CHAMBERS'S INFORMATION FOR THE PEOPLE. 



But 



BD = AC 



R, 



.'. side of regular decagon = ( \/5 i) 



(6). 



It is evident that the side of the regular hexagon 

 inscribed within the circle is equal to the radius. 



PROP. VI. Given the perimeter of a regular 

 polygon inscribed within a circle, and the perimeter 

 of a similar circumscribed polygon, to find the 

 perimeters of the regular inscribed and circum- 

 scribed polygons of double the number of sides. 



Let AB = side of inscribed polygon ; EF, the 

 tangent through the middle point C of arc AB, 

 meeting OA and OB produced in E and F ; then 



will EF = side of circumscribed figure. Join AC. 

 AC is the side of figure inscribed, with double the 

 number of sides. Through A and B draw tangents 

 AG, BH. GH is the side of the figure circum- 

 scribed, with double the number of sides. Join 

 OG. 



Since OE is the radius of the circle circum- 

 scribing the polygon whose perimeter is P, and 

 OC that of the circle circumscribing the similar 

 polygon whose perimeter is p, we have 

 JP _ OE 

 P ~ OC' 



But since OG bisects the angle EOC (Prop. 

 VII. 5, p. 612), we have (Euclid, VI. 3), 

 OE = EG 

 OC ~ GC '" 



. Z _ EG . 

 " p GC 5 



EG + GC_ EC 

 2GC GH' 



Now GH is a side of the polygon whose per- 

 imeter is P', and is contained in P' the same 

 number of times that EC is contained in P t or 



EC P 



GH P' ; 



Again, the triangles ACD and GKC are similar ; 

 AD KC 

 ' ' AC CG' 



But since AD is contained in P the same num- 

 ber of times that AC is contained in p', 



AD _ p 

 AC " / ; 



KC / 

 and similarly, pg = |p ; 



634 



. - 



' p 1 ~ F 



(2). 



PROP. VII. To find an expression for the ratio 

 of the circumference of a circle to the diameter. 



We shall take the diameter of the circle as 

 given, and determine the perimeters of an in- 

 scribed and similarly circumscribed polygon. We 

 shall then determine the perimeters of inscribed 

 and circumscribed polygons of double the number 

 of sides. Taking the last found perimeters as 

 given, we will then determine the perimeters of 

 polygons of double the number of sides by the 

 same method, and so on. As the number of sides 

 increases, it may be shewn that the perimeters 

 approach nearer and nearer to the circumference ; 

 hence their successively determined values will be 

 nearer and nearer approximations to the value of 

 the circumference. 



Taking the diameter of the circle as I, we shall 

 begin by inscribing and circumscribing a square, 

 and finding their perimeters. 



Here we have/ = 2 </ 2 = 2 8284271, 

 and P = 4. 



Having found P and p, we find P' and/' ; 



thus 



P' = 



= 3-3I3708S, 



and ff= >Jp X P'= 3-0614675. 



Then taking P' and p' as given quantities, we 



P = 3-3I37085, 

 and p 3-0614675 ; 



and find, by the same formulae, for the polygon of 

 1 6 sides, 



P'= 3-1825979, 



/= 3- 1 2 14452. 



Continuing this process, the results will be found 

 as in the following table : 



put 



From the last two numbers of this table, we 

 learn that the circumference of a circle whose 

 diameter is unity, is less than 



3-1415928, 



and greater than 3- 1415926. 



It follows therefore that 3-1415927 

 represents the circumference to within a unit of 



