2018-2019学年上海市杨浦区初三数学第一学期期末质量调研(参考答案)2019.01

来源：用户分享时间：2025/11/20 17:30:59 本文由

说明：文章内容仅供预览，部分内容可能不全，需要完整文档或者需要复制内容，请下载word后使用。下载word有问题请添加微信号:xxxxxxx或QQ：xxxxxx 处理（尽可能给您提供完整文档），感谢您的支持与谅解。

∵AD//BC，∴ADAE，即318-3x. x.∴

==BF=BFEBxBF6-x∵△AEF的面积为3，∴1. 18-3x鬃x=32x······························································································· （1分） ∴x=4. ·

∵AD=3，AB⊥AD，∴DE=5. ∵DE1，∴DC=10.

DC=2∵DF⊥DC，∴

S1. ·························································VDCE=2创510=25当点E在边AB延长线上，设AE=y，

∵AD//BC，∴ADAE，即3y.∴3y-18. BF=EBBF=y-6BF=y∵△AEF的面积为3，∴1.∴y=8. ······································2鬃y3y-18···y=3∵AD=3，AB⊥AD，∴DE=73.

联结CE，作CH⊥AD交AD的延长线于点H，同（1）可得DE1.

·················DC=2∴DC=273 ∵DF⊥DC，∴

S1. ················································VDCE=2创73273=73综上，当△AEF的面积为3时，△DCE的面积为25或73.

1分） 1分）

1分） 1分）（（（（

2018-2019学年上海市杨浦区初三数学第一学期期末质量调研(参考答案)2019.01.doc 将本文的Word文档下载到电脑，方便复制、编辑、收藏和打印