Question

（Ⅰ）（本小題滿分7分）
某服裝廠承攬一項生產夏涼小衫1600件的任務，計劃用t天完成．
（1）寫出每天生產夏涼小衫w（件）與生產時間t（天）（t＞4）之間的函數關系式;
（2）由于氣溫提前升高，商家與服裝廠商議調整計劃，決定提前4天交貨，那么服裝廠每天要多做多少件夏涼小衫才能完成任務？
(Ⅱ)（本小題滿分7分）
如圖，已知矩形ABCD中，E是AD上的一點，F是AB上的一點，EF⊥EC，且EF=EC，DE=4cm，矩形ABCD的周長為32cm，求AE的長．

Answer 1

(1)解：（1）························ 3分
（2）······························ 5分
 ··················· 6分
答：每天多做（或）件夏涼小衫才能完成任務．·········· 7分
(2)解：在Rt△AEF和Rt△DEC中，∵EF⊥CE，∴∠FEC=90°，
∴∠AEF+∠DEC=90°，而∠ECD+∠DEC=90°，
∴∠AEF=∠ECD．             ································································ 1分
又∠FAE=∠EDC=90°．EF=EC
∴Rt△AEF≌Rt△DCE．         ································································ 3分
AE=CD．                      ································································ 4分
AD=AE+4．
∵矩形ABCD的周長為32 cm，∴2（AE+AE+4）=32．          ······················· 6分
解得， AE="6" （cm）．          ······················ 7分解析:
略