七下期末复习卷(三)
- 格式:doc
- 大小:381.50 KB
- 文档页数:5
2023七年级下学期道德与法治期末高效复习卷一、单选题1.处于青春期的我们,往往更加关注自己的外表。
有时,一些正常的生理现象也可能给我们带来烦恼。
对此,我们()A.可以嘲弄长青春痘的同学B.因自己的某些生理变化而自卑是可以理解的C.要欣然接受青春花蕾的绽放D.不可以谈论青春期的生理变化2.人是能思想的苇草。
虽然生命有时像苇草一样脆弱,但思想使我们强大。
只有当思维日渐成熟,我们才能真正长大。
青春期思维的独立是指()A.一味追求独特,不人云亦云B.有自己独到的见解,同时接纳他人合理、正确的意见C.敢于和别人争辩,一切按照自己的想法去做D.有自己独到的见解,不倾听、接纳他人的意见3.青春的美丽既印在相纸上,也印在我们的心坎上。
在我们心灵的相册里,有男生也有女生。
打开青春相册()①男生女生各具特点,拥有各自的性别优势②男生女生应互相理解,互相帮助,相互学习③男生女生相互交往,优势互补益处多多④异性交往麻烦很多,容易招致闲言碎语⑤异性交往应内心坦荡、言谈得当举止得体A.①②③⑤B.①②③C.③④⑤D.②③④4.有人认为,男生和女生在语言表达、数学推理等方面存在一定的差异,各有优势。
面对这些差异我们()①不应因自己某一方面的优势而自傲②不应因自己某一方面的欠缺而自卑③要学会相互取长补短,共同进步④要学会认识性别个体差异A.①②③B.②③④C.①③④D.①②③④5.“当你不去旅行,不去冒险,不去拼一份奖学金,不过没试过的生活,整天挂着QQ,看微博。
逛着淘宝玩着网游,干着我80岁都能做的事,你要青春干什么?”这句网络流行语告诉我们()A.青春是美好的,要尽情享乐不要留下遗憾B.青春是矛盾的,要促进生理和心理相协调C.青春是短暂的,要努力探索青春,创造美好人生D.青春是梦想,梦想定会变成现实6.一般而言,“行己有耻”的养成路径是()①要有知耻之心②有所为有所不为③树立底线意识A.①→②→③B.①→③→②C.②→①→③D.③→①→②7.青春期是情绪发展的一个特殊时期。
2021学年部编版语文七年级下册期末复习专题训练:文言文阅读(三)部编人教版七年级下册期末复习专题训练:文言文阅读(三)2020-2021学年部编版语文七年级下册1.阅读下文,回答相关问题(一)初,权谓吕蒙曰:“卿今当涂掌事,不可不学!”蒙辞以军中多务。
权曰:“孤岂欲卿治经为博士邪!但当涉猎,见往事耳。
卿言多务,孰若孤?孤常读书,自以为大有所益。
”蒙乃始就学。
及鲁肃过寻阳,与蒙议论,大惊曰:“卿今者才略,非复吴下阿蒙!”蒙曰:“士别三日,即更刮目相待,大兄何见事之晚乎!”肃遂拜蒙母,结友而别。
(二)炳烛夜谈晋平公问于师旷曰:“吾年七十,欲学,恐已暮矣。
”师旷曰:“何不炳烛①乎?”平公曰:“安有为人臣而戏其君乎?”师旷曰:“盲臣②安敢戏君乎?臣闻之:少而好学,如日出阳;壮而好学,如日中之光;老而好学,如炳烛之明。
炳烛之明,孰与昧③行乎?”平公曰:“善哉!”。
【注释】①炳烛:点烛。
②盲臣:师旷为盲人,故自称盲臣。
③昧行:在黑暗中行走。
(1)解释下面句子中的加线词。
①卿今当涂掌事()②但当涉猎()③蒙辞以军中多务()④及鲁肃过寻阳()⑤恐已暮矣()⑥盲臣安敢戏君乎()(2)《孙权劝学》中有两个成语,请写出成语并解释含义(3)将下面的文言句子译成现代汉语。
①士别三日,即更刮目相待,大兄何见事之晚乎!②少而好学,如日出之阳。
(4)根据语段内容回答。
①“刮目相待”后,鲁肃有何举动?说明了什么?②吕蒙不愿学的理由和晋平公担心学不好的理由分别是什么?(5)两文都是谈学习的,共涉及四个人,读完后你得到什么启发?2.阅读下面的文言文,完成下面小题。
孙权劝学《资治通鉴》初,权谓吕蒙日:“卿今当涂掌事,不可不学!”蒙辞军中多务。
权日:“孤岂欲卿治经为博士邪!但当涉猎,往事耳。
卿言多务,孰若孤?孤常读书,自以为大有所益。
”蒙始就学。
及鲁肃寻阳,与蒙论议,大惊曰:“卿今者才略,非复吴下阿蒙!”蒙曰:“士别三日,即更刮目相,大兄何见事之晚乎!”肃遂拜蒙母,结友别。
期末复习测试卷一、选择题(每空1分,共20分)1. He likes singing very much and wants to be a ______ in the future.A. musicianB. writerC. workerD. doctor2.—Is this Lucy's exercise book?—Yes, it's ______ . She is looking for it everywhere.A. mineB. hisC. hersD. yours3. —Whose card is this?—It is ____. My brother Tom gave it to me on my birthday.A. myB. hersC. oursD. mine4. A robot show will be held in July, but ____ knows the date for sure.A. nobodyB. everybodyC. anybodyD. somebody5. In our hometown, _______ villagers leave for big cities to look for jobs.A. two hundreds ofB. hundred ofC. hundreds ofD. two hundreds6. T—Ken, could you please ______ the chairs? You can't put them here. —Sorry. I'm coming.A. knockB. waveC. touchD. move7. It usually ____ Mum about an hour to cook supper.A. paysB. takesC. spendsD. costs8. So many robots are worki ng in the factory. I simply can’t _______ my eyes.A. seeB. holdC. sellD. believe9. —Where did you go last winter holiday?—I ____ to London with my family.A. goB. am goingC. wentD. was going10. I'm looking forward to ______ the Palace Museum.A. visitedB. visitC. visitingD. visits11. —Tony, ______ read in bed. It's bad for your eyes.—OK, Mom.A. don'tB. doesn'tC. didn'tD. can't12. —The dress is too small for me. May I ______ that red one? —Certainly.A. turn onB. try onC. pick upD. put up13. The exam is coming. You’d better _______ your lessons carefully.A. worry aboutB. put awayC. go overD. pick up14. She saw the cup of coffee and _______.A. picks it upB. picked up itC. picked it upD. picks up it15. There _______ any schools in the future. Students will study at home.A. is going to haveB. will beC. aren’t going to haveD. won’t be16. Mr Green was happy and _______ to see Shakespeare’s plays in Beijing.A. strictB. excitedC. boredD. relaxed17. —Ann and her sister are ______ .—Yeah! Ann is friendly but her sister is a little difficult.A. sameB. quietC. differentD. tidy18. —____ hard-working boy Zhou Bin is!—Yes, I'm sure he'll pass the test.A. What aB. WhatC. How aD. How19. —The final exam is over. Will you stay at home or visit your grandparents? —______ . I miss them very much.A. Yes, I willB. I'll stay at homeC. No, I won'tD. I'll visit my grandparents20. —Tony, I will go to Dashu Mountain with my parents tomorrow.—______ .A. You're welcomeB. That's all rightC. Don't be sillyD. Have a nice day二、完形填空(每空1分,共10分)What did you do last weekend? Let me tell you how I __1____ my weekend.On Saturday I drove to Dallas to see my sister Alice. Alice is a(n) __2____. She gave a concert (音乐会) that day and it was excellent. After that, we walked to a __3____ for dinner with her friends. We had a good time together. After I went back home, I found I __4____ my bag in the restaurant. It was __5____ and I was really tired. __6____ I had to drive back to Dallas. Bad __7____!The Sunday was relaxing. __8____ I got up, it was half past nine. After having a quick breakfast, I __9____ to clean the house. Then I found a __10____ under my bed. I bought it last year and it still looked nice. So in the afternoon I took it to the park and flew it.1. A. made B. visited C. studied D. spent2. A. actress B. farmer C. singer D. teacher3. A. restaurant B. hospital C. bank D. park4. A. stopped B. left C. sold D. blew5. A. dark B. short C. cheap D. easy6. A. If B. Or C. When D. But7. A. girl B. fun C. luck D. time8. A. What B. Who C. Where D. When9. A. forgot B. started C. learned D. flew10. A. watch B. bike C. candle D. kite三、阅读理解(每题2分,共30分)AWei Hua had a busy day yesterday. She got up before 7 o’clock in the morning and quickly washed her hands and face. She got to school early. She had a rest after lunch. She played basketball after school, and then she walked home.Wei Hua’s pen was broken, so she needed a new one. On her way home, she bought a new pen at a shop. When she reached home, she had a short rest and drank a glass of water. After that she helped her parents. She quickly cooked the supper, and cleaned the house.She watched TV for half an hour after supper. Then she started doing her homework. She finished it at half past nine. She went to bed at a quarter to ten.1. Yesterday Wei Hua was _______.A. illB. busyC. sadD. happy2. How did she go home yesterday?A. By bus.B. By bike.C. On foot.D. By car.3. What did she buy on her way home?A. A book.B. Some food.C. Some drinks.D. A pen.4. She finished her homework at _______ in the evening.A. 9:30B. 9:00C. 10:00D. 4:305. She went to bed at _______.A. 10:15B. 9:45C. 10:30D. 9:30BWe spent a day in the country and picked a lot of flowers. Our car was full of flowers inside! On the way home we had to stop at traffic lights , and there my wife saw the bookshelf.It stood outside a furniture (家具)shop. "Buy it," she said at once. "We'll carry it home on the roof-rack (车顶架) . I've always wanted one like that . "What could I do? Ten minutes later I was twenty dollars poorer, and the bookshelf was tied on to the roof-rack. It was tall and narrow, quite heavy too .As it was getting darker, I drove slowly . Other drivers seemed more polite than usual that evening. The police even stopped traffic to let us through . Carrying furniture was a good idea.After a time my wife said, "There's a long line of cars behind . Why don't they overtake (超车) ?"Just at that time a police car did overtake. The two officers (警官)inside looked at us seriously when they went past . But then , with a kind smile they asked us to follow their car through the busy traffic. The police car stopped at our village church (教堂) . One of the offices came to me."Right, sir, " he said. "Do you need any more help now?"I didn't quite understand . "Thanks, officer, " I said . "You've been very kind. I lived just down the road. "He was looking at our things: first at the flowers, then at the bookshelf. "Well, Well, " he said and laughed . "It's a bookshelf you've got there ! We thought it was-er, something else . "My wife began to laugh . Suddenly I understood why the police drove here .I smiled at the officer. "Yes, it's a bookshelf, but thanks again." I drove home as fast as I could .6 . From the story we know that ___________.A . the writer was poor and didn't buy the bookshelf for his wifeB . the writer's wife didn't like the bookshelf at allC . the writer was always glad to buy something for his wifeD . the writer was not very dad to buy the bookshelf for his wife7 . What made the writer think that carrying furniture was "a good idea"?A . He could drive slowly and it was safe.B . Other drivers would let him go first .C . His wife could use a new bookshelf.D . He could save a lot of money and time.8 . Why were the police and other drivers so kind to the writer?A . Because they thought the writer liked studying very much and needed a bookshelf.B . Because they didn't think it was polite to overtake a car with a bookshelf on it .C . Because they thought somebody in the writer's family had died and he needed help .D . Because they thought it was dangerous to carry a bookshelf on a car.9. Why did the writer's wife begin to laugh?A . Because now she knew what mistake the police had made.B . Because at last her husband understood why the police had driven to the church .C . Because the officer was always looking at the flowers and the bookshelf.D . Because the police had helped them a lot .10 . When did the officers begin to realize (意识到)they had made a mistake?A . Before they arrived at the church.B . Before they overtook the writer's carC . After one of them looked at the flowers and the bookshelf carefully at the church.D . After the writer's family left the church.四、词汇运用(每空1分,共10分)A)根据句意及汉语提示写单词,完成句子。
人教版七年级下册数学期末复习试卷及答案一、选择题1.下列图形中,1∠与2∠是同旁内角的是( )A .B .C .D .2.在以下现象中,属于平移的是( )①在荡秋千的小朋友的运动;②坐观光电梯上升的过程;③钟面上秒针的运动;④生产过程中传送带上的电视机的移动过程. A .①② B .②④ C .②③ D .③④ 3.已知点P 的坐标为P (3,﹣5),则点P 在第( )象限.A .一B .二C .三D .四4.下列命题:①平面内,垂直于同一条直线的两直线平行;②经过直线外一点,有且只有一条直线与这条直线平行;③垂线段最短;④同旁内角互补.其中,正确命题的个数有( ) A .3个B .2个C .1个D .0个5.把一块直尺与一块含30的直角三角板如图放置,若134∠=︒,则2∠的度数为( )A .114︒B .126︒C .116︒D .124°6.下列说法正确的是( )A .a 2的正平方根是aB .819=±C .﹣1的n 次方根是1D .321a --一定是负数7.如图,已知直线//AB CD ,点F 为直线AB 上一点,G 为射线BD 上一点.若:2:1HDG CDH ∠∠=,:2:1GBE EBF ∠∠=,HD 交BE 于点E ,则E ∠的度数为( )A .45°B .55°C .60°D .75°8.如图,点A (0,1),点A 1(2,0),点A 2(3,2),点A 3(5,1)…,按照这样的规律下去,点A 100的坐标为( )A .(101,100)B .(150,51)C .(150,50)D .(100,53)九、填空题9.已知 325.6≈18.044,那么± 3.256≈___________.十、填空题10.平面直角坐标系中,点(3,2)A -关于x 轴的对称点是__________.十一、填空题11.如图,DB 是ABC 的高,AE 是角平分线,26BAE ∠=,则BFE ∠=______.十二、填空题12.如图,已知直线EF ⊥MN 垂足为F ,且∠1=138°,则当∠2等于__时,AB ∥CD .十三、填空题13.如图,将长方形ABCD 沿DE 折叠,使点C 落在边AB 上的点F 处,若45EFB ∠=︒,则DEC ∠=________°十四、填空题14.如图,在纸面上有一数轴,点A 表示的数为﹣1,点B 表示的数为3,点C 表示的数为3B 为中心折叠,然后再次折叠纸面使点A 和点B 重合,则此时数轴上与点C 重合的点所表示的数是_______.十五、填空题15.()2260a b ++-=,则(),a b 在第_____象限.十六、填空题16.如图,在平面直角坐标系中,点()10,0A ,点()22,1A ,点()34,2A ,点()46,3A ,,按照这样的规律下去,点2021A 的坐标为__________.十七、解答题17.计算:(1)3981++- (2)23427(3)+--- (3)2(23)+ (4)353325-++十八、解答题18.求下列各式中x 的值: (1)(x +1)3﹣27=0 (2)(2x ﹣1)2﹣25=0十九、解答题19.完成下列证明过程,并在括号内填上依据.如图,点E 在AB 上,点F 在CD 上,∠1=∠2,∠B =∠C ,求证AB ∥CD .证明:∵∠1=∠2(已知),∠1=∠4 ∴∠2= (等量代换), ∴ ∥BF ( ),∴∠3=∠ ( ). 又∵∠B =∠C (已知), ∴∠3=∠B ∴AB ∥CD ( ).二十、解答题20.在平面直角坐标系中,△ABC 三个顶点的坐标分别是A (﹣2,2)、B (2,0),C (﹣4,﹣2).(1)在平面直角坐标系中画出△ABC ;(2)若将(1)中的△ABC 平移,使点B 的对应点B ′坐标为(6,2),画出平移后的△A ′B ′C ′;(3)求△A ′B ′C ′的面积.二十一、解答题21.已知23|49|7a b a a -+-+=0,求实数a 、b 的值并求出b 的整数部分和小数部分.二十二、解答题22.如图,用两个面积为2200cm 的小正方形拼成一个大的正方形.(1)则大正方形的边长是 ;(2)若沿着大正方形边的方向裁出一个长方形,能否使裁出的长方形纸片的长宽之比为4:3,且面积为2360cm ?二十三、解答题23.(1)(问题)如图1,若//AB CD ,40AEP ∠=︒,130PFD ∠=︒.求EPF ∠的度数; (2)(问题迁移)如图2,//AB CD ,点P 在AB 的上方,问PEA ∠,PFC ∠,EPF ∠之间有何数量关系?请说明理由;(3)(联想拓展)如图3所示,在(2)的条件下,已知EPFα∠的平分线和∠=,PEA∠的平分线交于点G,用含有α的式子表示GPFC∠的度数.二十四、解答题24.如图,已知AM∥BN,∠A=64°.点P是射线AM上一动点(与点A不重合),BC、BD分别平分∠ABP和∠PBN,分别交射线AM于点C,D.(1)①∠ABN的度数是;②∵AM∥BN,∴∠ACB=∠;(2)求∠CBD的度数;(3)当点P运动时,∠APB与∠ADB之间的数量关系是否随之发生变化?若不变化,请写出它们之间的关系,并说明理由:若变化,请写出变化规律;(4)当点P运动到使∠ACB=∠ABD时,∠ABC的度数是.二十五、解答题25.如图,△ABC和△ADE有公共顶点A,∠ACB=∠AED=90°,∠BAC=45°,∠DAE=30°.(1)若DE//AB,则∠EAC=;(2)如图1,过AC上一点O作OG⊥AC,分别交A B、A D、AE于点G、H、F.①若AO=2,S△AGH=4,S△AHF=1,求线段OF的长;②如图2,∠AFO的平分线和∠AOF的平分线交于点M,∠FHD的平分线和∠OGB的平分线交于点N,∠N+∠M的度数是否发生变化?若不变,求出其度数;若改变,请说明理由.【参考答案】一、选择题1.A【分析】根据同旁内角的定义去判断【详解】∵A选项中的两个角,符合同旁内角的定义,∴选项A正确;∵B选项中的两个角,不符合同旁内角的定义,∴选项B错误;∵C选项中的两个角,不符合同旁内角的定义,∴选项C错误;∵D选项中的两个角,不符合同旁内角的定义,∴选项D错误;故选A.【点睛】本题考查了同旁内角的定义,结合图形准确判断是解题的关键.2.B【分析】平移是指在平面内,将一个图形上的所有点都按照某个方向作相同距离的移动,这样的图形运动叫作图形的平移运动,简称平移.平移不改变图形的形状和大小.平移可以不是水平的.据此解答.【详解】解析:B【分析】平移是指在平面内,将一个图形上的所有点都按照某个方向作相同距离的移动,这样的图形运动叫作图形的平移运动,简称平移.平移不改变图形的形状和大小.平移可以不是水平的.据此解答.【详解】①在荡秋千的小朋友的运动,不是平移;②坐观光电梯上升的过程,是平移;③钟面上秒针的运动,不是平移;④生产过程中传送带上的电视机的移动过程.是平移;故选:B.【点睛】本题考查了图形的平移,图形的平移只改变图形的位置,而不改变图形的形状和大小,学生易混淆图形的平移与旋转或翻转而误选.3.D【分析】直接利用第四象限内的点横坐标大于0,纵坐标小于0解答即可.解:∵点P 的坐标为P (3,﹣5), ∴点P 在第四象限. 故选D . 【点睛】本题主要考查了点的坐标,各象限坐标特点如下:第一象限(+,+),第二象限(-,+)第三象限(-,-)第一象限(+,-). 4.A 【分析】根据垂直的性质、平行公理、垂线段的性质及平行线的性质逐一判断即可得答案. 【详解】平面内,垂直于同一条直线的两直线平行;故①正确, 经过直线外一点,有且只有一条直线与这条直线平行,故②正确 垂线段最短,故③正确,两直线平行,同旁内角互补,故④错误, ∴正确命题有①②③,共3个, 故选:A . 【点睛】本题考查了命题与定理:判断一件事情的语句,叫做命题.许多命题都是由题设和结论两部分组成,题设是已知事项,结论是由已知事项推出的事项,一个命题可以写成“如果…那么…”形式.有些命题的正确性是用推理证实的,这样的真命题叫做定理. 5.D 【分析】根据角的和差可先计算出∠AEF ,再根据两直线平行同旁内角互补即可得出∠2的度数. 【详解】解:由题意可知AD//BC ,∠FEG=90°, ∵∠1=34°,∠FEG=90°, ∴∠AEF=90°-∠1=56°, ∵AD//BC ,∴∠2=180°-∠AEF=124°, 故选:D . 【点睛】本题考查平行线的性质.熟练掌握两直线平行,同旁内角互补并能正确识图是解题关键. 6.D 【分析】根据平方根、算术平方根、立方根的定义判断A 、B 、D ,根据乘方运算法则判断C 即可. 【详解】A :a 2的平方根是a ±,当0a ≥时,a 2的正平方根是a ,错误;B 9,错误;C :当n 是偶数时,()1=1n - ;当n 时奇数时,()1=-1n-,错误;D :∵210a --< ,∴【点睛】本题考查平方根、算术平方根、立方根的定义以及乘方运算,掌握相关的定义与运算法则是解题关键. 7.C 【分析】利用180ABG GBF ∠+∠=︒,及平行线的性质,得到180CDG GBF ∠+∠=︒,再借助角之间的比值,求出120BDE GBE ∠+∠=︒,从而得出E ∠的大小. 【详解】 解://AB CD ,ABG CDG ∴∠=∠, 180ABG GBF ∠+∠=︒,180CDG GBF ∴∠+∠=︒,:2:1HDG CDH ∠∠=,:2:1GBE EBF ∠∠=,2222()1801203333HDG GBE CDG GBF CDG GBF ∴∠+∠=∠+∠=∠+∠=⨯︒=︒,BDE HDG ∠=∠,120BDE GBE ∴∠+∠=︒,180()18012060E BDE GBE ∴∠=︒-∠+∠=︒-︒=︒,故选:C . 【点睛】本题考查了平行线的性质的综合应用,涉及的知识点有:平行线的性质、邻补角、三角形的内角和等知识,体现了数学的转化思想、见比设元等思想.8.B 【分析】观察图形得到偶数点的规律为,A2(3,2),A4(6,3),A6(9,4),…,A2n (3n ,n+1),由100是偶数,A100的横坐标应该是100÷2×3,纵坐标应该是100÷2+1解析:B 【分析】观察图形得到偶数点的规律为,A 2(3,2),A 4(6,3),A 6(9,4),…,A 2n (3n ,n +1),由100是偶数,A 100的横坐标应该是100÷2×3,纵坐标应该是100÷2+1,则可求A 100(150,51). 【详解】解:观察图形可得,奇数点:A 1(2,0),A 3(5,1),A 5(8,2),…,A 2n -1(3n -1,n -1),偶数点:A 2(3,2),A 4(6,3),A 6(9,4),…,A 2n (3n ,n +1),∵100是偶数,且100=2n,∴n=50,∴A100(150,51),故选:B.【点睛】本题考查点的坐标规律;熟练掌握平面内点的坐标,能够根据图形的变化得到点的坐标规律是解题的关键.九、填空题9.±1.8044【详解】∵,∴,即.故答案为±1.8044解析:±1.8044【详解】∵,∴,即 1.8044±.故答案为±1.8044十、填空题10.【分析】根据平面直角坐标系中,关于坐标轴对称的点的坐标特征,即可完成解答. 【详解】解:点关于轴的对称点的坐标是(3,2).【点睛】本题考查了根据平面直角坐标系中关于坐标轴对称的点的坐标特3,2解析:()【分析】根据平面直角坐标系中,关于坐标轴对称的点的坐标特征,即可完成解答.【详解】A-关于x轴的对称点的坐标是(3,2).解:点(3,2)【点睛】本题考查了根据平面直角坐标系中关于坐标轴对称的点的坐标特征,即关于x轴对称的点的坐标横坐标不变,纵坐标变为相反数;关于y轴对称的点的坐标纵坐标不变,横坐标变为相反数;十一、填空题 11.【分析】由角平分线的定义可得,∠FAD=∠BAE=26°,而∠AFD 与∠FAD 互余,与∠BFE 是对顶角,故可求得∠BFE 的度数. 【详解】∵AE 是角平分线,∠BAE=26°, ∴∠FAD=∠B 解析:64【分析】由角平分线的定义可得,∠FAD=∠BAE=26°,而∠AFD 与∠FAD 互余,与∠BFE 是对顶角,故可求得∠BFE 的度数. 【详解】∵AE 是角平分线,∠BAE=26°, ∴∠FAD=∠BAE=26°, ∵DB 是△ABC 的高,∴∠AFD=90°−∠FAD=90°−26°=64°, ∴∠BFE=∠AFD=64°. 故答案为64°. 【点睛】本题考查了三角形内角和定理,三角形的角平分线、中线和高,熟练掌握三角形内角和定理是解题的关键.十二、填空题 12.48° 【分析】先假设,求得∠3=∠4,由∠1=138°,根据邻补角求出∠3,再利用即可求出∠2的度数. 【详解】 解:若AB//CD , 则∠3=∠4,又∵∠1+∠3=180°,∠1=138°,解析:48° 【分析】先假设//AB CD ,求得∠3=∠4,由∠1=138°,根据邻补角求出∠3,再利用EF MN 即可求出∠2的度数. 【详解】 解:若AB //CD , 则∠3=∠4,又∵∠1+∠3=180°,∠1=138°,∴∠3=∠4=42°;∵EF⊥MN,∴∠2+∠4=90°,∴∠2=48°;故答案为:48°.【点睛】本题主要考查平行线的性质,两直线垂直,平角定义,解题思维熟知邻补角、垂直的角度关系.十三、填空题13.5【分析】根据翻折的性质,可得到∠DEC=∠FED,∠BEF与∠DEC、∠FED三者相加为180°,求出∠BEF的度数即可.【详解】解:∵△DFE是由△DCE折叠得到的,∴∠DEC=∠FE解析:5【分析】根据翻折的性质,可得到∠DEC=∠FED,∠BEF与∠DE C、∠FED三者相加为180°,求出∠BEF的度数即可.【详解】解:∵△DFE是由△DCE折叠得到的,∴∠DEC=∠FED,又∵∠EFB=45°,∠B=90°,∴∠BEF=45°,∴∠DEC=1(180°-45°)=67.5°.2故答案为:67.5.【点睛】本题考查角的计算,熟练掌握翻折的性质,找到相等的角是解决本题的关键.十四、填空题14.4+或6﹣或2﹣.【分析】先求出第一次折叠与A重合的点表示的数,然后再求两点间的距离即可;同理再求出第二次折叠与C点重合的点表示的数即可.【详解】解:第一次折叠后与A重合的点表示的数是:3+解析:62【分析】先求出第一次折叠与A重合的点表示的数,然后再求两点间的距离即可;同理再求出第二次折叠与C点重合的点表示的数即可.【详解】解:第一次折叠后与A重合的点表示的数是:3+(3+1)=7.与C重合的点表示的数:3+(36第二次折叠,折叠点表示的数为:12(3+7)=5或12(﹣1+3)=1.此时与数轴上的点C重合的点表示的数为:5+(5﹣11)=2故答案为:62【点睛】本题主要考查了数轴上的点和折叠问题,掌握折叠的性质是解答本题的关键.十五、填空题15.二【分析】根据非负数的性质列方程求出a、b的值,再根据各象限内点的坐标特征解答.【详解】解:由题意得,a+2=0,b-6=0,解得a=-2,b=6,所以,点(-2,6)在第二象限;故答解析:二【分析】根据非负数的性质列方程求出a、b的值,再根据各象限内点的坐标特征解答.【详解】解:由题意得,a+2=0,b-6=0,解得a=-2,b=6,所以,点(-2,6)在第二象限;故答案为:二【点睛】本题考查了各象限内点的坐标的符号特征,记住各象限内点的坐标的符号是解决的关键,四个象限的符号特点分别是:第一象限(+,+);第二象限(-,+);第三象限(-,-);第四象限(+,-).十六、填空题16.【分析】观察点,点,点,点点的横坐标为,纵坐标为,据此即可求得的坐标;【详解】,,,,,故答案为:【点睛】本题考查了坐标系中点的规律,找到规律是解题的关键.解析:(4040,2020)【分析】观察点()10,0A ,点()22,1A ,点()34,2A ,点()46,3A ,,点的横坐标为22n -,纵坐标为1n -,据此即可求得2021A 的坐标;【详解】()10,0A ,()22,1A ,()34,2A ,()46,3A ,,(22,1)n A n n --,∴2021(4040,2020)A故答案为:(4040,2020)【点睛】本题考查了坐标系中点的规律,找到规律是解题的关键.十七、解答题17.(1)6;(2)-4;(3);(4).【分析】(1)利用算术平方根和立方根、绝对值化简,再进一步计算即可;(2)利用算术平方根和立方根化简,再进一步计算即可;(3)类比单项式乘多项式展开计算解析:(1)6;(2)-4;(3)2+;(4)【分析】(1)利用算术平方根和立方根、绝对值化简,再进一步计算即可;(2)利用算术平方根和立方根化简,再进一步计算即可;(3)类比单项式乘多项式展开计算;(4)利用绝对值的性质化简,再进一步合并同类二次根式.【详解】解:(11-=3+2+1=6;(2=2-3-3=-4;(33)=2+;(4+=故答案为(1)6;(2)-4;(3)2+4)【点睛】本题考查立方根和算术平方根,实数的混合运算,先化简,再进一步计算,注意选择合适的方法简算.十八、解答题18.(1)x=2;(2)x=3或x=-2.【分析】(1)根据立方根的定义进行求解即可;(2)根据平方根的定义进行求解,即可得出答案.【详解】解:(1)(x+1)3-27=0,(x+1)3=2解析:(1)x=2;(2)x=3或x=-2.【分析】(1)根据立方根的定义进行求解即可;(2)根据平方根的定义进行求解,即可得出答案.【详解】解:(1)(x+1)3-27=0,(x+1)3=27,x+1=3,x=2;(2)(2x-1)2-25=0,(2x-1)2=25,2x-1=±5,x=3或x=-2.【点睛】本题考查了立方根和平方根,熟练掌握立方根和平方根的定义是解题的关键.十九、解答题19.∠4;CE;同位角相等,两直线平行;C;两直线平行,同位角相等;内错角相等,两直线平行【分析】根据平行线的判定和性质解答.【详解】解∵∠1=∠2(已知),∠1=∠4(对顶角相等),∴∠2=解析:∠4;CE;同位角相等,两直线平行;C;两直线平行,同位角相等;内错角相等,两直线平行【分析】根据平行线的判定和性质解答.【详解】解∵∠1=∠2(已知),∠1=∠4(对顶角相等),∴∠2=∠4(等量代换),∴CE∥BF(同位角相等,两直线平行),∴∠3=∠C(两直线平行,同位角相等).又∵∠B=∠C(已知),∴∠3=∠B(等量代换),∴AB∥CD(内错角相等,两直线平行).故答案为:对顶角相等;CE∥BF;同位角相等,两直线平行;两直线平行,同位角相等;内错角相等,两直线平行.【点睛】此题考查平行线的判定和性质,关键是根据平行线的判定和性质解答.二十、解答题20.(1)见解析;(2)见解析;(3)10【分析】(1)根据点A、B、C的坐标描点,从而可得到△ABC;(2)利用点B和B′的坐标关系可判断△ABC先向右平移4个单位,再向上平移2个单位得到△A′解析:(1)见解析;(2)见解析;(3)10【分析】(1)根据点A、B、C的坐标描点,从而可得到△ABC;(2)利用点B和B′的坐标关系可判断△ABC先向右平移4个单位,再向上平移2个单位得到△A′B′C′,利用此平移规律写出A′、C′的坐标,然后描点即可得到△A′B′C′;(3)用一个矩形的面积分别减去三个三角形的面积去计算△A′B′C′的面积.【详解】解:(1)如图,△ABC为所作;(2)如图,△A′B′C′为所作;(3)△A′B′C′的面积=111 6426244210 222⨯-⨯⨯-⨯⨯-⨯⨯=.【点睛】本题考查了平移变换:确定平移后图形的基本要素有两个:平移方向、平移距离.作图时要先找到图形的关键点,分别把这几个关键点按照平移的方向和距离确定对应点后,再顺次连接对应点即可得到平移后的图形.二十一、解答题21.4,【分析】根据分母不等于0,以及非负数的性质列式求出a、b的值,再根据根据被开方数估算无理数的大小即可得解.【详解】解:根据题意得,3a-b=0,a2-49=0且a+7>0,解得a=7,解析:4214【分析】根据分母不等于0,以及非负数的性质列式求出a 、b 的值,再根据根据被开方数估算无理数的大小即可得解.【详解】解:根据题意得,3a -b =0,a 2-49=0且a +7>0,解得a =7,b =21,∵16<21<25, ∴44.【点睛】本题考查了绝对值非负数,算术平方根非负数的性质,根据几个非负数的和等于0,则每一个算式都等于0列式是解题的关键.二十二、解答题22.(1);(2)无法裁出这样的长方形.【分析】(1)先计算两个小正方形的面积之和,在根据算术平方根的定义,即可求解; (2)设长方形长为cm ,宽为cm ,根据题意列出方程,解方程比较4x 与20的大小解析:(1)20;(2)无法裁出这样的长方形.【分析】(1)先计算两个小正方形的面积之和,在根据算术平方根的定义,即可求解;(2)设长方形长为4x cm ,宽为3x cm ,根据题意列出方程,解方程比较4x 与20的大小即可.【详解】解:(1)由题意得,大正方形的面积为200+200=400cm 2,∴cm ;()2根据题意设长方形长为4x cm ,宽为3x cm ,由题:43360x x ⋅= 则230x =0xx ∴=∴长为43020>∴无法裁出这样的长方形.【点睛】本题考查了算术平方根,根据题意列出算式(方程)是解决此题的关键.二十三、解答题23.(1)90°;(2)∠PFC=∠PEA+∠P ;(3)∠G=α【分析】(1)根据平行线的性质与判定可求解;(2)过P点作PN∥AB,则PN∥CD,可得∠FPN=∠PEA+∠FPE,进而可得∠PF 解析:(1)90°;(2)∠PFC=∠PEA+∠P;(3)∠G=12α【分析】(1)根据平行线的性质与判定可求解;(2)过P点作PN∥AB,则PN∥CD,可得∠FPN=∠PEA+∠FPE,进而可得∠PFC=∠PEA+∠FPE,即可求解;(3)令AB与PF交点为O,连接EF,根据三角形的内角和定理可得∠GEF+∠GFE=1 2∠PEA+12∠PFC+∠OEF+∠OFE,由(2)得∠PEA=∠PFC-α,由∠OFE+∠OEF=180°-∠FOE=180°-∠PFC可求解.【详解】解:(1)如图1,过点P作PM∥AB,∴∠1=∠AEP.又∠AEP=40°,∴∠1=40°.∵AB∥CD,∴PM∥CD,∴∠2+∠PFD=180°.∵∠PFD=130°,∴∠2=180°-130°=50°.∴∠1+∠2=40°+50°=90°.即∠EPF=90°.(2)∠PFC=∠PEA+∠P.理由:过P点作PN∥AB,则PN∥CD,∴∠PEA=∠NPE,∵∠FPN=∠NPE+∠FPE,∴∠FPN=∠PEA+∠FPE,∵PN∥CD,∴∠FPN=∠PFC,∴∠PFC=∠PEA+∠FPE,即∠PFC=∠PEA+∠P;(3)令AB与PF交点为O,连接EF,如图3.在△GFE中,∠G=180°-(∠GFE+∠GEF),∵∠GEF=12∠PEA+∠OEF,∠GFE=12∠PFC+∠OFE,∴∠GEF+∠GFE=12∠PEA+12∠PFC+∠OEF+∠OFE,∵由(2)知∠PFC=∠PEA+∠P,∴∠PEA=∠PFC-α,∵∠OFE+∠OEF=180°-∠FOE=180°-∠PFC,∴∠GEF+∠GFE=12(∠PFC−α)+12∠PFC+180°−∠PFC=180°−12α,∴∠G=180°−(∠GEF+∠GFE)=180°−180°+12α=12α.【点睛】本题主要考查平行线的性质与判定,灵活运用平行线的性质与判定是解题的关键.二十四、解答题24.(1)① ②;(2);(3)不变,,理由见解析;(4)【分析】(1)①由平行线的性质,两直线平行,同旁内角互补可直接求出;②由平行线的性质,两直线平行,内错角相等可直接写出;(2)由角平分线的解析:(1)①116,︒②CBN;(2)58︒;(3)不变,:2:1APB ADB∠∠=,理由见解析;(4)29.︒【分析】(1)①由平行线的性质,两直线平行,同旁内角互补可直接求出;②由平行线的性质,两直线平行,内错角相等可直接写出;(2)由角平分线的定义可以证明∠CBD=12∠ABN,即可求出结果;(3)不变,∠APB:∠ADB=2:1,证∠APB=∠PBN,∠PBN=2∠DBN,即可推出结论;(4)可先证明∠ABC=∠DBN,由(1)∠ABN=116°,可推出∠CBD=58°,所以∠ABC+∠DBN=58°,则可求出∠ABC的度数.【详解】解:(1)①∵AM//BN,∠A=64°,∴∠ABN=180°﹣∠A=116°,故答案为:116°;②∵AM//BN,∴∠ACB=∠CBN,故答案为:CBN;(2)∵AM//BN,∴∠ABN+∠A=180°,∴∠ABN=180°﹣64°=116°,∴∠ABP+∠PBN=116°,∵BC平分∠ABP,BD平分∠PBN,∴∠ABP=2∠CBP,∠PBN=2∠DBP,∴2∠CBP+2∠DBP=116°,∴∠CBD=∠CBP+∠DBP=58°;(3)不变,∠APB:∠ADB=2:1,∵AM//BN,∴∠APB=∠PBN,∠ADB=∠DBN,∵BD平分∠PBN,∴∠PBN=2∠DBN,∴∠APB:∠ADB=2:1;(4)∵AM//BN,∴∠ACB=∠CBN,当∠ACB=∠ABD时,则有∠CBN=∠ABD,∴∠ABC+∠CBD=∠CBD+∠DBN∴∠ABC=∠DBN,由(1)∠ABN=116°,∴∠CBD=58°,∴∠ABC+∠DBN=58°,∴∠ABC=29°,故答案为:29°.【点睛】本题考查了角平分线的定义,平行线的性质等,解题关键是能熟练运用平行线的性质并能灵活运用角平分线的定义等.二十五、解答题25.(1)45°;(2)①1;②是定值,∠M+∠N=142.5°【分析】(1)利用平行线的性质求解即可.(2)①利用三角形的面积求出GH,HF,再证明AO=OG=2,可得结论.②利用角平分线的定解析:(1)45°;(2)①1;②是定值,∠M+∠N=142.5°【分析】(1)利用平行线的性质求解即可.(2)①利用三角形的面积求出GH,HF,再证明AO=OG=2,可得结论.②利用角平分线的定义求出∠M,∠N(用∠FAO表示),可得结论.【详解】解:(1)如图,∵AB∥ED∴∠E=∠EAB=90°(两直线平行,内错角相等),∵∠BAC=45°,∴∠CAE=90°-45°=45°.故答案为:45°.(2)①如图1中,∵OG⊥AC,∴∠AOG=90°,∵∠OAG=45°,∴∠OAG=∠OGA=45°,∴AO=OG=2,∵S△AHG=12•GH•AO=4,S△AHF=12•FH•AO=1,∴GH=4,FH=1,∴OF=GH-HF-OG=4-1-2=1.②结论:∠N+∠M=142.5°,度数不变.理由:如图2中,∵MF,MO分别平分∠AFO,∠AOF,∴∠M=180°-12(∠AFO+∠AOF)=180°-12(180°-∠FAO)=90°+12∠FAO,∵NH,NG分别平分∠DHG,∠BGH,∴∠N=180°-12(∠DHG+∠BGH)=180°-12(∠HAG+∠AGH+∠HAG+∠AHG)=180°-12(180°+∠HAG)=90°-12∠HAG=90°-12(30°+∠FAO+45°)=52.5°-12∠FAO,∴∠M+∠N=142.5°.【点睛】本题考查平行线的性质,角平分线的定义,三角形内角和定理,三角形外角的性质等知识,最后一个问题的解题关键是用∠FAO表示出∠M,∠N.。
华师大版数学2023年七年级下册第二学期期末复习检测卷一、选择题(每题3分,共30分)1.下列图形中,是轴对称图形的有( )(第1题)A .4个B .3个C .2个D .1个2.若x =1是方程ax +2x =1的解,则a 的值是( )A .-1B .1C .2D .-123.下列等式变形不一定正确的是( )A .若x =y ,则x -5=y -5B .若x =y ,则ax =ayC .若x =y ,则3-2x =3-2yD .若x =y ,则=xc yc4.若关于x 的方程x +k =2x -1的解是负数,则k 的取值范围是( )A .k >-1B .k <-1C .k ≥-1D .k ≤-15.已知三角形三边为a 、b 、c ,其中a 、b 两边满足|a -3|+(b -7)2=0,那么这个三角形的最大边c 的取值范围是( )A .c >7 B .7<c <10 C .3<c <7D .4<c <106.如图,已知长方形的长为10 cm ,宽为4 cm ,则图中阴影部分的面积为( )A .20 cm 2B .15 cm 2C .10 cm 2D .25 cm2(第6题) (第7题) (第8题)7.如图,将△ABC 绕点A 逆时针旋转90°能与△ADE 重合,点D 在线段BC 的延长线上,若∠BAC =20°,则∠AED 的大小为( )A .135°B .125°C .120°D .115°8.如图,桐桐从A 点出发,前进3 m 到点B 处后向右转20°,再前进3 m 到点C 处后又向右转20°,…,这样一直走下去,她第一次回到出发点A 时,一共走了( )A .100 mB .90 mC .54 mD .60m9.小虎、大壮和明明三人玩飞镖游戏,各投5支镖,规定在同一环内得分相同,中靶和得分情况如图,则大壮的得分是( )A .20分B .22分C .23分D .25分(第9题) (第10题)10.如图,把△ABC 纸片沿DE 折叠,当点A 落在四边形BCED 的外面时,此时测得∠1=112°,∠A =40°,则∠2的度数为( )A .32°B .33°C .34°D .38°二、填空题(每题3分,共15分)11.若一个正多边形的每个外角都等于45°,则用这种多边形能铺满地面吗?答:________.(填“能”或“不能”)12.如图,在△ABC 中,点D 在BC 的延长线上,点F 是AB 边上一点,延长CA 到E ,连结EF ,则∠1、∠2、∠3的大小关系是________.(第12题) (第15题)13.若代数式3x +2与代数式x -10的值互为相反数,则x =________.14.二元一次方程组的解x ,y 的值相等,则k =________.{3x +2y =10,kx +(k +2)y =6)15.如图,l 1∥l 2,五边形ABCDE 是正五边形,那么∠1-∠2的度数为________.三、解答题(共75分)316.(8分)解方程(组):(1)-+=1; (2)2x -12x -24{34 x +y =12,4x -2y =10.)17.(9分)解不等式组:然后把它的解集在数轴上表示出来,{2x +3≥x +11,3x -105<4,)并求出x 的整数解.18.(8分)在图①,图②的网格纸中,△ABC 与△DEF 的三个顶点都在格点上.(1)在图①中,以点A 为对称中心画一个与△ABC 成中心对称的图形;(2)在图②中,将△DEF 绕点D 顺时针方向旋转90°,画出旋转后的图形.(第18题)19.(9分)如图,一条直线分别交△ABC的边及延长线于点D、E、F,∠A=20°,∠CED=100°,∠D=35°,求∠B的度数.(第19题)20.(9分)如图,∠1、∠2、∠3、∠4是四边形ABCD的四个外角.用两种方法说明∠1+∠2+∠3+∠4=360°.(第20题)21.(10分)如图,将△ABC沿射线AB的方向移动2 cm到△DEF的位置.5(1)找出图中所有平行的直线;(2)找出图中与AD 相等的线段,并写出其长度;(3)若∠ABC =65°,求∠BCF的度数.(第21题)22.(11分)如图,在△ABC 中,∠C =40°.将△ABC 绕点A 按逆时针方向旋转得到△ADE ,连结BD .当DE ∥AC 时,求∠ABD 的度数.(提示:在一个三角形中,若两条边相等,则它们所对的角也相等)(第22题)23.(11分)夕阳红街道办事处为给社区干净整洁的社区环境,加入环境保洁队伍,需要购置一批保洁用具,已知1把扫帚和3把拖把共需26元;3把扫帚和2把拖把共需29元.(1)求一把扫帚和一把拖把的售价各是多少元;(2)办事处准备购进这两种保洁工具共50把,并且扫帚的数量不多于拖把数量的3倍,不少于拖把数量的2倍,哪种方案最省钱?请说明理由.7答案一、1.C 2.A 3.D 4.B 5.B 6.A 7.D 8.C 9.C 10.A 点拨:设A ′D 与AC 交于点O .∵∠A =40°,∴∠A ′=∠A =40°.∵∠1=∠DOA +∠A ,∠1=112°,∴∠DOA =∠1-∠A =112°-40°=72°.∵∠DOA =∠2+∠A ′,∴∠2=∠DOA -∠A ′=72°-40°=32°.二、11.不能 12.∠1>∠2>∠3 13.2 14.1215.72° 点拨:如图,延长AB 交l 2于点M.(第15题)∵五边形ABCDE ∴正五边形ABCDE 的每个外角相等.∴∠MBC ==72°.360°5∵l 1∥l 2,∴∠2=∠BMD .∵∠1=∠BMD +∠MBC ,∴∠1-∠2=∠1-∠BMD =∠MBC =72°.三、16.解:(1)-+=1,2x -12x -24去分母,得-2(2x -1)+(x -2)=4,去括号,得-4x +2+x -2=4,移项,得-4x +x =4+2-2,合并同类项,得-3x =4,系数化为1,得x =-.43(2){34x +y =12,①4x -2y =10.②)①×2+②,得x =11,解得x =2.112把x =2代入②,得8-2y =10,解得y =-1,故方程组的解为{x =2,y =-1.)17.解:解2x +3≥x +11,得x ≥8;解<4,得x <10,3x -105∴不等式组的解集是8≤x <10.在数轴上表示为:(第17题)∴x 的整数解是8、9.18.解:(1)如图①,△AB ′C ′即为所求;(第18题)(2)如图②,△DE ′F ′即为所求.19.解:∵∠CED =100°,∠D =35°,∴∠BCD =180°-∠CED -∠D =180°-100°-35°=45°.∵∠BCD 是△ABC 的外角,∴∠B =∠BCD -∠A =45°-20°=25°.920.解:方法1:∵∠1+∠BAD =180°,∠2+∠ABC =180°,∠3+∠BCD =180°,∠4+∠CDA =180°,∴∠1+∠BAD +∠2+∠ABC +∠3+∠BCD +∠4+∠CDA =180°×4=720°.∵∠BAD +∠ABC +∠BCD +∠CDA =360°,∴∠1+∠2+∠3+∠4=360°.方法2:如图,连结BD,(第20题)∵∠1=∠ABD +∠ADB ,∠3=∠CBD +∠CDB ,∴∠1+∠2+∠3+∠4=∠ABD +∠ADB +∠2+∠CBD +∠CDB +∠4=180°×2=360°.21.解:(1)AE ∥CF ,AC ∥DF ,BC ∥EF .(2)AD =CF =BE =2 cm.(3)∵AE ∥CF ,∠ABC =65°,∴∠BCF =∠ABC =65°.22.解:∵将△ABC 绕点A 按逆时针方向旋转得到△ADE ,∴∠BAD =∠EAC ,△ADE ≌△ABC ,∴∠C =∠E =40°,AB =AD .∵DE ∥AC ,∴∠E =∠EAC .∴∠BAD =∠C =40°.∵AB =AD ,∴∠ABD =∠ADB ,∴∠ABD =(180°-∠BAD )=70°.1223.解:(1)设一把扫帚的售价是x 元,一把拖把的售价是y 元.由题意,可得解得{x +3y =26,3x +2y =29,){x =5,y =7.)答:一把扫帚的售价是5元,一把拖把的售价是7元.(2)设扫帚买了m 把,共花费W 元,则拖把买了(50-m )把.由题意得,W =5m +7(50-m )=-2m +350.∵扫帚的数量不多于拖把数量的3倍,不少于拖把数量的2倍,∴2(50-m )≤m ≤3(50-m ),解得≤m ≤.1003752∵m 为正整数,∴m 可以取34,35,36,37,∴共有四种方案:方案一:扫帚34把,拖把16把,共花费:-2×34+350=282(元).方案二:扫帚35把,拖把15把,共花费:-2×35+350=280(元).方案三:扫帚36把,拖把14把,共花费:-2×36+350=278(元).方案四:扫帚37把,拖把13把,共花费:-2×37+350=276(元).∵282>280>278>276,∴方案四最省钱.11。
安徽合肥市庐阳区2022-2023学年下学期七年级语文期末综合复习卷一、句子默写1.根据提示补写名句或填写课文原句。
(1)海日生残夜,_______。
(2)我寄愁心与明月,_______。
(3)_______,崔九堂前几度闻。
(4)______,切问而近思,仁在其中矣。
(5)牛背上牧童的短笛,______。
(6)肥胖的黄蜂伏在菜花上,轻捷的叫天子(云雀)______。
(7)诗文中蕴藏着四季之景,“山朗润起来了,________,太阳的脸红起来了”是朱自清笔下春日的温暖;“红莲被那密集的雨点,打得左右攲斜”是冰心笔下夏雨的滂沱;“_________,洪波涌起”是曹操笔下秋风的肃杀;“未若柳絮因风起”是谢道韫笔下冬雪的灵动。
二、基础知识综合2.根据语境,完成下面小题。
文学反yìng生活,承载着人们对生活的思考。
阅读文学作品可以让你拥有内在的丰沛,因为书中有辽阔的世界、有趣的灵魂、。
与书籍相伴,你可以与圣贤对话,叩问自己的内心;你可以与雅士相和.,陶冶自己的情趣……一路走来,你会发现自己变得阅历丰富、眼界开阔、心态平和、生活优雅。
你,便会遇见更好的自己!(1)从下面两幅字中任选一幅临写。
眼界开阔心态平和(2)看拼音写汉字:反yìng()(3)给加点字注音:相和.()(4)在横线处补写一个结构相同的短语。
三、综合性学习3.复兴学校开展以“美丽校园”为主题的综合实践活动,请你参与。
(1)小刚同学为本次活动拟写了一则宣传标语,请用对偶手法写出下句,要求符合主题。
上句:创建美丽校园;下句:(2)小刚同学准备在“美丽校园”主题班会上演讲,下面他的演讲稿中的一段话,请你帮他补充和修改。
四、现代文阅读阅读下面的文章,完成下面小题。
竹笋印象仇士鹏①我对竹笋的印象,一直是励志和积极的。
②在黑暗中积累了一季的力量,竹笋用根握紧了大地的脉动。
正所谓厚积而薄发,当春雨洒落,它便猛地抬头,捅破大地,一跃而上。
七年级下学期期末复习阅读理解语文试卷含答案一、七年级语文下册阅读理解1.阅读下文,完成下列小题。
紫藤萝瀑布宗璞①我不由得停住了脚步。
②从未见过开得这样盛的藤萝,只见一片辉煌的淡紫色,像一条瀑布,从空中垂下,不见其发端,也不见其终极。
只是深深浅浅的紫,仿佛在流动,在欢笑,在不停地生长。
紫色的大条幅上,泛着点点银光,就像迸溅的水花。
仔细看时,才知道那是每一朵紫花中的最浅淡的部分在和阳光互相挑逗。
③这里春红已谢,没有赏花的人群,也没有蜂围蝶阵。
有的就是这一树闪光的、盛开的藤萝。
花朵儿一串挨着一串,一朵接着一朵,彼此推着挤着,好不活泼热闹!④“我在开花!”它们在笑。
⑤“我在开花!”它们嚷嚷。
⑥每一穗花都是上面的盛开,下面的待放。
颜色便上浅下深,好像那紫色沉淀下来了,沉淀在最嫩最小的花苞里。
每一朵盛开的花就像是一个小小的张满了的帆,帆下带着尖底的舱。
船舱鼓鼓的,又像一个忍俊不禁的笑容,就要绽开似的。
那里装的是什么仙露琼浆?我凑上去,想摘一朵。
⑦但是我没有摘。
我没有摘花的习惯。
我只是伫立凝望,觉得这一条紫藤萝瀑布不只在我眼前,也在我心上缓缓流过。
流着流着,它带走了这些时一直压在我心上的关于生死的疑惑,关于疾病的痛楚。
我浸在这繁密的花朵的光辉中,别的一切暂时都不存在,有的只是精神的宁静和生的喜悦。
⑧这里除了光彩,还有淡淡的芳香,香气似乎也是浅紫色的,梦幻一般轻轻地笼罩着我。
忽然记起十多年前家门外也曾有过一大株紫藤萝,它依傍一株枯槐爬得很高,但花朵从来都是稀落的,东一穗西一串伶仃地挂在树梢,好像在察颜观色,试探什么。
后来索性连那稀零的花串也没有了。
园中别的紫藤花架也都拆掉,改种了果树。
那时的说法是,花和生活腐化有什么必然关系。
我曾遗憾地想:这里再看不见藤萝花了。
⑨过了这么多年,藤萝又开花了,而且开得这样盛,这样密,紫色的瀑布遮住了粗壮的盘虬卧龙般的枝干,不断地流着,流着,流向人的心底。
⑩花和人都会遇到各种各样的不幸,但是生命的长河是无止境的。
Unit 3 复习总结自测卷一一、根据句意及汉语提示完成单词。
1. Tom usually plays soccer for sixty ________ (分钟) after school.2. There are ________ (九十) students in the room.3. It's about two ___________ (千米) from my home to school.4. Lu Hu never ________ (骑) his bike to school.5. I want a ________ (新的) skirt for my daughter.6. My father always takes the ________ (火车) to Beijing.7. The boy can't spell the word ________ (七十).8. It's about two ____________ (公里) from here to the bus station.9. It's ________ (真的) that a new bridge will be built over the river.10. There is a big river ________ (介于……之间) their school and the village.二、单项选择。
1. — ____does it take to get there by bus?—About thirty minutes.A. How oftenB. How farC. How longD. How much2. —Do you get to school by ____bus?—No, I don't. I take ____subway.A. a;B. a; theC. the; theD. /; the3. —Have a good day at school.— ____.A. You, tooB. I'm sorryC. Yes, sureD. You're welcome4. There are ____of students in our school but only ____ students are girls.A. hundreds; two hundredB. hundred; two hundredsC. hundreds; two hundredsD. hundred; two hundred5. — ____does your mother go to work?—She walks to work.A. WhereB. HowC. WhenD. What6. It takes ____forty minutes ____there by bus.A. his; getB. him; getsC. his; gettingD. him; to get7. My brother and I ____about half an hour to get to school.A. needB. getC. takeD. has8. It usually takes me about twenty minutes ____to school.A. to rideB. ridingC. ridesD. ride9. —Can you ____a car?—No, but my father can. So I go to school in his car.A. takeB. rideC. driveD. get10. For the villagers, it's difficult ____the river.A. crossB. crossingC. crossesD. to cross11. Jingjing is a ____ girl.A. five year oldB. five-year-oldC. five years oldD. five-years-old12. —What do you want to be after you finish college(大学)?—My ____is to be a teacher.A. lifeB. habitC. dreamD. work13. —What do you think of math?—It's difficult ____me ____.A. to; to learnB. for; to learnC. for; learnD. to; learn14. —I want to be a scientist in the future(将来).—I hope your dream can ____.A. come inB. come outC. come trueD. come over15. —How is your English teacher?—She ____a mother to me.A. isB. likeC. likesD. is like16. It ____me half an hour ____the train station.A. takes; get toB. take; to get toC. takes; to get toD. take; getting to17. — ____does he ____the trip?—It's great.A. What; likeB. How; think ofC. What; think ofD. How; like of18. My uncle lives ____Beijing. He lives ____there with his son.A. in; inB. /; /C. in; /D. /; in19. —How far is it from the subway station to your home?—It's ____walk.A. five minuteB. five minute'sC. five minutesD. five minutes'20. Mr. Smith will arrive at our school next week.The underlined part “arrive at” means ________.A. getB. goC. comeD. reach三、根据汉语意思完成句子,每空一词。
期终检测卷AI. 词汇。
A. 根据句意, 首字母或中文提示完成句子。
1. They heard something ________ (not common, unusual) in ________ (灌木从).2. Suddenly,I heard a ________ (a quiet sound) behind the tree.3. The cat was very ________ (not strong).It made a sound like a whisper.4. Pete r’s dream is to ________(进入)a good university.5. She comes from Jiangsu ________ .6. If Chen Tao have the ________ (奖品), we will all be very happy.7. When the mouse saw the cat, it ________ (ran quickly) into the hole.8. .Simon c ________ play football when he was 5 yeas old.9. Dogs are very ________ to people. They are our good friends.10. ________ comes before Thursday.1. This morning Andy got up early as ________ .2. I have ________ money than you.3. The party is for my father’s ________ birthday.4. They ran away _______ .5. Our neighbors are all ________ and helpful.6. It’s co1d outside, so you should keep the door ________ .C.根据句意用所给动词的适当形式填空1. Look! Some students ________ (sit) under the big tree.2. ________ your mother ________ ( take) you to the zoo last weekend ?3. Miss Chen ________ (nod) to us after we said hello to her yesterday .4. My brother often ________ (watch ) TV on Saturday .5. Listen ! How nice the music ________ (sound) .6. I ________ (not go) to the library with Sandy yesterday afternoon .7. I heard my friend ________ (sing) an English song just now.8. Simon ________ (lose) his way on his way to the park a moment ago .9. Tom ________ (not be) late for school again now.10. Hi, you two! ________ (not shout) in the reading room.Ⅱ. 单项选择题.( ) 1. Can you tell me ________ that in English?A. how to callB. what to sayC. how to sayD. what to speak( ) 2. ________ a cat. ________ color is black.A. It’s; It’sB. Its; ItsC. Its; It’sD. It’s; Its( ) 3. Yesterday I met Amy ________.A. on my way homeB. in my way homeC. on my way to homeD. in my way to home( ) 4. Tom, ________ Jack, ________ to school by bus every morning.A. likes, goB. likes, goesC. like, goesD. like, go( ) 5. Mrs. Chen often ________ TV for two hours last year.A. watchB. watchesC. watchedD. is watched( ) 6. There isn’t ________ in today’s newspaper.A. nothing interestingB. interesting somethingC. anything interestingD. interesting anything( ) 7. He ________ his mother.A. decided not to callB. decided not callingC. decided didn’t callD. decided don’t call( ) 8. The little boy is not ________ to go to school. He is only 2 years old!A. old enoughB. enough oldC. young enoughD. enough young( ) 9. You shouldn’t be so ________. You should be more polite to your friends.A. unhappyB. rudeC. unhelpfulD. careless( ) 10, The buildings were on the fire last night. The firemen came and _______A. put out itB. put it outC. put them outD. put them off( ) 11. —Is that your best friend Simon?—No, it ________ be him. He flew to Hong Kong yesterday.A. mustn’tB. may notC. can’tD. needn’t( ) 12. ________ beautiful the picture is!A. WhatB. HowC. What aD. How a( ) 13. “Bubble” rhymes with ________.A. wideB. fightC. troubleD. edge( ) 14. Her mother looked ___ at that moment when she heard the good news.A. happyB. happilyC. unhappyD. sadly( ) 15. They all know that it isn’t difficult ________ cats.A. take careB. take care ofC. to look afterD. look afterⅢ. 句型转换。
1
七下期末数学复习题
姓名
一.选择题
1. 1010可以写成 ( )
A.521010 B.521010 C.52)10( D.55)10(
2.若m+n=3,则代数式624222nmnm的值为( )
A.12 B. 3 C. 4 D. 0
3. 若3x3-kx2+4被3x-1除后余3,则k的值为( )
A.2 B.4 C.9 D.10
4.要在二次三项式2x□6x的□中填上一个数,使它能按abxbax)(2型分解为))((bxax的
形式,那么这些数只能是 ( )
A.1,1 B.5,5 C.5,5,1,1 D.以上答案都不对。
5.如图,一块正方形铁皮的边长为a,如果一边截去6,另一边截去5,则所剩长方形铁皮的面积(阴影部
分)表示成:①);6)(5(aa②);5(652aaa③);6(562aaa ④;30652aaa其中正确
的有( )
A.1个 B.2个 C.3个 D.4个
6.已知1yx,则化简11()()xyxy的结果是( )
A.22x B.22y C.22xy D.222xy
7.方程21111xx的解为( )
A.0 B.1 C.-1 D.1或-1
8.若x2-x-2=0,则22121(1)1xxxx的值等于( )
A.32 B.0 C.32或0 D.12
9. 已知a>0>b>c,a+b+c=1,,,bcbcabMNPabc,则M,N,P之间的大小关系是 ( )
A.M>N>P B.N>P>M C.P>M>N D.M>P>N
10.同时使分式 8652xxx有意义,又使分式9)1(322xxx 无意义的x的取值范围是( )
A.x≠-4且x≠-2 B.x=-4或x=2 C.x=-4 D.x=2
2
11. 如图,在梯形ABCD中,AD⊥AB,BC⊥AB,P是AB的中点.已知AD=a+1,BC=2a-1,AB=2a,若梯形ABCD的面
积为S,则△CPD的面积为( )
A.13S B.12S C.23S D.34S
12.甲乙两人分别从相距s(km)的两地同时出发,若同向而行,则t1(h)后,快者追上
慢者.若相向而行,则t2(h)后,两人相遇,那么快者速度是慢者速度的( )
A.112ttt倍 B.121ttt 倍 C.1212tttt倍 D.1212tttt倍
二.填空题(每题3分,共30分)
13.已知5x2-3x-5=0,则22152525xxxx=
14.化简:22221369xyxyxyxxyy= ; 方程1233kxx有增根,则_________k
15.当a= ,b= 时关于x,y的二元一次方程组byxyax4232有无数个解
16.对于下列说法:①2xy是分式;②3232xx;③2111yx是分式方程;④式子
(1)yxyxx
从左到右的过程是因式分解;⑤214xx是完全平方式; 错误的是
17.在实数范围内进行因式分解:4244xx=
18.代数式2212011xx的最小值是 ;
19.如果11srs,将分式1rr用s表示,则1rr=
20.已知1192()abab,则baab
21.设a>b>0,a2-6b2-ab=0,则abba的值等于
22. 某商场在一楼和二楼之间安装了一自动扶梯,以均匀的速度向上行驶,一男孩和一女孩同时从自动扶梯上
走到二楼(扶梯行驶,两人也走梯).如果两人上梯的速度都是均匀的,每次只跨一级,且男孩每分钟走动的级数
是女孩的两倍.已知男孩走了27级到达扶梯顶部,而女孩走了18级到达扶梯顶部.则扶梯露在外面的部分级数
是
三.解答题(共46分)
21.因式分解: (每题4分,共8分)
(1)39xx (2)1-x2+6xy-9y
2
P
A
D
CB
3
21.(5分)解方程2714111xx
22.(5分)先化简,再求值:35(2)482yyyy,其中y=5
23.(6分)(1)如图①,加油站A和商店B在马路MN的同一侧,A到MN的距离大于B到MN的距离,AB=7m,
一个行人P在马路MN上行走.当P到A的距离PA与P到B的距离PB的差最大时,这个差等于 米.
(2)请结合第(1)题的解决方法,在图②的直线MN上确定一点Q,使QAQB最大.(不限画图工具)
24.(6分)列方程解应用题
一项工作,甲乙两人合作,4天可以完成.他们合做了3天后, 乙另有任务,甲单独又用了112天才全部完成.
问甲乙两人单独做,各需几天完成?
M
N
A
B
图①
图②
MN
A
B
4
25.描述证明(本小题满分6分)
海宝在研究数学问题时发现了一个有趣的现象:
(1)请你用数学表达式补充完整海宝发现的这个有趣现象;
已知a>0,b>0
如果
那么
(2)请你证明海宝发现的这个有趣的现象;
如图,长方形ABCD中放置9个形状、大小都相同的小长方形,小长方形的长为x,宽为y(尺寸如图)
(1)写出两个关于x,y 的关系式.
(2)求图中阴影部分的面积.
某电脑公司现有A,B,C三种型号的甲品牌电脑和D,E两种型号的乙品牌电脑.希望中学要从甲、乙两种
品牌电脑中各选购一种型号的电脑.
(1) 写出所有选购方案(利用树状图或列表方法表示);
(2) 如果(1)中各种选购方案被选中的可能性相同,那么A型号电脑被选中的概率是
多少?
(3) 现知希望中学购买甲、乙两种品牌电脑共36台(价格如图所示),恰好用了10万
元人民币,其中甲品牌电脑为A型号电脑,求购买的A型号电脑有几台.
C
B
D
22
7
A
(第29题)
5