题目

题型：不详难度：来源：

（Ⅰ）（本小题满分7分）
某服装厂承揽一项生产夏凉小衫1600件的任务，计划用t天完成．
（1）写出每天生产夏凉小衫w（件）与生产时间t（天）（t＞4）之间的函数关系式;
（2）由于气温提前升高，商家与服装厂商议调整计划，决定提前4天交货，那么服装厂每天要多做多少件夏凉小衫才能完成任务？
(Ⅱ)（本小题满分7分）
如图，已知矩形ABCD中，E是AD上的一点，F是AB上的一点，EF⊥EC，且EF=EC，DE=4cm，矩形ABCD的周长为32cm，求AE的长．

答案

(1)解：（1）························ 3分
（2）······························ 5分
 ··················· 6分
答：每天多做（或）件夏凉小衫才能完成任务．·········· 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的周长为32 cm，∴2（AE+AE+4）=32．          ······················· 6分
解得， AE="6" （cm）．          ······················ 7分

解析

略