2022英语周报高一第14期答案

9.(本小题满分12分)解:(1)性质1:DE⊥平面ABD.证明如下:翻折前,DE⊥D4,DE⊥BC,翻折后仍然有DE⊥D,DE⊥DB,且D∩DB=D则DE⊥平面ABD性质2:DE⊥AB,证明如下:与性质1证明方法相同,得到DE又因ABc平面ABD,则DEMB性质3DE与平面ABD一直线都垂直.证明如下与性质1证明方得到DE⊥平面ABD,从而DE与平面ABD内任一直线都垂直性质4:直线DE与平面ABE所成角等于一.证明如下如图5,取AB的中点F,连接DF,EF由DA=DB,得DF⊥AB,与性质2证明相同,得DE⊥AB,DE⊥DF再因DE∩DF=D,则AB⊥平面DEF,进而平面DEF⊥平面ABE作DH⊥EF于H,则DH⊥平面ABE,∠DEF就是直线DE与采面ABE所成的角DE=l, EF=cos∠DEF、DE22说明:写出一条并且只需写出一条正确的(允许在以上4条之外),给3分,完成正确的证明后合计给5分(2)解法一:AD=BD=√2,AB=2则△ABD是等腰直角三角形,如图6,取AB的中点F,则F是△ABD的外心图6设几何体E-ABD外接球的球心是O,则OF⊥平面ABD作OM⊥DE于M,则M是DE的中点,OFDM是矩形,OF=DMB=1几何体E-ABD的外接球半径R=√OF+FD2=+1=,则外接球的体积v=-R(12分)解法二:证明D,DB,DE两两垂信后,几何体E-ABD外接球就是以D4,DB,DE相邻的棱的长方体的外接球、(√5D4-+DB-+DE=2+2+1=5,R(12分)

第一节One possible versionDear Ms JenkinsI'm writing to ask you for help. Our school is going to sponsor an exhibition of Chinese paintings in theschool library next month. As president of the student council, i have been assigned the job of drafting an arnouncement in English as there aredreds of international students in our schoohave already written a rough draft of the above-mentioned announcement. But I'm afraid that there aresome mistakes in the English language. So I am hoping you would correct my possible mistakes in the announce-ment that i have attachedId appreciate it if you could do me the favocerely yourLIH

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