Problemata Geometrica varia.

3

Of the ratio that is between the Diameter pf a

Circle, and the circumference.

PROPOSITIO I.

The circumference of a circle is leise then the perimeter ofany ordinate Polygon that circumscribes the circle : butgrcater then the diameter of any ordinate Polygon inferi*hed within the circle.

^claration.^^S^^^Ht’ arh^ cb e is leffe then db f, great-

er then ege: that is , leffe then thetangent of the arh^ grcater then thefHbtenfe.

Demonstrat. I. Grcater then thesubtense , lecaufe between the twopoints cande the subtensec ge isfborter then any other curvcd line , and therefore thenarl^c b e drawn throngh the fame terms , by the 4 th. Der fJ*• Euclid. ™

2 Leffe then the tangent df. For by that which Archimoproves in his ßrß Propoßtion (ofthe Quadrature of n^irclc^ a triangle whof ? fide is a c 5 whof ? bafe is equal to theHeb e will be equal to the feBor c bea. But the 'Trianglebofe fide , or altitucTe is ab equal to a c, and whofe bafeis thetan gciit db ()is grcater then the forenamed fe&or by the exte-fi gure c b e f b d, and y et the altitude, or radius a b, or a c* 111 both equal , therefore it will follow , that d f is greater thentj s ark^c b e.

PROPOSITIO II.

^he diameter of a circle be of 1 0000.00000. ooooo.parts,

Fig. 1 .

31415.92653.57748

31415.92653.61440

The peripherysgrcater

fliallbe Riesle

L Et the arh c b or b e be 1

Tae

berfforeis the tadiusK^b bc ioooo.ooooO.oooo^ooooo.

n

a c

B or g c the fine of 1"

^ n d b f,orb c fio ali bc the tang. of 1

04848.13681.1076404848.1 3681.11333Double