8{>xGSP!4 > 0^4f>s9 44@>s4489 44>s4]D!( QppPossible SolutionmD!(DStHH"ynB'! Qpp& DaIn solving the open -ended part of the question, a student would be correct if he/she chose any point within the shaded region. To find the maximum distance, the student would have to find how far away a parallel chord of length 1 would be on either side of the central line. Using special right triangle properties, this "time" would be (sqrt 3)/2, which would then have to be converted to distance for the final answer.HmHz.UO?<?=G.N^NtREALLYCLNVH(n<. =F FfJ-gZJ-g~RWl BQppCB AQppC3CLkC 1QppC3CB?5%F?5%F\a CQpp CbBF'# jQppCbBFC3C?H EQpp CesCJ \a DQpp CbBR)5\ 6Qpp @eڥPDistance(E to D) = 4Y4P44D4>t4>>s44RJ4P 2.00 inchesB   DQppCy"C"6q [  mQppC3CCbBR? [Dd  kQppCbBRCesCJ? NSuld  DQppCkBp qN&_ 1Qpp @MAngle(DAD) = D) = 4Y4P44D4>t4>>s44RJ4P60 inchesB    DQppCICX=H   mQppC3CCy"C"6q? joD0 DQppBѼCɏ )5y"_XO 3Qpp @N-{Angle(DAD) = = 4Y4P44D4>t4>>s44RJ4P60 inchesB (tJx ` 2Qpp @M+Angle(EAD) = = 4Y4P44D4>t4>>s44RJ4P60 inchesB  nQppCICX=HCy"C"6q?/   mQppC3CCICX=H?Mi. T  nQppBѼCɏCkBp? (gP 5Qpp ?eڥPLength(Segment n) = 4Y4P44D4>t4>>s44RJ4P 1.00 inchesBM$  FQppCaC=96Bp2< 4Qpp @V&Angle(AFD) = n) = 4Y4P44D4>t4>>s44RJ4P90 inchesB v  pQppCaC=9C3C?