题目内容
(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.
(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
(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
练习册系列答案
相关题目