-{ GSP!lz6$ (P( 6$ vD60 zBBBT3(|R3(ޘ8(޺!h! p(P4d @The distance from the central path is 1/2r*3, because it's the height of an equilateral triangle with an edgelength of r, where r is the radius of the circle.>d<LM M E$en?< Bp(P4dClCM}L. Ap(P4dCBCOf!  5p(P4d @M(W radius = HH! ! D)PHH3.87 cm! Nُ'&!  kp(P4dCBClCM4H4H1H 1p(P4dCBB5)?5%F?5%FP\ 6p(P4d  @I3^ 10/radius = H! ! D)PHH 2.59 per cm! NP /P4U Gp(P4 dBBDډً'' mp(P4dCBBCL?)i 7p(P4d  @$ radius = = H! !  miles D)PHH10.00per cm! N(!  2p(P4d @FDistance(G to B) = ! ! miles D)PHH7.73 cmr cm! NOT  Ep(P4 dBCL 05 Dp(P4 dCmBH v 9p(P4d  @31E Diameter = to B) = ! !  miles D)PHH20.00cmr cm! N ?  Eՠp(P4dC5Ch |N  qՠp(P4dCBBCL? )&' pՠp(P4dCmBHC ? '& nՠp(P4dBCL@B? af !  E՛p(P4dCfBơ '&/, r՛p(P4dC5ChCN5CW?38 F՛p(P4 dBSC2  1 minute view4dC5ChBSC? ! Fminute view4dC< ACE 1minute view4d @fDistance(F to E) = ! ! miles D)PHH3.87 cmr cm! N)5~ 11minute view4d  @$h1 minute view = ) = ! !  miles D)PHH10.00cmr cm! NPډً'' u1minute view4dCBCmBG?f {{{{  1 minute view4dCfBơC< AC?H4H5)5 3 minute view4d  ?ӣ%Distance(G to B)/Distance(F to E) = ! miles D)PHH2.000cmr cm! N Y^H H minute view4dBiCA|X  distance view4dBiCACB?6Bl 4istance view4d @ ۼ[.distance from central path = o E) = ! miles D)PHH3.35 cmr cm! N 8istance view4d  @!R ֣)distance from central path*(10/radius) = ! miles D)PHH8.66 cmr cm! N