■SO llNES IN A CIRCLE. 



Propofitions dccoutit they ought to be made as correft as poflible; whicll 



refpcaing the h^s not been as yet done, either by Mr. Bofwell, or by his 



Circlets area, &c. _ . ._ -^ ' . ^ . ,- , r . 



more Icientihc commentator in your number for the prelent 



month. This declaration in a manner compels me to under- 

 take the following inveftigation oF the fiJ^jedt, in which I 

 IhaU refer to Fig. 4, Plate Xf. of your Journal for April, re- 

 quefting the reader to place the letter T at the upper extre- 

 mity of the vertical diameter, and C at the oppofite end. 



Theorem \Jt. Let the circle ITFC have unity for its dia- 

 meter; draw the diameters I F, T C at right angles to each 

 other; bife<5t the radius T O in W; join I W, and produce it 

 until it meets the circle again in B: thefe things being done, 

 the fquare upon I B is equal to .8000; which exceeds .7854, 

 or the common exprellion for the area ITFC by the fradtion 

 .014^. 



Demonftralion. Put the radius I O = R = .3 ; then O W 

 = ^R; lince the triangle I O W is right angled at O, by hy- 



pothefis, IW*=R*+iR^ = ^^> Eiic. 47.1. Now|the 

 4 4 



triangles O I W, B I ¥ are equiangular; becaufe the angle at 



I is common to both ; and the angle F B I is equal to W O I, 



being right, Lac. Si. Ill ; confequently, as W I : I O : : I F^ 



(=2lO):IB; hence as W P ( = — ) : 1 0*(= R*) : : 

 IF* (=4 R*): I B^- ( = 1^-5-!]; but R2=:.25; therefore 



I B2=:l£ii£l= I6x .05 = .8000. But .8000-.7854 = 



.0146. Q.E. D. 



Lemma. The area of the circle ITFC is equal to the re6i- 

 angle under the radius O I, and the femi-circumference I TF; 

 for this area is equal to a triangle having I O for its altitude, 

 and the whole circumference ITFC for its bafe. (Arclimi' 

 des de Circuh Prop. \(k.) 



Theorem 2nd, If the right line I B be the fide of a fquare, 

 which is equal to the area ITFC; and B G be drawn per- 

 pendicular to the diameter I F; the fegment I G of the dia- 

 meter I F, cutoff by B G, is equal to l of the circumference 

 ITFC or thearcTBF. 



Demonjiration, The tri^ingle I B F is right angled at B, 

 31 £.111. and BG is perpendicular to I F by hypothefis 



therefore 



