广东省深圳市宝安中学(集团)初中部2019-2020学年第二学期八年级第 10 周周测试卷(PDF版无答案)

2020年广东省深圳市宝安区宝安中学初二第一学期开学考英语试卷1.选择填空。

1．—Congratulations!—Thanks. I couldn’t achieve it without your support.A．advice B．plan C．help2．—I’m afraid I will give up learning English. It’s too difficult for me.—Nothing is difficult if you put your heart into it.A．stop B．enjoy C．keep3．—Do you still hear from Lisa?—Of course, although we’re not together, we remain friends.A．are only B．are very C．are still4．—When can we return the books to the library?—Maybe next week. It’s not open yet.A．go back B．give up C．give back5．—Didn’t you sleep well last night?—No, a terrible noise woke me up in the middle of the night.A．called me up B．made me stop sleeping C．gave me support6．—How long will the meeting last?—About half an hour.A．become B．achieve C．continue7．—Remember not to keep the computer on when you are not using it.—Thanks for reminding me. We must save energy.A．turning B．working C．connecting8．—Our car can’t go through that street. It is narrow.－Yes, so we have to walk there.A．not high B．not long C．not wide9．“I’m sorry, we’ve sold out,” the girl said. “What a pity!” Susan exclaimed.A．explained B．excused C．shouted10．To their surprise, the precious parcel was full of stones and sand!A．valuable B．pretty C．ordinary11．—Everything is ________ if we try our best.－Yes. you’re right.A．successful B．impossible C．possible12．—Can I have a ________ with you this Sunday afternoon? I want to talk about your green project for our school.—OK. See you then.A．joke B．conversation C．holiday13．—How big your school is!—Yes, it is ________ 66 classes.A．made of B．made up of C．made in14．—You look sad, Kate. What’s the matter?—I have made ________ mistakes in my report.A．a little B．a bit C．a few15．—You must give up fishing! It’s a ________ of time.－I’m not really interested in fishing. I am only interested in sitting in a boat and doing nothing at all!A．packet B．waste C．crowd16．The porter could not understand me. So I ________ my questions several times and at last he understood. A．answered B．requested C．repeated17．The writer had lost his money. He felt upset. He must have been ________.A．patient B．worried C．tired18．Some fishermen are unlucky. ________ catching fish, they catch old boots and rubbish.A．Instead of B．Because of C．In spite of19．I am ________ to stay here, which means that I make up my mind to stay here.A．doubted B．determined C．honest20．The writer lost his money and couldn’t do anything. So he started to ________ the wicked (惹人厌的) world. A．complain B．explain C．complete2. 完型填空In life, there is one thing or person that changes us. I still remember this person 21 . She was my English teacher when I came into junior high school. English was always 22 for me. I often felt sad and wanted to23 . But I changed my mind because of her. She was patient with me 24 encouraged me a lot. She planted hope in my heart. I began to work 25 than ever at learning English.In the beginning, it was so hard. But my English leacher often said, “There aren’t any shortcuts in school. 26you want to get high scores, you have wo do your best.” I agreed 27 her. I decided to overcome futureShe was also a friend. I could tell her anything She 28 me lots of good advice. She was always there when I was under pressure. As a result, I am now 29 enough to challenge myself and lead a better life.Maybe you will also have 30 that you can’t solve. But don’t worry. Look at the people around you———you will find someone who can help you solve the problems.21．A．quickly B．clearly C．nearly22．A．difficult B．interesting C．wonderful23．A．give them up B．give it up C．give up it24．A．so B．or C．and25．A．hard B．harder C．more hardly26．A．When B．Unless C．If27．A．with B．to C．about28．A．took B．asked C．offered29．A．stupid B．nervous C．brave30．A．problems B．pressure C．people3. 阅读理解AIf you had an emergency（紧急情况），who would you think of first? What would you do if your parents had an emergency?In 1988, a serious earthquake hit Armenia, killing about 25,000 people in less than four minutes. A father rushed to the school where his son was, only to discover that the building was flattened（变平）.But he remembered the promise he had made to his son: “No matter what, I’ll always be there for you!” Tears began to fill his eyes. He rushed to the place where his son’s classroom had been and started digging.Other parents tried to pull him off what was left out of the school. To each parent be asked, “Are you going to help me?” And then he went back to his digging, stone by stone.A firefighter showed up and tried to pull him away. The police came as well. To each, he asked, “Are you going to help me?”He dug for 8 hours…18 hours…28 hours…then, in the 38th hour, he pulled beck a large stone and heard his son, He called his son’s name: “ARMAND!” He heard back: “Dad? It’s me, Dad! I told the other kids not to worry. I told them that you would come to save us.”“What’s going on in there?” the father asked.“There are 14 of us left out of 33, Dad.” his son answered.“Let the other kids out first. I know you’ll get me! No matter what. I know you’ll always be there for me!”31．This story happened in ________.A．America B．Algeria C．Armenia D．Argentina32．How did the father feel when he heard his son’s voice?A．Excited.B．Calm.C．Hopeless.D．Worried.33．The son wanted other kids to get out first because ________A．there was a large stone on his body.B．many of the kids didn’t have parents.C．he promised the others that they could get out first.D．he believed that his father would still be there for him.34．What can we infer from the article?A．The father is lazy.B．The father knew his son was alive before digging.C．The father is in good relationship with his son.D．There were 34 students in the son’s classroom.BWhat does your dream school look like? John Hardy, a man from Canada, designs and builds his own dream school. He calls it the Green School.The Green School sits on BaliIsland (巴匣岛). The students are aged from 3 to 19 years old. It doesn't look like a school, but it is green. Everything in the school is made of bamboo. The classrooms have no walls or doors. The teachers write on the bamboo blackboard. The desks are not square. Everything is so beautiful.At school, there is a library, a computer room, a few teachers’ offices and some classrooms. Children learn usual subjects, such as reading, writing and Maths. They also learn how to grow vegetables in the school’s vegetable garden. There is a swimming pool for the kids to play in. There is also a small farm where children learn to keep goats, chickens and pigs. It is so much fun to go to the Green School.More and more people are living around the Green school. It becomes a community.35．Children between ________ and ________ years old can go to the Green School.A．1;3B．3;19C．6;20D．16;2236．The students won’t learn how to ________ in the Green School.A．grow vegetables in the gardenB．make a bamboo deskD．work out Maths problems37．Which of the following is TRUE?A．All the desks are square in the school.B．Children don’t need to write in the school.C．You can see animals in the school.D．There is a music room in the school.38．The best title for the passage may be “________ ”.A．The Green School on BaliIslandB．How to be the Groen SchoolC．Life story of John HardyD．John Hardy’s dream bouseCGiving children music lessons won’t just introduce them to a world of rhythm and melody — it could also improve their language skills.A study of 74 Chinese children before primary school was led by scientist Robert Desimone from MIT. In the study, the 4 to 5-year-old children were divided into three groups. One group received a 45-minute piano lesson three times a week, while another received extra reading instruction(授课) lessons. The third group took no extra lessons.The lessons lasted for six months. Then the children were tested on their abilities of discriminate (辨别) words based on differences in tone and so on. The test results showed that the children who had taken piano lessons performed better at discriminating between words, when compared with the children taking extra reading lessons. But both groups did better than the third group.The researchers think that the musical education helped these children perform better in language tests, even better than the kids taking extra reading lessons. This should make educators pay close attention. “That means if schools spend more time and energy on music education, there will be an improvement in children’s studies.” Desimone said.Nowadays, many schools give extra reading to kids instead of the arts education. In fact, giving children music lessons is no worse than extra reading.39．Where was Robert Desimone from?A．China.B．Germany.C．A primary school.D．MIT.40．How many groups of children took part in the study?A．74.B．45.C．2.D．3.41．Which of the following statements is TRUE?B．The third group performed best in discriminating words.C．The researchers think that reading helped the children perform better in the tests.D．The researchers think schools should add more music class instead of extra reading class.42．According to the passage, what lessons should be paid more attention to in schools?A．Reading.B．Art.C．PE.D．Science.DPaper-cutting is one of the traditional folk arts(民间传统艺术) in China. It has a long history of about 1,500 years. Artists use paper, scissors and knives to make paper-cuttings. People find happiness and luck in paper-cuttings. In Spring Festival, they paste paper-cuttings like “Fu” on windows to bring good luck for the new year. And parents may paste “Xi” inside the houses when their sons or daughters get married.In many places, the paper-cuttings are red. In old times, people worked on the farm and respected the sun. Red is the colour of the sun. They see red as the symbol of hope and life. Now, you can see red everywhere in China, the walls of old palaces, lanterns, weddings and so on.Do you know some of the paper-cuttings are black in Shanzhou? Shanzhou is in Henan province(省). Thepaper-cuttings there are black. Black is the best colour there. Since Shanzhou is a dry place, people make blackpaper-cuttings and wish for rain.Moreover, paper-cuttings are colourful in Y uxian(蔚县), Hebei province. Artists put five pieces of paper-cuttings together. They just paint the first piece with brushes. And the colour goes through. Soon a beautiful paper-cutting is made. 43．The underlined word “symbol” means “________” in Chinese.A．象征B．暗示C．出现D．记录44．Why people in Shanzhou make black paper-cuttings?A．Because they see it as the symbol of hope.B．Because black is their favourite colour.C．Because they wish for rain.D．Because they think black is lucky.45．Which of the following isn’t mentioned in the passage?A．Red paper-cuttings.B．Green paper-cuttings.C．Black paper-cuttings.D．Colourful paper-cuttings.46．What may the writer continue to talk in paragraph 5?B．When do people paste paper-cuttings.C．Paper-cutting museums around us.D．How do people in Yuxian make paper-cuttings.EIn many, even most space -themed movies, whenever Earth faces a disaster, the solution is always fleeing (逃出) the planet in spaceships.But the latest Chinese science-fiction movie, The Wandering Earth (流浪地球), offers a different and more special idea to audiences.The movie is based on a short story by Chinese sci-fi writer Liu Cixin. In the movie, Earth is in danger of being destroyed by the dying sun. In order to save it, humans around the world work together to build a giant engine system that will push Earth away from the sun. Instead of leaving Earth again this time we’re taking it with us.This “ambition” didn’t come nowhere. For thousands of years, “homeland” has had a soft spot in the hearts and minds of Chinese people. One old idiom is “Luoye Guigen”, which means returning to one’s homeland in old age, like fallen leaves returning to the roots of their tree. This special cultural background probably shows The Wandering Earth is different from Hollywood-style space movies.“What is Chinese sci-fi?” Guo Fan, the film’s director, said in an interview. “A vehicle (方式) that can really express our cultural and spiritual core can be called Chinese sci-fi. Or we’re just imitating (模仿) others and telling the same American stories.”And the makers of The Wandering Earth may have chosen the best time to tell its Chinese sci-fi story. The movie was on show on Feb.5, the first day of Chinese New Year. It was a time when many people had just made the hard journey back to their hometowns.So to them, there is only one possible way to tell the story: Earth goes wherever humans go, because it’s our home. 47．What’s the solution in most space-themed movies when Earth faces a disaster?A．People usually stay with the Earth.B．People usually fight for the Earth.C．The Earth usually becomes a better place.D．People usually escape from the Earth.48．Paragraph 4 mainly explains that ________.A．the movie The Wandering Earth shows special cultural background “homeland”B．the movie The Wandering Earth shows the Chinese great ambitionC．most Chinese people especially prefer the traditional Chinese culture1．C【详解】句意：——祝贺！——谢谢。

宝安中学⼋年级下数学第9周周末测试出题：张洪涛审题：王红霞学校：班级：姓名：学号：家⻓签名：⼀.选择题（每题3分，共36分）1.若a＞b，则下列各式中⼀定成⽴的是（）A．a+2＜b+2B．a-2＜b-2C．＞D．-2a＞-2b2.下列图案中是中⼼对称图形，但不是轴对称图形的是（）A．B．C．D．3.下列各组数中，以它们为边⻓的线段能构成直⻆三⻆形的是（）A．2，3，4B．4，5，6C．3，5，7D．1，1，4.下列多项式能⽤完全平⽅公式分解因式的是（）A．m2－mn＋n2B．x2＋4x–4 C.x2－4x＋4 D.4x2－4x＋45.要使分式有意义，那么x的取值范围是（）A．x≠3B．x≠－3C．x≠0且x≠3D．x≠3且x≠-36.平⾯直⻆坐标系内,点P（m+3,m-5）在第四象限，则的取值范围是（）A.-5＜m＜3B.-3＜m＜5C.3＜m＜5D.-5＜m＜-37.如右图，△ABC中，∠C=90°，点E在AC上，且∠1=∠2，DE垂直平分AB，垂⾜是D，如果EC=cm，则AE等于（）A．B．C．D．128．已知实数a，b满⾜，则以a，b的值为两边⻓的等腰三⻆形的周⻓是（）A．17或7B．31C．41D．31或419.对于⾮零实数x、y，规定若，则x的值为（）A．4B．－4C．0D．⽆解10.如图，将△ABC绕C点顺时针旋转⾄△DEC，使得A、C、E三点共线，此时点D恰好在AB延⻓线上，若∠A=25°，则∠BCD的度数是（）A.80° B.90° C.100° D.110°11.如图，在Rt△ABC中，∠C=90°，∠A=60°，以顶点B为圆⼼，适当⻓为半径画弧，分别交AB，BC于点M、N，再分别以点M、N为圆⼼，⼤于MN的⼀半⻓为半径画弧，两弧交于点P，作射线BP交边AC于点D，作DE∥AB，交BC于点E，若DE=8，则点D到AB的距离是（）A.2 B.3 C.4 D.12.在平⾯直⻆坐标系xOy中，有⼀个等腰直⻆三⻆形AOB，∠OAB＝90°，直⻆边AO在x轴上，且AO＝1．将Rt△AOB绕原点O顺时针旋转90°后各边增⾄原来的两倍得到等腰直⻆三⻆形A1OB1，再将Rt△A1OB1绕原点O顺时针旋转90°后各边增⾄原来的两倍得到等腰三⻆形A2OB2……，依此规律，得到等腰直⻆三⻆形A2020OB2020．则点B2020的坐标是() A．(22020，－22020)B．(－22020，22020)C．(－22020，－22020)D．(22020，22020)第12题第14题第16题⼆、填空题（共4题，每题3分，共12分）13．分解因式：=；14．⼀次函数y=－3x+b和y=kx+1的图象如图所示，其交点为P（3，4），则不等式－3x+b≤kx+1的解集是；15.若关于x的不等式组有解，则实数a的取值范围是；16．如图，Rt△ABC中，∠ABC＝90º，AB＝BC＝，将△ABC绕点A逆时针旋转60º，得到△ADE，作EF⊥BA，交BA延⻓线于点F，则EF的⻓是；三、解答题（本题共7⼩题，共52分）17．因式分解：（每⼩题4分，共8分）①②18．分式化简（每⼩题4分，共8分）①②19．①解不等式组（4分）②解分式⽅程：（4分）20．（5分）化简求值：，其中x是不等式组的整数解．21．（6分）（应⽤分式⽅程解应⽤题）某商店经销⼀种纪念品，9⽉的销售额为2000元，为扩⼤销售，10⽉该店对这种纪念品打九折销售，结果销量增加20件，销售额增加700元．（1）求这种纪念品9⽉的销售单价？（4分）（2）若9⽉销售这种纪念品获利800元，问10⽉销售这种纪念品获利多少元？（2分）22．（8分）如图，矩形ABCO中，点C在x轴上，点A在y轴上，点B的坐标是（-3，4）．矩形ABCO 沿直线BD折叠，使得点A落在对⻆线OB上的点E处，折痕与OA、x轴分别交于点D、F．(1)直接写出线段BO的⻓度，BO=；（1分）(2)求点D的坐标.（4分）(3)若点M是y轴上的动点，若△MDF是以DF为腰的等腰三⻆形，请直接写出点M的坐标..（3分）23.（9分）在Rt△ABC中，∠ACB=90°，∠A=30°，点D是AB的中点，DE⊥BC，垂⾜为点E，连接CD．（1）如图1，DE与BC的数量关系是；（3分）（2）如图2，若P是线段CB上⼀动点（点P不与点B、C重合），连接DP，将线段DP绕点D逆时针旋转60°，得到线段DF，连接BF，猜想DE、BF、BP三者之间的数量关系，并证明你的结论；(4分)（3）若点P是线段CB延⻓线上⼀动点，按照（2）中的作法，可在图3中补全图形，直接写出DE、BF、BP三者之间的数量关系．（2分）。

宝安区2021-2022学年第一学期学情调查问卷八年级数学第一部分（选择题）一、选择题1. 下列各数中，是无理数的是（ ）A.3.14B. πC. 38D. 【答案】B【解析】【分析】根据无理数的定义，“无限不循环的小数是无理数”逐项分析即可．【详解】解：A. 3.14 是有理数，故该选项不符合题意；B. π是无理数，故该选项符合题意；C.38是有理数，故该选项不符合题意； D. 3=是有理数，故该选项不符合题意；故选B【点睛】本题考查了无理数，解答本题的关键掌握无理数的三种形式：①开方开不尽的数，②无限不循环小数，③含有π的数．2. 若点(,2)P m −在第三象限内，则m 的值可以是（ ）A. 2B. 0C. 2−D. 2±【答案】C【解析】【分析】根据第三象限内点的特点可知横纵坐标都为负，据此判断即可．【详解】解：∵点(,2)P m −在第三象限内，∴0m < ∴m 的值可以是2−故选C【点睛】本题考查了第三象限内点的坐标特征，掌握各象限内点的坐标特征是解题的关键．平面直角坐标系中各象限点的坐标特点：①第一象限的点：横坐标>0，纵坐标>0；②第二象限的点：横坐标<0，纵坐标>0；③第三象限的点：横坐标<0，纵坐标<0；④第四象限的点：横坐标>0，纵坐标<0． 3. 下列计算中，正确的是（ ）A.B. 3+C.D. 2−【答案】C【解析】【分析】根据二次根式的加减法以及二次根式的乘法运算进行计算即可．【详解】A.A 选项不正确；B. 3与B 选项不正确；C. ,计算正确，故C 选项正确D. 与2不同类二次根式不能合并，故D 选项不正确；故选C【点睛】本题考查了二次根式的加减法以及二次根式的乘法，掌握二次根式的运算法则是解题的关键． 4. 下列各组数中，不能作为直角三角形的三边的是（ ）A. 3，4，5B. 2，3C. 8，15，17D. 23，24，25【答案】D【解析】【分析】由题意直接根据勾股定理的逆定理即如果三角形有两边的平方和等于第三边的平方，那么这个三角形是直角三角形，如果没有这种关系，这个就不是直角三角形进行分析判断即可．【详解】解：A 、32+42=52，符合勾股定理的逆定理，故选项错误；B、22223+=，符合勾股定理的逆定理，故选项错误；C 、82+152=172，符合勾股定理的逆定理，故选项错误；D 、∵（32）2+（42）2=81+256=337，（52）2=625，∴（32）2+（42）2≠（52）2，不符合勾股定理的逆定理即此时三角形不是直角三角形，故选项正确. 故选：D.【点睛】本题考查勾股定理的逆定理，注意掌握在应用勾股定理的逆定理时，应先认真分析所给边的大小关系，确定最大边后，再验证两条较小边的平方和与最大边的平方之间的关系，进而作出判断． 5. 如图，将一副三角板平放在一平面上（点D 在BC 上），则1∠的度数为（ ） 是A. 60°B. 75°C. 90°D. 105°【答案】B【解析】 【分析】根据三角尺可得45,30EDB ABC ∠=°∠=°，根据三角形的外角性质即可求得1∠【详解】解： 45,30EDB ABC ∠=°∠=°175EDB ABC ∴∠=∠+∠=°故选B【点睛】本题考查了三角形的外角性质，掌握三角形的外角性质是解题的关键．6. 生活中常见的探照灯、汽车大灯等灯具都与抛物线有关．如图，从光源P 点照射到抛物线上的光线,PA PB 等反射以后沿着与直线PF 平行的方向射出，若CAP α∠=°，DBP β∠=°，则APB ∠的度数为（ ）°A. 2αB. 2βC. αβ+D. 5()4αβ+ 【答案】C【解析】 【分析】根据平行线的性质可得,EPA PAC EPB PBD ∠=∠∠=∠，进而根据APB APE BPE ∠=∠+∠即可求解【详解】解： ,PF AC PF BD ∥∥∴,EPA PAC EPB PBD ∠=∠∠=∠∴APB APE BPE ∠=∠+∠αβ=+【点睛】本题考查了平行线的性质，掌握平行线的性质是解题的关键．7. 下列命题正确的是（ ）A. 数轴上的每一个点都表示一个有理数B. 甲、乙两人五次考试平均成绩相同，且20.9S =甲，21.2S =乙，则乙的成绩更稳定C. 三角形的一个外角大于任意一个内角D. 在平面直角坐标系中，点(4,2)−与点(4,2)关于x 轴对称【答案】D【解析】【分析】根据数轴上的点与实数一一对应即可判断A ；根据平均数相同的情形下，方差越小，成绩越稳定即可判断B ；根据三角形的外角与内角的关系即可判断C ；根据关于x 轴对称的点的坐标特征即可判断D【详解】A. 数轴上的每一个点都表示一个实数，故该选项不正确，不符合题意；B. 甲、乙两人五次考试平均成绩相同，且20.9S =甲，2 1.2S =乙，则甲的成绩更稳定，故该选项不正确，不符合题意；C. 三角形的一个外角不一定大于任意一个内角，故该选项不正确，不符合题意；D. 在平面直角坐标系中，点(4,2)−与点(4,2)关于x 轴对称，故该选项正确，符合题意；故选D【点睛】本题考查了实数与数轴，方差的意义，三角形的外角的性质，关于x 轴对称的点的坐标特征，掌握以上知识是解题的关键． 8. 如图，已知点(1,2)B 是一次函数(0)y kx b k =+≠上的一个点，则下列判断正确的是（ ）A. 0,0k b >>B. y 随x 的增大而增大C. 当0x >时，0y <D. 关于x 的方程2kx b +=的解是1x =【答案】D【分析】根据已知函数图象可得0,0k b <>，是递减函数，即可判断A 、B 选项，根据0x >时的函数图象可知y 的值不确定，即可判断C 选项，将B 点坐标代入解析式，可得2k b +=进而即可判断D【详解】A.该一次函数经过一、二、四象限∴ 0,0k b <>， y 随x 的增大而减小，故A,B 不正确；C. 如图，设一次函数(0)y kx b k =+≠与x 轴交于点(,0)C c ()0c >则当x c >时，0y <，故C 不正确D. 将点(1,2)B 坐标代入解析式，得2k b +=∴关于x 的方程2kx b +=的解是1x =故D 选项正确故选D【点睛】本题考查了一次函数的图象与性质，一次函数与二元一次方程组的解的关系，掌握一次函数的图象与性质是解题的关键．9. 某学校体育场的环形跑道长250m ，甲、乙分别以一定的速度练习长跑和骑自行车，同时同地出发，如果反向而行，那么他们每隔20s 相遇一次．如果同向而行，那么每隔50s 乙就追上甲一次，设甲的速度为x m/s ，乙的速度为y m/s ，则可列方程组为（ ）A. ()()2025050250x y y x += −=B. ()()2025050250y x x y −= +=C. ()()2050050250x y x y −= +=D. ()()2025050500x y y x += −=【答案】A【解析】 【分析】利用路程＝速度×时间，结合“如果反向而行，那么他们每隔20s 相遇一次；如果同向而行，那么每隔50s 乙就追上甲一次”，即可得出关于x ，y 二元一次方程组，此题得解．【详解】解：∵如果反向而行，那么他们每隔20s 相遇一次，∴20（x +y ）＝250；∵如果同向而行，那么每隔50s 乙就追上甲一次，∴50（y ﹣x ）＝250．∴所列方程组为()()2025050250x y y x += −=． 故选：A ．【点睛】本题考查了由实际问题抽象出二元一次方程组，找准等量关系，正确列出二元一次方程组是解题的关键．10. 如图，直线443y x =−+与x 轴交于点B ，与y 轴交于点C ，点(1,0)E ，D 为线段BC 的中点，P 为y 轴上的一个动点，连接PD 、PE ，当PED V 的周长最小时，点P 的坐标为（ ） A. 40,5B. (0,1)C. (1,0)D. 30,2【答案】A【解析】 【分析】作点E 关于y 轴的对称点F ，连接DF ，交y 轴于点Q ，则QE QF =，进而根据对称性求得当点P 与Q 重合时，PED V 的周长最小，通过求直线DF 的解析式，即可求得P 点的坐标【详解】解：如图，作点E 关于y 轴的对称点F ，连接DF ，交y 轴于点Q ，则QE QF =，连接PF ，的PED V 的周长PD PE DE PF PE PD DF DE =++=++≥+，点,D E 是定点，则DE 的长不变， ∴当PQ 重合时，PED V 的周长最小， 由443y x =−+，令0,x =4y =，令0y =，则3x = (3,0),(0,4)B C ∴D Q 是BC 的中点3(,2)2D ∴ (1,0)E ,点F 是E 关于y 轴对称的点(1,0)F ∴−设直线DF 的解析式为：y kx b =+，将3(,2)2D ，(1,0)F −代入， 0322k b k b =−+ =+解得4545k b = =∴直线DF 的解析式为：44+55y x =令0x =，则45y =即4(0,)5P故选A【点睛】本题考查了轴对称的性质求最值，求一次函数解析式，求直线与坐标轴的交点，求线段中点坐标，掌握根据轴对称的性质求线段和的最值是解题的关键． 第二部分（非选择题）二、填空题11. 9的算术平方根是 ．【答案】3【解析】【分析】根据一个正数的算术平方根就是其正的平方根即可得出．【详解】∵239=，∴9算术平方根为3．故答案为：3．【点睛】本题考查了算术平方根，熟练掌握算术平方根的概念是解题的关键．12. 为了庆祝中国共产党成立100周年，某校举行“歌唱祖国”班级合唱比赛，评委将从“舞台造型、合唱音准和进退场秩序”这三项进行打分，各项成绩均按百分制计算，然后再按舞台造型占40%，合唱音准占40%，进退场秩序占20%计算班级的综合成锁．七（1）班三项成绩依次是95分、90分、95分，则七（1）班的综合成绩为________．【答案】93【解析】【分析】根据题意求这组数据的加权平均数即可.【详解】解：七（1）班的综合成绩为9540%9040%9520%93×+×+×=分故答案为：93 【点睛】本题考查了求加权平均数，掌握加权平均数的计算是解题的关键，加权平均数计算公式为：1122()1k k x x f x f x f n++…+，其中12k f f f …，，，代表各数据的权. 13. 如图，长方形ABCD 的边AB 落在数轴上，A 、B 两点在数轴上对应的数分别为1−和1，1BC =，连接BD ，以B 为圆心，BD 为半径画弧交数轴于点E ，则点E 在数轴上所表示的数为_________．【答案】11+【解析】【分析】根据勾股定理求得BD ，进而根据数轴上的两点距离即可求得点E 在数轴上所表示的数．【详解】解： 四边形ABCD 是长方形，A 、B 两点在数轴上对应的数分别为1−和1，1BC =，1,2AD BC AB ∴===依题意BE BD =.设点E 在数轴上所表示的数为x ，则1x −解得1x =−故答案为：1【点睛】本题考查了勾股定理，实数与数轴，掌握勾股定理求得BD 是解题的关键．14. 如图，若一次函数3y kx =+与正比例函数2y x =的图象交于点(1,)m ，则方程组320kx y x y −=− −=的解为_________．【答案】12x y = =【解析】【分析】先将点(1,)m 代入正比例函数2y x =求得m ，则交点的坐标即为方程组的解．【详解】解：将点(1,)m 代入正比例函数2y x =，得2m =∴点()1,2为一次函数3y kx =+与正比例函数2y x =的图象的交点∴320kx y x y −=− −=的解为12x y = = 故答案为：12x y = =【点睛】本题考查了两直线交点与二元一次方程组的关系，理解交点的坐标即为方程组的解是解题的关键．15. 如图，将长方形纸片ABCD 沿MN 折叠，使点A 落在BC 边上点A ′处，点D 的对应点为D ¢，连接A D ′′交边CD 于点E ，连接CD ′，若9AB =，6AD =，A ′点为BC 的中点，则线段ED ′的长为________．【答案】94##2.25 【解析】【分析】连接NA ′，勾股定理求得DN ，进而证明A D N NCA ′′′ ≌，设,EC a A E b ′==，根据6NC =,以及Rt A EC ′ 三边关系建立方程组，解方程组求解即可．【详解】解：如图，连接NA ′，折叠DN D N ′∴=，AD A D ′′=，A D N D ′′∠=∠四边形ABCD 是长方形，9AB =，6AD =，9DC AB ∴==,6BC AD ==，90D BCD ∠=∠=°设DN x =则9NC DC DN x =−=−A ′是BC 的中点，6BC AD == ∴132CA BC ′== 在Rt A CN ′ 中， 222A N CN A C ′′=+在Rt A D N ′′ 中，222A N ND AD ′′′=+∴22CN A C ′+22ND AD ′′+ 即()2222936x x −+=+解得3x =ND ND A C ′′∴==3=,6NC A D ′′==又∵90ND A A CD ′′′∠=∠=°A D N NCA ′′′∴ ≌NA D A NC ′′′∴∠=∠A E NE ′∴=A D CN ′′=CE ED ′∴=设,EC a A E b ′== 在Rt A EC ′ 中222A E EC A C ′′−=即2223b a −=①又6CE EN CN +==6EC A E EC EN a b ′∴+=+=+=②由①可得()()9b a b a +−=③ 将②代入③得32b a −=④②-④得922a =解得94a =即94EC = 94ED CE ′∴== 故答案为：94【点睛】本题考查了勾股定理，折叠问题，因式分解，三角形全等的性质与判定，解二元一次方程组，掌握折叠的性质是解题的关键． 三、解答题16.计算：++【答案】3+【解析】【分析】根据二次根式的乘法运算以及求一个数的立方根进行计算即可【详解】解：++3=+63=+−=3+【点睛】本题考查了二次根式的乘法运算以及求一个数的立方根，掌握二次根式的乘法运算是解题的关键．17. 解方程组：22263x yx y−=−=【答案】91015xy==−【解析】【分析】根据加减消元法解二元一次方程组即可详解】解：22263x yx y−=−=①②①-②得：623y y−+=−解得15y=−将15y=−代入①1225x=−解得910x=【∴原方程组的解为：91015 xy==−【点睛】本题考查了加减消元法解二元一次方程组，掌握加减消元法是解题的关键．18. 深圳市教育局印发的《深圳市义务教育阶段学校课后服务实施意见》明确中小学课后延时服务从2021年3月5日开始实施某校积极开展课后延时服务活动，提供了“有趣的生物实验、经典影视欣赏、虚拟机器人竞赛、趣味篮球训练、国际象棋大赛……”等课程供学生自由选择．一个学期后，该校现为了解学生对课后延时服务的满意情况，随机对部分学生进行问卷调查，并将调查结果按照“A．非常满意；B．比较满意；C．基本满意；D．不满意”四个等级绘制成了如图所示的两幅不完整的统计图：请你根据图中信息，解答下列问题：（1）该校抽样调查的学生人数为人，请补全条形统计图；（2）样本中，学生对课后延时服务满意情况的“中位数”所在等级为，“众数”所在等级为；（填“A、B、C或D”）（3）若该校共有学生2100人，据此调查估计全校学生对延时服务满意（包含A、B、C三个等级）的学生有人．【答案】（1）50；统计图见解析（2）B;A（3）1890【解析】【分析】（1）根据A:非常满意的人数除以所占的百分比即可求得总人数，进而根据总人数减去,,A B D等级的人数即可求得C等级的人数，进而补全统计图；（2）根据中位数的定义可知该组数据的中位数为第25和26个的平均数，根据条形统计图即可求得位于B 等级，根据条形统计图即可求得人数最多的位于A等级；（3）根据总人数乘以20151050++即可求得答案 【小问1详解】 解：该校抽样调查的学生人数为：2040%50÷=（人）C 等级的人数为502015510−−−=（人） 补全统计图如图，【小问2详解】按满意度从大到小排列，根据中位数的定义可知，中位数为第25和26个的平均数，则样本中，学生对课后延时服务满意情况的“中位数”所在等级为BA 的人数最多，则“众数”所在等级为A故答案为：,B A【小问3详解】该校共有学生2100人，据此调查估计全校学生对延时服务满意（包含A 、B 、C 三个等级）的学生有2015102100189050++×=（人） 故答案为：1890【点睛】本题考查了扇形统计图与条形统计图信息关联，根据样本估计总体，求中位数和众数，求条形统计图，从统计图中获取信息是解题的关键．19. 列方程组解应用题：全自动红外体温检测仪是一种非接触式人体测温系统，通过人体温度补偿、温度自动校正等技术实现准确、快速测温工作，具备人体非接触测温、高温报警等功能．为了提高体温检测效率，某医院引进了一批全自动红外体温检测仪．通过一段时间使用发现，全自动红外体温检测仪的平均测温用时比人工测温快2秒，全自动红外体温检测仪检测60个人的体温的时间比人工检测40个人的体温的时间还少50秒，请计算全自动红外体温检测仪和人工测量测温的平均时间分别是多少秒？【答案】全自动红外体温检测仪和人工测量测温的平均时间分别是1.5秒和3.5秒【解析】【分析】设全自动红外体温检测仪的平均测温用时为x 秒，人工测量的平均测温用时为y 秒，根据“全自的动红外体温检测仪检测60个人的体温的时间比人工检测40个人的体温的时间还少50秒”列出方程组，解方程求组解即可【详解】解：设全自动红外体温检测仪的平均测温用时为x 秒，则人工测量的平均测温用时为y 秒，则6050402x y x y += +=解得 1.53.5x y = =答：全自动红外体温检测仪和人工测量测温的平均时间分别是1.5秒和3.5秒．【点睛】本题考查了二元一次方程组的应用，根据题意列出等量关系是解题的关键．20. 如图，1l 表示星月公司某种电子产品的销售收入与销售量之间的关系，2l 表示该电子产品的生产成本与销售量之间的关系．（1）当销售量为 件时，销售收入等于生产成本．（2）当6x =时，生产成本= 万元．（3）若星月公司要想获得不低于22万元的利润，那么销售量至少为多少件？【答案】（1）3 （2）4（3）36【解析】【分析】（1）根据函数图象可知12,l l 的交点的横坐标即为所求；（2）根据函数图象分别求得2l 对应的解析式2y ，令6x =即可求解；（3）根据销售收入减去生产成本即可求得利润，根据题意列一元一次不等式求解即可．【小问1详解】解：根据函数图象可知12,l l 的交点的横坐标为3，此时，销售收入等于生产成本，故答案为：3【小问2详解】解：设2l 的解析式为2y 22k x b +，将点()()0,2,3,3代入得222233b k b = =+解得22132k b = = 2123y x ∴=+ 令6x =，2224y =+=故答案为：4【小问3详解】解：设直线1l 的解析式为1y kx =，将点()3,3代入得1y x =根据题意，1222y y −≥ 即12223x x −+≥解得36x ≥x 为正整数，36x ∴=答：若星月公司要想获得不低于22万元的利润，那么销售量至少为36件．【点睛】本题考查了一次函数的应用，一元一次不等式的应用，根据函数图象获取信息是解题的关键．21. 定义：我们把一次函数(0)y kx b k =+≠与正比例函数y x =的交点称为一次函数(0)y kx b k =+≠的“不动点”．例如求21y x =−的“不动点”；联立方程21y x y x =− = ，解得11x y = =，则21y x =−的“不动点”为(1,1)． （1）由定义可知，一次函数32y x =+“不动点”为 ． （2）若一次函数y mx n =+的“不动点”为(2,1)n −，求m 、n 的值． 的（3）若直线3(0)y kx k =−≠与x 轴交于点A ，与y 轴交于点B ，且直线3y kx =−上没有“不动点”，若P 点为x 轴上一个动点，使得3ABP ABO S S = ，求满足条件的P 点坐标．【答案】（1）()1,1−−（2）1,32m n =−= （3）(6,0)−或()12,0【解析】 【分析】（1）联立一次函数解析式32y x =+与正比例函数y x =，解二元一次方程组即可； （2）将“不动点”为(2,1)n −，代入y x =求得n ，进而代入y mx n =+求得m 即可； （3）根据题意可得1k =，进而设(,0)P x ，根据三角形面积公式求解即可．【小问1详解】解：由定义可知，一次函数32y x =+的“不动点”为一次函数解析式32y x =+与正比例函数y x =的交点，即32y x y x=+ = 解得11x y =− =−∴一次函数32y x =+的“不动点”为()1,1−−【小问2详解】解：根据定义可得，点(2,1)n −在y x =上，12n ∴−=解得3n =点(2,1)n −又在y mx n =+上， 12n m n ∴−=+，又3n =3123m ∴−=+ 解得12m =−123m n =− ∴ = 【小问3详解】直线3y kx =−上没有“不动点”，∴直线3y kx =−与y x =平行1k ∴=∴3y x =−，令0x =，3y =−令0y =，则3x =()()3,0,0,3A B ∴−3,3OA OB ∴==设(,0)P x3ABP ABO S S =11322AP OB OA OB ∴⋅⋅=×⋅⋅ ∴3AP OA =333x ∴−=×即39x −=或39x −=−解得6x =−或12x =()6,0P ∴−或()12,0【点睛】本题考查了一次函数的性质，一次函数与坐标轴围成的三角形的面积，两直线交点问题，掌握一次函数的性质是解题的关键．22. 【问题背景】学校数学兴趣小组在专题学习中遇到一个几何问题：如图1，已知等边ABC ，D 是ABC 外一点，连接AD 、CD 、BD ，若30ADC ∠=°，3AD =，5BD =，求CD 的长．该小组在研究如图2中OMN OPQ ≅ 中得到启示，于是作出如图3，从而获得了以下的解题思路，请你帮忙完善解题过程．解：如图3所示，以DC 为边作等边CDE △，连接AE ．∵ABC ，DCE 是等边三角形，∴BC AC =，DC EC =，60BCA DCE ∠=∠=°．∴BCA ACD ∠+∠= ACD +∠，∴BCD ACE ∠=∠，∴ ，∴5AE BD ==，∵30ADC ∠=°，60CDE ∠=°，∴90ADE ADC CDE ∠=∠+∠=°．∵3AD =，∴CD DE == ．【尝试应用】如图4，在ABC 中，45ABC ∠=°，AB =，4BC =，以AC 为直角边，A 为直角顶点作等腰直角ACD △，求BD 的长．【拓展创新】如图5，在ABC 中，4AB =，8AC =，以BC 为边向往外作等腰BCD △，BD CD =，120BDC ∠=°，连接AD ，求AD 的最大值．【答案】[问题背景]DCE ∠；BCD ACE ≌；4；[尝试应用][拓展创新]【解析】【分析】[问题背景]根据等式的性质，三角形全等的判定与性质，勾股定理填空即可；[尝试应用]以AB 为直角边，A 为直角顶点作等腰Rt ABF ，连接,,AF BF CF ，进而证明BAD FAC △≌△，根据勾股定理求得FC ，即可求得BD 的长；[拓展创新] 以DA 为腰，作等腰DAG △，DA DG =，120ADG ∠=°，过点D 作DH AG ⊥，同理证明ABD GCD ≌，进而根据含30度角的直角三角形的性质，勾股定理求得,DH AH ，根据三角形三边关系确定AD 最大值时，,,A C G 三点共线，进而即可求得AD 的最大值．【详解】[问题背景] 解：如图3所示，以DC 为边作等边CDE △，连接AE ．∵ABC ，DCE 是等边三角形，∴BC AC =，DC EC =，60BCA DCE ∠=∠=°．∴BCA ACD ∠+∠=DCE ∠ACD +∠，∴BCD ACE ∠=∠，∴BCD ACE ≌，∴5AE BD ==，∵30ADC ∠=°，60CDE ∠=°，∴90ADE ADC CDE ∠=∠+∠=°．∵3AD =，∴CD DE ==4．[尝试应用] 解：如图4所示，以AB 为直角边，A 为直角顶点作等腰Rt ABF ，连接,,AF BF CF ．∵DAC △，FAB 是等腰直角三角形，∴AF AB =，AD AC =，90FAB DAC ∠=∠=°．∴BAF FAD CAD FAD ∠+∠=∠+∠，∴FAC BAD ∠=∠，∴BAD FAC △≌△，∴AF AB ==，2FB ∴=∵45ABC ∠=°，45ABF ∠=°， ∴90FBC ABF ABC ∠=∠+∠=°．∵4BC =，∴BD FC ==[拓展创新]解：如图，以DA 为腰，作等腰DAG △，DA DG =，120ADG ∠=°，过点D 作DH AG ⊥，90,30DHA HAD ∴∠=°∠=°，12AH HG AG == 12HD AD ∴=AH AD ∴=即AD =AG = ∵DBC △，DAG △是等腰三角形，,DC DB DG DA ∴==∴GDA CDA CDB CDA ∠−∠=∠−∠ GDC ADB ∴∠=∠∴ABD GCD ≌∴==CG AB4=AD=AG则当AG取得最大值时，AD取得最大≤+=+=AG CG AC AB AC12A C G三点共线时，AD取得最大值，如图，当,,∴ADAG【点睛】本题考查了等腰三角形的性质与判定，三角形全等的性质与判定，勾股定理，线段最值问题，从题干部分理解作等腰三角形辅助线是解题的关键．。

宝安中学（集团）初中部2019-2020学年度第二学期八年级期中考试数学试卷一、选择题（每题3分，共36分）1.已知a<b ，则下列不等式一定成立的是（ ） A.33+>+b aB.b a 22>C.b a 33-<-D.0<-b a2.下列图形中，中心对称图形个数是（ ）A.1个B.2个C.3个D.4个3.下列从左到右的变形中，因式分解正确的是（ ） A.1)2(21422+-=+-x x x xB.)2(22-=-x x x xC.1)1)(1(2-=-+x x xD.22)2(42+=++x x x4.一个多边形从一个顶点可引对角线3条，这个多边形内角和等于（ ） A.360°B.540°C.720°D.900°5.已知点P (3-m ，1-m )在第二象限，则m 的取值范围在数轴上表示正确的是（ ）A. B. C. D.6.等腰三角形的一个角是80°，则它的顶角的度数是（ ） A.80°B.80°或20°C.80°或50°D.20°7.下列各组线段中，不能构成直角三角形的是（ ） A.321、、B.532、、C.532、、D.321、、8.已知mn n m =-22，则nmm n -的值等于（ ） A.1B.0C.1-D.41-9.下列命题为真命题的是（ ） A.若ab >0，则a >0，b>0B.两个锐角分别相等的两个直角三角形全等C.在一个角的内部，到角的两边距离相等的点在这个角的平分线上D.一组对边平行，另一组对边相等的四边形是平行四边形10.学校计划选购甲，乙两种图书为“初中数学分享学习课堂之生讲生学”初赛的奖品，已知甲图书的单价是乙图书单价的1.5倍，用600元单独购买甲种图书比单独购买乙种图书少10本，设乙种图书的价格为x 元，依据题意列方程正确的是（ ） A.105.1600600=-xx B.106005.1600=-xx C.5.160010600=-+x x D.5.110600600=+-x x 11.如图，在△ABC 中，∠C =90°，以点B 为圆心，任意长为半径画弧，分别AB 、BC 于点M 、N .分别以点M 、N 为圆心，以大于21MN 的长度为半径画弧，两弧相交于点P ，过点P 作线段BD ，交AC 于点D ，过点D 作DE ⊥AB 宇2点E ，则下列结论：①CD =ED ；②∠ABD =21∠ABC ；③BC =BE ；④AE =BE 中，一定正确的是（ ） A.①②③ B.①②④C.①③④D.②③④12.如图，为一幅重叠放置的三角板，其中∠ABC =∠EDF =90°，BC 与DF 共线，将△DEF 沿CB 方向平移，当EF 经过AC 的中点O 时，直线EF 交AB 于点G ，若BC =3，则此时OG 的长度为（ ） A.223B.323C.23 D.23323- 二、填空题（每题3分，共12分） 13.要使分式41-+x x 有意义，则x 的取值应满足 . 14.如图，一次函数3+-=x y 与一次函数m x y +=2图像交于点（2-，n ），则关于x 的不等式32+-<+x m x 的解集为 .15.如图，口ABCD 中，对角线AC 、BD 交于点O ，OE ⊥AC 于点E ，已知△DCE 的周长为14.则口ABCD 的周长为 .16.如图，在直角坐标系中，正方形OABC 的顶点B 的坐标为（3，3），直线CD 交直线OA 于点D ，直线OE 交线段AB 于E ，且CD ⊥OE ，垂直为点F ，若图中阴影部分的面积是正方形OABC 的面积的31，则△OFC 的周长为 .三、解答题（共52分）17.（8分）因式分解：（1）2422+-x x （2）)(16)3y x y x ---（18.（6分）解不等式组⎪⎩⎪⎨⎧+≥++≤-33221)12x x x x （，并求出不等式组的整数解之和.19.（6分）先化简再求值：a a a a a a 44822222-÷⎪⎭⎫- ⎝⎛+-+，其中a 满足方程0142=++a a .20.（6分）如图，在平面直角坐标系中，已知△ABC 的三个顶点的坐标分别为A （4-，3），B （3-，1），C （1-，3） （1）请按下列要求画图：①平移△ABC ，使A 的对应点A 1 的坐标为（4-，3-），请画出平移后的△111C B A ；②△222C B A 与△ABC 关于原点中心对称，画出△222C B A ；（2）若将△111C B A 绕点M 旋转可得到△222C B A ，请直接写出旋转中心M 点的坐标 .21.（8分）端午节前夕，小东妈妈准备购买若干个粽子和咸鸭蛋（每个粽子的价格相同，每个咸鸭蛋的价格相同）.已知某超市粽子的价格比咸鸭蛋的价格贵1.8元，小东妈妈发现，花30元购买粽子的个数与花12元购买的咸鸭蛋个数相同. （1）求该超市粽子与咸鸭蛋的价格各是多少元？（2）小东妈妈计划购买粽子与咸鸭蛋共18个，她的一张购物卡上还有余额40元，若只用这张购物卡，她最多能买粽子多少个？22.（9分）如图，在△ABC 中，AB 、AC 边的垂直平分线相交于点O ，分别交BC 边于点M 、N ，连接AM ，AN .(1)若△AMN 的周长为6，求BC 的长； (2)若∠MON =30°，求∠MAN 的度数；(3)若∠MON =45°，BM =3，BC =12，求MN 的长度.23.（9分）已知：直线643+=x y 与x 轴、y 轴分别相交于点A 和点B ，点C 在线段AO 上.将△ABO 沿BC 折叠后，点O 恰好落在AB 边上点D 处. （1）求出OC 的长？（2）如图，点E 、F 是直线BC 上的两点，若△AEF 是以EF 为斜边的等腰直角三角形，求点F 的坐标；（3）取AB 的中点M ，若点P 在y 轴上，点Q 在直线AB 上，是否存在以C 、M 、P 、Q 为顶点的四边形为平行四边形？若存在，请求出所有满足条件的Q 点坐标；若不存在，请说明理由.参考答案一、选择题：二、填空题：三、解答题17．（1）2)1(2-x （2）)4)(4)((--+--y x y x y x18.30≤≤x ，整数解之和为6 19．化简为2)2(1+a ，代入求值得3120. （1）画图略（2）M （0，3-）21.（1）咸鸭蛋价格为1.2元，粽子价格为3元（2）最多购买粽子10个 22.（1）BC =6（2）∠MAN =120°（3）MN =5 23.（1）CO =3（2）F （6-，6-）或（2-，2） （3）Q （1-，421）或（1，427）或（7-，43）。

单元测试卷（满分：100分）一．积累与运用（28分）1. 下列各组词语中划线字注音无误的一项是（）A .优雅（yǎ）掠过（lüè）翘首而望（qiào）B .瞬间（shùn）酷似（kù）屏息敛声（píng）C .沸腾（fèi）一刹那（chà）悄然不惊（qiāo）D .旋风（xuàn）赢得（yíng）惊讶不已（yà）2. 下列词语中没有错别字的一项是（）A .锐不可挡督战封锁业已B .负隅顽抗拒绝协定功势C .排山倒海策略占据管辖D .横度长江豫北惨败企图3. 下面有关新闻评论的说法错误的一项（）A .新闻评论是现代新闻传播工具经常采用的社论、评论、评论员文章、短评、编者按、专栏评论和评述等的总称。

B .新闻评论是议论文体裁中重要的一类，由论点、论据、论证三要素组成，具有政策性、针对性和准确性。

C .新闻评论代表一定的机构、组织对当前重要问题和事件的态度、观点、看法，可以指导受众的意见走向、行为走向。

D .新闻评论在舆论监督中处于一种显要的地位，在弘扬先进思想和精神的同时，还要不断揭露和抨击各种腐败现象和不正之风，对不良之风和现象形成强大的舆论压力。

4. 下列对划线词语的含义理解有误的一项是（）A .瑞典国王和挪威诺贝尔基金会今天首次颁发了诺贝尔奖。

（颁发：发布（命令、指示、政策等）。

）B .根据诺贝尔的遗嘱，诺贝尔奖由4个机构（瑞典3个，挪威1个）颁发。

（遗嘱：人在生前或临死时对自己身后事如何处理用口头或书面形式所做的嘱咐。

）C .他在血清疗法的研究方面卓有成就。

（卓有成就：有突出的成绩。

）D .他在诗歌创作方面颇有建树。

（建树：建立的功绩。

）5. 下列对病句修改不正确的一项是（）A .要问最近最火的电视剧是哪部，无疑确定就是《欢乐颂2》。

（在句末加上“的剧情”）B .因为售卖富氧空气是新鲜事物，目前还没有统一的检测标准，今后这些都会逐步落实。

宝安中学（集团）初中部八年级（数学）学科寒假作业调研一、选择题（10×3=30分）1. 36的算术平方根是（ ）A. 6± B. 6C.D.2. 若a b <，则下列各式中一定成立的是（ ）A. 11a b -<- B.33a b > C. a b -<- D.ac bc<3. 已知ABC 三边为a 、b 、c ，下列条件不能判定ABC 为直角三角形的是（ ）A. 222b ac =- B. A B C =+∠∠∠C. ::3:4:5A B C ∠∠∠= D. 222::1:2:3a b c =4. 目前新冠变异毒株“奥密克戎“肆虐全球，疫情防控形势严峻．体温超过37.3℃的必须如实报告，并主动到发热门诊就诊．体温T “超过37.3℃”用不等式表示为（ ）A 37.3T >℃B. 37.3T <℃C. 37.3T ≤℃D.37.3T ≥℃5. 一个不等式组解集为12x -<≤，那么在数轴上表示正确的是（ ）A. B. C.D.6. 下列四个命题中，真命题是（ ）A. 两个无理数的和还是无理数B. 体积为8的正方体，边长是无理数C. 两直线被第三条直线所截，内错角相等D.有意义，则3x ≥7. 已知：如图，在ABCD Y 中，4AB =，7AD =，ABC ∠的平分线交AD 于点E ，交CD 的延长线于点F ，则DF 的长为（ ）的.的A. 6B. 5C. 4D. 38. 如图，一次函数y kx b =+的图像经过()0,2A 、()1,0B 两点，则不等式0kx b +<的解集是（ ）A. 0x <B. 01x <<C. 1x <D. 1x >9. 如图，用大小形状完全相同的长方形纸片在直角坐标系中摆成如图图案，已知A (﹣2，6)，则点B 的坐标为( )A. (﹣6，4)B. (203-，143) C. (﹣6，5) D. (203-，4)10. 如图，90ABC ADB ∠=∠=︒，DA DB =，若2BC =，4AB =，则点D 到AC 的距离是（ ）A.B.C.D.二、填空题（本题共5小题，每小题3分，共15分）11._______3． （选填“>”、“<”或“=”）12 若(,3),(2,)A a B b 关于x 轴对称，则ab =_________．13. 如图，直线2y x =与y kx b =+相交于点(),2P m ，则关于x 的方程2x kx b =+的解是______．14. 如图，把两块大小相同的含45︒的三角板ACF 和三角板CFB 如图所示摆放，点D 在边AC 上，点E 在边BC 上，且12,33CFE CFD ∠=︒∠=︒，则DEC ∠的度数为________．15. 如图，在Rt △ABC 中，∠ACB ＝90°，AC ＝BC ＝4，点E 在AC 上，且AE ＝1，连接BE ，∠BEF ＝90°，且BE ＝FE ，连接CF ，则CF 的长为____________.三、解答题（共55分）16. 计算题（1）(32)+；（2）2(3+17. 解下面的不等式（组），并把解集在数轴上表示出来：（1）2152146x x -+-≥-；（2）()3351462233x x x x ⎧+>-⎪⎨--≥⎪⎩．18. 如图，在平面直角坐标系中，ABC 各顶点分别为()3,3A -，()4,2B --，()1,1C --．（1）在图中作A B C ''' ，使A B C ''' 和ABC 关于y 轴对称；（2）直接写出点B 关于x 轴对称的点的坐标______；（3）在x 轴上存在一点Q ，使得QB QC +的值最小，QB QC +的最小值为______；请直接写出点Q 的坐标______．19. 如图，ABC 中，,AC BC CD =平分ACB ∠，交AB 于点D ，延长BC 至点E ，使CE BC =，连接AE ．（1）求证：CD AE ∥；（2）连接DE ，若5,6AC AB ==，求DCE △的面积．20. 某教育科技公司销售A ，B 两种多媒体，这两种多媒体的进价与售价如表所示：A B进价（万元/套）3 2.4售价（万元/套）3.32.8（1）若该教育科技公司计划购进两种多媒体共50套，共需资金132万元，该教育科技公司计划购进A ，B 两种多媒体各多少套？（2）若该教育科技公司计划购进两种多媒体共50套，其中购进A 种多媒体m 套()1020m ≤≤，当把购进的两种多媒体全部售出，求购进A 种多媒体多少套时，能获得最大利润，最大利润是多少万元？21. 如图①所示，在A 、B 两地之间有一车站C ，甲车从A 地出发经C 站驶往B 地，乙车从B 地出发经C 站驶往A 地，两车同时出发，匀速行驶，图②是甲、乙两车行驶时离C 站的路程，()km y 与行驶时间()h x 之间的函数图像．图① 图②（1）填空：a 值为____________，m 的值为____________，AB 两地的距离为____________km ．（2）请直接写出乙车离C 站的路程()km y 与行驶时间()h x 之间的函数关系式．（3）请求出乙车到达A 地前，两车与车站C 的路程之和等于300km 时行驶时间x 的值．22. 如图，在平面直角坐标系中，直线1:2AB y x m =+与x 轴交于点A ，与y 轴交于点()0,2B ，直线AC 经过y 轴负半轴上的点C ，且45ACO ∠=︒．（1）求直线AC 的函数表达式；（2）直线AC 向上平移9个单位，平移后的直线与直线AB 交于点D ，连结DC ，求ACD 面积；（3）在（2）的条件下，平移后的直线与x 轴交于点E ，点M 为x 轴上的一点，直线DE 上是否存在点N （不与点D 重合），使以点E ，M ，N 为顶点的三角形与ADE V 全等，若存在，请直接写出点N的坐标；若不存在，请说明理由．的宝安中学（集团）初中部八年级（数学）学科寒假作业调研一、选择题（10×3=30分）1. 36的算术平方根是（ ）A. 6±B. 6C. D.【答案】B 【解析】【分析】算术平方根的定义：一个非负数的正的平方根，即为这个数的算术平方根，利用定义即可求出结果．【详解】解：∵6的平方为36，∴36算术平方根为6．故选B ．【点睛】此题主要考查了算术平方根的概念，算术平方根易与平方根的概念混淆而导致错误．2. 若a b <，则下列各式中一定成立的是（ ）A. 11a b -<- B. 33a b> C. a b -<- D.ac bc<【答案】A 【解析】【分析】本题主要考查了不等式的基本性质．不等式的基本性质：不等式两边加（或减）同一个数（或式子），不等号的方向不变；不等式两边乘（或除以）同一个正数，不等号的方向不变；不等式两边乘（或除以）同一个负数，不等号的方向改变．再逐一分析本题即可得到答案.【详解】解：A 、不等式的两边都减1，不等号的方向不变，故A 符合题意；B 、不等式的两边都除以3，不等号的方向不变，故B 不符合题意；C 、不等式的两边都乘以1-，不等号的方向改变，故C 不符合题意；D 、当0c >时，ac bc <；当0c <时，ac bc >，故D 不符合题意；故选：A ．3. 已知ABC 的三边为a 、b 、c ，下列条件不能判定ABC 为直角三角形的是（ ）A. 222b ac =- B. A B C=+∠∠∠C. ::3:4:5A B C ∠∠∠=D. 222::1:2:3a b c =【答案】C 【解析】【分析】根据勾股定理的逆定理及三角形内角和定理对各选项进行逐一判断即可．【详解】解：A .∵222b a c =-，∴222c b a +=，∴此三角形是直角三角形，故本选项不符合题意；B .∵180A B C A B C ∠+∠+∠=︒∠=∠+∠，，∴90A ∠=︒，∴此三角形是直角三角形，故本选项不符合题意；C .设3A x ∠=，则∠45B x C x =∠=，，∵180A B C ∠+∠+∠=︒，∴345180x x x ++=︒，解得15x =︒，∴51575C ∠=⨯︒=︒，∴此三角形不是直角三角形，故本选项符合题意；D .∵222::1:2:3a b c =，设2a k =，则222,3b k c k ==，∴222+=a b c ，∴此三角形是直角三角形，故本选项不符合题意；故选：C ．【点睛】本题考查的是勾股定理的逆定理：如果三角形的三边长a b c ，，满足222+=a b c ，那么这个三角形就是直角三角形．也考查了三角形内角和定理，熟知以上知识是解答此题的关键．4. 目前新冠变异毒株“奥密克戎“肆虐全球，疫情防控形势严峻．体温超过37.3℃的必须如实报告，并主动到发热门诊就诊．体温T “超过37.3℃”用不等式表示为（ ）A. 37.3T >℃B. 37.3T <℃C. 37.3T ≤℃D.37.3T ≥℃【答案】A 【解析】【分析】“超过37.3℃”的意思就是“大于37.3℃”由此即可得到答案．【详解】解：由题意得：“超过37.3℃”列不等式为37.3T >℃，故选：A ．【点睛】本题主要考查了列不等式，正确理解题意是解题的关键．5. 一个不等式组的解集为12x -<≤，那么在数轴上表示正确的是（ ）A. B.C. D.【答案】A 【解析】【分析】根据数轴上的点表示的数，右边的总是大于左边的数，这个解集就是不等式1x >-和2x ≤的解集的公共部分．【详解】解：数轴上12x -<≤表示1-与2之间的部分，并且包含2，不包含1-，在数轴上可表示为：故选：A【点睛】此题考查了数轴上表示不等式的解集，解题关键是掌握用数轴上的点表示数．6. 下列四个命题中，真命题是（ ）A. 两个无理数的和还是无理数B. 体积为8的正方体，边长是无理数C. 两直线被第三条直线所截，内错角相等D. 有意义，则3x ≥【答案】D 【解析】【分析】本题考查了命题与定理的知识，利用二次根式有意义的条件、无理数的定义、无理数的应用，平行四边形的性质分别判断后即可确定正确的选项．【详解】解：A. 两个无理数的和不一定是无理数，原说法是假命题；B. 体积为8的正方体，边长是2，是有理数，原说法是假命题；C. 两平行线被第三条直线所截，内错角相等，原说法是假命题；D. 有意义，则3x ≥，是真命题；故选D ．7. 已知：如图，在ABCD Y 中，4AB =，7AD =，ABC ∠的平分线交AD 于点E ，交CD 的延长线于点F ，则DF 的长为（ ）A. 6B. 5C. 4D. 3【答案】D 【解析】【分析】利用DE 平分ABC ∠，可得ABE CBE ∠∠=，利用平行四边形性质得AB CF ，7BC AD ==，4CD AB ==，进而利用等腰三角形的判定即可求得DF ．【详解】解：∵DE 平分ABC ∠，∴ABE CBE ∠∠=，∵四边形ABCD 是平行四边形，7AD =，4AB =，∵ABCF ，7BC AD ==，4CD AB ==，∴F ABE ∠∠=，∴F CBE ∠∠=，∴7BC CF ==，∴3DF =．故选：D ．【点睛】本题主要考查的是平行四边形的性质的应用，以及等腰三角形的判定，熟练掌握平行四边形的性质是解题的关键．8. 如图，一次函数y kx b =+的图像经过()0,2A 、()1,0B 两点，则不等式0kx b +<的解集是（）的A. 0x < B. 01x << C. 1x < D. 1x >【答案】D【解析】【分析】不等式0kx b +<的解集为直线y kx b =+落在x 轴下方的部分对应的x 的取值范围．【详解】解：由题意知一次函数y kx b =+的图像经过点(1,0)B ，并且函数值y 随x 的增大而减小，因而不等式0kx b +<的解集是：1x >．故选：D ．【点睛】本题考查了一次函数与一元一次不等式的关系，解题的关键是从函数的角度看，就是寻求使一次函数y ax b =+的值大于（或小于）0的自变量x 的取值范围；从函数图像的角度看，就是确定直线y kx b =+在x 轴上（或下）方部分所有的点的横坐标所构成的解集．9. 如图，用大小形状完全相同的长方形纸片在直角坐标系中摆成如图图案，已知A (﹣2，6)，则点B 的坐标为( )A. (﹣6，4)B. (203-，143)C. (﹣6，5)D. (203-，4)【答案】B【解析】【分析】设长方形纸片的长为x ，宽为y ，根据题意列方程组解答即可.【详解】设长方形纸片的长为x ，宽为y ，根据题意得：226x y x y -=⎧⎨+=⎩， 解得：10343x y ⎧=⎪⎪⎨⎪=⎪⎩，∴﹣2x =﹣203，x +y =143，∴点B 坐标为(﹣203，143).故选：B.【点睛】此题考查几何图形类二元一次方程组的实际应用，正确理解图形中线段的大小关系列得方程组是解题的关键.10. 如图，90ABC ADB ∠=∠=︒，DA DB =，若2BC =，4AB =，则点D 到AC 的距离是（ ）A.B.C.D. 【答案】B【解析】【分析】过点D 作DF AC ⊥，垂足为F ，过点D 作DG CB ⊥，交CB 的延长线于点G ，在Rt ABC △中，利用勾股定理可求出AC 的长，再利用等腰直角三角形的性质可得45DBA DAB ∠=∠=︒，AD BD ==，然后在Rt DBG △中，利用锐角三角函数的定义求出DG 的长，最后根据ADC △的面积ABC = 的面积ADB + 的面积DBC - 的面积进行计算即可解答．【详解】解：过点D 作DFAC ⊥，垂足为F ，过点D 作DG CB ⊥，交CB 的延长线于点G ，的90ABC ∠=︒ ，2BC =，4AB =，AC ∴===，90ADB ∠=︒ ，DA DB =，45DBA DAB ∴∠=∠=︒，AD BD ====，90ABC ∠=︒ ，18090ABG ABC ∠=︒-∠=︒∴，9045DBG DBA ∴∠=︒-∠=︒，在Rt DBG △中，DB =，sin 452DG DB ∴=⋅︒==，ADC ∴ 的面积ABC = 的面积ADB + 的面积DBC - 的面积，11112222AC DF AB BC AD DB BC DG ∴⋅=⋅+⋅-⋅111142222222DF ∴⨯=⨯⨯+⨯-⨯⨯，442=+-，DF ∴=，∴点D 到AC ，故选：B ．【点睛】本题考查了等腰直角三角形，点到直线的距离，利用了勾股定理，锐角三角函数，根据题目的已知条件结合图形添加适当的辅助线是解题的关键．二、填空题（本题共5小题，每小题3分，共15分）11. _______3． （选填“>”、“<”或“=”）【答案】<【解析】【分析】本题考查了对有理数的大小比较法则的应用，注意：3=>．根据实数的大<，即可求出答案．【详解】解： 79<<，3<，故答案为：<．12. 若(,3),(2,)A a B b 关于x 轴对称，则ab =_________．【答案】6-【解析】【分析】本题考查关于x 轴对称的点的坐标特点，掌握关于x 轴对称的点的坐标规律:横坐标相同，纵坐标互为相反数是解题的关键．【详解】解：∵(,3),(2,)A a B b 关于x 轴对称，∴23a b ==-，，∴2(3)6ab ⨯=-=-，故答案为：6-．13. 如图，直线2y x =与y kx b =+相交于点(),2P m ，则关于x 的方程2x kx b =+的解是______．【答案】1x =【解析】【分析】求出m 的值，利用图象法解方程即可．【详解】解：∵直线2y x =与y kx b =+相交于点(),2P m ，∴22m =，∴1m =，∴()1,2P ，∴由图象可知：方程2x kx b =+的解是1x =．故答案为：1x =．【点睛】本题考查图象法解方程．熟练掌握两条直线交点的横坐标为联立两个函数形成的一元一次方程的解，是解题的关键．14. 如图，把两块大小相同的含45︒的三角板ACF 和三角板CFB 如图所示摆放，点D 在边AC 上，点E 在边BC 上，且12,33CFE CFD ∠=︒∠=︒，则DEC ∠的度数为________．【答案】66︒##66度【解析】【分析】本题考查三角形内角和定理、等腰直角三角形的性质、全等三角形的判定和性质等知识，解题的关键是学会添加常用辅助线，构造全等三角形解决问题．作FH FE ⊥交AC 于H ，证明58DEF DHF FEB ∠=∠=︒=∠即可解决问题．【详解】作FH FE ⊥交AC 于H ，∵90AFC EFH ∠=∠=︒，∴12AFH CFE ∠=∠=︒，∵45,A FCE FA FC ∠=∠=︒=，∴FAH FCE ≌，∴FH FE =，∵123345DFE CFE DFC ∠=∠+∠=︒+︒=︒，∴45DFH DFE ∠=∠=︒,∵DF DF =，∴DFE DFH ≌，∴57DEF DHF A AFH ∠=∠=∠+∠=︒∵57FEB CFE FCE ∠=∠+∠=︒，∴180575766DEC ∠=︒-︒-︒=︒，故答案为66︒．15. 如图，在Rt △ABC 中，∠ACB ＝90°，AC ＝BC ＝4，点E 在AC 上，且AE ＝1，连接BE ，∠BEF ＝90°，且BE ＝FE ，连接CF ，则CF 的长为____________.【解析】【分析】过点F 作FM ⊥AC 交AC 延长线于M ，根据∠BEF =90°且BE =EF ，可以得到△EFM ≌△BEC ，从而可以计算出CM 、FM 的长，再利用勾股定理即可得到CF 的长.详解】解：∵∠ACB ＝90°，AC ＝BC ＝4，，AE ＝1∴CE ＝3∵FM ⊥AC ，∠BEF ＝90°【∴∠ACB＝∠BEF =∠FME =90°∴∠FEM+∠EFM=90°=∠BEC+∠FEM∴∠EFM=∠BEC又∵BE=FE∴△EFM≌△BEC∴BC＝EM＝4，CE＝FM＝3∴CM=EM-EC=1∴CF==.【点睛】此题主要考查了全等三角形的性质与判定，勾股定理的运用，解题的关键在于能够熟练掌握相关知识进行求解.三、解答题（共55分）16. 计算题+；（1）(32)(3+（2）2-【答案】16. 117. 7+【解析】【分析】本题考查二次根式的混合运算，在运算过程中，有同类二次根式的要合并；相乘的时候，被开方数简单的直接让被开方数相乘，再化简；较大的也可先化简，再相乘，灵活对待，结果化为最简二次根式．也考查了二次根式的性质，完全平方公式和平方差公式．掌握二次根式的运算法则、性质和乘法公式是解题的关键．（1）运用二次根式的乘法运算，然后合并即可；（2）先用完全平方公式和和二次根式的约分，然后再进行加减运算．【小问1详解】解：(32)+65=-+-1=-【小问2详解】解：2(3+935=++-7=+17. 解下面的不等式（组），并把解集在数轴上表示出来：（1）2152146x x -+-≥-；（2）()3351462233x x x x ⎧+>-⎪⎨--≥⎪⎩．【答案】（1）54x ≤；数轴见解析 （2）24x ≤<；数轴见解析【解析】【分析】本题主要考查了解不等式组，把不等式（组）的解集表示在数轴上时，需注意：“圆点”和“圆圈”的使用区别，当解集中的不等号是“≥”及“≤”符号时，用“圆点”；当解集中的不等号是“>”及“<”符号时，用“圆圈”．（1）按解一元一次不等式的一般步骤解答，并把解集规范的表示在数轴上即可；（2）按解一元一次不等式组的一般步骤解答，并把解集规范的表示在数轴上即可．【小问1详解】解：2152146x x -+-≥-，去分母得：()()32125212x x --+≥-，去括号得：6310412x x ---≥-，移项、合并同类项得：45x -≥-，系数化为1得：54x ≤，解集表示在数轴上为：【小问2详解】解：()3351462233x x x x ⎧+>-⎪⎨--≥⎪⎩，解不等式()3351x x +>-得：4x <，解不等式462233x x --≥得：2x ≥，把两个不等式的解集表示在数轴上为：∴不等式组的解集为：24x ≤<．18. 如图，在平面直角坐标系中，ABC 各顶点分别为()3,3A -，()4,2B --，()1,1C --．（1）在图中作A B C ''' ，使A B C ''' 和ABC 关于y 轴对称；（2）直接写出点B 关于x 轴对称的点的坐标______；（3）在x 轴上存在一点Q ，使得QB QC +的值最小，QB QC +的最小值为______；请直接写出点Q 的坐标______．【答案】（1）见解析（2）()4,2-（3）()2,0-【解析】【分析】本题主要考查坐标与图形，熟练掌握关于坐标轴对称图形的画法，关于坐标轴对称的两点坐标关系，两点之间线段最短和勾股定理是解题的关键．（1）分别找出、、A B C 关于y 轴时称的点，画出图形即可；（2）根据x 轴对称的点的坐标写出答案即可；（3）作点C 关于x 轴对称的点的坐标C ''，连接BC ''，QB QC +的值最小值为BC ''的长；设BC ''的解析式为：y kx b =+，求出解析式，令0y =，即可得到答案．【小问1详解】解：分别找出、、A B C 关于y 轴对称的点，画出图形即可； 小问2详解】解：x 轴对称的点的坐标横坐标不变，纵坐标变成相反数，故点B 关于x 轴对称的点的坐标为()4,2-；【小问3详解】解：作点C 关于x 轴对称的点的坐标(1,1)C ''-，连接BC ''，故QB QC +的值最小值为BC ''的长，BC ''==QB QC +的最小值为 ()4,2B --，(1,1)C ''-，【设BC ''的解析式为：y kx b =+，故241k b k b-=-+⎧⎨=-+⎩，解得12k b =⎧⎨=⎩，∴BC ''的解析式为：2y x =+，令0y =，故2x =-，∴点Q 的坐标为()2,0-．19. 如图，ABC 中，,AC BC CD =平分ACB ∠，交AB 于点D ，延长BC 至点E ，使CE BC =，连接AE ．（1）求证：CD AE ∥；（2）连接DE ，若5,6AC AB ==，求DCE △的面积．【答案】（1）见详解（2）6【解析】【分析】本题考查等腰三角形的判定和性质，勾股定理，平行线的判定，掌握等腰三角形的判定和性质是解题的关键．（1）根据等边对等角得到CAE CEA ∠=∠，由外角的性质得到2ACB CEA ∠=∠，再根据角平分线的定义得到2ACB BCD ∠=∠，既可以证得CEA BCD ∠=∠，进而得到结论；（2）根据三线合一得到CD AB ⊥，12DB AB =，然后根据勾股定理得到4CD =，然后利用BDC EDC S S = 解题即可．【小问1详解】证明：∵AC BC CE BC ==，，∴AC CE =，∴CAE CEA ∠=∠，∵2ACB CAE CEA CEA ∠=∠+∠=∠，∵CD 平分ACB ∠，∴2ACB BCD ∠=∠，∴CEA BCD ∠=∠，∴CD AE ∥；【小问2详解】解：∵AC BC CD =，平分ACB ∠，∴CD AB ⊥，132DB AB ==，5AC BC ==，∴4CD ===，又∵CE BC =，∴1134622BDC EDC S S BD CD ==⨯=⨯⨯= ．20. 某教育科技公司销售A ，B 两种多媒体，这两种多媒体的进价与售价如表所示：A B进价（万元/套）3 2.4售价（万元/套） 3.3 2.8（1）若该教育科技公司计划购进两种多媒体共50套，共需资金132万元，该教育科技公司计划购进A ，B 两种多媒体各多少套？（2）若该教育科技公司计划购进两种多媒体共50套，其中购进A 种多媒体m 套()1020m ≤≤，当把购进的两种多媒体全部售出，求购进A 种多媒体多少套时，能获得最大利润，最大利润是多少万元？【答案】（1）购进A 种多媒体20套，B 种多媒体30套（2）购进A 种多媒体10套时，能获得最大利润，最大利润是19万元【解析】【分析】本题考查二元一次方程组的应用、 一次函数的应用，解答本题的关键是明确题意，列出相应的方程组，写出相应的函数解析式，利用一次函数的性质求最值．（1）根据题意和表格中的数据，可以列出相应的二元一次方程组，然后求解即可；（2）根据题意可以写出利润与m 的函数关系式，然后根据m 的取值范围和一次函数的性质，可以求得利润的最大值．【小问1详解】设A 种多媒体a 套，B 种多媒体b 套，由题意可得：503 2.4132a b a b +=⎧⎨+=⎩，解得 2030a b =⎧⎨=⎩，答：购进A 种多媒体20套，B 种多媒体30套；【小问2详解】设利润为w 元,由题意可得：()()()3.33 2.8 2.4500.120w m m m =-+-⨯-=-+，∴w 随m 的增大而减小，1020m ≤≤ ，∴当 10m =时，w 取得最大值，此时 19w =，答：购进A 种多媒体10套时，能获得最大利润，最大利润是19万元．21. 如图①所示，在A 、B 两地之间有一车站C ，甲车从A 地出发经C 站驶往B 地，乙车从B 地出发经C 站驶往A 地，两车同时出发，匀速行驶，图②是甲、乙两车行驶时离C 站的路程，()km y 与行驶时间()h x 之间的函数图像．图① 图②（1）填空：a 的值为____________，m 的值为____________，AB 两地的距离为____________km ．（2）请直接写出乙车离C 站的路程()km y 与行驶时间()h x 之间的函数关系式．（3）请求出乙车到达A 地前，两车与车站C 的路程之和等于300km 时行驶时间x 的值．【答案】21. 120,1.5,48022. 80120y x =-23 97x =或3x =【解析】【分析】本题考查了一次函数的应用，理解图象，求出甲，乙速度是本题的关键．（1）先求出甲的速度，利用路程=速度×时间，可求a 的值, m 的值, AB 的距离；（2）利用待定系数法可求解析式；（3）分两种情况讨论，由题意列出不等式，即可求解．【小问1详解】∵甲的速度()36060km /h 6==，∴BC 的距离 ()602120km a =⨯=，()360120480km AB ∴=+=，∴乙车速度 ()48080km /h 6==， ()120 1.5h 80m ∴==，故答案为:120,1.5,480；.【小问2详解】设1.5小时后, 乙车离C 站的路程()km y 与行驶时间()h x 之间的函数关系式 y kx b =+，则36060 1.5k b k b =+⎧⎨=+⎩，解得： 80120k b =⎧⎨=-⎩，∴函数关系式为80120y x =-；【小问3详解】当0 1.5x ≤≤时，3606012080300x x -+-=， 97x ∴=，当 1.56x <≤时，3606080120300x x -+-=，.3x = ，综上所述：当97x =或.3x =时两车与车站C 的路程之和等于300km .22. 如图，在平面直角坐标系中，直线1:2AB y x m =+与x 轴交于点A ，与y 轴交于点()0,2B ，直线AC 经过y 轴负半轴上的点C ，且45ACO ∠=︒．（1）求直线AC 的函数表达式；（2）直线AC 向上平移9个单位，平移后的直线与直线AB 交于点D ，连结DC ，求ACD 面积；（3）在（2）的条件下，平移后的直线与x 轴交于点E ，点M 为x 轴上的一点，直线DE 上是否存在点N （不与点D 重合），使以点E ，M ，N 为顶点的三角形与ADE V 全等，若存在，请直接写出点N 的坐标；若不存在，请说明理由．【答案】（1）4y x =--（2）18（3）()8,3-或5⎛+ ⎝⎭或5⎛- ⎝【解析】【分析】（1）由点B 的坐标可求得m 的值，然后根据直线AB 的解析式可以求得A 的坐标，再结合OA OC =得到C 的坐标，然后用待定系数法求出直线AC 的解析式；（2）根据直线的平移规律得到直线DE 的解析式，从而求得D 的坐标，然后根据ACD ABC BCD S S S =+△△△即可求解；（3）先根据直线DE 的解析式求出点E ，根据勾股定理以及平行四边形的性质，分三种情况可得到点N 的坐标．【小问1详解】解：将点()0,2B 代入直线1:2AB y x m =+，得到2m =，∴直线1:22AB y x =+，令0y =，得到4x =-，∴()4,0A -，4OA =，∵45ACO ∠=︒，∴OA OC =，∴()0,4C -，设直线AC 的表达式为y kx b =+，则404k b b -+=⎧⎨=-⎩，解得14k b =-⎧⎨=-⎩，∴直线AC 的表达式为4y x =--，【小问2详解】解：∵直线AC 向上平移9个单位，∴直线DE 的解析式为495y x x =--+=-+，∵平移后的直线与直线AB 交于点D ，∴1225y x y x ⎧=+⎪⎨⎪=-+⎩，解得23x y =⎧⎨=⎩，∴()2,3D ，∵()0,2B ，()0,4C -，∴6BC =，∴ACD ABC BCDS S S =+△△△=11222BC OA BC ⋅+⨯=1164621822⨯⨯+⨯⨯=；【小问3详解】解：∵直线DE ：5y x =-+与x 轴交于点E ，∴点()5,0E ，∴9AE =，当9EN AE ==时，过点N 作x 轴的垂线交x 轴于一点F ，如图所示：设(),5N a a -+，则5NF a =-+，5FE a =-，在NEF 中，222NF EF NE +=，即()()222559a a -++-=，解得5a =±，∴5N ⎛+ ⎝⎭或5N ⎛ ⎝⎭；当DE EN =时，如图所示：∵ADE MNE △≌△，∴,ADE MNE DE NE ∠=∠=，∴点E 为DN 的中点，∵()2,3D ，()5,0E ，∴()8,3N -，综上，存在，此时点N 的坐标为()8,3-或5⎛⎝⎭或5⎛ ⎝⎭．【点睛】本题考查了一次函数的图像、待定系数法求一次函数解析式、一次函数图像的平移、平面直角坐标系中求图形的面积、求两直线交点坐标，分类讨论，数形结合是解答本题的关键．。

红岭教育公司第二学期初二年级第一阶段测试物理试卷( 说明 : 本试卷考试时间为80分钟,满分为100分)一、选择题（本大题共27 小题，每题 2 分，共54 分）1、对于力的观点，以下说法中错误的选项是（）A．力是物体对物体的作用B．物体受力的同时也必定在施力C．力是改变物体运动状态的原由D．只有接触的物体才能产生力的作用2、小红和小明站在冰面上静止．小明在后边推了小红一下，使小红向前滑去，同时，小明后退，这个现象不可以说明（）A．力能够改变物体的运动状态B．力的作用成效与力的方向相关C．力的作用成效与力的作用点相关D．物体间力的作用是互相的3、用手经过一根绳索提起一重物时，会感得手遇到一个向下的力的作用，这个力的施力物体是 ( )A．重物B．地球C．绳索D．绳索和重物4、以下各力中不属于弹力的是（）A．悬挂重物的细绳，对重物的拉力B．水平桌面对放在其上的书籍的支持力C．按图钉时，手对图钉的压力D．地球对人造卫星的吸引力5、以下对于弹力的说法中，正确的选项是()A．互相接触的物体之间必定存在弹力作用B．只有受弹簧作用的物体才遇到弹力作用C．只有互相接触并发生弹性形变的物体间才存在弹力作用D．弹簧的弹力老是跟弹簧的长度成正比6、小明和小忠想把一条弹性绳拉开，使弹性绳两端的拉环能分别套在相隔一段距离的两根柱子上，用来晒衣服。

现有两种方法：一种是按图甲所示的方法做；另一种是按图乙所示的方法做。

对于这两种做法，以下说法正确的选项是()A．图甲中每一个人所用的力比图乙中每一个人所使劲要小B．图乙中每一个人所用的力比图甲中每一个人所使劲要小C．图甲中每一个人所用的力与图乙中每一个人所使劲同样D．条件不足，没法比较图甲中每一个人所用的力与图乙中每一个人所用力的大小7、在实验时，小明将一个正常的铁质外壳测力计的挂钩挂在铁架台上，静止时有以下图的示数。

接着，他把这个测力计像如图中乙那样，上下各挂一个 50 g的钩码，并挂到甲测力计下，则甲、乙两测力计的示数分别是()A．1.0 N 和 1.5 N B． 1.0 N 和 0.5 N C.2.0 N和 1.0 N D.2.0 N 和 1.5 N8、用天平易弹簧测力计分别在地球和月球上测同一物体，丈量的结果是（）A．天平弹簧测力计测的都同样B．天平测的同样，弹簧测力计测的不一样C．天平测的不一样，弹簧测力计测的同样D．天平、弹簧测力计测的都不一样9、对于重力，以下说法中正确的选项是（）A． G=mg表示物体遇到的重力跟它的质量成正比B． m=表示物体的质量跟它遇到的重力成正比C．由于g ＝ 9.8 N/kg ，因此 1 kg就等于9.8 ND．物体遇到的重力是由物体质量产生的10、一辆汽车停在水平路面上，汽车与路面之间的互相作使劲为（）A．汽车的重力和路面对汽车的支持力B．路面对汽车的支持力和汽车对路面的压力C．汽车的重力和汽车对路面的压力D．以上说法均不对11、以下图，在竖直平面内用轻质细线悬挂一个小球，将小球拉至 A 点，使细线处于拉直状态，由静止开始开释小球，不计摩擦，小球可在 A、B 两点间往返摇动．当小球摆到 B 点时，细线恰巧断开，则小球将（）A ．在B点保持静止B．沿BE方向运动 C．沿 BC方向运动D．沿 BD方向运动状况是（）A.向左匀速运动 B.向右匀速运动C. 向右加快运动或向左减速运动D. 向右减速运动或向左加快运动13、我国于2013 年 6 月 11 日 17 ︰ 38 在酒泉发射中心成功发射“神州十号”飞船，并创始了中国载人应用型飞翔的先河。

深圳市2019-2020学年第二学期八年级下第1-6单元综合测试Class __________________ Name____________________NO._______________笔试部分（75分）I.选择填空（15分）。

ⅰ. 从下面每小题的A、B、C三个选项中选出可以替换划线部分的最佳选项。

（共7小题，每小题1分）( )1. — I require someone to help me. Could you give me a hand?— Of course I can. What’s wrong?A.adviseB. allowC. need( )2. — The speech is starting. You need to get into the hall with your student card separately.—We see, sir. We will stand in a line.A.at the same timeB. in publicC. one after another( )3. — Jim, what do you think of the questions in the last exam?—They were simple so I got full marks.A. strangeB. wonderfulC. easy( )4. — Everything appears perfect. I think you did a good job.— Thank you. I hope the party will be a success.A.turnsB.looksC. gets( )5. — I’m worried about my future.—Take it easy. Although you don’t know what will take place in the future, you can try your best to make preparations (准备) for it.A.improveB.happenC. lose( )6. It is cruel of people to kill wild animals. They should protect them.A. foolishB. seriousC. unkind( )7. My grandma is nearly ninety years old, but he is healthy.A. overB. hardlyC. almostii. 根据句子意思，从下面每小题的A、B、C三个选项中选出恰当的词语完成句子。

2019-2020学年广东省深圳外国语学校初中部八年级（下）期中数学试卷一、选择题（每题3分，共36分）1．（3分）在下列四个汽车标志图案中，是中心对称图形的是（）A．B．C．D．2．（3分）下列各式从左到右的变形中，属于因式分解的是（）A．a（x+y）＝ax+ay B．x2﹣2x+1＝x（x﹣2）+1C．6ab＝2a•3b D．x2﹣8x+16＝（x﹣4）23．（3分）若分式在实数范围内有意义，则x的取值范围为（）A．x＞3B．x≠0且x≠3C．x≥0D．x≠34．（3分）下列运算正确的是（）A．（a+b）2＝a2+b2B．5a﹣a＝5C．+＝1D．（﹣2a2b）3＝﹣6a6b35．（3分）在平面直角坐标系中，线段A′B′是由线段AB经过平移得到的，已知点A（﹣2，1）的对应点为A′（3，4），点B的对应点为B′（4，0），则点B的坐标为（）A．（9，3）B．（﹣1，﹣3）C．（3，﹣3）D．（﹣3，﹣1）6．（3分）“五一”江北水城文化旅游节期间，几名同学包租一辆面包车前去旅游，面包车的租价为180元，出发时又增加了两名同学，结果每个同学比原来少摊了3元钱车费，设实际参加游览的同学共x人，则所列方程为（）A．B．C．D．7．（3分）一个多边形的每个内角都相等，并且它的一个外角与一个内角的比为1：3，则这个多边形为（）A．三角形B．四边形C．六边形D．八边形8．（3分）如图，把△ABC绕点C顺时针旋转某个角度θ得到△A′B′C，∠A＝30°，∠1＝70°，则旋转角θ可能等于（）A．40°B．50°C．70°D．100°9．（3分）如图，在△ABC中，D是AC边的中点，且BD⊥AC，ED∥BC，ED交AB于点E，若AC＝4，BC＝6，则△ADE的周长为（）A．6B．8C．10D．1210．（3分）如图，平行四边形ABCD的对角线AC与BD相交于点O，AE⊥BC，垂足为E，AB＝2，AC＝4，BD＝8，则点D到BC的距离为（）A．B．3C．D．11．（3分）如图，在三角形ABC中，∠ABC＝90°，将三角形ABC沿AB方向平移AD的长度得到三角形DEF，已EF＝8，BE＝3，CG＝3，则图中阴影部分的面积是（）A．12.5B．19.5C．32D．45.512．（3分）如图，分别以Rt△ABC的直角边AC，斜边AB为边向外作等边三角形△ACD和△ABE，F为AB的中点，连接DF，EF，∠ACB＝90°，∠ABC＝30°．则以下4个结论：①AC⊥DF；②四边形BCDF为平行四边形；③DA+DF＝BE；④其中，正确的是（）A．只有①②B．只有①②③C．只有③④D．①②③④二、填空题（每题3分，共12分）13．（3分）因式分解：a2b﹣25b＝．14．（3分）一个平行四边形的一边长是3，两条对角线的长分别是4和，则此平行四边形的面积为．15．（3分）如图，Rt△ABC中，∠C＝90°，BD＝2CD，AD是∠BAC的角平分线，∠CAD＝度．16．（3分）如图，Rt△ABC中，∠ACB＝90°，BC＝AC＝3，点D是BC边上一点，∠DAC＝30°，点E是AD边上一点，CE绕点C逆时针旋转90°得到CF，连接DF，则AD+DF的最小值是．三、解答题（共52分）17．解方程：．18．先化简，再求值：﹣÷，其中m＝2020．19．若x为整数，且的值也为整数，则所有符合条件的x的值之和．20．如图，在平行四边形ABCD中，点O是AB的中点，且OC＝OD．（1）求证：平行四边形ABCD是矩形；（2）若AD＝3，∠COD＝60°，求矩形ABCD的面积．21．某汽车销售公司经销某品牌A款汽车，随着汽车的普及，其价格也在不断下降，今年5月份A款汽车的售价比去年同期每辆降价1万元，如果卖出相同数量的A款汽车，去年销售额为90万元，今年销售额只有80万元．（1）求今年5月份A款汽车每辆售价多少万元；（2）为了增加收入，汽车销售公司决定再经销同品牌的B款汽车，已知B款汽车每辆进价为7.5万元，每辆售价为10.5万元，A款汽车每辆进价为6万元，若卖出这两款汽车共15辆后，获利不低于39万元，求B款汽车至少卖出多少辆？22．如图所示，平行四边形ABCD和平行四边形CDEF有公共边CD，边AB和EF在同一条直线上，AC⊥CD且AC＝AF，过点A作AH⊥BC交CF于点G，交BC于点H，连接EG．（1）若AE＝2，CD＝5，则△BCF的面积为；△BCF的周长为；（2）求证：BC＝AG+EG．23．已知：在△ABC中，∠BAC＝90°，AB＝AC，点D为直线BC上一动点（点D不与B、C重合）．以AD为边作正方形ADEF，连接CF．（1）如图1，当点D在线段BC上时，请直接写出线段BD与CF的数量关系：；（2）如图2，当点D在线段BC的延长线上时，其它条件不变，若AC＝2，CD＝1，则CF＝；（3）如图3，当点D在线段BC的反向延长线上时，且点A、F分别在直线BC的两侧，其它条件不变：①请直接写出CF、BC、CD三条线段之间的关系：；②若连接正方形对角线AE、DF，交点为O，连接OC，探究△AOC的形状，并说明理由．2019-2020学年广东省深圳外国语学校初中部八年级（下）期中数学试卷参考答案与试题解析一、选择题（每题3分，共36分）1．【解答】解：A、是轴对称图形，但不是中心对称图形，故本选项错误；B、是中心对称图形，故本选项正确；C、不是中心对称图形，故本选项错误；D、不是中心对称图形，故本选项错误．故选：B．2．【解答】解：A、等式右边不是整式积的形式，故不是因式分解，故本选项错误；B、等式右边不是整式积的形式，故不是因式分解，故本选项错误；C、等式右边是分式积的形式，故不是因式分解，故本选项错误；D、等式右边是整式积的形式，故是因式分解，故本选项正确．故选：D．3．【解答】解：∵分式在实数范围内有意义，∴x﹣3≠0，∴x≠3故选：D．4．【解答】解：（A）原式＝a2+2ab+b2，故A错误．（B）原式＝4a，故B错误．（D）原式＝﹣8a6b3，故D错误．故选：C．5．【解答】解：横坐标从﹣2到3，说明是向右移动了3﹣（﹣2）＝5，纵坐标从1到4，说明是向上移动了4﹣1＝3，求原来点的坐标，则为让新坐标的横坐标都减5，纵坐标都减3．则点B的坐标为（﹣1，﹣3）．6．【解答】解：设实际参加游览的同学共x人，根据题意得：﹣＝3．故选：D．7．【解答】解：设这个多边形的边数为n，依题意得（n﹣2）×180°＝3×360°，解得n＝8，∴这个多边形为八边形，故选：D．8．【解答】解：∵△ABC绕点C顺时针旋转某个角度θ得到△A′B′C，∴∠A＝∠A′＝30°，又∵∠1＝∠A′+∠ACA′＝70°，∴∠θ＝∠ACA′＝40°，故选：A．9．【解答】解：∵D是AC的中点，且BD⊥AC，∴AB＝BC＝6，AD＝AC＝2，∵ED∥BC，∴AE＝BE＝AB＝3，ED＝BC＝3，∴△AED的周长＝AE+ED+AD＝8．故选：B．10．【解答】解：∵AC＝4，BD＝8，四边形ABCD是平行四边形，∴AO＝AC＝2，BO＝BD＝4，∵AB＝2，∴AB2+AO2＝BO2，∴∠BAC＝90°，∵在Rt△BAC中，BC＝，S△BAC＝×AB×AC＝×BC×AE，∴2×4＝2AE，∴AE＝，即点D到BC的距离为，故选：D．11．【解答】解：△ABC沿AB的方向平移AD的长度得到△DEF，∴△DEF≌△ABC，∴EF＝BC＝8，S△DEF＝S△ABC，∴S△ABC﹣S△DBG＝S△DEF﹣S△DBG，∴S四边形ACGD＝S梯形BEFG，∵CG＝3，∴BG＝BC﹣CG＝8﹣3＝5，∴图中阴影部分的面积＝S梯形BEFG＝×（5+8）×3＝19.5，故选：B．12．【解答】解：∵∠ACB＝90°，∠ABC＝30°，∴∠BAC＝60°，AC＝AB，∵△ACD是等边三角形，∴∠ACD＝60°，∴∠ACD＝∠BAC，∴CD∥AB，∵F为AB的中点，∴BF＝AB，∴BF∥CD，CD＝BF，∴四边形BCDF为平行四边形，②正确；∵四边形BCDF为平行四边形，∴DF∥BC，又∠ACB＝90°，∴AC⊥DF，①正确；∵DA＝CA，DF＝BC，AB＝BE，BC+AC＞AB∴DA+DF＞BE，③错误；设AC＝x，则AB＝2x，S△ACD＝x2，S△ACB＝x2，S△ABE＝x2，＝＝，④错误，故选：A．二、填空题（每题3分，共12分）13．【解答】解：a2b﹣25b＝b（a2﹣25）＝b（a﹣5）（a+5），故答案为：b（a﹣5）（a+5）．14．【解答】解：∵平行四边形两条对角线互相平分，∴它们的一半分别为2和，∵22+（）2＝32，∴两条对角线互相垂直，∴这个四边形是菱形，∴S＝4×2＝4．故答案为：4．15．【解答】解：过点D作DE⊥AB于E点，∵AD是∠BAC的角平分线，DC⊥AC，DE⊥AB，∴DC＝DE．∵BD＝2CD，∴BD＝2DE．∴∠B＝30°．∵∠C＝90°，∴∠CAB＝60°．∴∠CAD＝×60°＝30°．故答案为30．16．【解答】解：由旋转可得，FC＝EC，∠ECF＝90°，又∵∠ACB＝90°，BC＝AC＝3，∴∠CAE＝∠CBF，∴△ACE≌△BCF（SAS），∴∠CBF＝∠CAE＝30°，∴点F在射线BF上，∵Rt△ACD中，∠CAD＝30°，AC＝3＝BC，∴CD＝，AD＝2CD＝2∴BD＝3﹣，∴AD+DF＝2+DF，即DF的值最小时，AD+DF有最小值，如图，当DF⊥BF时，DF最小，∵∠DBF＝30°，∴DF＝BD＝，∴AD+DF的最小值＝2+＝，故答案为．三、解答题（共52分）17．【解答】解：去分母得：2＝x2+2x﹣x2+4，解得：x＝﹣1，经检验x＝﹣1是分式方程的解．18．【解答】解：﹣÷＝﹣×＝+＝＝＝．∵m＝2020，∴原式＝＝．19．【解答】解：＝＝，∵x为整数，且的值也为整数，∴x﹣2的值为﹣4，﹣2，﹣1，1，2或4．∴x的值为：﹣2，0，1，3，4或6，经检验，当x＝﹣2时，原式分母为0，不符合题意，故舍去．∴0+1+3+4+6＝14．∴所有符合条件的x的值之和为14．20．【解答】（1）证明：∵四边形ABCD是平行四边形，∴AD＝BC，AD∥BC，∴∠A+∠B＝180°，∵点O是AB的中点，∴OA＝OB，在△AOD和△BOC中，，∴△AOD≌△BOC（SSS），∴∠A＝∠B＝90°，∴平行四边形ABCD是矩形；（2）解：由（1）得：△AOD≌△BOC，∴∠AOD＝∠BOC，∵∠COD＝60°，∴∠AOD＝∠BOC＝60°，∵∠A＝90°，∴∠ADO＝30°，∴OA＝AD＝，∴AB＝2OA＝2，∴矩形ABCD的面积＝AB×AD＝2×3＝6．21．【解答】（1）解：设今年5月份A款汽车每辆售价为x万元，根据题意，得．解得x＝8．经检验，x＝8是原方程的解．答：今年5月份A款汽车每辆售价为8万元；（2）解：设B款汽车卖出a辆，根据题意，得a（10.5﹣7.5）+（15﹣a）×（8﹣6）≥39，解得a≥9．答：B款汽车至少卖出9辆．22．【解答】（1）解：∵四边形ABCD，四边形CDEF是平行四边形，∴AB＝CD＝5，CD＝EF，AB∥CD，∴AB＝EF＝5，∴AE＝BF＝2，∴AF＝AC＝3，∵AB∥CD，AC⊥CD∴AB⊥AC，∴CF＝＝3，BC＝＝＝，∴△BCF的面积＝BF•AC＝×2×3＝3，△BCF的周长＝BF+BC+CF＝2+3+；（2）证明：如图，在AD上取一点M，使得AM＝AG，连接CM．∵四边形ABCD，四边形EFCD都是平行四边形，∴AB＝CD＝EF，AD＝BC，AD∥BC，AB∥CD，∵AH⊥BC，∴AH⊥AD，∵AC⊥AB，∴∠BAC＝∠GAM＝90°，∴∠F AG＝∠CAM，∵AF＝AC，AG＝AM，∴△F AG≌△CAM（SAS），∴∠ACM＝∠AFG＝45°，FG＝CM．∵∠ACD＝∠BAC＝90°，∴∠MCD＝45°＝∠EFG，∵EF＝CD，FG＝CM，∴△EFG≌△DCM（SAS），∴EG＝DM，∴AG+EG＝AM+DM＝AD＝BC．即BC＝AG+EG．故答案为：3；．23．【解答】（1）证明：∵∠BAC＝90°，AB＝AC，∴∠ABC＝∠ACB＝45°，∵四边形ADEF是正方形，∴AD＝AF，∠DAF＝90°，∵∠BAC＝∠BAD+∠DAC＝90°，∠DAF＝∠CAF+∠DAC＝90°，∴∠BAD＝∠CAF，在△BAD和△CAF中，，∴△BAD≌△CAF（SAS），∴BD＝CF；故答案为：BD＝CF；（2）∵∠BAC＝90°，AB＝AC，∴∠ABC＝∠ACB＝45°，∵四边形ADEF是正方形，∴AD＝AF，∠DAF＝90°，∵∠BAC＝∠BAD+∠DAC＝90°，∠DAF＝∠CAF+∠DAC＝90°，∴∠BAD＝∠CAF，在△BAD和△CAF中，，∴△BAD≌△CAF（SAS），∴BD＝CF，∵AC＝2，∴BC＝2，∵CD＝1，∴CF＝BD＝BC+CD＝2+1；故答案为：2+1；（3）①与（1）同理可得，BD＝CF，所以，CF＝CD﹣BC，故答案为：CF＝CD﹣BC；②∵∠BAC＝90°，AB＝AC，∴∠ABC＝∠ACB＝45°，则∠ABD＝180°﹣45°＝135°，∵四边形ADEF是正方形，∴AD＝AF，∠DAF＝90°，∵∠BAC＝∠BAF+∠CAF＝90°，∠DAF＝∠BAD+∠BAF＝90°，∴∠BAD＝∠CAF，在△BAD和△CAF中，，∴△BAD≌△CAF（SAS），∴∠ACF＝∠ABD＝180°﹣45°＝135°，∴∠FCD＝∠ACF﹣∠ACB＝90°，则△FCD为直角三角形，∵正方形ADEF中，O为DF中点，∴OC＝DF，∵在正方形ADEF中，OA＝AE，AE＝DF，∴OC＝OA，∴△AOC是等腰三角形．。

2024-2025学年第一学期期中学情调查八年级数学考试时间：90分钟一、选择题（每小题3分，共24分）1.8的算术平方根是（ ）A.4B. C. D.22.在一次“寻宝”游戏中，寻宝人已经找到了和两个标志点，并且知道藏宝地点的坐标为，如图，藏宝地点可能是（ ）A.M 点B.N 点C.P 点D.Q 点3.下列计算正确的是（ ）4.在中，a 、b 、c 分别是、、的对边，则下列条件不能判断是直角三角形的是（ ）A.，B.C.D.5.正比例函数（）的图象经过第一、三象限，则一次函数的图象大致是（ ）A. B. C. D.6.如图，正方形的边长为2，在数轴上，C 点为原点，以中点M 为圆心，线段的长为半径画弧，交数轴的正半轴于点F ，则F 点表示的实数为（ ）±(3,2)A (3,2)B -(0,0)-=+=3=-4=ABC △A ∠B ∠C ∠ABC △35A ∠=︒55B ∠=︒::3:4:5a b c =222::3:4:5a b c =A B C∠+∠=∠y kx =0k ≠y x k =+ABCD BC BC MDD.37.已知一次函数与的自变量x 与因变量，的部分对应数值如下表，则关于x ，y 的二元一次方程组的解为（ ）x…-1012……-3159……-7-315…A. B. C. D.无解8.如图，15只空油桶堆在一起，每只油桶底面的直径均为45，要给他们盖一个遮雨棚，遮雨棚的最低高度为（ ）.A.225B. C. D.二、填空题（每小题3分，共15分）9.点与点关于x 轴对称，则的值为_________.10.若，则_________.11.小亮、小红和笑笑三个人玩飞镖游戏，各投6支飞镖，规定在同一圆环内得分相同，三人中靶和得分情况如图，则小红得分为_________分.12.某市去年居民用水按照4.6元/吨收取费用，为提倡居民节约用水，自今年1月1日起对居民用水实行阶梯水费，规定：若用水超过a 吨，超过a吨的部分每吨增加2元.图中，分别表示去年、今年水费y （元）1-1+111y k x b =+222y k x b =+1y 2y 12y k x by k x b =+⎧⎨=+⎩1y 2y 13x y =-⎧⎨=-⎩01x y =⎧⎨=⎩15x y =⎧⎨=⎩cm cm 45225-(1,)P a -(,4)Q b a b +4y =+y x -=1l 2l与用水量x （吨）之间的关系.实行阶梯水费后，若用水超过a 吨，则超过a 吨的部分每吨水费为_______元.13.如图，在中，，M 为上一点，且，N 为边上一点，连接，将沿翻折，使点C 的对应点D 落在延长线上，交于点E ，若，则的长为__________.三、解答题（本题共7小题，共61分）14.（9（215.（5分）解方程组：16.（7分）如图，在平面直角坐标系中，已知的顶点坐标分别是，，.（1）画出关于y 轴对称的，其中点A 的对应点是点，点B 的对应点是点；（2）请直接写出点的坐标为_________，点的坐标为_________，点的坐标为_________；（3）P 为x 轴上一点，最小时，点P 的坐标为_________.17.（8分）为测量学校旗杆的高度，八年级1班的学习小组设计了多种方案，请结合下面表格的信息，完成任务问题：测量工具含45°角的直角三角板、足够长的皮尺Rt ABC △90ABC ∠=︒AC 2CM AM ==BC MN CMN △MN AB MD BC ADM C ∠=∠NC -2433x y x y +=-⎧⎨+=⎩②①ABC △(1,1)A (4,2),B (3,4)C ABC △A B C '''△A 'B 'A 'B 'C 'PA PB +方案一方案二方案三测量方案示意图设计方案及测量数据在地面确定点C ，并测得小明站在距离旗杆2.4m 的点D 处，眼睛距离地面1.6m ，视线沿着三角板的一直角边落在旗杆顶部A 处，小亮沿着直线垂直移动一高为4m 的竹竿，直到小明视线沿着三角板的另一直角边恰好落在竹竿顶部E 处，此时测得竹竿距离旗杆12.8m.如图，旗杆顶端的绳子垂落地面后还多出1m ，将绳子斜拉直后，使得绳子底端C 刚好接触地面，此时测得.任务一判断分析（1）在方案一中，要确定旗杆的高度应测量_________的长度，请说明理由：__________________；任务二推理计算（2）请在方案二或方案三中任选一个方案，并根据测量数据，求旗杆的高度.18.（8分）某网购平台开展“爱心助农”活动，准备在平台推送两种特色水果.经过对往年情况的调查，这两种水果的进价和售价如下表所示：种类进价（元/）售价（元/）甲x 12乙y14（1）（3分）购进甲种水果5和乙种水果10需要160元；购进甲种水果12和乙种水果5需要156元.求x ，y 的值；（2）（5分）该平台决定每天对甲、乙两种水果共1000进行销售，其中甲种水果的数量不超过200，平台每天售完1000水果能获利2500元吗？19.（12分）我们发现，用不同的方式表示同一图形的面积可以解决线段长度之间的数量关系的相关问题.我国汉代数学家赵爽（公元3-4世纪）就通过一幅“弦图”，证明出勾股定理，后人称之“赵爽弦图”.如图1，“赵爽弦图”是由4个全等的直角三角形拼成的一个大正方形，记为“正方形”，设直角三角形较短的直角边为a ，较长的直角边为b ，面积为.图1图2图3图445ACB ∠=︒BD EF 5m BC =AB kg kg kg kg kg kg kg kg kg ABCD S △（1）（2分）小桐用这4个直角三角形拼出图2所示的正方形，发现：若a、b的值确定，则正方形的面积、正方形的面积、直角三角形的面积的值都唯一确定，当，时，_________，_________；（2）（10分）小桐进一步思考，并提出问题：已知、、中的任意两个量可否求出a、b的值？于是给出以下条件，并进行探索：条件（I）条件（II）条件（III）①（2分）选择条件（I）（II），则_________，_________；②（3分）选择条件（I）（III），请你帮小桐计算出a，b的值；③（5分）【探索发现】选择条件（II）（III），由（II）得：，由（III）得：，进而得出关于a的方程：，小桐尝试从“形”的角度来确定a的值，将看作是长为，宽为a的长方形，且长方形面积为48，根据“赵爽弦图”的构图思路，小桐用4个这样的长方形构造“空心”大正方形（如图3），则图3中大正方形的面积为：，也可以表示为：，于是：，因此，所以或（舍去），故，.这正是赵爽在《勾股圆方图注》中记载的一类方程的几何解法.【类比迁移】小桐继续根据以上解法求解方程，请将其解答过程补充完整.第一步：利用四个全等的长方形构造“空心”大正方形；第二步：根据大正方形的面积可得新的方程__________________，解得原方程的一个正根为_________；【拓展应用】一般地，对于关于x的方程可以构造图4求解.已知图4是由四个面积为5的相同矩形构成，中间围成的正方形面积为16，那么，此方程中的_________，求得方程的正根为_________.20.（12分）阅读材料，若点M到直线a，b的距离相等，则称点M为直线a，b的关联点.图1备用图例如：如图1，在平面直角坐标系中，点到x轴和y轴的距离相等，故是x轴和y轴的关联EFGHABCDABCDS正方形EFGHEFGHS正方形S△1a=4b=ABCDS=正方形EFGHS=正方形ABCDS正方形EFGHS正方形S△256ABCDS=正方形4EFGHS=正方形24S=△a=b=2b a-=1242ab=(2)48a a+=(2)48a a+=(2)a+4844196⨯+=22()(2)a b a a+=++2(2)196a a++=2214a+=±6a=8a=-6a=8b=(3)18x x-=()x x m n+=n=(3,3)-(3,3)-点.在平面直角坐标系中，已知，直线：（）交x 轴于点，交y 轴于点C ，点D 为x 轴上一个点；（1）（5分）直线经过点A ，①_________.，若在直线上，则比较t 与6的大小：t ________6；②当点D 坐标为时，点B 恰好为、的关联点，求直线的解析式；（2）（7分）若（），D 为中点，点P 为线段上一点，且为x 轴和y 轴的关联点，将绕点P 逆时针旋转90°至，①求证：点E 为直线：与直线：的关联点；②对于直线：上任意两点M 、N ，始终有，直接写出m 的值.(0,6)A 1l 4y kx m =+0k <(,0)B n 1l m =(1,)t 1l (8,0)CO CD 1l 8n m =0m >OB BC PD PE 1l 4y kx m =+2l 4y kx m =-+2l 4y kx m =-+AMN EMN S S =△△2024-2025学年第一学期期中学情调查八年级数学参考答案与评分标准一、选择题（每小题3分，共24分）BDACA BDB二、填空题（每小题3分，共15分）9.-5；10.1；11.33；12.5.5；13.；三、解答题（本题共7小题，共61分）14.（1）解：原式∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙3分∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙4分（2）解：原式4分5分15.解：①，得： ③∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙1分③②，得：∴∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙3分将代入①，得：∴∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙4分∴原方程组的解是∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙5分16.（1）如图，即为所求；∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙2分10312=-⨯+==+-4=+-4=-3⨯6312x y +=--515x =-3x =-3x =-2(3)4y ⨯-+=-2(3)4y ⨯-+=-32x y =-⎧⎨=⎩A B C '''△（2）；；；∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙5分（3）∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙7分17.（8分）为测量学校旗杆的高度，八年级1班的学习小组设计了多种方案，请结合下面表格的信息，完成任务问题：（1）；∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙1分为等腰直角三角形，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙2分（2）选择方案二：过点C 作分别交于点M ，交于点N ，则∴∵∴∴∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙3分由题可知，米，米，米，米，∴米，米∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙5分∵，∴∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙6分∴∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙7分∴米故旗杆的高度为12米......................8分选择方案三：由题可知，，，设米，则米∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙3分在中，即∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙5分解得：∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙7分故旗杆的高度为12米∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙8分18.（1）∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙2分(1,1)-(4,2)-(3,4)-(2,0)BC ABC △AB CB =MN BF ∥AB EF 90AMC CNE ∠=∠=︒2390∠+∠=︒90ACE ∠=︒1390∠+∠=︒12∠=∠ 1.6MB NF CD ===4EF = 2.4MC BD ==12.8MN BF ==10.4CN = 2.4EN =CM =90AMC CNE ∠=∠=︒12∠=∠()AMC CNE ASA △≌△10.4AM CN ==12AB AM MB =+=AB 1AC AB =+5BC =90ABC ∠=︒AB x =(1)AC x =+Rt ABC △222AB BC AC+=2225(1)x x +=+12x =AB 510160,125156;x y x y +=⎧⎨+=⎩解得：∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙3分∴x ，y 的值分别为8，12.（2）设甲种水果售出m ，则乙种水果售出，该平台利润为w 元，则，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙5分∵，∴w 随m 增大而增大∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙.6分∵∴当时，w 最大，且最大值为2400元.∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙7分故每天售完1000kg 水果获利无法达到2500元......................8分法二：设甲种水果售出m ，则乙种水果售出，若要获利2500元，则，故每天售完1000水果获利无法达到2500元.19.（1），；∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙2分（2）①，；∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙4分②由题，，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙5分∴∵∴∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙6分∵，a ，b 均为正数∴联立，得解得：∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙7分③【类比迁移】∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙8分；∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙9分【拓展应用】，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙10分8,12;x y =⎧⎨=⎩kg (1000)kg m -(128)(1412)(1000)w m m =-+--22000m =+20k =>200m ≤200m =kg (1000)kg m -(128)(1412)(1000)2500m m -+--=250200m =>kg 25ABCD S =正方形9EFGH S =正方形7a =9b =2()256a b +=48ab =22()()464a b a b ab -=+-=b a>8b a -=2()256a b +=16a b +=168a b b a +=⎧⎨-=⎩412a b =⎧⎨=⎩2(3)1849x x +-=⨯+6x =5n =方程的正根为1或5.∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙12分20.（1）①，<；.....................2分②由①得：，则直线的解析式为∵在中，作于点H ，∵点B 恰好为、的关联点则∵，∴∴∴，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙4分将代入，得：∴直线的解析式为∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙5分（2）①∵（），D 为中点则将代入：中，得：∴∴∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙6分∵点P 为线段上一点，且为x 轴和y 轴的关联点，设则326OC =1l 6y kx =+8OD =Rt COD △10CD =BH CD ⊥CO CD BO BH =12BOC S BO CO =⋅△1122BDC S CD BH BD CO =⋅=⋅△BOC BCD S BO CO BO CO S BD CO CD BH⋅⋅==⋅⋅△△35BO CO BD CD ==3OB =(3,0)B (3,0)B 6y kx =+2k =-1l 26y x =-+8n m =0m >OB (4,0)D m (8,0)B m 1l 4y kx m =+084mk m=+12k =-142y x m =-+BC (,)P a a 142a a m =-+∴即过点P 作轴于点M ，过点E 作交的延长线于点N 则解得，故，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙7分连接，则∴，由题∴，∴E 为直线：与直线：的关联点∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙.8分②m 的值为，简要思路：法一：由题可知，则，∴解得或法二：由题可知，A 点为直线与直线的关联点可知或83a m =88,33P m m ⎛⎫ ⎪⎝⎭PM x ⊥EN PM ⊥MP ENP PMD≌△△83NE PM m ==43NP MD m ==16,43E m m ⎛⎫ ⎪⎝⎭E C y y =CE CE OB∥13∠=∠24∠=∠34∠=∠12∠=∠1l 4y kx m =+2l 4y kx m =-+192m =2910m =QE AC =203Q y m =|64|AC m =-204|64|3Q E y y m m m -=-=-910m =921l 2l 1AE l ∥2AE l ∥故或将代入得，或（其他解法不赘述）162AE y x =-+162AE y x =+16,43E m m ⎛⎫ ⎪⎝⎭910m =92。

2019-2020学年广东省深圳市宝安区八年级下学期期中考试数学试卷（带解析）一、选择题1.若x＞y，则下列式子中错误的是（）A．x-3＞y-3 B．x+3＞y+3 C．-3x＞-3y D．＞【答案】C．【解析】试题分析：A、不等式的两边都减3，不等号的方向不变，故A正确；B、不等式的两边都加3，不等号方向不变，故B正确；C、不等式的两边都乘-3，不等号的方向改变，故C错误；D、不等式的两边都除以3，不等号的方向改变，故D正确；故选C．考点：不等式的性质2.函数中自变量x的取值范围在数轴上表示正确的是（）A．B．C．D．【答案】A．【解析】试题分析：由函数，得到3x+6≥0，解得：x≥﹣2，表示在数轴上，如图所示：故选A．考点：在数轴上表示不等式的解集；函数自变量的取值范围．3.如图，OP是∠AOB的平分线，点C，D分别在角的两边OA，OB上，添加下列条件，不能判定△POC≌△POD的选项是（）A．PC⊥OA，PD⊥OBB．OC=ODC．∠OPC=∠OPDD．PC=PD【答案】D．【解析】试题分析：A．PC⊥OA，PD⊥OB得出∠PCO=∠PDO=90°，根据AAS判定定理成立，B．OC=OD，根据SAS判定定理成立，C．∠OPC=∠OPD，根据ASA判定定理成立，D．PC=PD，根据SSA无判定定理不成立，故选D．考点：角平分线的性质；全等三角形的判定．二、单选题1.若分式的值为0，则x的值为( )A．±2 B．2 C．﹣2 D．4【答案】C【解析】依据题意列出方程，然后把分式方程转化为整式方程，解出解后，再去检验，最后作答.解：由题意可得：=0，方程两边同乘x-2得：，则有x2=4，解得：经检验可得：x=2不合题意，故舍去.所以原方程的解为x=-2.故选C.“点睛”本题属于中等题，考查了分式方程的解法和直接开平方法解一元二次方程，关键是分式方程的检验不能忘了.2.下列汽车标志中既是轴对称图形又是中心对称图形的是( )A． B． C． D．【答案】C【解析】根据轴对称图形与中心对称图形的概念求解．解：A、是轴对称图形，不是中心对称图形．故本选项错误；B、不是轴对称图形，也不是中心对称图形．故本选项错误；C、是轴对称图形，也是中心对称图形．故本选项正确；D、是轴对称图形，不是中心对称图形．故本选项错误．故选C．“点睛”考查了中心对称图形与轴对称图形的概念：轴对称图形的关键是寻找对称轴，图形两部分沿对称轴折叠后可重合；中心对称图形是要寻找对称中心，旋转180度后与原图重合．3.下列从左边到右边的变形，是因式分解的是( )A．（a+3）（a﹣3）=a2﹣9B．x2+x﹣5=x（x+1）﹣5C．x2+1=x（x+）D．x2+4x+4=（x+2）2【答案】D【解析】根据分解因式就是把一个多项式化为几个整式的积的形式的定义，利用排除法求解．解：A、是多项式乘法，不是因式分解，错误；B、x2+x﹣5=x（x+1）﹣5，右边不是积的形式，错误；C、不是因式分解，错误；D、是因式分解，右边是积的形式，正确；故选D．“点睛”这类问题的关键在于能否正确应用分解因式的定义来判断．4.如图，边长为a，b的矩形的周长为14，面积为10，则a2b+ab2的值为( )A．140 B．70 C．35 D．24【答案】B【解析】由矩形的周长和面积得出a+b=7，ab=10，再把多项式分解因式，然后代入计算即可．解：根据题意得：a+b==7，ab=10，∴a2b+ab2=ab（a+b）=10×7=70；故选B．“点睛”本题考查了矩形的性质、分解因式、矩形的周长和面积的计算；熟练掌握矩形的性质，并能进行推理计算是解决问题的关键．5.明明准备用自己节省的零花钱充值共享单车“摩拜”，他现在已存有45元，计划从现在起以后每个月节省30元，直到他至少有300元．设x个月后他至少有300元，则可以用于计算所需要的月数x的不等式是( )A．30x﹣45≥300 B．30x+45≥300 C．30x﹣45≤300 D．30x+45≤300【答案】B【解析】此题中的不等关系：现在已存有45元，计划从现在起以后每个月节省30元，直到他至少有300元．至少即大于或等于．解：x个月可以节省30x元，根据题意，得30x+45≥300．故选B．“点睛”本题主要考查简单的不等式的应用，解题时要注意题目中的“至少”这类的词．6.下列命题中，逆命题是假命题的是( )A．全等三角形的对应角相等B．直角三角形两锐角互余C．全等三角形的对应边相等D．两直线平行，同位角相等【答案】A【解析】把一个命题的条件和结论互换就得到它的逆命题，再进行判断即可．解：A、全等三角形的对应角相等的逆命题是对应角相等的三角形全等，是假命题；B、直角三角形两锐角互余的逆命题是两锐角互余的三角形是直角三角形，是真命题；C、全等三角形的对应边相等的逆命题是对应边相等的三角形全等，是真命题；D、两直线平行，同位角相等的逆命题是同位角相等，两直线平行，是真命题；故选A．“点睛”此题考查了命题与定理，两个命题中，如果第一个命题的条件是第二个命题的结论，而第一个命题的结论又是第二个命题的条件，那么这两个命题叫做互逆命题．其中一个命题称为另一个命题的逆命题．7.如图，Rt△ABC中，∠ACB=90°，AC=3，BC=4，将边AC沿CD翻折，使点A落在AB上的点E处；再将边BC沿CF翻折，使点B落在CE的延长线上的点B′处，两条折痕与斜边AB分别交于点D、F，则线段B′F的长为( )A． B． C． D．【答案】B【解析】首先根据折叠可得CD=AC=3，BC=4，∠ACE=∠DCE，∠BCF=∠B/CF，CE⊥AB，然后求得△BCF是等腰直角三角形，进而求得∠B/GD=90°，CE-EF=，ED=AE=，从而求得B/D=1，DF=，在Rt△B/DF中，由勾股定理即可求得B/F的长.解：根据首先根据折叠可得CD=AC=3，B/C=B4，∠ACE=∠DCE，∠BCF=∠B/CF，CE⊥AB，∴BD=4-3=1，∠DCE+∠B/CF=∠ACE+∠BCF，∴∠ACB=90°，∴∠ECF=45°，∴△ECF是等腰直角三角形，∴EF=CE，∠EFC=45°，∴∠BFC=∠B/FC=135°，∴∠B/FD=90°，∵S=AC×BC=AB×CE，△ABC∴AC×BC=AB×CE，∵根据勾股定理求得AB=5，∴CE=，∴EF=，ED=AE==∴DE=EF-ED=，∴B/F==.故答案为：“点睛”此题主要考查了翻折变换，等腰三角形的判定和性质，勾股定理的应用等，根据折叠的性质求得相等的角是解本题的关键.三、填空题1.多项式3x2﹣6x的公因式为____；【答案】3x【解析】根据公因式的定义求解即可．解：多项式3x2﹣6x的公因式为3x．“点睛”此题主要考查了公因式概念的应用，掌握定义是解题的关键．2.如图，在△ABC中，AB=4，BC=6，∠B=60°，将△ABC沿射线BC的方向平移2个单位后，得到△A′B′C′，连接A′C，则△A′B′C的周长为____；【答案】12【解析】根据平移性质，判定△A′B′C为等边三角形，然后求解．解：由题意，得BB′=2，∴B′C=BC-BB′=4．由平移性质，可知A′B′=AB=4，∠A′B′C=∠ABC=60°，∴A′B′=B′C，且∠A′B′C=60°，∴△A′B′C为等边三角形，∴△A′B′C的周长=3A′B′=12．故答案为：12．“点睛”本题考查的是平移的性质，熟知图形平移后新图形与原图形的形状和大小完全相同是解答此题的关键．3.已知一次函数y=ax+b的图象如图，根据图中信息请写出不等式ax+b≥2的解集为________；【答案】【解析】观察函数图形得到当x≤0时，一次函数y=ax+b的函数值不小于2，即ax+b≥2．解：根据题意得当x≤0时，ax+b≥2，即不等式ax+b≥2的解集为x≤0．故答案为x≤0．“点睛”本题考查了一次函数与一元一次不等式：从函数的角度看，就是寻求使一次函数y=ax+b的值大于（或小于）0的自变量x的取值范围；从函数图象的角度看，就是确定直线y=kx+b在x轴上（或下）方部分所有的点的横坐标所构成的集合．4.如图，在Rt △ABC 中，已知∠C =90°，∠A =60°，AC =3cm ，以斜边AB 的中点P 为旋转中心，把这个三角形按逆时针方向旋转90°得到Rt △A′B′C′，则旋转前后两个直角三角形重叠部分的面积为______．【答案】【解析】根据已知及勾股定理求得DP 的长，再根据全等三角形的判定得到△B′PH ≌△BPD ，从而根据直角三角形的性质求得GH ，BG 的长，从而不难求得旋转前后两个直角三角形重叠部分的面积． 解：如图所示，在直角△DPB 中，BP=AP=AC=3， ∵∠A=60°， ∴DP 2+BP 2=BD 2， ∴x 2+32=（2x ）2， ∴DP=x=，∵B′P=BP ，∠B=∠B′，∠B′PH=∠BPD=90°， ∴△B′PH ≌△BPD ， ∴PH=PD=，∵在直角△BGH 中，BH=3+，∴GH=，BG=， ∴S △BGH =××=，S △BDP =×3×=，∴S DGHP ==cm 2．“点睛”此题考查勾股定理，三角形的全等的判定及性质，旋转的性质等知识的综合运用． 四、解答题1.（1）因式分解：﹣2a 3+12a 2﹣18a ．（2）因式分解：a2（x﹣y）+4（y﹣x）．【答案】(1)；（2）【解析】解答分解因式的问题要先分析是否可以提取公因式，再分析是否可以采用公式法.（1）原式 ==（2）原式 ===“点睛”本题属于基础应用题，只需学生熟练掌握分解因式的方法，即可完成.2.（1）解不等式组：，并写出整数解．（2）解不等式组：，并把它的解集在所示的数轴上表示出来．【答案】（1），整数解为2；（2）【解析】分别解两个不等式得到x≥1和x＜5，根据大于小的小于大的取中间得到不等式的解集，然后利用数轴表示，再写出整数解．解：（1）解不等式①得：，解不等式②得：，∴原不等式组的解集是，∴原不等式组的整数解为2．（2）解不等式①得：解不等式②得：在同一数轴上分别表示出它们的解集为∴原不等式组的解集是“点睛”本题考查了解一元一次不等式组：先分别解两个不等式，然后根据“同大取大，同小取小，大于小的小于大的取中间，大于大的小于小的无解”确定不等式组的解集．也考查了数轴表示不等式的解集．3.先化简，再求值：其中x=2017．【答案】-1，-1【解析】先根据分式混合运算的法则把原式进行化简，再把x 的值代入进行计算即可． 原式 ==" -1"∴当x =2017时，原式 =-1“点睛”本题考查的是分式的化简求值，熟知分式混合运算的法则是解答此题的关键． 4.如图，在平面直角坐标系中，△ABC 的三个顶点都在格点上，点A 的坐标为（2，4），请解答下列问题：（1）AB 的长等于 ；（结果保留根号）（2）把△ABC 向下平移5个单位后得到对应的△A 1B 1C 1，画出△A 1B 1C 1，点A 1的坐标是___ ； （3）画出△ABC 绕原点O 旋转180°后得到的△A 2B 2C 2，并写出点A 2的坐标 ；【答案】（1）；（2）画图见解析，（2,-1）；（3）画图见解析，（-2,-4）【解析】（1）分别找出A 、B 、C 三点关于x 轴的对称点，再顺次连接，然后根据图形写出A 点坐标；（2）将△A 1B1C 1中的各点A 1、B 1、C 1绕原点O 旋转180°后，得到相应的对应点A 2、B 2、C 2，连接各对应点即得△A 2B2C 2． 解：①AB=；②画出△A 1B1 C 1， 点A 1的坐标是（2,-1）； ③画出旋转后的△A 2B 2C 2， 点A 2的坐标是（-2,-4）“点睛“图形与变换空间与坐标：中考试题中分值约为3-4分，题型以选择，填空为主，难易度属于易。

深圳市宝安区八年级下学期第一次月考语文试卷姓名:________ 班级:________ 成绩:________一、选择题 (共8题；共16分)1. （2分）下列词语中加横线的字，每对读音都相同的一项是（）。

A . 绯红／芳菲引擎／风驰电掣数一数二／历历可数B . 砌成／贯彻侥幸／饶有兴味相提并论／无与伦比C . 怄气／讴歌粗犷／旷日持久转弯抹角／一平如抹D . 震撼／遗憾遐想／目不暇接不省人事／独处自省2. （2分）下列句子没有语病的一项是（）A . 有没有坚定的意志，是一个人在事业上能够取得成功的关键。

B . 中学生是学习的重要阶段。

C . 不管严寒酷暑，我们都坚持清晨的锻炼。

D . 我们听到的是欢乐的歌声和愉快的笑容。

3. （2分） (2017七下·萧山期末) 下列句子中划线的词语使用恰当的一项是（）。

A . 盛开的紫藤花吸引很多驻足鉴赏，一片辉煌的淡紫色，像一条瀑布，从空中垂下。

B . 家风，是一个家庭或家族长期以来形成并传承的道德操守和处世方法。

C . 在学校上课认真听讲，有疑难就不耻下问，老师一定会帮你耐心解答，让你豁然开朗。

D . 神舟八号像一支离弦之箭，气冲斗牛，直上云天，与天宫一号成功对接，成为我国载人航天发展史上新的里程碑。

4. （2分）下列句子中标点符号使用有误的一项是（）A . 奶奶是会说故事的，说了一个，还要再说一个……奶奶突然说：“月亮进来了！”B . 它走了，它是匆匆的；你们快出去寻月吧。

C . 菜市场上正在出售菠菜、黄瓜、韭菜……等蔬菜。

D . “吹面不寒杨柳风”，不错的，像母亲的手抚摸着你。

5. （2分） (2017七上·重庆月考) 下列对课文内容理解有误的一项是（）A . 《春》一文的作者用诗的笔调，描绘了大地回春、万物复苏、生机勃勃的景象，抒发了热爱春天、赞美春天的感情。

B . 读史铁生的《秋天的怀念》，我们读出了母爱的艰辛与伟大，同时也读出了面对生命的残缺遗恨，我们应勇敢镇定地“好好儿活”。

2019-2020学年广东省深圳市宝安区八年级（下）期末物理试卷一、选择题（在下列1-20小题中，每小题只有一个选项符合题意，请将正确答案用2B铅笔涂到答题卡上，每小题2分，共40分）A．静止不动的物体一定没有受到力的作用B．两个物体没有接触就一定没有力的作用C．人可以左手拍右手，说明一个物体也可以有力的作用D．力是物体对物体的作用，没有物体就没有力的作用A．拍篮球时，篮球上下跳动B．高尔夫球被球杆推动C．射箭瞄准时，用力把弓拉弯D．滑冰运动员急速拐弯A．对电动自行车限速，是因为速度越大惯性越大B．限制电动自行车的质量，是因为质量越大惯性越大C．电动自行车在行驶过程中，外力突然全部消失，车将马上停下来D．刹车后，电动自行车不能立即停下来，说明车受到惯性力的作用A．在地球上，1kg等于9.8NB．质量是1kg的物体所受的重力是9.8NC．物体受到的重力是它质量的9.8倍D．物体的质量是它所受重力的9.8倍•5．如图所示，一物体在推力F的作用下沿水平方向做匀速直线运动，下列说法正确的是（）A．物体受到的重力和物体受到的推力是一对平衡力B．物体受到的重力和地面对物体的支持力是一对平衡力C．物体受到的推力和地面对物体的摩擦力是一对相互作用力D．物体对地面的压力和地面对物体的支持力是一对平衡力•6．如图所示，重10N的磁铁被吸附在竖直放置的铁板上，当对磁铁施加一个竖直向上大小为15N的拉力时，磁铁沿着铁板匀速向上运动，则此时磁铁受到的摩擦力大小和方向分别是（）A．15N，竖直向下B．15N，竖直向上C．10N，竖直向上D．5N，竖直向下A．鞋底的花纹B．浴室的脚垫做成凹凸不平C．在门的合页轴上加润滑油D．矿泉水瓶盖上的条纹•8．如图所示，甲、乙两个实心均匀正方体分别放在水平地面上，若它们质量相等，则它们对水平地面的压强（）A．甲大B．乙大C．一样大D．无法•9．如图，水平桌面上放着底面积和质量都相同的甲、乙两容器，分别装有深度相同、质量相等的不同种液体。

2019-2020学年深圳市宝安中学初中部高三英语第一次联考试题及答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项ANew events and changes of junior golf competition calendarNew eventsNotah BegayⅢJunior Golf National ChampionshipWhat does a junior golfer aim to pursue? One thing is to be noticed, ideally by a college coach. A remarkable opportunity will be offered by the Notah BegayⅢJunior Golf National Championship to its participants: an event broadcast by Golf Channel. Players aged between eight and 18 can compete in the new event; information about where and when it will be held will be released later.Barbasol Junior ChampionshipBeginning the career in the PGA Tour is something that a junior golfer tends to dream of. The Barbasol Junior Championship, which is scheduled to take place between June 29 and July 2 at Keene Trace Golf Club inNicholasville,Kentucky, will provide such an opportunity. Boys under 19 years old will qualify for this new 54-hole event, and the winner will be awarded a spot at the PGA Tour's Barbasol Championship in July.Changed eventsThunderbird International JuniorThe dates of the AJGA's Thunderbird International Junior have to be changed since the NCAA Championships move to Grayhawk Golf Club inScottsdale,Arizona, for the next three years. Generally, the Thunderbird is played at the end of May. However, this year it is scheduled on different dates for the first time, from April 9 to 12, which means, of course, that the finish date is on Masters Sunday.Gator InvitationalJunior golf intends to prepare for the following college golf. If this is the case, then it is crucial to simulate the higher-level experience as much as possible. Because of that, the Gator Invitational, as a junior boys' event, has made a significant decision on becoming a 54-hole event by adding a round this year. The new version will be played from March 13 to 15 at The Country Club of Jackson inJackson,Mississippi.1. Which event can be watched on TV?A. Notah BegayⅢJunior Golf National Championship.B. Barbasol Junior Championship.C. Thunderbird International Junior.D. Gator Invitational.2. When will the Thunderbird International Junior be played?A. At the end of May.B. From April 9 to 12.C. Between June 29and July 2.D. From March 13 to 15.3. What has been changed about the Gator Invitational?A. The award given to the winnerB. The place where it is played.C. The required age of the players.D. The number of rounds it has.BJohn Montefiore's path to graduation from theUniversityofTorontowas a little unusual. He recently completed his bachelor's degree(学士学位) which he started in1995.Montefiore left university in 1996 and tried to tell himself that a degree wasn't necessary for personal success. But it remained unfinished businessuntil he made the decision to return in 2018 at the age 42. At that time, his job development stalled, so he made up his mind to go back to school in order to improve himself and work well in future.The second time around, Montefiore never missed a class and always sat in the front row. He found many courses invaluable and he received the Award of Excellence twice. He found support services played an important role in his successful return. He said, “I hadn't written a paper for years, so I found the college writing centre to be of great use. Before I handed in a paper, they would give me feedback, which was really great. The university has so much built in to help students succeed and I'm so thankful for that.” But he also met difficulties this time. As a student much older than others, he found it harder to make friends with classmates.However, no matter what had happened before, he finally got his bachelor's degree. He said, “After all these years, I had thought it wouldn't mean anything, but it means a lot. I totally understand the value of education now. It's not necessarily the value that other people see in it, but the value it gives me as a person, for my confidence and my self-esteem. It also helps me prepare well before I re-enter the workplace.”4. Why was Montefiore's graduation unusual?A. He had achieved personal success before.B. He understood the importance of a degree.C. He finished his degree at a much older age.D. He began his university study at a young age.5. What does the underlined word "stalled" in paragraph 2 probably mean?A. Stopped.B. Started.C. Survived.D. Succeeded.6. What can we learn about Montefiore from the third paragraph?A. He got good grades very easily.B. He took his study seriously this timeC. He was thankful for his classmates' helpD. He had difficulty using support services.7. What did Montefiore learn from his experience?A. Confidence is important in one's life.B. Others' opinions on education matter a lot.C. Higher education is a must for personal success.D. Education makes one feel better about himself.CAccording to a survey published by the American Institutes for Research last year, a total of 57 colleges were operating some form of CBE programs and about 85 percent of all the higher education officials said they were either designing a CBE program at their school or were considering doing so.Students in a CBE program choose a central field of study, just as they would at a traditional college or university. Yet instead of attending a series of classes led by professors or teaching assistants at schools, the students study online and direct themselves.CBE programs require students to show their understanding of a given set of sills Students must prove their mastery of skills that relate to their field of choice by taking related exams. Once they have met all the requirements of their study programs, the students will get their degrees.CBE programs have made use of many new technologies, especially internet and online media. This helps reduce barriers for nontraditional and other students by bringing higher education to them. And programs that permit students to work at their own speed may save students' money by reducing the time it takes for them to earn a degree.But some educators have concerns about the value of the education that CBE programs offer. Johann Neem atWesternWashingtonUniversityargues that the purpose of higher education is not simply to help students master certain skills. It should teach students how to think critically (批判性地) understand the subjects they are studying more deeply and see how they are connected to other subjects. Only that way can they put the knowledge to better use.He said, “You need to explore, think .. get shaken, have a conversation and struggle. Andthose things taketime.”Instead of supporting CBE, he adds, policy makers and educators should look for ways to improve access and reduce costs for traditional higher education.8. How are CBE programs different from traditional college education?A. They require students to choose their subjects.B. They offer shorter curricula and are less expensive.C. They heavily rely on the information technologies.D. They allow students to take easier examinations.9. What can we learn from Johann Neem's words?A Free access to traditional education should be provided.B. Higher education just focuses on critical thinking skills.C. Students should spend longer time completing the degree courses.D. College students should be challenged to explore around their subjects.10. How does Johann Neem's attitude toward CBE programs?A. Supportive.B. Disapproving.C. Sympathetic.D. Uncaring.11. What is the author's purpose in writing the text?A. To press policy-makers to provide more affordable education.B. To show the disadvantages of the traditional college education.C. To introduce a new controversial trend in the higher education.D. To encourage educators to improve the quality of CBE programs.DNew Yorkis among the slowest cities during rush hour in the world, according to a report published in January. Crossing midtown by car is soul-destroying. The average speed is 4.7 miles per hour, not much quicker than a quick walk. But relief is in sight. On April 1st, state lawmakers agreed to introduce road charges, makingNew Yorkthe first big American city to do so. By next year vehicles will have to pay to enterManhattansouth of60th Street.The details of the new rule, including how much drivers will have to pay, how they will pay and how often they will pay, haveyet to be decided. A “traffic mobility review board" will be set up to work all this out. New Yorkers living in the fee zone who make less than $60,000 a year will be exempt (获豁免) . Other drivers, including motorcyclists, the city' s civil servants, disabled drivers and the trucking industry, all want discounts or exemptions,which might not be a good sign.If done right, road pricing could be expanded beyondManhattan.New Yorkcan learn from other cities.Singapore, for instance, which has had pricing for decades, adjusts prices regularly. It can also learn from mistakes.London, which rolled out its pricing in 2003, is only starting to charge on-demand car hires like Uber. Stockholm exempted too many vehicles, which caused a drop in revenues (收入) .Other cities considering road charges, includingLos Angeles, Philadelphia Portland,San FranciscoandSeattle, are watchingNew York. "We really have t1o make a good example," says Nicole Gelinas of the Manhattan Institute, aNew Yorkthink tank.12. What does the report find?A.New Yorkhas terrible road traffic.B.New Yorkoften introduces new laws.C. New Yorkers prefer walking to driving.D. New Yorkers face an increased cost of living.13. What can be inferred about the new rule from paragraph 2?A. It faces some potential problems.B. It has clear and detailed fee standards.C. It favors New Yorkers living in the fee zone.D. It fails to win a traffic mobility review board s approval.14. What doLondonandStockholmhave in common?A. They learn a lot fromSingapore.B. They have greatly increased revenues.C. They charge on-demand car hires heavily.D. They are bad examples of placing road charges.15. What is the best title for the text?A. How much does it cost to drive intoManhattan?B.New Yorkapproves road pricing forManhattanC. Drivers fear crossingManhattansouth of60th StreetD. Who will be exempt from road charges inManhattan?第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

2019-2020学年深圳市宝安中学初中部高三英语下学期期末考试试卷及参考答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项ASome of the world’s most talented musicians have played concerts atLincolnCenter. On September 10, the center hosted a unique class of star musicians: kids and teens. The young players are part of the World Peace Orchestra, or WPO. The group held its first Music for Peace concert inNew York City, with 134 students from more than 60 countries taking part. Musicians were chosen from all over the world.The World Peace Orchestra began in 2013. The nonprofit group brings kids together using the common language of music. To be considered for the orchestra, students first had to be recommended by their teachers and then audition online for a group of judges. Once chosen, professional musicians and teachers then worked with the players to improve their skills.Some of the young musicians played classical instruments, such as violins and flutes. Others played instruments unique to their home country. For example, Amold Mugo, 16, fromKenya, played an instrument called the djembe. The drum is originally fromWest Africa. Mugo said he was shocked when he learned he was chosen for the orchestra. “I can’t express how I felt. I was rolling on the floor. It’s a once in a lifetime opportunity,” he said.Adomas Hendrixson, 13, fromLithuania, played piano for the WPO. Before theNew York Cityperformance, Hendrixson talked about what he hoped to take away from the event. “Fun and joy-- people smiling and clapping,” he said.“This is one of the only times in your life this could happen and I'm very excited.”Mugo says his favorite part of the WPO is making new friends from around the world. “I hope when I go home, I just take a little bit of every friend that I made here-- Brazil, Portugal, Queens,” Mugo said, “I hope that I take part of their culture home with me so I understand them more.”1. Who is most likely to have played for the WPO according to this passage?A. Li Ming, 16, a student fromChina.B. Daniel, 44, a judge fromAmerica.C. Edward, 21, a clerk fromAustralia.D. Catherine, 32, a teacher fromEngland.2. We can learn from the last paragraph that Mugo ________.A.performed best for the WPO.B. has learned some foreign culture.C. was eager to be chosen for a second time.D. received a great award for his performance.3. What would be the best title for this passage?A. Kids were good at playing music.B. Mugo liked making foreign friends.C. Concerts were played atLincolncentre.D. Young musicians played for peace.BAlaska—The American city Anchorage is recovering from a powerful earthquake Friday that damaged public buildings, homes and roads.The 7.0 earthquake caused buildings to slake. But there have been no reports of deaths, serious injuries or damage. Officials say the quake has not affected transportation of food and her supplies. “The ships are coming in on schedule, the supply lines are at this point working well,” the government told reporters Sunday.The Glenn Highway was probably the road hit hardest by the earthquake. It connects the state's largest city to other parts in the north. Traffic has been heavy and slow—moving since the quake. Drivers are being guided. Groups of workers are trying to rebuild areas where the quake left large holes in the road.People who are still nervous after the major quake have been more upset by more than 1, 700 aftershocks. “Anything that moves, you feel terrified”said David, whose home suffered structural(结构)damage, including a sunken foundation(地基). Actually, Alaska came up with strict building rules after a 9. 2 earthquake in 1964. That was the second most powerful earthquake on record.Government officials said a public health center promises that moneyfor medical treatment will continue to come. Mental healthy service(心理健康服务)is also available for people hurt by the disaster.Earthquake experts say there is a 4 percent chance of another 7. 0 earthquake or greater in the following week. "The chance is very small, but its not impossible, ” said the expert, Paul Caruso.4. What was the result of the earthquake?A. Buildings were damaged.B. Food supply was cut off.C. Many people were killedD. The ships could not come in.5. Why is the traffic slow on the Glenn Highway?A. Because small quakes hit the city.B. Because falling rocks are a danger.C. Because the highway is badly damaged.D. Because drivers are misled.6. What can you learn from Paul Caruso?A. Another greater earthquake is on the way.B. Chances still exists of another earthquake.C. It will be safe in the 1th week after the quake.D. There is no possibility for more quakes.7. Where can your possibly read the passage?A. Ina story book.B. In a travel journal.C. In a poster.D. In a newspaper.CPopularization has in some cases changed the original meaning of emotional (情感的) intelligence. Many people now misunderstand emotional intelligence as almost everything desirable in a person's makeup that cannotbe measured by an IQ test, such as character, motivation, confidence, mental stability, optimism and “people skills.” Research has shown that emotional skills may contribute to some of these qualities, but most of them move far beyond skill-based emotional intelligence.We prefer to describe emotional intelligence as a specific set of skills that can be used for either good or bad purposes. The ability to accurately understand how others are feeling may be used by a doctor to find how best to help her patients, while a cheater might use it to control potential victims. Being emotionally intelligent does not necessarily make one a moral person.Although popular beliefs regarding emotional intelligence run far ahead of what research can reasonably support, the overall effects of the publicity have been more beneficial than harmful. The most positive aspect of this popularization is a new and much needed emphasis (重视) on emotion by employers, educators and others interested in promoting social well-being. The popularization of emotional intelligence has helped both the public and researchers re-evaluate the functionality of emotions and how they serve people adaptively in everyday life.Although the continuing popular appeal of emotional intelligence is desirable, we hope that such attention will excite a greater interest in the scientific and scholarly study of emotion. It is our hope that in coming decades, advances in science will offer new perspectives (视角) from which to study how people manage their lives. Emotional intelligence, with its focus on both head and heart, may serve to point us in the right direction.8. What is a common misunderstanding of emotional intelligence?A. It can be measured by anIQ test.B. It helps to exercise a person’s mind.C. It includes a set of emotional skills.D. It refers to a person’s positive qualities.9. Why does the author mention “doctor” and “cheater” in paragraph 2?A. To explain a rule.B. To clarify a concept.C. To present a fact.D. To make a prediction.10. What is the author’s attitude to the popularization of emotional intelligence?A. Favorable.B. Intolerant.C. Doubtful.D. Unclear.11. What does the last paragraph mainly talk about concerning emotional intelligence?A. Its appeal to the public.B. Expectations for future studies.C. Its practical application.D. Scientists with new perspectives.DThe headmaster of a primary school showed on television to support her idea that parents should “dress appropriately in daywear” when they drop off and pick up their kids from school.Kate Chisholm, head teacher atSkerneParkAcademyin Darlington, theU.K., sent a letter home asking parents to set a better example for their children.“I have noticed there has been an increasing tendency for parents to drop off and pick up their kids from school while still wearing their pajamas (睡衣),” Chisholm wrote.“Could I please ask that when you are sending your children, you take the time to dress appropriately in daywear that is suitable for the weather conditions?”Kate Chisholm wants parents at her school to dress nicer. She appeared on British television station ITV to further explain her decision, saying she had started noticing the pajama trend had been picked up by “30 or 40” parents at school.Despite her determination to make school a nicer place to be, Chisholm admits that she can't demand that parents dress up-such as Karen Routh, 49, who wore pajamas to drop off her 8-ycar-old daughter Holly, because she was running late and didn't feel well.“I imagine there might be some people who keep up wearing pajamas for the next six months to prove a point,” Chisholm said. “I can't force people to get dressed but I will keep sending letters home in the hope that they decide to put on a pair of jeans.”Wearing pajamas in public has also become a hot issue for some schools and States in theU.S.In 2015, aFloridaschool board member insisted on a dress rule for parents who showed up in the school in sleepwear.12. The headmaster asks parents to pay attention to ______.A. the way they dressB. the relations with teachersC. the way they treat their kidsD. the clothes they buy for their kids13. How does Chisholm try to change this situation?A. Asking kids to set examples.B. Keeping them out of school.C. Sending letters to persuade them.D. Forcing them to change by laws.14. Why did Kate Chisholm appear on ITV?A. She wanted to force Karen to dress properly.B. Parents spent less time on their clothes.C. She wanted to explain her decision about the parents' dress.D. She wanted to tell us more and more parents wear pajamas to school.15. It can be inferred from the last paragraph that ______.A. strict laws should be passed to stop pajamasB. aFloridaschool will force parents to wear jeansC. people wearing pajamas in public will be punishedD. more and more people are concerned about dressing properly in public第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

2019-2020学年深圳市宝安中学初中部高三英语第一次联考试卷及答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AWashingtonDCBusToursDC Highlights TourThis is their base tour. It begins at 10 am daily and lasts for 5 hours. This part-bus, part-walking tour includes guided stops in theUS. All of their tours include a driver as well as a tour guide, and yourtour guide will get off the bus and give you walking tours of each stop, while your bus and driver wait for you. Capitol Building, the White House, Washington Monument as well as the Lincoln and MLK Memorials and the Vietnam War and Korean War Veterans Memorials.$54—Adult I $44—Child (3—12)Discover DC TourIf you want pretty much to explore every famous monument and landmark in DC and take a 1 -hour cruise on thePotomac River, then consider the Discover DC Tour. This 6-hour tour will take you to all locations (景点) listed on their DC Highlights Tour as well as the World WarⅡMemorial, the Franklin Delano Roosevelt Memorial, and the Thomas Jefferson Memorial.TIP: If you are planning on visiting NYC, you will get 30% off the Discover NYC Tour (normally $100) if you buy it at the same time as your Discover DC Tour.$74—Adult I $54—Child (3—12)VIP ExclusiveWashingtonDCCity TourThis 8-hour tour is actually the Discover DC Tour above with a VIP add-on at the beginning and the end. You * 11 meet your guide early for reserved (预留)tickets to tour inside the US Capitol Building. The 45 -minute tour and film have reserved tickets so you don't have to worry about it being sold out. After your day of sightseeing, you 'll be dropped off at the National Archives, again with reserved time tickets so you don't have to worry about waiting in line.$125 for Adults and ChildrenSkyview Changeable Bus TourOn this changeable mini bus, you get to experience a guided tour with panoramic views (全景)without the glare of a window in the photos! A guided bus tour takes you not only through all locations listed on the DC Highlights Tour, but also Old Town Alexandria andNationalHarbor.$69—Adult I $59—Child (3—12)Time: 9 am—4 pm1. What's special about the DC Highlights Tour?A. It uses mini buses.B. It has the fewest locations.C. It has the most tour guides.D. It provides reserved tickets.2. How much should a man pay if he buys the Discover DC Tour and the Discover NYC Tour at the same time?A. $128.B. $174.C. $104.D. $144.3. Which tour lasts the longest?A. DC Highlights Tour.B. Discover DC Tour.C. VIP ExclusiveWashingtonDCCity Tour.D. Skyview Changeable Bus Tour.BI dropped out of college after my first year. Three years later, I returned to college after having been stuck in a dead-end job, working at a department store. I saw school as my way out. But I quickly found myself up against the same problems that had caused me to give up before. I was in over my head with college-level algebra (代数) and a heavy workload of reading and writing homework. In addition, I was still unsure of my career (职业) direction。

2019-2020学年深圳市宝安中学初中部高三英语下学期期末试卷及参考答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项ABook reading is certainly one of the most absorbing habits. For young adults who love to read, finding some good books to read is very essential. Writing a book review can help you to improve your language and writing skills.The Book ThiefListed onThe New York Times Children’s Best Seller List for over 100 weeks, The Book Thief by Markus Zusak is the story of a young girl in the Nazi camps set during World War II. So, if you love history and wish to learn how the life was during Adolf Hitler’s time, read this historic book.The Diary of Young GirlEven Anne Frank can not have imagined that her personal diary written during World War II would become such a popular book. It’s a must read that describes the situation of a family in the evils of wars through the eyes of a teenager.Animal FarmAnimal Farm is one of the most popular books by George Orwell. It is just a reflection of the Stalin and World War II period that has been so creatively presented in this book. It is an interesting example of how literature can be used to present conditions common in the society.Adventures of Huckleberry FinnMark Twain’s Adventures of Huckleberry Finn is one of the great American novels in history, and is certainly a great pick for young adults. Young Huck Finn and his mischief along with the color1 ful description of people around theMississippi Rivermake this novel a great book to read.1.Which book describes the author’s own experiences according to this passage?A.The Book ThiefB.The Diary of Young GirlC.Animal FarmD.Adventures of Huckleberry Finn2.What do the first three books have in common?A.All of them are about wars.B.All of them are about farms.C.All of them are intended for history lovers.D.All of them were written during World War II.3.The purpose of this passage is to _________.A.instruct youngsters how to improve skillsB.tell youngsters some wonderful reading habitsC.introduce several good books to youngstersD.give youngsters advice on writing a book reviewBPreparations for the Tokyo Olympics have suffered another challenge after a survey found that 60% of people in Japan want them to be cancelled,less than three months before the Games are scheduled to open.Japan has extended a state of emergency in Tokyo and several other regions until the end of May as it struggles to control a fast increase in COVID-19 cases caused by new, more catching variants（变异体）with medical staff warning that health services in some areas are on the edge of breaking down.The Olympics, which were delayed by a year due to the pandemic, are set to open on 23 July, with the International Olympic Committee（IOC）and organizers insisting that measures will be put in place to ensure the safety of athletes and other visitors, as well as a nervous Japanese public.The survey, conducted between 7 and 9 May by the conservative Yomiuri Shimbun, showed 60% wanted the Games cancelled as opposed to 39% who said they should be held. “Postponement” — an option abandoned by the IOC — was not offered as a choice.Of those who said the Olympics should go ahead, 23% said they should take place without audience. Foreign audience have been banned but a final decision on native attendance will be made in June.Another poll conducted at the weekend by TBS News found 65% wanted the Games cancelled or postponed again, with 37% voting to give up the event altogether and 28% calling for another delay. A similar poll in April conducted by Kyodo news agency found 70% wanted the Olympics cancelled or postponed.The IOC's vice president, John Coates, said that while Japanese sentiment about the Games “was a concern”, he could foresee no situation under which the sporting events would not go ahead.4. How many Japanese wish the Olympics would not be held in Tokyo according to the survey?A. 60%.B. 28%.C. 37%.D. 70%.5. What should be put into consideration if the Olympics open?A. The economic crisis.B. The urban transport.C. The safety of athletes.D. The health condition of citizens.6. What is some people's attitude towards foreign spectators in Paragraph 5?A. Welcome.B. Unfriendly.C. Cold.D. Unsupported.7. What can we conclude from John Coates'words?A. The Olympics will be stopped this year.B. The Olympics will be put off.C. The Olympics will be held normally.D. The Olympics will take place in other place.CThink ofJapanin the spring and the image that comes to mind is likely the country’s famous cherry blossoms, also known as “Sakura” — white and pink flowers, blooming across cities and mountains.The flowers, which experience a “peak bloom” that only lasts a few days, have been loved inJapanfor more than a thousand years. Crowds celebrate with viewing parties,flockingto the most popular locations to take photos and have picnics underneath the branches.But this year, cherry blossom season has come and gone in the blink of an eye, in one of the earliest blooms on record. Scientists warn it’s a symptom of the larger climate crisis threatening ecosystems everywhere.Yasuyuki Aono, a researcher atOsakaPrefectureUniversity, has gathered records fromKyotoback to 812 AD from historical documents and diaries. In the central city ofKyoto, cherry blossoms peaked on March 26, the earliest in more than 1,200 years, Aono said. And in the capitalTokyo, cherry blossoms reached full bloom on March 22, the second-earliest date on record.The peak bloom dates shift every year, depending on numerous factors including weather and rainfall, but have shown a general trend of moving earlier and earlier. InKyoto, the peak date stayed around mid-April for centuries, but began moving into early April during the 1800s. The date has only dipped into late March a handful of times in recorded history.“Sakura blooms are very temperature sensitive,” said Aono. “Flowering and full bloom could be earlier or later depending on the temperature alone,” he said. “The temperature was low in the 1820s, but it has risen by about 3.5 degrees Celsius to this day.”This year’s seasons in particular influenced the blossom dates, he added. The winter was very cold, but the spring came fast and unusually warm.8. What is the best title of the passage?A. Cherry blossom celebrations.B. Warning of a climate crisis.C. A strong love for cherry blossom.D. Cherry blossom season coming earlier.9. What does the underlined word “flocking” mean?A. Blocking.B. Flooding.C. Running.D. Following.10. What can we infer from paragraph 5?A. The peak blossom dates fall on a fixed date.B. The cherry blossom rarely peaks in March.C. The peak bloom dates mainly depend on weather and rainfall.D. Cherry blossom peaks around mid-April inTokyo.11. What is the author’s purpose in writing the passage?A. To inform people the date of cherry blossom.B. To show a study on cherry blossom dates.C. To present a Japanese tradition of cherry blossom celebration.D. To make people aware of the influence of climate change on cherry blossom.DWhen the COVID-19 hit and supermarket shelves were empty, Chris Hall and Stefanny Lowey decided they no longer wanted to rely on others for food. The couple, who live on Pender Island in BritishColumbia, Canada, decided to start a year-long challenge where they wouldn't buy a single thing to eat. Instead they would grow, raise or catch everything—right down to sugar, salt and flour. Now, five months in, they say the challenge has changed their lives.Chris, 38, said, “It has always been something that we have wanted to do. We have had a garden and grown vegetables for a long time already. When the COVID-19 hit, it gave us that extra push that we needed to do it. We were both out of work when we started, and with the reality check of grocery stores running out of items, it gave us even more motivation to see if we could look after ourselves.”The pair spent the months before building a house for chickens, ducks and turkey as well as studying as much as possible to figure out where they would get all the things they needed. Chris adds, “We had to learn so many new things like how to grow mushrooms, process our Stevia plants, and harvest salt from the ocean. We spent alot of time reading and studying online to figure out all the things we were going to need to do.”Now after five months, they both feel its been going well but Chris admits the first few weeks were difficult. “The first three weeks were very challenging as our bodies adjusted to cutting out coffee, wine and sugar all on the same day,” he says. “After three weeks our energy levels balanced out and our wishes reduced and now we feel great.” Now February has ended. As they come through winter, they feel positive about continuing with this way of living, with their challenge officially ending in August.12. Why did the pair decide to produce foods on their own?A. They were isolated by Pender Island.B. They couldn't afford to buy them because they were out of work.C. They believed it's good for their health.D. They could hardly buy them in shops.13. Which words can be used to describe the couple?A. Rich and generous.B. Helpful and positive.C. Optimistic and self-dependent.D. Motivated and brave.14. What can we learn from the last paragraph?A. Their challenge may last about eleven months in total.B. They were discouraged by the difficulty at first.C. They had difficulty because they wanted more.D. They couldn't adjust their bodies to the hard work after three weeks.15. In which column may you read such a passage?A. Sports.B. Agriculture.C. Lifestyle.D. Business.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。