Elements of geometry and trigonometry . other as the polygons A and A, nC which theyform part : hence A : A : : CI) : CM. Again, the trianglesCAM, CME, having the cf)mm()n v(;rtex M, are to each otheras their bases CA, CK ; tiny an; lik<;wisc to each other as thepolygons A and B of which tlicy form part ; hence A : B : :CA : CK. But since AD and ME are parall«l, we haveCD : CM : : CA : CE; hence A : A : : A : B ; hence thepfjlygon AlOnc of those rccjuircd, is a mean pro])ortional betweenihc two given polygons A and B,and consctjuenllyA - VAx B. 124 GEOMETRY. Secondly- The altitude CM be-ing
Elements of geometry and trigonometry . other as the polygons A and A, nC which theyform part : hence A : A : : CI) : CM. Again, the trianglesCAM, CME, having the cf)mm()n v(;rtex M, are to each otheras their bases CA, CK ; tiny an; lik<;wisc to each other as thepolygons A and B of which tlicy form part ; hence A : B : :CA : CK. But since AD and ME are parall«l, we haveCD : CM : : CA : CE; hence A : A : : A : B ; hence thepfjlygon AlOnc of those rccjuircd, is a mean pro])ortional betweenihc two given polygons A and B,and consctjuenllyA - VAx B. 124 GEOMETRY. Secondly- The altitude CM be-ing common, the triangle CPM isto the triangle CPE as PM is toPE ; but since CP bisects the an-gle MCE, we have PM : PE : :CM : CE (Book IV. ) ::CD : CA : : A : A:hence CPM : CPE : : A : A ;and consequently CPM : CPM +CPE or CME : : A : A-f A. ButCMPA, or 2CMP, and CME areto each other as the polygons B ^ and B, of which they form part : hence B : B : : 2A : A + A has been already determined ; this new proportion will serve for determining B, and give us B: - ; and thus by A+A ^ means of the polygons A and B it is easy to find the polygonsA and B, which shall have double the number of sides. PROPOSITION XIV. PROBLEM. To find the approximate ratio of the circumference to thediameter. Let the radius of the circle be 1 ; the side of the inscribedsquare will be \/2 (Prop. III. Sch.), that of the circumscribedsquare will be equal to the diameter 2 ; hence the surface ofthe inscribed square is 2, and that of the circumscribed squareis 4. Let us therefore put Ar:=2, and B=4 ; by the last pro-position we shall find the inscribed octagon Am V 8:=, and the circumscribed octagon B=j——/ô = The inscribed and the circumscribed octagons being thus deter-mined, we shall easily, by means of them, determine the poly-gons having twice the number of sides. We have only in thiscase to put A=, B = ; we shall find A = 2 A B V
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