Ex. 935. A square constructed on a diameter of a circle is equivalent to twice the area of the inscribed square. PROPOSITION VI. PROBLEM 400. To inscribe a regular decagon in a given circle. Construction. In the given circle AB, draw any radius 0A, and divide it into extreme and mean ratio, so that OA: OC = OC: CA. Then OC is the side of the required decagon, and by applying it ten times as a chord, a regular decagon ADEF, etc., is formed. Proof. Draw DO and DC. AO:OC = OC: CA, or since OC= AD, (Why ?) (Why ?) But AOAD being isosceles, the similar triangle DAC must be isosceles, or, CD= AD= CO, whence 20=2CDO. (Why ?) But 20= LCDA. .:. 2 20= LODA, (Ax. 2.) and 220= 2 DAO. But 20+ZODA + ZDAO = 2 rt. Is. ..520= 2 rt. , 20={rt. Z, or of 4 rt. 6. .. arc AD is of the circumference, and AD is the side of an inscribed regular decagon. or Q.E.F. 401. Cor. 1. By joining the alternate vertices A, E, G, etc., an inscribed regular pentagon is formed. 402. Cor. 2. Polygons of 5, 10, 20, etc., sides may be inscribed in and circumscribed about a given circle. PROPOSITION VII. PROBLEM 403. To construct the side of a regular polygon of fifteen sides inscribed in a given circle. D B Construction. In the given circle ABD, draw the chord AB equal to the radius, and chord AC equal to the side of the inscribed regular decagon. Then CB is the side of the inscribed regular polygon of fifteen sides. Proof. Arc AB = d of the circumference. Arc AC = b of the circumference. Arc CB=\-, or t of the circumference, and CB is the side of the required polygon. Q.E.F. 404. Cor. Polygons of 15, 30, 60, etc., sides may be inscribed in, and circumscribed about a given circle. Ex. 936. To construct a regular decagon, having given a side. Ex. 940. The diagonals of a regular pentagon divide each other into extreme and mean ratio. Ex. 941. The area of a regular inscribed hexagon is a mean proportional between the areas of the inscribed and circumscribed equilateral triangles. Ex. 942. Any radius of a regular polygon bisects an angle of the polygon. Ex. 943. The diagonals drawn from a vertex of a regular decagon divide that angle into eight equal parts. Ex. 944. To construct a regular pentagon, having given a side. Ex. 945. An equiangular polygon inscribed in a circle is regular if the number of its sides is odd. PROPOSITION VIII. THEOREM 405. Regular polygons of the same number of sides are similar. B B' E E E D' Hyp. ABCDE and A'B'C'D' E' are regular polygons of n sides. ABCDE ~ A'B'C'D'E'. To prove n n 2 Proof. Every angle of both polygons is equal to st. Ls. (Why?) .: polygons ABCDE and A'B'C'D'E' are mutually equiangular. Since AB= BC, etc., and A'B' = B'C', etc., (Why?) AB: AB BC: B'C' = etc. Q.E.D. Ex. 946. To construct a regular hexagon equivalent to one-half of a given regular hexagon. Ex. 947. Construct a regular pentagon equivalent to the sum of two given regular pentagons. THEOREM PROPOSITION IX. 406. The perimeters of regular polygons of the same number of sides have the same ratio as their radii or as their apothems. To prove Hyp. P and P' are the perimeters of two regular polygons having the radii OA and O'A', and the apothems OD and O'D', respectively. P: P = QA: 0'A' = OD: O'D'. (398) .:. Δ Ο ADΝΔΟ' Α'D'. '. (Why ?) Hence AD; A'D' OD: O'D' = AO: A'O'. (Why?) But P: P = AB: A'B' = AD: A'D'. (Why?) .:: P: P = OD: O'D' = AO: A'O'. Q.E.D. |