八年级下试题—2016_2017学年广东广州越秀区广州大学附属中学初二下学期六校联考期末物理试卷

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2016-2017学年广东省广州市越秀区八年级(下)期末英语试卷

2016-2017学年广东省广州市越秀区八年级(下)期末英语试卷

2016-2017学年广东省广州市越秀区八年级(下)期末英语试卷一、语法选择(共1小题;每小题15分,满分15分)1.(15分)阅读下面短文,按照句子结构的语法性和上下文连贯的要求,从1~15各题所给的A、B、C和D项中选出最佳选项,并在答题卡上将该项涂黑.There would be a history exam and Phillip looked more and more worried.He (1)the subject important since they began to study it this term.In class he always (2)about the football matches (3)Class Three and Class Two and tried to find out a way (4)Class Two.Of course he didn't hear (5)Mr.Beckman said.His friend Sally could help him with his homework and hint how he answered Mr.Beckman.But bad luck!The girl was ill in hospital and (6)would help him next Monday!Phillip thought hard and at last he found a way.He copied all the main points on his wristbands (袖口)of his shirt and then went to play football with his friends.(7)was Sunday yesterday.Even he didn't open his students'book,he was sure he would pass the exam.So he went on (8)football and watching TV.But this morning,when the boy was going to dress(9),he couldn't find his shirt.He jumped(10)the bed and began to look for it everywhere,(11)he failed.His mother brought his breakfast in and found his bedroom was in a fearful mess.She asked in(12),"What's the matter,son?""I can't find my shirt!""You wore it for a week and I washed it when you fell (13)last night." "Who let you(14)it?"Phillip shouted (15).1.A.doesn't think B.didn't think C.hasn't thought D.hadn't thought 2.A.think B.thought C.thinking D.thinks 3.A.about B.inside C.among D.between 4.A.to beat B.beating C.beat D.beats5.A.when B.how C.what D.who 6.A.anybody B.nobody C.everybody D.somebody 7.A.It B.This C.That D.There 8.A.play B.plays C.to play D.playing 9.A.it B.him C.himself D.them 10.A.into B.off C.up D.inside 11.A.when B.and C.so D.but 12.A.surprise B.surprised C.surprising D.more surprised 13.A.sleep B.asleep C.sleeping D.sleepless 14.A.wash B.to wash C.washed D.washing 15.A.angry B.angrier C.angrily D.more angrily.二、完形填空(共1小题;每小题1.5分,满分1.5分)16.(15分)阅读下面短文,掌握其大意,然后从16~25各题所给的A、B、C 和D项中,选出最佳选项,并在答题卡上将该项涂黑.I was cute,active and happy until that day when I was 15months old.I had a bad fall.I fell on a glass rabbit and it cut my eye badly enough to blind it.The doctor said that if the eye was (16),my face would become out of shape,so my scarred (有疤痕的),blind and grey eye lived on with me.People often asked uncomfortable questions about my eye or (17)played jokes on me.When the kids played games,I was always the"(18)guy."I knew that everyone looked down on me.Yet every time Mom noticed I was unhappy,she would say to me:"Hold your head up high and face the world."It helped me move on in life.As a (19),I thought Mom meant:"Be careful or you will fall down or knock into something because you are not looking."As a teenager,I always looked down to hide my shame,but I found that people (20)me when sometimes I held my head up high.In high school I was even made monitor,but outside of school I still felt like a freak(怪物).All I really wanted was to look like everyone else.When things got really bad,I would cry to my mom and she would look at me with loving eyes and (21),"Hold your head up high and face the world.Let them see the (22)inside you."That is how Mom taught me pride.Mom's love and encouragement gave me the confidence to remove my own doubt.(23),I was able to look people in the eye.I learned not only to appreciate (欣赏)myself but to have deep love for others."Hold your head up high"has been (24)many times in my home.Each of my children has felt its (25).The gift she gave me lives on in my children.16.A.cared for B.taken care C.taken out D.brought back 17.A.still B.even C.almost D.hardly18.A.bad B.good C.kind D.active19.A.student B.person C.pioneer D.child20.A.helped B.liked C.enjoyed D.hated21.A.cry B.shout C.say D.speak22.A.face B.fear C.anger D.love23.A.Luckily B.Really C.Finally D.Suddenly 24.A.shouted B.repeated C.recorded D.discussed 25.A.power B.promise C.energy D.secret.三、阅读(共5小题;满分40分)第一节阅读理解(共5小题;每小题10分,满分40分)26.(10分)阅读下列短文,从每题所给的A、B、C和D项中,选出最佳选项并在答题卡上将该项涂黑.In the past ten years,many scientists have studied the differences between men and women.And they all got the same answer.The sexes are different,because their brains are different.And this,the scientists say,makes men and women see the world in different ways.Boys,for example,generally are better than girls at mathematical (数学的)ideas.Boys are also better at the kind of hand and eye movements necessary for ball sports.Girls,on the other hand generally start speaking earlier than boys.And they generally see better in the dark than boys and are better at learning foreign languages.What makes men and women better at one thing or another?The answer is the brain.The brain has two sides connected by nerve tracks.The left side generally is used for maths,speech and writing.The right side is used for artistic creation and the expression for emotions (情感).Girls develop the right side of the brain faster than boys,while boys develop the left side faster.Another interesting difference is that the two sides of a man's brain are connected by a smaller number of nerves (神经)than the two sides of a woman's brain are.26.Which of the following tells the main idea of this passage?A.Research on the BrainB.Who Are Better,Boys or GirlsC.People's Different BrainsD.Differences Between Men and Women27.make(s)men and women think differently.A.The different waysB.The different brainsC.The different experiencesD.The influences (影响)of the world28.According to the article,girls are generally better than boys at learning.A.mathsB.physicsC.EnglishD.chemistry29.Which of the following is TRUE according to the passage?A.Men are better than women in all things.B.The area in girls used for language skills develops better.C.The left side of brain in girls develops faster.D.Girls can see better in the dark because they are better at eye movements.30.The number of nerves connecting the two sides of the brain in men women.A.is different fromB.is the same asC.is larger thanD.is twice as large as.31.(10分)Computer programmer David Jones gets 35,000a year designing new computer games.Yet he cannot find a bank ready to let him have a credit card (信用卡).Instead he has been told to wait another two years until he is 18.The 16﹣year﹣old boy works for a small company in Liverpool where the problem of most young people of his age is finding a job.David's company releases (推出)two new games for the fast growing computer market each month.But David's biggest headache is what to do with his money.Even though he makes a lot money,he cannot drive a car,take out a mortgage (抵押贷款),or get credit cards,David got his job with the Liverpool﹣based company four months ago,a year after leaving school with six O﹣levels and working for a time in a computer shop."I got the job because they knew I had already written some programs."David said he spends some of his money on records and clothes and gives his mother 50pounds a week.But most of his free time is spent working."Unfortunately,computing was not part of our studies at school"he said."But I had been studying it in books and magazines for four years in my free time.I knew what I wanted to do and never thought about staying on at school.Most people in this business are very young,anyway."David added,"I would like to get a million and I may retire (退休)early.You never know when the market might disappear." 31.How is David different from people of his age?A.He often goes out with friends.B.He lives with his mother.C.He can make a lot of money.D.He graduated with six O﹣levels.32.Which of the following problems does David need to solve?A.He has a lot of free time.B.He will soon lose his job.C.He has no time to learn driving.D.He is too young to get a credit card.33.Why was David able to get the job in the company?A.He had done well in all his exams.B.He had written some computer programs.C.He was good at playing computer games.D.He had learnt to use computers at school.34.He left school because.A.he did not enjoy schoolB.he wanted to make a lot of moneyC.he was afraid of getting too young to start computingD.he wanted to work with computers and staying at school did not help him 35.Why does David think he may retire early?A.Because you have to be young to write computer programs.B.Because he wants to stop working when he has a million dollars.C.Because he thinks some day computer market may disappear.D.Because he thinks his company might be closed soon.36.(6分)There was once a man called Mr.Flowers,and flowers were his only joy in life.He spent all his free time in one of his four glass﹣houses and grew flowers of every color,with long and.He tried to grow a rose of a new color to win the silver cup Mr.Flowers did his best to be in or near his glass﹣houses at the beginning and end of the school day.But it was not always possible to be on watch (看守)at those times.Mr.Flowers had tried many ways to protect his glass﹣houses,but nothing had been useful.Hehad been to school to report to the teachers;but this had not done any good.He had tried to drive away the boys that threw stones into his garden;but the boys could run faster than he could,and they laughed at him from far away.He had even picked up all the stones around his garden,so that the boys would have nothing to throw;but they soon found others.At last Mr.Flowers had a good idea.He put up a large notice made of good,strong wood,some meters away from the glass﹣houses.On it he had written the words:DO NOT THROW STONES AT THIS NOTICE.After this,Mr.Flowers had no further trouble;the boys were much more tempted to throw stones at the notice than at the glass﹣houses.36.Mr.Flowers'hoped to.A.build glass﹣houses in his free timeB.grow the Rose of the Year in a silver cupC.win a silver cup for growing a rose of a new colorD.grow a rose with the longest name37.Boys were often tempted to.A.throw stones at Mr.Flowers'glass﹣housesB.throw stones at Mr.Flowers'from their schoolC.be in or near Mr.Flowers'roses in the glass﹣housesD.play with Mr.Flowers near his glass﹣houses38.Mr.Flowers stayed in or near his glass﹣houses.A.in all his free timeB.at the weekendsC.when he could not see the boysD.at the beginning and the end of the school day.41.(10分)with modern furniture (家具).Near main railway station.Buses pass front door.Phone 332087.Mr.Brown wants to live near his workplace.Mrs.Brown wants a house with four bedrooms.The Brown children don't care what the house is like as long as (只要)the garden is big enough.Ann,their daughter has always wanted to work in a restaurant.She thinks it is an exciting place.It's hard work but she isn't lazy.You have to work on Saturdays and Sundays,but you get other days off.It's not very well﹣paid.But who cares about money?Tom is a middle school student.He is twelve years old.He stops school at half past three,so he has plenty of time to finish his homework.Jack,Mr.Brown's brother,is looking for a room to rent.He doesn't care what colour the walls are or how old the furniture is.He is getting ready for his drawing.So the house must be quiet.41.Why didn't the Browns buy the house in the Advertisement 1?A.It was on the north side of town.B.There weren't enough bedroom.C.There was no dinning﹣room.D.There was a garden in the house.42.Why does Jimmy want to get the job in the restaurant?A.It's an easy job.B.The place is good and the food is free.C.He thinks it will be exciting.D.He can get long holidays.43.Why can't Tom play for the new football team?A.He's not old enough.B.It's in a different village.C.School stops too late.D.He has to do his homework.44.What's wrong with the room for Jack in Advertisement 4?A.It's too small.B.The walls are the wrong color.C.The man likes old furniture better.D.It's too noisy.45.What does the underlined sentence"All meals free."Mean?A.Waiters in the restaurant can have the meals at any time.B.Waiters can have the meals in the restaurant without paying money.C.Customers can have the meals without paying.D.Customers can eat anywhere at any time.46.(4分)阅读短文及文后选项,选出可以填入空白处的最佳选项,把答案写在答题卷相应的空格中.At the shopping mall,you probably think about spending,not saving.(46)Think about all the water faucets (水龙头)in the toilet.Think about all the air conditioning in summer.Saving energy is a hot topic for such a big place!(47)Electric lights,infrared (感应的)faucets and air﹣conditioners need electricity to work.(48)More and more dust and dirt make the lights less bright.When malls keep lights clean,they can use fewer lights and keep the area bright.(49)A.Many things at the mall use energy.B.Using fewer lights means using less energy.C.Sometimes people forget to turn the water off after they wash.D.Think about all the lights that keep the mall bright.E.One way to save energy is by keeping lights clean.四、写作(共三节,满分30分)根据下列句子及所给单词的首字母写出所缺单词.在填写答卷时,要求写出完整单词.(每空限填一词)50.(1分)Forever Green Houses are so good that I will c recommend them to my friends.51.(1分)I never feel l because I've made many good friends here.52.(1分)Benny had a t dream last night.It has made him frightened.53.(1分)She is still angry and refuses to a my apology.54.(1分)A to my mum's idea,keeping pets helps us become more responsible people.55.(1分)She keeps doing sports everyday,so she has lost a lot of w recently.56.(1分)如果我们还不采取行动,这个世界上将不再有大熊猫.If we do nothing,there will be giant pandas in the world.57.(1分)我考虑了一下,最后选定了那条红裙子.I thought it over and finally that red skirt.58.(1分)没有什么可以阻止我们拯救濒危动物.Nothing can the endangered animals.59.(1分)为了满足每一位客人的要求,酒店不断改善自己的服务质量.every guest,the hotel keeps improving its service.60.(1分)他跑得飞快,足以抓住那小偷.He ran to catch the thief.61.(2分)杂志不能被带出阅览室.The magazines out of the reading room.62.(2分)当我开着车时,我看到了那场事故.While I my car,I saw the accident.63.(15分)假如你有两位同学﹣﹣芳芳和明明,参加了上个星期二学校组织的英语演讲比赛,他们的不同表现使你深有感触.请你根据下列表格中的内容提示写一篇英语短文.参考词汇:英语演讲比赛English Speech Contest要求:1.词数:80~100个.2.短文开头已给出,不计入总词数.3.要点齐全,行文连贯,可适当发挥.Last Tuesday Fangfang and Mingming entered the English Speech Contest in our school.What they did made me think deeply.Before the contest.Fangfang thought .2016-2017学年广东省广州市越秀区八年级(下)期末英语试卷参考答案与试题解析一、语法选择(共1小题;每小题15分,满分15分)1.(15分)阅读下面短文,按照句子结构的语法性和上下文连贯的要求,从1~15各题所给的A、B、C和D项中选出最佳选项,并在答题卡上将该项涂黑.There would be a history exam and Phillip looked more and more worried.He (1)C the subject important since they began to study it this term.In class he always (2)B about the football matches (3)D Class Three and Class Two and tried to find out a way (4)A Class Two.Of course he didn't hear (5)C Mr.Beckman said.His friend Sally could help him with his homework and hint how he answered Mr.Beckman.But bad luck!The girl was ill in hospital and (6)B would help him next Monday!Phillip thought hard and at last he found a way.He copied all the main points on his wristbands (袖口)of his shirt and then went to play football with his friends.(7)A was Sunday yesterday.Even he didn't open his students'book,he was sure he would pass the exam.So he went on (8)D football and watching TV.But this morning,when the boy was going to dress(9)C,he couldn't find his shirt.He jumped(10)B the bed and began to look for it everywhere,(11)D he failed.His mother brought his breakfast in and found his bedroom was in a fearful mess.She asked in(12)A,"What's the matter,son?""I can't find my shirt!""You wore it for a week and I washed it when you fell (13)B last night." "Who let you(14)A it?"Phillip shouted (15)C.1.A.doesn't think B.didn't think C.hasn't thought D.hadn't thought 2.A.think B.thought C.thinking D.thinks 3.A.about B.inside C.among D.between4.A.to beat B.beating C.beat D.beats 5.A.when B.how C.what D.who 6.A.anybody B.nobody C.everybody D.somebody 7.A.It B.This C.That D.There 8.A.play B.plays C.to play D.playing 9.A.it B.him C.himself D.them 10.A.into B.off C.up D.inside 11.A.when B.and C.so D.but 12.A.surprise B.surprised C.surprising D.more surprised 13.A.sleep B.asleep C.sleeping D.sleepless 14.A.wash B.to wash C.washed D.washing 15.A.angry B.angrier C.angrily D.more angrily.【解答】1.C 考查动词,A.doesn't think一般现在时B.didn't think一般过去时C.hasn't thought现在完成时D.hadn't thought过去完成时,根据since they began to study it this term,可知since引导的句子通常用于现在完成时,故选C.2.B 考查动词,A.think思考,原形B.thought过去式C.thinking动名词D.thinks 三单式,根据后半句中的动词形式,可知是一般过去时,故选B.3.D 考查介词,A.about关于B.inside里面的C.among在…之中(两者以上)D.between 在…和…之间(两者之间),根据Class Three and Class Two ,可知是短语between…and…在…和…之间,这是固定,故选D.4.A 考查动词,A.to beat击打,不定式B.beating动名词C.beat原形D.beats 三单式,根据tried to find out a way,可知此处是不定式表目的,努力找到打败二班的方法,故选A.5.C 考查代词,A.when什么时候B.how如何C.what什么D.who谁,根据Mr.Beckman said,可知是句型what Mr.Beckman said贝克曼先生所说的,故选C.6.B 考查代词,A.anybody任何人B.nobody没人C.everybody每个人D.somebody 某人,根据The girl was ill in hospital,可知这个女孩生病住院了,所以没人帮助他了,故选B.7.A 考查代词,A.It它B.This这个C.That那个D.There那儿,根据was Sunday yesterday,可知是指时间,时间用it作主语,故选A.8.D 考查动词,A.play玩,动词原形B.plays三单式C.to play不定式D.playing 动名词,根据went on ,可知介词后接动名词,故选D.9.C 考查代词,A.it它B.him他C.himself他自己D.them他们,根据dress,可知是短语dress oneself给自己穿衣服,故选C.10.B 考查介词,A.into进入B.off离开C.up往上D.inside里面的,根据the bed and began to look for it everywhere,可知是从床上跳下来到处寻找衬衣,jump off从…跳下来,故选B.11.D 考查连词,A.when什么时候B.and和C.so因此D.but但是,根据he failed,可知这跟前文是转折关系,故选D.12.A 考查形容词,A.surprise惊讶B.surprised感到惊讶的,修饰人C.surprising 惊讶的,修饰物D.more surprised更加惊讶,根据She asked in,可知是指她惊讶的问,in surprise惊讶的,故选A.13.B 考查形容词,A.sleep睡觉,动词B.asleep睡着的C.sleeping昏昏欲睡的D.sleepless失眠的,根据fell ,结合选项,可知是短语fall asleep入睡,这是固定搭配,故选B.14.A 考查动词,A.wash洗,动词原形B.to wash不定式C.washed过去式D.washing动名词,根据let you,可知是句型let sb do sth让某人做某事,故选A.15.C 看副词,A.angry生气的B.angrier更生气C.angrily生气地,副词D.more angrily更加生气地,根据shouted ,可知此处是副词修饰动词shouted,此处是原级,故选C.二、完形填空(共1小题;每小题1.5分,满分1.5分)16.(15分)阅读下面短文,掌握其大意,然后从16~25各题所给的A、B、C 和D项中,选出最佳选项,并在答题卡上将该项涂黑.I was cute,active and happy until that day when I was 15months old.I had a bad fall.I fell on a glass rabbit and it cut my eye badly enough to blind it.The doctor saidthat if the eye was (16)C,my face would become out of shape,so my scarred (有疤痕的),blind and grey eye lived on with me.People often asked uncomfortable questions about my eye or (17)B played jokes on me.When the kids played games,I was always the"(18)A guy."I knew that everyone looked down on me.Yet every time Mom noticed I was unhappy,she would say to me:"Hold your head up high and face the world."It helped me move on in life.As a (19)D,I thought Mom meant:"Be careful or you will fall down or knock into something because you are not looking."As a teenager,I always looked down to hide my shame,but I found that people (20)B me when sometimes I held my head up high.In high school I was even made monitor,but outside of school I still felt like a freak (怪物).All I really wanted was to look like everyone else.When things got really bad,I would cry to my mom and she would look at me with loving eyes and (21)C,"Hold your head up high and face the world.Let them see the (22)D inside you."That is how Mom taught me pride.Mom's love and encouragement gave me the confidence to remove my own doubt.(23)C,I was able to look people in the eye.I learned not only to appreciate (欣赏)myself but to have deep love for others."Hold your head up high"has been (24)B many times in my home.Each of my children has felt its (25)A.The gift she gave me lives on in my children.16.A.cared for B.taken care C.taken out D.brought back 17.A.still B.even C.almost D.hardly18.A.bad B.good C.kind D.active19.A.student B.person C.pioneer D.child20.A.helped B.liked C.enjoyed D.hated21.A.cry B.shout C.say D.speak22.A.face B.fear C.anger D.love23.A.Luckily B.Really C.Finally D.Suddenly24.A.shouted B.repeated C.recorded D.discussed 25.A.power B.promise C.energy D.secret.【解答】16:C 考查短语,A.cared for关心B.taken care关心C.taken out 取出D.brought back拿回来,根据my face would become out of shape,so my scarred (有疤痕的),blind and grey eye lived on with me.可知眼睛被取出,我的脸会变形,我的眼睛会瞎,故选C.17:B 考查副词.A仍然;B甚至;C几乎;D几乎不;根据前面句子People often asked uncomfortable questions about my eye人们经常问我不舒服的问题,后面应该是or(18)played jokes on me,人们甚至开我玩笑故选择B.18:A 考查形容词.A糟糕的;B好的,C友好的;D活跃的;根据前面眼睛摘了,看起来会很凶狠,所以后面玩游戏时I was always the"(19)guy,可知我总是坏家伙故选择A.19.D 考查名词,A.student学生B.person人C.pioneer先锋D.child孩子,根据后文As a teenager,可知我是个孩子,故选D.20:B 考查动词.A帮助;B喜欢;C喜欢;D讨厌;根据前面说I always looked down to hide my shame(自卑)我总是低头掩饰我的羞愧,but I found that people (22)me when sometimes I held my head up high.应该是但我发现人们喜欢我,有时我高昂着头,故选择B.21.C 考查动词,A.cry哭B.shout大喊C.say说D.speak讲话,根据后文,"Hold your head up high and face the world…",可知这是陈述的内容,故选C.22.D 考查名词,A.face脸B.fear害怕C.anger愤怒D.love 爱,根据That is how Mom taught me pride,可知是指内心的爱,故选D.23.C 考查副词,A.Luckily幸运地B.Really真正地C.Finally最终地D.Suddenly 突然地,根据I was able to look people in the eye,前面说我妈妈鼓励我,经过她的鼓励,有了自信,后面应该是最后能直视人们的眼睛,故选C.24:B 考查动词.A大喊;B重复;C广播;D讨论;根据前面他妈妈鼓励他说Hold your head up high and face the world,所以这里应该是抬起你的头,在我家里已经重复了很多次,故选B.25:A 考查名词.A权力,能量,B许诺;C精神,活力;D秘密;根据后面说The gift she gave me lives on in my children.她给我的礼物继续用于我的孩子们,前面Each of my children has felt its 应该是我的每一个孩子都感受到了它的力量,故选A.三、阅读(共5小题;满分40分)第一节阅读理解(共5小题;每小题10分,满分40分)26.(10分)阅读下列短文,从每题所给的A、B、C和D项中,选出最佳选项并在答题卡上将该项涂黑.In the past ten years,many scientists have studied the differences between men and women.And they all got the same answer.The sexes are different,because their brains are different.And this,the scientists say,makes men and women see the world in different ways.Boys,for example,generally are better than girls at mathematical (数学的)ideas.Boys are also better at the kind of hand and eye movements necessary for ball sports.Girls,on the other hand generally start speaking earlier than boys.And they generally see better in the dark than boys and are better at learning foreign languages.What makes men and women better at one thing or another?The answer is the brain.The brain has two sides connected by nerve tracks.The left side generally is used for maths,speech and writing.The right side is used for artistic creation and the expression for emotions (情感).Girls develop the right side of the brain faster than boys,while boys develop the left side faster.Another interesting difference is that the two sides of a man's brain are connected by a smaller number of nerves (神经)than the two sides of a woman's brain are.26.Which of the following tells the main idea of this passage?DA.Research on the BrainB.Who Are Better,Boys or GirlsC.People's Different BrainsD.Differences Between Men and Women27.B make(s)men and women think differently.A.The different waysB.The different brainsC.The different experiencesD.The influences (影响)of the world28.According to the article,girls are generally better than boys at learning C.A.mathsB.physicsC.EnglishD.chemistry29.Which of the following is TRUE according to the passage?BA.Men are better than women in all things.B.The area in girls used for language skills develops better.C.The left side of brain in girls develops faster.D.Girls can see better in the dark because they are better at eye movements.30.The number of nerves connecting the two sides of the brain in men A women.A.is different fromB.is the same asC.is larger thanD.is twice as large as.【解答】26.D.主旨大意题.根据语境:In the past ten years,many scientists have studied the differences between men and women.可知本文主要讲了男人和女人的不同之处.故选D.27.B.细节理解题.根据语境:The sexes are different,because their brains are different.And this,the scientists say,makes men and women see the world in different ways.可知因为男女的大脑结构不同,所以他们的想法也不同.故选B.28.C.细节理解题.根据语境:And they generally see better in the dark than boys and are better at learning foreign languages.可知女孩比男孩更擅长学外语,如英语等.故选C.29.B.是非判断题.根据语境:And they generally see better in the dark than boys and are better at learning foreign languages.可知女孩在语言领域发展的比男孩更好.故选B.30.A.推理判断题.根据语境:Another interesting difference is that the two sides of a man's brain are connected by a smaller number of nerves (神经)than the two sides of a woman's brain are.可知男女大脑两侧的神经数不同,故选A.31.(10分)Computer programmer David Jones gets 35,000a year designing new computer games.Yet he cannot find a bank ready to let him have a credit card (信用卡).Instead he has been told to wait another two years until he is 18.The 16﹣year﹣old boy works for a small company in Liverpool where the problem of most young people of his age is finding a job.David's company releases (推出)two new games for the fast growing computer market each month.But David's biggest headache is what to do with his money.Even though he makes a lot money,he cannot drive a car,take out a mortgage (抵押贷款),or get credit cards,David got his job with the Liverpool﹣based company four months ago,a year after leaving school with six O﹣levels and working for a time in a computer shop."I got the job because they knew I had already written some programs."David said he spends some of his money on records and clothes and gives his mother 50pounds a week.But most of his free time is spent working."Unfortunately,computing was not part of our studies at school"he said."But I had been studying it in books and magazines for four years in my free time.I knew what I wanted to do and never thought about staying on at school.Most people in this business are very young,anyway."David added,"I would like to get a million and I may retire (退休)early.You never know when the market might disappear." 31.How is David different from people of his age?CA.He often goes out with friends.B.He lives with his mother.C.He can make a lot of money.D.He graduated with six O﹣levels.32.Which of the following problems does David need to solve?DA.He has a lot of free time.B.He will soon lose his job.C.He has no time to learn driving.D.He is too young to get a credit card.33.Why was David able to get the job in the company?BA.He had done well in all his exams.B.He had written some computer programs.C.He was good at playing computer games.D.He had learnt to use computers at school.34.He left school because D.A.he did not enjoy schoolB.he wanted to make a lot of moneyC.he was afraid of getting too young to start computingD.he wanted to work with computers and staying at school did not help him 35.Why does David think he may retire early?CA.Because you have to be young to write computer programs.B.Because he wants to stop working when he has a million dollars.C.Because he thinks some day computer market may disappear.D.Because he thinks his company might be closed soon.【解答】31.答案:C.考查细节理解题,由第一段首句可知,大卫设计了一种新的电脑游戏而获得$35.000.故此可推断在这个年龄段,他可以挣很多钱.故选:C32.答案:D.考查细节理解题,根据But David's biggest headache is what to do with his money.Even though he makes a lot money,he cannot drive a car,take out a mortgage (抵押贷款),or get credit cards.可知:,由于他没有达到法定年龄,没有资格申请信用卡.故选:D.33.答案:B.考查细节理解题,根据I got the job because they knew I had already written some programs.可知:David写了一些。

广州大学附属中学2016-2017学年第二学期期末考试(六校联考) 初二年级 物理 试卷及参考答案

广州大学附属中学2016-2017学年第二学期期末考试(六校联考) 初二年级 物理 试卷及参考答案

力的作用,该力的方向是

(2)“玉兔”月球车的轮子做得宽大,能
才不至于陷入松软的月壤中,而不使用充气轮胎
是因为月球表面大气压较
(选填“大”或“小”),充气轮胎容易爆胎.
(3)宇航员向前抛出一个小球,小球离开手后仍能继续向前运动,这是由于小球具有

13 2016~2017学年广东广州越秀区广州大学附属中学初二下学期期末第14题
16 2016~2017学年广东广州越秀区广州大学附属中学初二下学期期末第17题 为了鉴别两个烧杯中哪个装的是清水哪杯是盐水.
(1) 小明将压强计的金属盒先后浸入到两杯液体中,如图甲和乙所示.请指出他实验操作不妥
之处:

(2) 小红用一长方形小木块,先后放入两杯液体中,小木块静止时的情况如图丙和丁.
,则吸盘对玻璃的压力是

18 2016~2017学年广东广州越秀区广州大学附属中学初二下学期期末第19题
( )托里拆利测量大气压强值实验如图甲所示,当时的大气压强等于
高水银柱所产
生的压强.
( )若在实验过程中不小心有气泡进入玻璃管内,则管内外水银液面高度差将
(选填“变
大”、“变小”或“不变”).
一个西瓜慢慢放入装满水的桶中时,排出 的水,最后漂浮如图所示.( 取



(1) 求西瓜漂浮时排开水的体积. (2) 求西瓜的重力. (3) 从水中缓慢托起西瓜,直到它离开水面的过程中,小明感到:竖直向上用的力越来越大.
请用学过的知识解释这种现象.
四、实验探究题
21 2016~2017学年广东广州越秀区广州大学附属中学初二下学期期末第22题 实验探究题 (1) 如图甲是小华同学探究二力平衡条件时的实验情景.

越秀区2016-2017初二下英语期末试卷及答案

越秀区2016-2017初二下英语期末试卷及答案
2016~2017学年广东广州越秀区初二下学期期末英语试卷
语法选择
1. There would be a history exam and Phillip looked more and more worried. He since they began to study it this term. In class he always and Class Two and tried to find out a way 4 2 1 the subject important 3 Class Three
last night."
D. hadn't thought D. thinks D. between D. beats D. who D. somebody D. There D. playing D. them D. inside D. but D. more surprised D. sleepless D. washing D. more angrily
完形填空
2. I was cute, active and happy until that day when I was 15 months old. I had a bad fall. I fell on a glass rabbit and it cut my eye badly enough to blind it. The doctor said that if the eye was out of shape, so my scarred (有疤痕的), blind and grey eye lived on with me. People often asked uncomfortable questions about my eye or played games, I was always the " 3 2 played jokes on me. When the kids 1 , my face would become

18八下春季数学课程第十八讲 广州市越秀区2016-2017学年八年级(下)期末数学试卷

18八下春季数学课程第十八讲  广州市越秀区2016-2017学年八年级(下)期末数学试卷

2016-2017学年广东省广州市越秀区八年级(下)期末数学试卷一、选择题(共10小题,每小题3分,满分30分)1.下列式子没有意义的是()A.B. C. D.2.下列计算中,正确的是()A.÷=B.(4)2=8 C. =2 D.2×2=23.刻画一组数据波动大小的统计量是()A.平均数B.方差 C.众数 D.中位数4.在暑假到来之前,某机构向八年级学生推荐了A,B,C三条游学线路,现对全级学生喜欢哪一条游学线路作调查,以决定最终的游学线路,下面的统计量中最值得关注的是()A.方差 B.平均数C.中位数D.众数5.关于正比例函数y=﹣2x,下列结论中正确的是()A.函数图象经过点(﹣2,1) B.y随x的增大而减小C.函数图象经过第一、三象限 D.不论x取何值,总有y<06.以下列各组数据中的三个数作为三角形的边长,其中能构成直角三角形的是()A.2,3,4 B.,,C.1,,2 D.7,8,97.若一个直角三角形的一条直角边长是5cm,另一条直角边比斜边短1cm,则斜边长为()cm.A.10 B.11 C.12 D.138.如图,在菱形ABCD中,对角线AC,BD相交于点O,AB=5,AC=6,则菱形ABCD的面积是()A.24 B.26 C.30 D.489.在下列命题中,是假命题的是()A.有一个角是直角的平行四边形是矩形B.一组邻边相等的矩形是正方形C.一组对边平行且相等的四边形是平行四边形D.有两组邻边相等的四边形是菱形10.已知平面上四点A(0,0),B(10,0),C(12,6),D(2,6),直线y=mx﹣3m+6将四边形ABCD分成面积相等的两部分,则m的值为()A.B.﹣1 C.2 D.二、填空题:本大题共6小题,每小题3分,满分18分.11.已知a=+2,b=﹣2,则ab= .12.一次函数y=kx+b(k≠0)中,x与y的部分对应值如下表:x ﹣2 ﹣1 0 1 2y ﹣6 ﹣4 ﹣2 0 2那么,一元一次方程kx+b=0的解是x= .13.如图是一次函数y=mx+n的图象,则关于x的不等式mx+n>2的解集是.14.一组数据:2017、2017、2017、2017、2017,它的方差是.15.考古学家们发现了几块大约完成于公元前2000年左右的古巴比伦的泥版书,据专家们考证,其中一块上面刻有如下问题:“一根长度为30个单位的棍子直立在墙上,当其上端垂直滑下6个单位时,请问其下端离开墙角有多远?”,这个问题的答案是:其下端离开墙角个单位.16.如图所示,在Rt△ABC中,∠A=90°,DE∥BC,F,G,H,I分别是DE,BE,BC,CD 的中点,连接FG,GH,HI,IF,FH,GI.对于下列结论:①∠GFI=90°;②GH=GI;③GI=(BC﹣DE);④四边形FGHI是正方形.其中正确的是(请写出所有正确结论的序号).三、解答题:本大题共9小题,满分72分.解答须写出文字说明、推理过程和演算步骤.17.计算:( +﹣)×.18.如图,在△ABC中,AD⊥BC,AB=5,BD=4,CD=.(1)求AD的长.(2)求△ABC的周长.19.如图在平行四边形ABCD中,AC交BD于点O,AE⊥BD,CF⊥BD,垂足分别为E、F,求证:四边形AECF为平行四边形.20.下表是某校八年级(1)班43名学生右眼视力的检查结果.视力 4.0 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 5.0人数 1 2 5 4 3 5 1 1 5 10 6(1)该班学生右眼视力的平均数是(结果保留1位小数).(2)该班学生右眼视力的中位数是.(3)该班小鸣同学右眼视力是4.5,能不能说小鸣同学的右眼视力处于全班同学的中上水平?试说明理由.21.如图,正方形ABCD的对角线相交于点O,BC=6,延长BC至点E,使得CE=8,点F是DE的中点,连接CF、OF.(1)求OF的长.(2)求CF的长.22.如图,在平面直角坐标系中,直线y=kx+b经过点A(﹣30,0)和点B(0,15),直线y=x+5与直线y=kx+b相交于点P,与y轴交于点C.(1)求直线y=kx+b的解析式.(2)求△PBC的面积.23.2016年下半年开始,不同品牌的共享单车出现在城市的大街小巷.现已知A品牌共享单车计费方式为:初始骑行单价为1元/半小时,不足半小时按半小时计算.内设邀请机制,每邀请一位好友注册认证并充值押金成功,双方骑行单价均降价0.1元/半小时,骑行单价最低可降至0.1元/半小时(比如,某用户邀请了3位好友,则骑行单价为0.7元/半小时).B 品牌共享单车计费方式为:0.5元/半小时,不足半小时按半小时计算.(1)某用户准备选择A品牌共享单车使用,设该用户邀请好友x名(x为整数,x≥0),该用户的骑行单价为y元/半小时.请写出y关于x的函数解析式.(2)若有A,B两种品牌的共享单车各一辆供某用户一人选择使用,请你根据该用户已邀请好友的人数,给出经济实惠的选择建议.24.下面我们做一次折叠活动:第一步,在一张宽为2的矩形纸片的一端,利用图(1)的方法折出一个正方形,然后把纸片展平,折痕为MC;第二步,如图(2),把这个正方形折成两个相等的矩形,再把纸片展平,折痕为FA;第三步,折出内侧矩形FACB的对角线AB,并将AB折到图(3)中所示的AD处,折痕为AQ.根据以上的操作过程,完成下列问题:(1)求CD的长.(2)请判断四边形ABQD的形状,并说明你的理由.25.如图,正方形ABCD中,AB=4,P是CD边上的动点(P点不与C、D重合),过点P作直线与BC的延长线交于点E,与AD交于点F,且CP=CE,连接DE、BP、BF,设CP═x,△PBF 的面积为S1,△PDE的面积为S2.(1)求证:BP⊥DE.(2)求S1﹣S2关于x的函数解析式,并写出x的取值范围.(3)分别求当∠PBF=30°和∠PBF=45°时,S1﹣S2的值.2016-2017学年广东省广州市越秀区八年级(下)期末数学试卷参考答案与试题解析一、选择题(共10小题,每小题3分,满分30分)1.下列式子没有意义的是()A.B. C. D.【考点】72:二次根式有意义的条件.【分析】根据二次根式中的被开方数是非负数进行分析即可.【解答】解:A、有意义,故此选项不合题意;B、没有意义,故此选项符合题意;C、有意义,故此选项不合题意;D、有意义,故此选项不合题意;故选:B.2.下列计算中,正确的是()A.÷=B.(4)2=8 C. =2 D.2×2=2【考点】75:二次根式的乘除法;73:二次根式的性质与化简.【分析】各项计算得到结果,即可作出判断.【解答】解:A、原式===3,不符合题意;B、原式=32,不符合题意;C、原式=|﹣2|=2,符合题意;D、原式=4,不符合题意;故选C.3.刻画一组数据波动大小的统计量是()A.平均数B.方差 C.众数 D.中位数【考点】WA:统计量的选择.【分析】根据方差的意义:体现数据的稳定性,集中程度,波动性大小;方差越小,数据越稳定.【解答】解:由于方差反映数据的波动情况,衡量一组数据波动大小的统计量是方差.故选B.4.在暑假到来之前,某机构向八年级学生推荐了A,B,C三条游学线路,现对全级学生喜欢哪一条游学线路作调查,以决定最终的游学线路,下面的统计量中最值得关注的是()A.方差 B.平均数C.中位数D.众数【考点】WA:统计量的选择.【分析】全级学生喜欢哪一条游学线路最值得关注的应该是喜欢哪条线路的人数最多,即众数.【解答】解:由于众数是数据中出现次数最多的数,故全级学生喜欢的游学线路最值得关注的应该是统计调查数据的众数.故选:D.5.关于正比例函数y=﹣2x,下列结论中正确的是()A.函数图象经过点(﹣2,1) B.y随x的增大而减小C.函数图象经过第一、三象限 D.不论x取何值,总有y<0【考点】F6:正比例函数的性质.【分析】根据正比例函数图象上的坐标特征,正比例函数图象的性质对各选项分析判断后利用排除法求解.【解答】解:A、当x=﹣2时,y=﹣2×(﹣2)=4,即图象经过点(﹣2,4),不经过点(﹣2,1),故本选项错误;B、由于k=﹣2<0,所以y随x的增大而减小,故本选项正确;C、由于k=﹣2<0,所以图象经过二、四象限,故本选项错误;D、∵x>0时,y<0,x<0时,y>0,∴不论x为何值,总有y<0错误,故本选项错误.故选:B.6.以下列各组数据中的三个数作为三角形的边长,其中能构成直角三角形的是()A.2,3,4 B.,,C.1,,2 D.7,8,9【考点】KS:勾股定理的逆定理.【分析】欲判断是否是直角三角形,则需满足较小两边平方的和等于最大边的平方.【解答】解:A、22+32≠42,故不是直角三角形;B、()2+()2≠()2,故不是直角三角形;C、12+()2=22,故是直角三角形;D、72+82≠92,故不是直角三角形;故选C.7.若一个直角三角形的一条直角边长是5cm,另一条直角边比斜边短1cm,则斜边长为()cm.A.10 B.11 C.12 D.13【考点】KQ:勾股定理.【分析】设斜边长为xcm,根据勾股定理列出方程,解方程即可.【解答】解:设斜边长为xcm,则另一条直角边为(x﹣1)cm,由勾股定理得,x2=52+(x﹣1)2,解得,x=13,则斜边长为13cm,故选:D.8.如图,在菱形ABCD中,对角线AC,BD相交于点O,AB=5,AC=6,则菱形ABCD的面积是()A.24 B.26 C.30 D.48【考点】L8:菱形的性质.【分析】根据菱形的对角线互相垂直,利用勾股定理列式求出OB,再根据菱形的对角线互相平分求出AC、BD,然后利用菱形的面积等于对角线乘积的一半列式进行计算即可得解.【解答】解:∵四边形ABCD是菱形,∴OA=OC=3,OB=OD,AC⊥BD,在Rt△AOB中,∠AOB=90°,根据勾股定理,得:OB=,=,=4,∴BD=2OB=8,∴S菱形ABCD=×AC×BD=×6×8=24.故选A.9.在下列命题中,是假命题的是()A.有一个角是直角的平行四边形是矩形B.一组邻边相等的矩形是正方形C.一组对边平行且相等的四边形是平行四边形D.有两组邻边相等的四边形是菱形【考点】O1:命题与定理.【分析】利用矩形、正方形、平行四边形、菱形的判定等知识分别判断后即可确定正确的选项.【解答】解:A、有一个角是直角的平行四边形是矩形,正确;B、一组邻边相等的矩形是正方形,正确;C、一组对边平行且相等的四边形是平行四边形,正确;D、有两组邻边相等且平行的四边形是菱形,错误;故选D.10.已知平面上四点A(0,0),B(10,0),C(12,6),D(2,6),直线y=mx﹣3m+6将四边形ABCD分成面积相等的两部分,则m的值为()A.B.﹣1 C.2 D.【考点】FI:一次函数综合题.【分析】根据点的坐标先判定出四边形ABCD是平行四边形,再根据过平行四边形中心的直线把平行四边形分成面积相等的两部分,求出中心点的坐标,然后代入直线解析式进行计算即可求解.【解答】解:如图,∵A(0,0),B(10,0),C(12,6),D(2,6),∴AB=10﹣0=10,CD=12﹣2=10,又点C、D的纵坐标相同,∴AB∥CD且AB=CD,∴四边形ABCD是平行四边形,∵12÷2=6,6÷2=3,∴对角线交点P的坐标是(6,3),∵直线y=mx﹣3m+6将四边形ABCD分成面积相等的两部分,∴直线y=mx﹣3m+6经过点P,∴6m﹣3m+6=3,解得m=﹣1.故选B.二、填空题:本大题共6小题,每小题3分,满分18分.11.已知a=+2,b=﹣2,则ab= 1 .【考点】76:分母有理化.【分析】将a与b的值代入原式计算即可得到结果.【解答】解:∵a=+2,b=﹣2,∴ab=(+2)(﹣2)=5﹣4=1,故答案为:112.一次函数y=kx+b(k≠0)中,x与y的部分对应值如下表:x ﹣2 ﹣1 0 1 2y ﹣6 ﹣4 ﹣2 0 2那么,一元一次方程kx+b=0的解是x= 1 .【考点】FC:一次函数与一元一次方程.【分析】此题实际上是求当y=0时,所对应的x的值.根据表格求解即可.【解答】解:根据上表中的数据值,当y=0时,x=1,即一元一次方程kx+b=0的解是x=1.故答案是:1.13.如图是一次函数y=mx+n的图象,则关于x的不等式mx+n>2的解集是x>0 .【考点】FD:一次函数与一元一次不等式.【分析】根据一次函数y=mx+n的图象经过点(0,2)以及函数的增减性,即可求出关于x 的不等式mx+n>2的解集.【解答】解:由题意,可知一次函数y=mx+n的图象经过点(0,2),且y随x的增大而增大,所以关于x的不等式mx+n>2的解集是x>0.故答案为x>0.14.一组数据:2017、2017、2017、2017、2017,它的方差是0 .【考点】W7:方差.【分析】根据方差的意义解答.【解答】解:该组数据一样,没有波动,方差为0,故答案为0.15.考古学家们发现了几块大约完成于公元前2000年左右的古巴比伦的泥版书,据专家们考证,其中一块上面刻有如下问题:“一根长度为30个单位的棍子直立在墙上,当其上端垂直滑下6个单位时,请问其下端离开墙角有多远?”,这个问题的答案是:其下端离开墙角18 个单位.【考点】KU:勾股定理的应用.【分析】如图,根据题意PC=AC=30,AC=24,再利用勾股定理求出CB的值即可求出其下端离开墙角有多远.【解答】解:∵PC=AB=30,PA=6,∴AC=24,∴BC===18,∴下端离开墙角18个单位.故答案为18.16.如图所示,在Rt△ABC中,∠A=90°,DE∥BC,F,G,H,I分别是DE,BE,BC,CD 的中点,连接FG,GH,HI,IF,FH,GI.对于下列结论:①∠GFI=90°;②GH=GI;③GI=(BC﹣DE);④四边形FGHI是正方形.其中正确的是①③(请写出所有正确结论的序号).【考点】LN:中点四边形.【分析】①正确.延长IF交AB于K,利用两直线平行同位角相等即可解决问题;②④错误.只要证明四边形FGHI是矩形,即可判断;③正确.延长EI交BC于N,则△DEI≌△CNI,推出DE=CN,EJ=JN,由EG=GB,EI=IN,推出GI=BHN=(BC﹣DE),故③正确;【解答】解:延长IF交AB于K,∵DF=EF,BG=GE,∴FG=BD,GF∥AB,同理IF∥AC,HI=BD,HI∥BD,∴∠BKI=∠A=90°,∴∠GFI=∠BKI=90°,∴GF⊥FI,故①正确,∴FG=HI,FG∥HI,∴四边形FGHI是平行四边形,∵∠GFI=90°,∴四边形FGHI是矩形,故②④错误,延长EI交BC于N,则△DEI≌△CNI,∴DE=CN,EJ=JN,∵EG=GB,EI=IN,∴GI=BHN=(BC﹣DE),故③正确,故答案为①③.三、解答题:本大题共9小题,满分72分.解答须写出文字说明、推理过程和演算步骤.17.计算:( +﹣)×.【考点】79:二次根式的混合运算.【分析】先把二次根式化为最简二次根式,然后把括号内合并后进行二次根式的乘法运算.【解答】解:原式=(6+﹣﹣3)×=×=7.18.如图,在△ABC中,AD⊥BC,AB=5,BD=4,CD=.(1)求AD的长.(2)求△ABC的周长.【考点】KQ:勾股定理.【分析】(1)根据勾股定理求出AD;(2)根据勾股定理求出AC,计算即可.【解答】解:(1)在Rt△ABD中,AD==3;(2)在Rt△ACD中,AC==2,则△ABC的周长=AB+AC+BC=5+4++2=9+3.19.如图在平行四边形ABCD中,AC交BD于点O,AE⊥BD,CF⊥BD,垂足分别为E、F,求证:四边形AECF为平行四边形.【考点】L7:平行四边形的判定与性质.【分析】由四边形ABCD是平行四边形,可得AB=CD,AB∥CD,又由AE⊥BD,CF⊥BD,即可得AE∥CF,∠AEB=∠CFD=90°,然后利用AAS证得△AEB≌△CFD,即可得AE=CF,由有一组对边相等且平行的四边形是平行四边形,即可证得四边形AECF是平行四边形.【解答】证明:∵四边形ABCD是平行四边形,∴AB=CD,AB∥CD,∴∠ABE=∠CDF,∵AE⊥BD,CF⊥BD,∴AE∥CF,∠AEB=∠CFD=90°,在△AEB和△CFD中,∵,∴△AEB≌△CFD(AAS),∴AE=CF,∴四边形AECF是平行四边形.20.下表是某校八年级(1)班43名学生右眼视力的检查结果.视力 4.0 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 5.0人数 1 2 5 4 3 5 1 1 5 10 6(1)该班学生右眼视力的平均数是 4.6 (结果保留1位小数).(2)该班学生右眼视力的中位数是 4.7 .(3)该班小鸣同学右眼视力是4.5,能不能说小鸣同学的右眼视力处于全班同学的中上水平?试说明理由.【考点】W4:中位数;W2:加权平均数.【分析】(1)根据加权平均数的定义求解可得;(2)由中位数的定义知中位数为第22个数据,据此可得;(3)根据小鸣同学右眼视力是4.5,小于中位数4.7,解答可得.【解答】解:(1)该班学生右眼视力的平均数是×(4.0+4.1×2+4.2×5+4.3×4+4.4×3+4.5×5+4.6+4.7+4.8×5+4.9×10+5.0×6)≈4.6,故答案为:4.6;(2)由于共有43个数据,其中位数为第22个数据,即中位数为4.7,故答案为:4.7;(3)不能,∵小鸣同学右眼视力是4.5,小于中位数4.7,∴不能说小鸣同学的右眼视力处于全班同学的中上水平.21.如图,正方形ABCD的对角线相交于点O,BC=6,延长BC至点E,使得CE=8,点F是DE的中点,连接CF、OF.(1)求OF的长.(2)求CF的长.【考点】LE:正方形的性质.【分析】(1)只要证明OF是△DBE的中位线即可解决问题;(2)在Rt△DCE中,利用勾股定理求出DE,再利用斜边中线的性质即可解决问题;【解答】解:(1)∵四边形ABCD是正方形,∴BC=CD=6,∠BCD=∠ECD=90°,OB=OD,∵CE=8,∴BE=14,∵OB=OD,DF=FE,∴OF=BE=7.(2)在Rt△DCE中,DE===10,∵DF=FE,∴CF=DE=5.22.如图,在平面直角坐标系中,直线y=kx+b经过点A(﹣30,0)和点B(0,15),直线y=x+5与直线y=kx+b相交于点P,与y轴交于点C.(1)求直线y=kx+b的解析式.(2)求△PBC的面积.【考点】FF:两条直线相交或平行问题.【分析】(1)根据点A、B的坐标,利用待定系数法求出直线AB的解析式即可;(2)联立两直线解析式成方程组,通过解方程组可得出点P的坐标,由一次函数图象上点的坐标特征可求出点C的坐标,进而可得出线段BC的长度,再利用三角形的面积公式,即可求出△PBC的面积.【解答】解:(1)将点A(﹣30,0)、B(0,15)代入y=kx+b,,解得:,∴直线y=kx+b的解析式为y=x+15.(2)联立两直线解析式成方程组,,解得:,∴点P的坐标为(20,25).当x=0时,y=x+5=5,∴点C的坐标为(0,5),∴BC=15﹣5=10,∴S△PBC=BC•x P=×10×20=100.23.2016年下半年开始,不同品牌的共享单车出现在城市的大街小巷.现已知A品牌共享单车计费方式为:初始骑行单价为1元/半小时,不足半小时按半小时计算.内设邀请机制,每邀请一位好友注册认证并充值押金成功,双方骑行单价均降价0.1元/半小时,骑行单价最低可降至0.1元/半小时(比如,某用户邀请了3位好友,则骑行单价为0.7元/半小时).B 品牌共享单车计费方式为:0.5元/半小时,不足半小时按半小时计算.(1)某用户准备选择A品牌共享单车使用,设该用户邀请好友x名(x为整数,x≥0),该用户的骑行单价为y元/半小时.请写出y关于x的函数解析式.(2)若有A,B两种品牌的共享单车各一辆供某用户一人选择使用,请你根据该用户已邀请好友的人数,给出经济实惠的选择建议.【考点】FH:一次函数的应用;9A:二元一次方程组的应用.【分析】(1)根据题意可以分段求出y关于x的函数解析式;(2)根据题意可以分段写出相应的不等式,从而可以解答本题.【解答】解:(1)由题意可得,当0≤x≤9且x为正整数时,y=1﹣0.1x,当x≥10且x为正整数时,y=0.1,即y关于x的函数解析式是y=;(2)由题意可得,当0≤x≤9时,1﹣0.1x>0.5,可得,x<5,则当x≤x<5且x为正整数时,选择B品牌的共享单车;当0≤x≤9时,1﹣0.1x=0.5,得x=5,则x=5时,选择A或B品牌的共享单车消费一样;当0≤x≤9时,1﹣0.1x<0.5,得x>5,则x>5且x为正整数,选择A品牌的共享单车;当x≥10且x为正整数时,0.1<0.5,故选项A品牌的共享单车.24.下面我们做一次折叠活动:第一步,在一张宽为2的矩形纸片的一端,利用图(1)的方法折出一个正方形,然后把纸片展平,折痕为MC;第二步,如图(2),把这个正方形折成两个相等的矩形,再把纸片展平,折痕为FA;第三步,折出内侧矩形FACB的对角线AB,并将AB折到图(3)中所示的AD处,折痕为AQ.根据以上的操作过程,完成下列问题:(1)求CD的长.(2)请判断四边形ABQD的形状,并说明你的理由.【考点】LO:四边形综合题.【分析】(1)先证明四边形MNCB为正方形,再利用折叠得:CA=1,AB=AD,所以CD=AD﹣AC,可得结论;(2)根据平行线的性质得折叠得:∠BAQ=∠BQA,由等角对等边得:AB=BQ,由一组对边平行且相等可得:四边形ABQD是平行四边形,再由AB=AD,可得四边形ABQD是菱形.【解答】解:(1)∵∠M=∠N=∠MBC=90°,∴四边形MNCB是矩形,∵MB=MN=2,∴矩形MNCB是正方形,∴NC=CB=2,由折叠得:AN=AC=NC=1,Rt△ACB中,由勾股定理得:AB==,∴AD=AB=,∴CD=AD﹣AC=﹣1;(2)四边形ABQD是菱形,理由是:由折叠得:AB=AD,∠BAQ=∠QAD,∵BQ∥AD,∴∠BQA=∠QAD,∴∠BAQ=∠BQA,∴AB=BQ,∴BQ=AD,BQ∥AD,∴四边形ABQD是平行四边形,∵AB=AD,∴四边形ABQD是菱形.25.如图,正方形ABCD中,AB=4,P是CD边上的动点(P点不与C、D重合),过点P作直线与BC的延长线交于点E,与AD交于点F,且CP=CE,连接DE、BP、BF,设CP═x,△PBF 的面积为S1,△PDE的面积为S2.(1)求证:BP⊥DE.(2)求S1﹣S2关于x的函数解析式,并写出x的取值范围.(3)分别求当∠PBF=30°和∠PBF=45°时,S1﹣S2的值.【考点】LO:四边形综合题.【分析】(1)如图1中,延长BP交DE于M.只要证明△BCP≌△DCE,推出∠BCP=∠CDE,由∠CBP+∠CPB=90°,∠CPB=∠DPM,即可推出∠CDE+∠DPM=90°,延长即可解决问题;(2)根据S1﹣S2=S△PBE﹣S△PDE计算即可解决问题;(3)分两种情形分别求出PC的长,利用(2)中结论计算即可;【解答】解:(1)如图1中,延长BP交DE于M.∵四边形ABCD是正方形,∴CB=CD,∠BCP=∠DCE=90°,∵CP=CE,∴△BCP≌△DCE,∴∠BCP=∠CDE,∵∠CBP+∠CPB=90°,∠CPB=∠DPM,∴∠CDE+∠DPM=90°,∴∠DMP=90°,∴BP⊥DE.(2)由题意S1﹣S2=(4+x)•x﹣•(4﹣x)•x=x2(0<x<4).(3)①如图2中,当∠PBF=30°时,∵∠CPE=∠CEP=∠DPF=45°,∠FDP=90°,∴∠PFD=∠DPF=45°,∴DF=DP,∵AD=CD,∴AF=PC,∵AB=BC,∠A=∠BCP=90°,∴△BAF≌△BCP,∴∠ABF=∠CBP=30°,∴x=PC=BC•tan30°=,∴S1﹣S2=x2=.②如图3中,当∠PBF=45°时,在CB上截取CN=CP,理解PN.由①可知△ABF≌△BCP,∴∠ABF=∠CBP,∵∠PBF=45°,∴∠CBP=22.5°,∵∠CNP=∠NBP+∠NPB=45°,∴∠NBP=∠NPB=22.5°,∴BN=PN=x,∴x+x=4,∴x=4﹣4,∴S1﹣S2=(4﹣4)2=48﹣32.。

八年级下试题—2016_2017学年广东广州海珠区中山大学附属中学初二下学期期末英语试卷

八年级下试题—2016_2017学年广东广州海珠区中山大学附属中学初二下学期期末英语试卷

15. A.timesB.sometimesC.at the same timeD.for the first time 6. A.plantedB.affordedC.endedD.went 7. A.SinceB.UnlessC.IfD.Although 8. A.theyB.heC.sheD.you 9. A.receivedB.expectedC.refusedD.got 10. A.sorryB.worriedC.encouragedD.sad 11. A.evenB.hardlyC.stillD.too 12. A.get used toB.worry aboutC.pay attention toD.look forward to 13. A.withB.withoutC.byD.for 14. A.foolB.volunteerC.heroD.boy 15. A.knock into B.dream about C.talk about plain about1. A.quietelessC.importantD.easy 2. A.clocksB.carsC.booksD.watches 3. A.wheneverB.howeverC.whateverD.wherever 4. A.atB.onC.forD.in 5. A.nothingB.everythingC.something elseD.some things 6. A.suchB.asC.soD.only 7. A.understandB.findC.forgetD.remember 8. A.WhileB.butC.thoughD.because 9. A.timeB.placeC.weatherD.news 10. A.put B.want C.let D.expectDifferent people have different ideas about time. People in the USA think that itis 1 to know the time. In cities in America, there are 2 in stations, factories and other buildings. Radio announcers give you the correct time during the day. Most Americans also have watches with them 3 they go. They want to do certain things 4 certain time. They don't like to be late. They think everyone is supposed to do 5 on time.But time is not 6 important to everybody in the world. When 7 that people there you visit a county in South America, you will don't like to rush. If you had an appointment with somebody,' he could probably be late 8 he might not want to arrive on time.In South America, even the radio programs may not begin on time. The men on the radio may not think it is important to tell the exact 9 . People in South America think that clocks or watches are just machines. They think that you 10 a clock or a watch control your life if you do everything on time.2二、完形填空第一节、阅读理解ASingles' day falls on every November 11th. No one is quite sure exactly who first thought it up, and the most widely-accepted story is that it emerged from the dorms of Nanjing University in 1993 when four single male students got together to discuss how to break free of the loneliness of single life. One suggested that because of the ones in the date, November 11would be a good day to organize activities for singles. Singles' Day, started by a small group of friends, gradually grew into something like the anti-Valentine's Day, a day China's single young people could use as an excuse to get together and have fun together.Originally about being with friends and having fun, now Singles Day is about shopping-mostly online shopping. How did this happen? It's a long story, but the short answer is Alibaba.By the late 2000s, most of China's Internet users were familiar with Singles' Day. There might have been some small shops online and off line offering sales on that day earlier, but no major company until Alibaba launched its first Singles' Day online sale in 2009.In that first year, Alibaba was the only major ecommerce company to offer a sale, and it featured just 27 brands offering discounts via its T-mall marketplace. The sale was definitely successful, but it wasn't enough to redefine the holiday on its own. In the following year,Alibaba went bigger, offering more brands and deeper discounts. But other companies had noticed the potential of the 2009 sales bonanza and decided to follow suit. E-commerce platforms like JD had their first major Singles' Day sales in 2010, and overnight Singles' Day went from a T-mall sale to something that was beginning to look like Cyber Monday in some western countries.Over the next few years, Alibaba. JD, and other Chinese e-commerce players all expanded their one-day discounts, and sales grew exponentially (以指数方式). On Singles Day in 2012, Alibaba's marketplaces, Taobao and T-mall, did about $3 billion in sales. In 2013that number nearly doubled, and Chinese shoppers had obliterated America's Cyber Monday spending records in just the first few hours of the sale. And it's only getting bigger.3 A.B.C.D.We can know that Singles' Day was fit set .by a large e-commerce company called Alibabafor the young people who want to have fun togetherto organize activities for singles to get rid of their lonelinessto express something against Valentine’s Day(1)Before 2009 on Singles' Day, single young people were most likely to .(2)三、阅读A.B.C.D.just spend the day alonehave fun with friendsdo some online shoppinggo back home to visit parentsA.B.C.D.Which of the following statements is TRUE according to the passage?By 2009, Singles' Day have already became a famous online-shopping day.Cyber Monday in some western countries is to celebrate Singles' Day.In 2013, Alibaba's marketplaces did about nearly $4 billion in sale.Alibaba was the first major e-commerce company to offer a sale on Singles' Day.(3)A.B.C.D.According to the last two paragraphs, we can know that .Alibaba played an important part in the new definition of Singles' Day.other Chinese e-commerce company like JD followed Cyber Monday.Alibaba offered more brands and deeper discounts to make it bigger.Alibaba, Taobao, T-mall and JD are different e-commerce companies.(4)A.B.C.D.What is the passage mainly about?The origin of Singles' Day.The life of modern young single people.The e-commerce company Alibaba.How Singles' Day become a shopping day.(5)BDear Lan Lan,I'm now writing to you at Hartsop, a village in the Lake District, a place to have most beautiful scenery in English couple here to spend Christmas with them. This was planned for overseas students to know British way of life. by living with British families.We came two days before Christmas and during our stay, we have had everything we expected. Christmas turkey, Christmas cake, Christmas puddings, Christmas parties and Christmas gifts under the Christmas tree. All are exciting and amusing, but above all these, we are deeply touched by the hospitality of the family, Roger and Anne Marie.Anne Marie was a nurse and all these days, she had been busy cooking meals, washing dishes and showing us around. Roger, who was a doctor, knows a lot about China and still wants to know more. He plays us Chinese music and it seems to me he does better in that than we do. In the evenings, we all sit around the fireplace with Tim, a lovely dog, by our side.Like long-time-no-see friends, we talk about all the things that have happened or we hope to happen in our lives. There is always so much to tell and to know .4After three months away from home, we are again feeling how sweet a home can be.Roger and Anne Marrie are not like most of the other people we have met who always make us feel we are foreigners in a foreign country. They show such kindness to us that they bring us a person-to-person feeling. instead of a British-to-Chinese feeling. They make us believe that though there are differences of languages and cultures, one may always expect to find in every corner of the world the feeling of love and being loved.How I wish Mum and Dad could meet them! How I wish you were here with me! How strange it is that the more I feel at home here, the more I miss my real home and all of you. We always talk about "when we go back home next year..." and soon we will turn this into "When we go back home THIS year..." for the New Year's Day is coming. Miss you.LoveCao WenA.B.C.D.The writer is now in England.a Chinese visitor to an English familya Chinese student studyinga Chinese teacher workinga Chinese student living(1)A.B.C.D.In Hartsop, Cao Wen felt .quite at homethe English home better than her homeshe was a foreigner in a foreign countryshe was one member of the family(2)A.B.C.D.It seems that Roger .can speak Chinese very wellknows something about Chinese for a long timeonce lived in China for a long time.can do everything better than the Chinese(3)A.B.C.D.When you are away from home, it is good for you to have the feeling .of being aloneof going home soonof love and being lovedas a foreigner(4)A.B.C.D. In the writer's opinion, the best home is .the home they stayed in Englandthe home that you left behindChinese homeone's own home(5)CSuper Typhoon Megi crashed into the Philippines on Monday, leaving a trail of destruction in its path. Thousands of residents fled to safety as heavy rains and strong winds caused Hooding, collapsed trees and cut off power. It was the strongest storm to hit the island nation in four years.At least seven people died in the storm, and at least six others were injured by falling trees, broken rooftops and glass, officials said More than 4,150 residents were forced lo take shelter in schools, town halls, churches and relatives' homes.A typhoon is a tropical cyclone, or hurricane, that occurs in the western Pacific or Indian Oceans. The storm is named a "super typhoon" when winds exceed 150 mph. Megi crashed ashore in the Isabela province of the northern Philippines with whipping winds that reached 162 mph. The storm created big waves in Palanan Bay. All ships and fishing boats were advised to stay in ports. Several airline flights were canceled.Luckily, the winds blew high from the ground, sparing many rice fields from damage, says Alvaro Antonio, the governor of Cagayan province. The region also took a hard hit from the storm.Thousands of military reserve officers and volunteers stood by to help with relief efforts.Evacuations and emergency preparations for the storm were made days in advance. Retired army Major General Benito Ramos heads the Philippines' disaster-preparedness agency. He said readying for the typhoon was like "preparing for war."Notes :①tropical cyclone 热带气旋 ②evacuation n.疏散 ③collapsed adj.倒塌的5 A.B.C.D.How many people were homeless in this natural disaster?At least six.At least seven.About 13.More than 4,150.(1)A.B.C.D. Which of the following description about typhoon is RIGHT?It is a tropical cyclone occurs in the eastern Pacific.It is a hurricane occurs in the Atlantic Ocean.When winds exceed 100 mph, the storm is named a "super typhoon" .It can bring a great destruction.(2)A.B.C.What does the underlined word mean?Avoiding.Loosing.Taking.(3)D.Protecting.A. B.C.D.Which of the following didn't get involved in this disaster?Military reserve officers.Volunteers.Army.Disaster-preparedness agency.(4)A.B.C.D.What can we infer from the major General's words?The preparing work is all done by soldiers.The preparing work is a hard and responsible task.The army should get into the relief efforts.They did the same thing with soldiers.(5)DNo one is sure how the ancient Egyptians built the pyramids near Cairo. But a new study suggests they used a little rock 'n' roll. Long-ago builders could have attached wooden poles to the stones and rolled them across the sand, the scientists say."Technically, I think what they're proposing is possible." physicist Daniel Bonn said.People have long puzzled over how the Egyptians moved such huge rocks. And there's no obvious answer. On average, each of the two million big stones weighed about as much as a large pickup truck. The Egyptians somehow moved the stone blocks to the pyramid site from about one kilometer away.The most popular view is that Egyptian workers slid the blocks along smooth paths. Many scientists suspect workers first would have put the blocks on sleds (滑板), Then they would have dragged them along paths. To make the work easier, workers may have lubricated the paths either with wet clay or with the fat from cattle. Bonn has now tested this idea by building small sleds and dragging heavy objects over sand.Evidence from the sand supports this idea. Researchers found small amounts of fat, as well as a large amount of stone and the remains of paths.However, physicist Joseph West thinks there might have been a simpler way, who led the new study. West said, "I was inspired while watching a television program showing how sleds might have helped with pyramid construction" , I thought, 'Why don't they just try rolling the things?' "A square could be turned into a rough sort of wheel by attaching wooden poles to its sides, he realized. That, he notes, should make a block of stone" a lot easier to roll than a square”.So he tried it .6He and his students tied some poles to each of four sides of a 30-kilogram stone block.That action turned the block into somewhat a wheel. Then they- placed the block on the ground,They wrapped one end of a rope around the block and pulled. The researchers found they could easily roll the block along different kinds of paths. They calculated that rolling the block required about as much force as moving it along a slippery (滑的) path.West hasn't tested his idea on larger blocks,but he thinks rolling has clear advantages over sliding. At least, workers wouldn't have needed to carry cattle fat or water to carry cattle fat or water to smooth the paths.A.B.C.D.It's widely believed that the stone blocks were moved to the pyramid site by .rolling them on roadspushing them over the sandsliding them on smooth pathsdragging them on some poles(1)A.B.C.D.The underlined part "lubricated the paths" in Paragraph 4 means .made the path wetmade the path hardmade the path widemade the path slippery(2)A.B.C.D.What does the underlined word "it" in Paragraph 7 refer to?Rolling the blocks with poles attached.Rolling the blocks on wooden wheels.Rolling poles to move the blocks.Roiling the blocks with fat.(3)A.B.C.D.Why is rolling better than sliding according to West?Because more force is needed for sliding.Because rolling work can' t be done by fewer cattle.Because sliding on smooth road is more dangerous.Because less preparation on path is needed for rolling.(4)A.B.C.D.What is the text mainly about?An experiment on ways, of moving blocks to the pyramid site.An application of the method of moving blocks to the pyramid site.An argument about different methods of moving blocks to the pyramid site.An introduction to .a possible new way of moving blocks to the pyramid site.(5)第二节、阅读填空A.B.C.D.E.Taking good notes is a time-saving skill that will help you to become a better student inseveral ways. 1 Second, your notes are excellent materials to refer to when you are studying for a test. Third, note-taking offers variety to your study time and helps you to hold your interest.You will want to take notes during classroom discussions and while reading a textbook or doing research for a report, 2 Whenever or however you take notes, keep in mind that note-taking is a process of choosing notes. 3The following methods may work best for you.● Read the text quickly to find the main facts and ideas in it.●Read the text carefully and watch for words that can show main points and supporting facts.● Write your notes in your own words.● 4● Note any questions or ideas you may have about what was said or written.As you take notes, you may want to use your own shorthand (速记). When you do it, be sure that you understand your symbols and you will use them all the time. 5Use words, not complete sentences.Otherwise, you may not be able to read your notes later.you will also want to develop your own method for taking notes.That means you must first decide what is important enough to include in your notes.First, the simple act of writing something down makes it easier for you to understand and remember it.7第一节、单词拼写单词拼写根据句子意思和所给的首字母写出所缺单词8When meeting something difficult or dangerous, he showed great c .(1)When she came in, she didn't s our hands or raise your hand.(2)The police asked the shopkeeper to d more about the thief, so that they couldknow more about the theft.(3)That's great! We have s another goal!(4)Some of the stones w twenty-six tons.(5)I prefer the quiet countryside to the n cities.(6)四、写作第二节、完成句子根据所给的汉语内容,用英语完成下列句子9妈妈正在为圣诞晚餐做准备。

广东省2016-2017学年八年级语文下学期第二周小测试题 新人教版(附答案)

广东省2016-2017学年八年级语文下学期第二周小测试题 新人教版(附答案)

第二周小测试卷一、默写或按要求填空(16分)(1)山随平野尽,。

(李白《渡荆门送别》)(2)予独爱莲之出淤泥而不染,。

(周敦颐《爱莲说》)(3),松间沙路净无泥,。

(苏轼《浣溪沙》)(4)八月湖水平,。

,波撼岳阳城。

(孟浩然《望洞庭湖赠张丞相》)(5)陶渊明《归园田居》(其三)一诗中最能表现诗人摒弃世俗、皈返自然的意愿的两句是:,。

二、根据拼音写出相应的词语。

(8分)(1)日落的景象和日出同样壮观、qǐlì( ),而且神秘、迷人。

(2)到徐州见着父亲,看见满院lánɡjí()的东西,又想起祖母。

(3)北雁南飞,活跃在田间草际的昆虫也都xiāo shēnɡnìjì()。

(4)设计者和匠师们yīn dìzhìyí(),自出心裁,修建成功的园林当然各个不同。

三、下列句中加点词语使用不恰当的一项是()(3分)A. 终于要砌新台阶了,父亲明明该高兴,却露出些尴尬..的笑。

B. 在自然看来,人类上下翻飞的这片巨大空间,不过是咫尺..之间而已。

C. 现在的电信诈骗案层出不穷,行骗者手段之隐蔽、蒙骗形式之多样,简直令人叹为观止....。

D. 学习知识不仅需要刻苦钻研的精神,更需要一丝不苟....的态度。

四、下列对病句的修改不正确的一项是()(3分)A. 我这次考不好的原因是因为平时没有扎扎实实学好基础知识。

(删除“因为”)B. 这次受检的十余种教辅资料合格率不足10%左右。

(“不足”和“左右”删除其中一个)C. 同学们要尽量选择无异味的橡皮擦、无刺激气味的涂改液和学习用品。

(把“学习用品”改成“文具”)D. 既然你能够认真地读完课程标准推荐阅读的名著,那么,一生将受益无穷。

(“既然”改成“如果”)五、根据语境,仿照前两个分句续写两个句子,使之构成语意相关的排比句。

(4分)美是飘荡在蓝天的几缕白云,美是挺立在峭壁的几棵小松,,。

只要我们善于观察和感悟,美无处不在。

广州市2016-2017学年八年级下学期期中考试数学试卷

广州市2016-2017学年八年级下学期期中考试数学试卷

2016-2017学年广州市八年级下学期期中考试数学试卷一.选择题(本大题10小题,每小题2分,共20分)1.式子3+x 有意义,则x 的取值范围是( )A 、x ≥3B 、x ≤3C 、x ≥−3D 、x ≤−32.在△ABC 中,三边长满足b 2−a 2=c 2,则互余的一对角是( )A 、∠A 与∠B B 、∠B 与∠CC 、∠A 与∠CD 、以上都不正确3.一直角三角形两边分别为3和5,则第三边为( )A 、4B 、34C 、4或34D 、24.下列二次根式中,最简二次根式是( )A 、5.0B 、a 4C 、8D 、105.如图,所有三角形都是直角三角形,所有四边形都是正方形,已知S 1=4,S 2=9,S 3=8,S 4=10,则S =( )A 、25B 、31C 、32D 、406.在平行四边形ABCD 中,∠A :∠B :∠C =2:3:2,则∠D=( )A 、36°B 、108°C 、72°D 、60°7.实数a ,b 在数轴上的位置如图所示,则化简2)1(-a −2)(b a -+b 的结果是( )A 、1B 、b +1C 、2aD 、1−2a8.在△ABC 中,∠ACB =90°,AC =40,CB =9,M 、N 在AB 上且AM =AC ,BN =BC ,则MN 的长为( )A 、6B 、7C 、8D 、99.下列命题中,真命题是( )A 、对角线相等且互相垂直的四边形是菱形B 、有一条对角线平分对角的四边形是菱形C 、菱形是对角线互相垂直平分的四边形D 、菱形的对角线相等10.下列命题:如图,正方形ABCD 中,E 、F 分别为AB 、AD 上的点,AF =BE ,CE 、BF 交于H ,BF 交AC 于M ,O 为AC 的中点,OB 交CE 于N ,连OH .下列结论中:①BF ⊥CE ;②OM =ON ;③OH =21CN ;④2OH +BH =CH .其中正确的命题有( ) A 、只有①② B 、只有①②④C 、只有①④D 、①②③④二.填空题(共6小题,每小题3分,共18分)11.215×5=__________.12.m ,n 分别是2−1的整数部分和小数部分,则2m −n =________.13.如图,台阶A 处的蚂蚁要爬到B 处搬运食物,它爬的最短距离是__________.14.如图,在Rt △ABC 中,∠C =90°,BC =3,AC =4,M 为斜边AB 上一动点,过M 作MD ⊥AC ,过M 作ME ⊥CB 于点E ,则线段DE 的最小值为________.15.如图,矩形ABCD 中,AB =15cm ,点E 在AD 上,AE =9cm ,连接EC ,将矩形ABCD 沿BE 翻折,点A 恰好落在EC 上的点A ′处,则BC=_________cm .16.如图,在平面直角坐标系中,边长为1的正方形OA 1B 1C 1的两边在坐标轴上,以它的对角线OB 1为边作正方形OB 1B 2C 2,再以正方形OB 1B 2C 2的对角线OB 2为边作正方形OB 2B 3C 3,以此类推…、则正方形OB 2015B 2016C 2016的顶点B 2016的坐标是________.三.解答题(共9小题,共62分)17.计算:(5−2)2014(5+2)2015−2|−25|−(1−2)0.18.先化简,再求值:1)(42--x x x +(x −2)2−692x ,其中,x =5+1.19.如图,四边形ABCD 中,AB =20,BC =15,CD =7,AD =24,∠B =90°, 求证:∠A +∠C =180°.20.如图,将平行四边形ABCD 的对角线BD 向两个方向延长至点E 和点F ,使BE =DF ,求证:四边形AECF 是平行四边形.21.如图,点C 在线段BD 上,AC ⊥BD ,CA =CD ,点E 在线段CA 上,且满足DE =AB ,连接DE 并延长交AB 于点F .(1)求证:DE ⊥AB ;(2)若已知BC =a ,AC =b ,AB =c ,设EF =x ,则△ABD 的面积用代数式可表示为;S △ABD =21c (c +x )你能借助本题提供的图形,证明勾股定理吗?试一试吧.22.如图,在菱形ABCD 中,F 是BC 上任意一点,连接AF 交对角线BD 于点E ,连接AF 交对角线于点E ,连接EC(1)求证:AE =EC ;(2)当∠ABC =60°,∠CEF =60°时,点F 在线段BC 的什么位置?说明理由.23.如图,已知四边形ABCD 为正方形,AB =22,点E 为对角线AC上一动点,连接DE ,过点E 作EF ⊥DE .交射线BC 于点F ,以DE 、EF 为邻边作矩形DEFG ,连接CG.①求证:矩形DEFG是正方形;②探究:CE+CG的值是否为定值?若是,请求出这个定值;若不是,请说明理由.24.如图:在△ABC中,CE、CF分别平分∠ACB与它的邻补角∠ACD,AE⊥CE于E,AF⊥CF于F,直线EF分别交AB、AC于M、N.(1)求证:四边形AECF为矩形;(2)试猜想MN与BC的关系,并证明你的猜想;(3)如果四边形AECF是菱形,试判断△ABC的形状,直接写出结果,不用说明理由.25.如图,四边形OMTN中,OM=ON,TM=TN,我们把这种两组邻边分别相等的四边形叫做筝形.(1)探究筝形对角线之间的位置关系,并证明你的结论;(2)在筝形ABCD中,已知AB=AD=10,BC=CD,BC>AB,BD、AC为对角线,BD=16.①若∠ABC=90°,求AC的长;②过点B作BF⊥CD于F,BF交AC于点E,连接DE.当四边形ABED为菱形时,求点F到AB的距离.。

广东省广州市越秀区2016-2017学年八年级下册数学期末考试试卷(解析版)

广东省广州市越秀区2016-2017学年八年级下册数学期末考试试卷(解析版)

广东省广州市越秀区2016-2017 学年八年级下册数学期末考试试卷(解析版)一、选择题1. 下列式子没有意义的是()A. B.C.D.2.下列计算中,正确的是()A.÷=B. (4)2=8 C.=2 D.2×2=23. 刻画一组数据波动大小的统计量是()A.平均数B.方差 C.众数 D.中位数4.在暑假到来之前,某机构向八年级学生推荐了A,B,C 三条游学线路,现对全级学生喜欢哪一条游学线路作调查,以决定最终的游学线路,下面的统计量中最值得关注的是()A.方差B.平均数 C.中位数 D.众数5. 关于正比例函数y=﹣ 2x ,下列结论中正确的是()A.函数图象经过点(﹣ 2,1) B. y 随 x 的增大而减小C.函数图象经过第一、三象限 D.不论x取何值,总有y <06. 以下列各组数据中的三个数作为三角形的边长,其中能构成直角三角形的是()A. 2,3,4B.,,C. 1,,2 D.7, 8,97. 若一个直角三角形的一条直角边长是5cm,另一条直角边比斜边短1cm,则斜边长为()cm.A. 10B.11 C.12 D. 13 8. 如图,在菱形ABCD中,对角线AC,BD相交于点O, AB=5, AC=6,则菱形 ABCD的面积是()A. 24B.26 C. 30 D. 489. 在下列命题中,是假命题的是()A.有一个角是直角的平行四边形是矩形B.一组邻边相等的矩形是正方形C.一组对边平行且相等的四边形是平行四边形D.有两组邻边相等的四边形是菱形10.已知平面上四点 A( 0, 0), B( 10,0), C( 12, 6), D( 2,6),直线 y=mx﹣ 3m+6将四边形 ABCD 分成面积相等的两部分,则m的值为()A. B.﹣1 C.2D.二、填空题11. 已知 a=+2, b=﹣2,则ab=________.12.一次函数 y=kx+b (k≠0)中, x 与 y 的部分对应值如下表:x ﹣ 2 ﹣ 1 0 1 2y ﹣ 6 ﹣ 4 ﹣ 2 0 2那么,一元一次方程kx+b=0 的解是 x=________.13.如图是一次函数 y=mx+n的图象,则关于 x 的不等式 mx+n> 2 的解集是 ________.14.一组数据: 2017、 2017、2017、 2017、 2017,它的方差是 ________.15.考古学家们发现了几块大约完成于公元前2000 年左右的古巴比伦的泥版书,据专家们考证,其中一块上面刻有如下问题:“一根长度为30 个单位的棍子直立在墙上,当其上端垂直滑下 6 个单位时,请问其下端离开墙角有多远?”,这个问题的答案是:其下端离开墙角________个单位.16.如图所示,在 Rt△ ABC中,∠ A=90°, DE∥ BC, F, G, H, I 分别是 DE, BE, BC, CD的中点,连接 FG,GH, HI ,IF , FH,GI.对于下列结论:①∠ GFI=90°;② GH=GI;③ GI=(BC﹣DE);④四边形FGHI 是正方形.其中正确的是________(请写出所有正确结论的序号).三、解答题17.计算:(+﹣)×.18.如图,在△ ABC中, AD⊥BC, AB=5, BD=4,CD=.(1)求 AD的长.(2)求△ ABC的周长.19. 如图在平行四边形ABCD中, AC交 BD于点 O, AE⊥ BD, CF⊥ BD,垂足分别为E、F,求证:四边形AECF为平行四边形.20. 下表是某校八年级(1)班 43 名学生右眼视力的检查结果.视力 4.0 4.14.24.34.44.54.64.7 4.84.95.0人数 125435115106( 1)该班学生右眼视力的平均数是________(结果保留 1 位小数).(2)该班学生右眼视力的中位数是________.(3)该班小鸣同学右眼视力是 4.5 ,能不能说小鸣同学的右眼视力处于全班同学的中上水平?试说明理由.21.如图,正方形 ABCD的对角线相交于点 O, BC=6,延长 BC至点 E,使得 CE=8,点 F 是 DE的中点,连接CF、 OF.(1)求 OF的长.(2)求 CF的长.22.如图,在平面直角坐标系中,直线y=kx+b 经过点 A(﹣ 30, 0)和点 B(0, 15),直线 y=x+5 与直线y=kx+b 相交于点P,与 y 轴交于点C.(1)求直线 y=kx+b 的解析式.(2)求△ PBC的面积.23.2016年下半年开始,不同品牌的共享单车出现在城市的大街小巷.现已知 A 品牌共享单车计费方式为:初始骑行单价为 1 元/ 半小时,不足半小时按半小时计算.内设邀请机制,每邀请一位好友注册认证并充值押金成功,双方骑行单价均降价0.1 元 / 半小时,骑行单价最低可降至0.1 元 / 半小时(比如,某用户邀请了 3 位好友,则骑行单价为0.7 元 / 半小时). B 品牌共享单车计费方式为:0.5 元 / 半小时,不足半小时按半小时计算.( 1)某用户准备选择 A 品牌共享单车使用,设该用户邀请好友x 名( x为整数, x≥0),该用户的骑行单价为y 元 / 半小时.请写出y 关于x 的函数解析式.(2)若有 A, B 两种品牌的共享单车各一辆供某用户一人选择使用,请你根据该用户已邀请好友的人数,给出经济实惠的选择建议.24.下面我们做一次折叠活动:第一步,在一张宽为 2 的矩形纸片的一端,利用图(1)的方法折出一个正方形,然后把纸片展平,折痕为 MC;第二步,如图(2),把这个正方形折成两个相等的矩形,再把纸片展平,折痕为FA;第三步,折出内侧矩形FACB的对角线AB,并将AB折到图(3)中所示的AD处,折痕为AQ.根据以上的操作过程,完成下列问题:(1)求 CD的长.(2)请判断四边形 ABQD的形状,并说明你的理由.25.如图,正方形 ABCD中, AB=4,P 是 CD边上的动点( P 点不与 C、D 重合),过点 P 作直线与 BC的延长线交于点积为 S2E,与.AD交于点F,且CP=CE,连接DE、 BP、 BF,设CP═x,△ PBF的面积为S1,△ PDE的面( 1)求证: BP⊥ DE.( 2)求 S1﹣S2关于 x 的函数解析式,并写出( 3)分别求当∠ PBF=30°和∠ PBF=45°时,x 的取值范围.S1﹣ S2的值.答案解析部分一、 <b > 选择题 </b>1.【答案】 B【考点】二次根式有意义的条件【解析】【解答】 A、有意义,A不合题意;B、没有意义,B符合题意;C、有意义,C不合题意;D、有意义,D不合题意;故答案为: B.【分析】依据二次根式被开放数为非负数求解即可.2.【答案】 C【考点】二次根式的性质与化简,二次根式的乘除法【解析】【解答】解: A、原式 ===3,A 不符合题意;B、原式 =32, B 不符合题意;C、原式 =| ﹣ 2|=2 ,C 符合题意;D、原式 =4,D不符合题意;故答案为: C.【分析】依据二次根式的除法法则可对 A 作出判断;依据二次根式的性质可对 B、 C 作出判断,依据二次根式的乘法法则可对 D 作出判断 .3.【答案】 B【考点】统计量的选择【解析】【解答】由于方差反映数据的波动情况,衡量一组数据波动大小的统计量是方差.故答案为: B.【分析】方差是反应一组数据波动大小的量.4.【答案】 D【考点】统计量的选择【解析】【解答】由于众数是数据中出现次数最多的数,故全级学生喜欢的游学线路最值得关注的应该是统计调查数据的众数.故答案为: D.【分析】决定最终的线路应改由多数人员的意见决定,故此可得到问题的答案.5.【答案】 B【考点】正比例函数的图象和性质【解析】【解答】解: A、当 x=﹣ 2 时, y=﹣2×(﹣ 2)=4,即图象经过点(﹣ 2, 4),不经过点(﹣ 2,1),故本选项错误;B、由于 k=﹣ 2< 0,所以 y 随 x 的增大而减小,故本选项正确;C、由于 k=﹣ 2< 0,所以图象经过二、四象限,故本选项错误;D、∵ x> 0 时, y< 0,x< 0 时, y> 0,∴不论 x 为何值,总有y< 0 错误,故本选项错误.故答案为: B.【分析】依据正比例函数的图像和性质可对 B、 C、 D 作出判断,将 x=-2 代入函数解析式可求得 y 的值,从而可对A作出判断 .6.【答案】 C【考点】勾股定理的逆定理222【解析】【解答】 A、 2 +3≠4,故不是直角三角形, A 不符合题意;B、()2+()2≠()2,故不是直角三角形, B 不符合题意;C、 12+()2=22,故是直角三角形, C 符合题意;222D、7 +8≠9 ,故不是直角三角形,D 不符合题意;故答案为: C.【分析】依据勾股定理的逆定理进行判断即可.7. 【答案】 D【考点】勾股定理【解析】【解答】设斜边长为xcm,则另一条直角边为(x﹣1) cm,222解得, x=13,则斜边长为13cm,故答案为: D..【分析】设斜边长为xcm,则另一条直角边为(x-1 ) cm,然后依据勾股定理列方程求解即可8. 【答案】 A【考点】菱形的性质【解析】【解答】∵四边形ABCD是菱形,∴OA=OC=3, OB=OD,AC⊥ BD,在 Rt △ AOB中,∠ AOB=90°,根据勾股定理,得:OB=,=,=4,∴ BD=2OB=8,∴ S 菱形ABCD=×AC×BD=×6×8=24.故答案为: A.BO 【分析】根据菱形的对角线互相垂直且互相平分可得到AC⊥BD,且 AO=OC=3,然后依据勾股定理可求得的长,从而可得到BD的长,最后依据菱形的面积等于对角线乘积的一半求解即可.9. 【答案】 D【考点】命题与定理【解析】【解答】 A、有一个角是直角的平行四边形是矩形,正确, A 不符合题意;B、一组邻边相等的矩形是正方形,正确, B 不符合题意;;C、一组对边平行且相等的四边形是平行四边形,正确,C不符合题意;D、有两组邻边相等且平行的四边形是菱形,错误, D 不符合题意 .故答案为: D.【分析】首先依据矩形的定义、正方形的判定定理、平行四边形的判定定理、菱形的判定定理判定命题的对错,从而可做出判断 .10.【答案】 B【考点】待定系数法求一次函数解析式【解析】【解答】解:如图,∵A( 0, 0), B( 10, 0), C( 12, 6), D( 2, 6),∴AB=10﹣ 0=10, CD=12﹣ 2=10,又点 C、 D的纵坐标相同,∴AB∥ CD且 AB=CD,∴四边形ABCD是平行四边形,∵12÷2=6,6÷2=3,∴对角线交点P 的坐标是( 6,3),∵直线 y=mx﹣ 3m+6将四边形ABCD分成面积相等的两部分,∴直线 y=mx﹣ 3m+6经过点 P,∴6m﹣ 3m+6=3,解得 m=﹣1.故答案为:B.【分析】首先依据各点的坐标可确定出四边形ABCD为平行四边形,然后可求得两对角线交点的坐标,然后由直线平分线四边形的面积可知直线经过点( 6, 3),最后将点( 6, 3)代入直线解析式求解即可 . 二、 <b > 填空题 </b>11.【答案】 1【考点】分母有理化【解析】【解答】解:∵a=+2,b=﹣2,∴ ab=(+2)(﹣2)=5﹣4=1,故答案为: 1【分析】依据平方差公式和二次根式的性质进行计算即可.12.【答案】 1【考点】一次函数与一元一次方程【解析】【解答】解:根据上表中的数据值,当 y=0 时, x=1,即一元一次方程 kx+b=0 的解是 x=1.故答案是: 1.【分析】依据表格找出当y=0 时,对应的x 的取值即可 .13.【答案】 x> 0【考点】一次函数与一元一次不等式【解析】【解答】解:由题意,可知一次函数y=mx+n 的图象经过点(0, 2),且 y 随 x 的增大而增大,所以关于x 的不等式mx+n>2 的解集是 x> 0.故答案为: x> 0.【分析】不等式的解集为当y>2 时,函数自变量的取值范围.14.【答案】 0【考点】方差【解析】【解答】解:该组数据一样,没有波动,方差为0,故答案为: 0.【分析】方差的意义或利用方差公式进行解答即可.15.【答案】 18【考点】勾股定理的应用【解析】【解答】解:∵PC=AB=30, PA=6,∴AC=24,∴ BC===18,∴下端离开墙角18 个单位.故答案为: 18.【分析】根据题意可得到PC=AB=30,AC=24,然后在Rt △ABC中利用勾股定理求出CB的长即可 .16.【答案】①③【考点】中点四边形【解析】【解答】解:延长 IF 交 AB于 K,∵DF=EF, BG=GE,∴FG= BD, GF∥AB,同理 IF ∥ AC, HI=BD, HI∥ BD,∴∠ BKI=∠A=90°,∴∠ GFI=∠BKI=90°,∴GF⊥ FI ,故①正确,∴FG=HI, FG∥ HI ,∴四边形 FGHI是平行四边形,∵∠ GFI=90°,∴四边形 FGHI是矩形,故②④错误,延长 EI 交 BC于 N,则△ DEI≌△ CNI,∴DE=CN, EJ=JN,∵EG=GB, EI=IN ,∴GI= BHN= ( BC﹣ DE),故③正确,故答案为①③.【分析】对于①,延长IF 交 AB 于 K,然后根据两直线平行同位角相等进行解答即可;对于②和④.只要证明四边形FGHI 是矩形即可判断;对于③,先延长EI 交 BC于 N,然后再证明△DEI≌△ CNI,依据全等三角形的性质可得到DE=CN, EJ=JN,然后再结合中点的定义可推出GI=HN= ( BC-DE) .三、 <b > 解答题 </b>17. 【答案】解:原式 =( 6+﹣3)×=×=7.【考点】二次根式的混合运算【解析】【分析】先将各二次根式化简为最简二次根式,然后再合并同类二次根式,最后,在依据二次根式的乘法法则进行计算即可 .18. 【答案】( 1)解:在Rt△ ABD中, AD==3( 2)解:在Rt △ ACD中, AC==2,则△ ABC的周长 =AB+AC+BC=5+4++2=9+3【考点】勾股定理【解析】【分析】( 1)在 Rt△ ABD中,依据勾股定理可求得AD的长;( 2)在 Rt △ ACD中,依据勾股定理可求得AC的长,然后再依据三角形的周长等于三边长度之和求解即可. 19.【答案】证明:∵四边形 ABCD是平行四边形,∴AB=CD, AB∥ CD,∴∠ ABE=∠CDF,∵AE⊥ BD,CF⊥ BD,∴AE∥ CF,∠ AEB=∠CFD=90°,在△ AEB和△ CFD中,∵,∴△ AEB≌△ CFD( AAS),∴AE=CF,∴四边形AECF是平行四边形.【考点】平行四边形的判定与性质【解析】【分析】首先依据四边形的性质可得AB=CD, AB∥CD,然后再证明AE∥CF,接下来,利用AAS证得△ AEB≌△ CFD,依据全等三角形的性质可得到 AE=CF,最后依据一组对边相等且平行的四边形是平行四边形进行证明即可 .20.【答案】( 1) 4.6(2) 4.7(3)解:不能,∵小鸣同学右眼视力是 4.5 ,小于中位数 4.7 ,∴不能说小鸣同学的右眼视力处于全班同学的中上水平.【考点】中位数、众数【解析】【解答】解:(1)该班学生右眼视力的平均数是×(4.0+4.1 ×2+4.2 ×5+4.3 ×4+4.4 ×3+4.5 ×5+4.6+4.7+4.8 ×5+4.9 ×10+5.0 ×6)≈ 4.6 ,故答案为: 4.6 ;( 2)由于共有43 个数据,其中位数为第22 个数据,即中位数为 4.7 ,( 3)不能,∵小鸣同学右眼视力是 4.5 ,小于中位数 4.7 ,∴不能说小鸣同学的右眼视力处于全班同学的中上水平.故答案为:(1) 4.6 ;( 2)4.7 ;( 3)不能 .【分析】( 1)根据加权平均数公式求解即可;( 2)首先将这组数据按照从小到大的顺序排列,中位数为第22 个数据;( 3)根据小鸣同学右眼视力是 4.5 ,小于中位数 4.7 ,故此可得到问题的答案.21.【答案】( 1)解:∵四边形 ABCD是正方形,∴ BC=CD=6,∠ BCD=∠ECD=90°, OB=OD,∵ CE=8,∴ BE=14,∵ OB=OD, DF=FE,∴OF= BE=7.( 2)解:在Rt △ DCE中, DE===10,∵ DF=FE,∴ CF=DE=5.【考点】正方形的性质【解析】【分析】( 1)由正方形的性质可知O为 BD的中点,故此OF是△ DBE的中位线,然后依据三角形中位线的性质解答即可;( 2)在 Rt△ DCE中,利用勾股定理求出DE,再利用直角三角形斜边上中线等于斜边的一半求解即可.22.【答案】( 1)解:将点 A(﹣ 30, 0)、 B( 0, 15)代入 y=kx+b ,,解得:,∴直线 y=kx+b 的解析式为y=x+15.( 2)解:联立两直线解析式成方程组,,解得:,∴点 P 的坐标为( 20, 25).当x=0 时, y=x+5=5,∴点 C 的坐标为( 0, 5),∴ BC=15﹣ 5=10,∴S△PBC= BC?x P= ×10×20=100.【考点】两条直线相交或平行问题k、b 【解析】【分析】( 1)将点 A 和点 B 的坐标代入直线的解析式得到关于k、b 的方程组,从而可求得的值,于是可得到直线AB 的解析式;( 2)联立两直线解析式成方程组,通过解方程组可得出点P 的坐标,由一次函数图象上点的坐标特征可求出点 C 的坐标,进而可得出线段BC的长度,最后利用三角形的面积公式求解即可.23.【答案】( 1)解:由题意可得,当0≤x≤9且 x 为正整数时, y=1﹣ 0.1x ,当x≥10 且 x 为正整数时, y=0.1 ,即 y 关于 x 的函数解析式是y=( 2)解:由题意可得,当 0≤x≤9时, 1﹣ 0.1x > 0.5 ,可得, x< 5,则当 x≤x< 5 且 x 为正整数时,选择 B 品牌的共享单车;当 0≤x≤9时, 1﹣ 0.1x=0.5 ,得 x=5,则 x=5 时,选择 A 或 B 品牌的共享单车消费一样;当 0≤x≤9时, 1﹣ 0.1x < 0.5 ,得 x>5,则 x>5 且 x 为正整数,选择 A 品牌的共享单车;当 x≥10 且 x 为正整数时, 0.1 < 0.5 ,故答案为:项 A 品牌的共享单车.【考点】二元一次方程组的应用,一次函数的应用【解析】【分析】( 1)可分为 0≤x≤9且 x 为正整数或x≥10 且 x 为正整数两种情况列出y 与 x 的函数关系式;( 2)分为 0≤x≤9;0≤x≤9;0≤x≤9;当x≥10 四种情况列出关于x 的方程或不等式,然后再进行求解即可 .24. 【答案】( 1)解:∵∠ M=∠N=∠MBC=90°,∴四边形MNCB是矩形,∵ MB=MN=2,∴矩形 MNCB是正方形,∴ NC=CB=2,由折叠得:AN=AC=NC=1,Rt △ACB中,由勾股定理得:AB==,∴ AD=AB=,∴ CD=AD﹣ AC=﹣ 1;( 2)解:四边形ABQD是菱形,理由是:由折叠得: AB=AD,∠ BAQ=∠QAD,∵BQ∥ AD,∴∠BQA=∠QAD,∴∠BAQ=∠BQA,∴AB=BQ,∴ BQ=AD, BQ∥ AD,∴四边形 ABQD是平行四边形,∵AB=AD,∴四边形ABQD是菱形.【考点】正方形的判定与性质【解析】【分析】( 1)首先证明四边形 MNCB为正方形,然后再依据折叠的性质得到: CA=1,AB=AD,最后再依据 CD=AD-AC求解即可;( 2)根据平行线的性质和折叠的性质可得到∠BAQ=∠ BQA,然后依据等角对等边的性质得到AB=BQ,接下来,依据一组对边平行且相等的四边形为平行四边形可证明四边形ABQD是平行四边形,再由AB=AD,可得四边形 ABQD是菱形 .25.【答案】( 1)解:如图 1 中,延长 BP交 DE于 M.∵四边形ABCD是正方形,∴CB=CD,∠BCP=∠DCE=90°,∵ CP=CE,∴△ BCP≌△ DCE,∴∠ BCP=∠CDE,∵∠ CBP+∠CPB=90°,∠ CPB=∠DPM,∴∠ CDE+∠DPM=90°,∴∠ DMP=90°,∴BP⊥ DE.( 2)解:由题意S1﹣ S2=(4+x)?x﹣?( 4﹣x) ?x=x2( 0< x<4).( 3)解:①如图 2 中,当∠ PBF=30°时,∵∠ CPE=∠CEP=∠DPF=45°,∠ FDP=90°,∴∠ PFD=∠DPF=45°,∴DF=DP,∵ AD=CD,∴AF=PC,∵ AB=BC,∠A=∠BCP=90°,∴△ BAF≌△ BCP,∴∠ ABF=∠CBP=30°,∴ x=PC=BC?tan30 ° =,∴S1﹣ S2=x2=.②如图 3 中,当∠ PBF=45°时,在CB上截取 CN=CP,理解 PN.由①可知△ ABF≌△ BCP,∴∠ ABF=∠CBP,∵∠ PBF=45°,∴∠ CBP=22.5°,∵∠ CNP=∠NBP+∠NPB=45°,∴∠ NBP=∠NPB=22.5°,∴BN=PN= x,∴x+x=4,∴x=4﹣4,∴ S1﹣ S2=( 4﹣4)2=48﹣32.【考点】正方形的性质【解析】【分析】( 1)首先延长 BP 交 DE于 M.然后依据 SAS可证明△ BCP≌△ DCE,依据全等三角形的性质可得到∠ BCP=∠ CDE,由∠ CBP+∠CPB=90°,∠ CPB=∠DPM,即可推出∠ CDE+∠DPM=90°;( 2)根据题意可得到S1-S 2=S△PBE-S △PDE,然后依据三角形的面积公式列出函数关系式即可;( 3)分当∠ PBF=30°和∠ PBF=45°两种情形分别求出PC的长,最后再利用(2)中结论进行计算即可.。

最新广东省广州市越秀区-2017学年八年级下册数学期末考试试卷(解析版)

最新广东省广州市越秀区-2017学年八年级下册数学期末考试试卷(解析版)

广东省广州市越秀区2016-2017学年八年级下册数学期末考试试卷(解析版)一、选择题1.下列式子没有意义的是()A. B.C.D.2.下列计算中,正确的是()A.÷ =B.(4 )2=8 C.=2 D. 2 ×2 =23.刻画一组数据波动大小的统计量是()A.平均数B.方差 C.众数 D.中位数4.在暑假到来之前,某机构向八年级学生推荐了A,B,C三条游学线路,现对全级学生喜欢哪一条游学线路作调查,以决定最终的游学线路,下面的统计量中最值得关注的是()A.方差B.平均数 C.中位数 D.众数5.关于正比例函数y=﹣2x,下列结论中正确的是()A.函数图象经过点(﹣2,1) B.y随x的增大而减小C.函数图象经过第一、三象限 D.不论x取何值,总有y <06.以下列各组数据中的三个数作为三角形的边长,其中能构成直角三角形的是()A.2,3,4B.,,C.1,,2 D.7,8,97.若一个直角三角形的一条直角边长是5cm,另一条直角边比斜边短1cm,则斜边长为()cm.A.10B.11 C.12 D.13 8.如图,在菱形ABCD中,对角线AC,BD相交于点O,AB=5,AC=6,则菱形ABCD的面积是()A.24B.26 C.30 D.489.在下列命题中,是假命题的是()A.有一个角是直角的平行四边形是矩形B.一组邻边相等的矩形是正方形C.一组对边平行且相等的四边形是平行四边形D.有两组邻边相等的四边形是菱形10.已知平面上四点A(0,0),B(10,0),C(12,6),D(2,6),直线y=mx﹣3m+6将四边形ABCD 分成面积相等的两部分,则m的值为()A. B.﹣1 C.2D.二、填空题11.已知a= +2,b= ﹣2,则ab=________.12.一次函数y=kx+b(k≠0)中,x与y的部分对应值如下表:那么,一元一次方程kx+b=0的解是x=________.13.如图是一次函数y=mx+n的图象,则关于x的不等式mx+n>2的解集是________.14.一组数据:2017、2017、2017、2017、2017,它的方差是________.15.考古学家们发现了几块大约完成于公元前2000年左右的古巴比伦的泥版书,据专家们考证,其中一块上面刻有如下问题:“一根长度为30个单位的棍子直立在墙上,当其上端垂直滑下6个单位时,请问其下端离开墙角有多远?”,这个问题的答案是:其下端离开墙角________个单位.16.如图所示,在Rt△ABC中,∠A=90°,DE∥BC,F,G,H,I分别是DE,BE,BC,CD的中点,连接FG,GH,HI,IF,FH,GI.对于下列结论:①∠GFI=90°;②GH=GI;③GI= (BC﹣DE);④四边形FGHI是正方形.其中正确的是________(请写出所有正确结论的序号).三、解答题17.计算:(+ ﹣)× .18.如图,在△ABC中,AD⊥BC,AB=5,BD=4,CD= .(1)求AD的长.(2)求△ABC的周长.19.如图在平行四边形ABCD中,AC交BD于点O,AE⊥BD,CF⊥BD,垂足分别为E、F,求证:四边形AECF 为平行四边形.20.下表是某校八年级(1)班43名学生右眼视力的检查结果.(1)该班学生右眼视力的平均数是________(结果保留1位小数).(2)该班学生右眼视力的中位数是________.(3)该班小鸣同学右眼视力是4.5,能不能说小鸣同学的右眼视力处于全班同学的中上水平?试说明理由.21.如图,正方形ABCD的对角线相交于点O,BC=6,延长BC至点E,使得CE=8,点F是DE的中点,连接CF、OF.(1)求OF的长.(2)求CF的长.22.如图,在平面直角坐标系中,直线y=kx+b经过点A(﹣30,0)和点B(0,15),直线y=x+5与直线y=kx+b相交于点P,与y轴交于点C.(1)求直线y=kx+b的解析式.(2)求△PBC的面积.23.2016年下半年开始,不同品牌的共享单车出现在城市的大街小巷.现已知A品牌共享单车计费方式为:初始骑行单价为1元/半小时,不足半小时按半小时计算.内设邀请机制,每邀请一位好友注册认证并充值押金成功,双方骑行单价均降价0.1元/半小时,骑行单价最低可降至0.1元/半小时(比如,某用户邀请了3位好友,则骑行单价为0.7元/半小时).B品牌共享单车计费方式为:0.5元/半小时,不足半小时按半小时计算.(1)某用户准备选择A品牌共享单车使用,设该用户邀请好友x名(x为整数,x≥0),该用户的骑行单价为y元/半小时.请写出y关于x的函数解析式.(2)若有A,B两种品牌的共享单车各一辆供某用户一人选择使用,请你根据该用户已邀请好友的人数,给出经济实惠的选择建议.24.下面我们做一次折叠活动:第一步,在一张宽为2的矩形纸片的一端,利用图(1)的方法折出一个正方形,然后把纸片展平,折痕为MC;第二步,如图(2),把这个正方形折成两个相等的矩形,再把纸片展平,折痕为FA;第三步,折出内侧矩形FACB的对角线AB,并将AB折到图(3)中所示的AD处,折痕为AQ.根据以上的操作过程,完成下列问题:(1)求CD的长.(2)请判断四边形ABQD的形状,并说明你的理由.25.如图,正方形ABCD中,AB=4,P是CD边上的动点(P点不与C、D重合),过点P作直线与BC的延长线交于点E,与AD交于点F,且CP=CE,连接DE、BP、BF,设CP═x,△PBF的面积为S1,△PDE的面积为S2.(1)求证:BP⊥DE.(2)求S1﹣S2关于x的函数解析式,并写出x的取值范围.(3)分别求当∠PBF=30°和∠PBF=45°时,S1﹣S2的值.答案解析部分一、<b >选择题</b>1.【答案】B【考点】二次根式有意义的条件【解析】【解答】A、有意义,A不合题意;B、没有意义,B符合题意;C、有意义,C不合题意;D、有意义,D不合题意;故答案为:B.【分析】依据二次根式被开放数为非负数求解即可.2.【答案】C【考点】二次根式的性质与化简,二次根式的乘除法【解析】【解答】解:A、原式= = =3,A不符合题意;B、原式=32,B不符合题意;C、原式=|﹣2|=2,C符合题意;D、原式=4 ,D不符合题意;故答案为:C.【分析】依据二次根式的除法法则可对A作出判断;依据二次根式的性质可对B、C作出判断,依据二次根式的乘法法则可对D作出判断.3.【答案】B【考点】统计量的选择【解析】【解答】由于方差反映数据的波动情况,衡量一组数据波动大小的统计量是方差.故答案为:B.【分析】方差是反应一组数据波动大小的量.4.【答案】D【考点】统计量的选择【解析】【解答】由于众数是数据中出现次数最多的数,故全级学生喜欢的游学线路最值得关注的应该是统计调查数据的众数.故答案为:D.【分析】决定最终的线路应改由多数人员的意见决定,故此可得到问题的答案.5.【答案】B【考点】正比例函数的图象和性质【解析】【解答】解:A、当x=﹣2时,y=﹣2×(﹣2)=4,即图象经过点(﹣2,4),不经过点(﹣2,1),故本选项错误;B、由于k=﹣2<0,所以y随x的增大而减小,故本选项正确;C、由于k=﹣2<0,所以图象经过二、四象限,故本选项错误;D、∵x>0时,y<0,x<0时,y>0,∴不论x为何值,总有y<0错误,故本选项错误.故答案为:B.【分析】依据正比例函数的图像和性质可对B、C、D作出判断,将x=-2代入函数解析式可求得y的值,从而可对A作出判断.6.【答案】C【考点】勾股定理的逆定理【解析】【解答】A、22+32≠42,故不是直角三角形,A不符合题意;B、()2+()2≠()2,故不是直角三角形,B不符合题意;C、12+()2=22,故是直角三角形,C符合题意;D、72+82≠92,故不是直角三角形,D不符合题意;故答案为:C.【分析】依据勾股定理的逆定理进行判断即可.7.【答案】D【考点】勾股定理【解析】【解答】设斜边长为xcm,则另一条直角边为(x﹣1)cm,由勾股定理得,x2=52+(x﹣1)2,解得,x=13,则斜边长为13cm,故答案为:D.【分析】设斜边长为xcm,则另一条直角边为(x-1)cm,然后依据勾股定理列方程求解即可.8.【答案】A【考点】菱形的性质【解析】【解答】∵四边形ABCD是菱形,∴OA=OC=3,OB=OD,AC⊥BD,在Rt△AOB中,∠AOB=90°,根据勾股定理,得:OB= ,= ,=4,∴BD=2OB=8,∴S菱形ABCD= ×AC×BD= ×6×8=24.故答案为:A.【分析】根据菱形的对角线互相垂直且互相平分可得到AC⊥BD,且AO=OC=3,然后依据勾股定理可求得BO 的长,从而可得到BD的长,最后依据菱形的面积等于对角线乘积的一半求解即可.9.【答案】D【考点】命题与定理【解析】【解答】A、有一个角是直角的平行四边形是矩形,正确,A不符合题意;B、一组邻边相等的矩形是正方形,正确,B不符合题意;;C、一组对边平行且相等的四边形是平行四边形,正确,C不符合题意;D、有两组邻边相等且平行的四边形是菱形,错误,D不符合题意.故答案为:D.【分析】首先依据矩形的定义、正方形的判定定理、平行四边形的判定定理、菱形的判定定理判定命题的对错,从而可做出判断.10.【答案】B【考点】待定系数法求一次函数解析式【解析】【解答】解:如图,∵A(0,0),B(10,0),C(12,6),D(2,6),∴AB=10﹣0=10,CD=12﹣2=10,又点C、D的纵坐标相同,∴AB∥CD且AB=CD,∴四边形ABCD是平行四边形,∵12÷2=6,6÷2=3,∴对角线交点P的坐标是(6,3),∵直线y=mx﹣3m+6将四边形ABCD分成面积相等的两部分,∴直线y=mx﹣3m+6经过点P,∴6m﹣3m+6=3,解得m=﹣1.故答案为:B.【分析】首先依据各点的坐标可确定出四边形ABCD为平行四边形,然后可求得两对角线交点的坐标,然后由直线平分线四边形的面积可知直线经过点(6,3),最后将点(6,3)代入直线解析式求解即可.二、<b >填空题</b>11.【答案】1【考点】分母有理化【解析】【解答】解:∵a= +2,b= ﹣2,∴ab=(+2)(﹣2)=5﹣4=1,故答案为:1【分析】依据平方差公式和二次根式的性质进行计算即可.12.【答案】1【考点】一次函数与一元一次方程【解析】【解答】解:根据上表中的数据值,当y=0时,x=1,即一元一次方程kx+b=0的解是x=1.故答案是:1.【分析】依据表格找出当y=0时,对应的x的取值即可.13.【答案】x>0【考点】一次函数与一元一次不等式【解析】【解答】解:由题意,可知一次函数y=mx+n的图象经过点(0,2),且y随x的增大而增大,所以关于x的不等式mx+n>2的解集是x>0.故答案为:x>0.【分析】不等式的解集为当y>2时,函数自变量的取值范围.14.【答案】0【考点】方差【解析】【解答】解:该组数据一样,没有波动,方差为0,故答案为:0.【分析】方差的意义或利用方差公式进行解答即可.15.【答案】18【考点】勾股定理的应用【解析】【解答】解:∵PC=AB=30,PA=6,∴AC=24,∴BC= = =18,∴下端离开墙角18个单位.故答案为:18.【分析】根据题意可得到PC=AB=30,AC=24,然后在Rt△ABC中利用勾股定理求出CB的长即可.16.【答案】①③【考点】中点四边形【解析】【解答】解:延长IF交AB于K,∵DF=EF,BG=GE,∴FG= BD,GF∥AB,同理IF∥AC,HI= BD,HI∥BD,∴∠BKI=∠A=90°,∴∠GFI=∠BKI=90°,∴GF⊥FI,故①正确,∴FG=HI,FG∥HI,∴四边形FGHI是平行四边形,∵∠GFI=90°,∴四边形FGHI是矩形,故②④错误,延长EI交BC于N,则△DEI≌△CNI,∴DE=CN,EJ=JN,∵EG=GB,EI=IN,∴GI= BHN= (BC﹣DE),故③正确,故答案为①③.【分析】对于①,延长IF交AB于K,然后根据两直线平行同位角相等进行解答即可;对于②和④.只要证明四边形FGHI是矩形即可判断;对于③,先延长EI交BC于N,然后再证明△DEI≌△CNI,依据全等三角形的性质可得到DE=CN,EJ=JN,然后再结合中点的定义可推出GI=HN=(BC-DE).三、<b >解答题</b>17.【答案】解:原式=(6 + ﹣3 )×= ×=7.【考点】二次根式的混合运算【解析】【分析】先将各二次根式化简为最简二次根式,然后再合并同类二次根式,最后,在依据二次根式的乘法法则进行计算即可.18.【答案】(1)解:在Rt△ABD中,AD= =3(2)解:在Rt△ACD中,AC= =2 ,则△ABC的周长=AB+AC+BC=5+4+ +2 =9+3【考点】勾股定理【解析】【分析】(1)在Rt△ABD中,依据勾股定理可求得AD的长;(2)在Rt△ACD中,依据勾股定理可求得AC的长,然后再依据三角形的周长等于三边长度之和求解即可.19.【答案】证明:∵四边形ABCD是平行四边形,∴AB=CD,AB∥CD,∴∠ABE=∠CDF,∵AE⊥BD,CF⊥BD,∴AE∥CF,∠AEB=∠CFD=90°,在△AEB和△CFD中,∵,∴△AEB≌△CFD(AAS),∴AE=CF,∴四边形AECF是平行四边形.【考点】平行四边形的判定与性质【解析】【分析】首先依据四边形的性质可得AB=CD,AB∥CD,然后再证明AE∥CF,接下来,利用AAS证得△AEB≌△CFD,依据全等三角形的性质可得到AE=CF,最后依据一组对边相等且平行的四边形是平行四边形进行证明即可.20.【答案】(1)4.6(2)4.7(3)解:不能,∵小鸣同学右眼视力是4.5,小于中位数4.7,∴不能说小鸣同学的右眼视力处于全班同学的中上水平.【考点】中位数、众数【解析】【解答】解:(1)该班学生右眼视力的平均数是×(4.0+4.1×2+4.2×5+4.3×4+4.4×3+4.5×5+4.6+4.7+4.8×5+4.9×10+5.0×6)≈4.6,故答案为:4.6;(2)由于共有43个数据,其中位数为第22个数据,即中位数为4.7,(3)不能,∵小鸣同学右眼视力是4.5,小于中位数4.7,∴不能说小鸣同学的右眼视力处于全班同学的中上水平.故答案为:(1)4.6;(2)4.7;(3)不能.【分析】(1)根据加权平均数公式求解即可;(2)首先将这组数据按照从小到大的顺序排列,中位数为第22个数据;(3)根据小鸣同学右眼视力是4.5,小于中位数4.7,故此可得到问题的答案.21.【答案】(1)解:∵四边形ABCD是正方形,∴BC=CD=6,∠BCD=∠ECD=90°,OB=OD,∵CE=8,∴BE=14,∵OB=OD,DF=FE,∴OF= BE=7.(2)解:在Rt△DCE中,DE= = =10,∵DF=FE,∴CF= DE=5.【考点】正方形的性质【解析】【分析】(1)由正方形的性质可知O为BD的中点,故此OF是△DBE的中位线,然后依据三角形中位线的性质解答即可;(2)在Rt△DCE中,利用勾股定理求出DE,再利用直角三角形斜边上中线等于斜边的一半求解即可. 22.【答案】(1)解:将点A(﹣30,0)、B(0,15)代入y=kx+b,,解得:,∴直线y=kx+b的解析式为y= x+15.(2)解:联立两直线解析式成方程组,,解得:,∴点P的坐标为(20,25).当x=0时,y=x+5=5,∴点C的坐标为(0,5),∴BC=15﹣5=10,∴S△PBC= BC•x P= ×10×20=100.【考点】两条直线相交或平行问题【解析】【分析】(1)将点A和点B的坐标代入直线的解析式得到关于k、b的方程组,从而可求得k、b 的值,于是可得到直线AB的解析式;(2)联立两直线解析式成方程组,通过解方程组可得出点P的坐标,由一次函数图象上点的坐标特征可求出点C的坐标,进而可得出线段BC的长度,最后利用三角形的面积公式求解即可.23.【答案】(1)解:由题意可得,当0≤x≤9且x为正整数时,y=1﹣0.1x,当x≥10且x为正整数时,y=0.1,即y关于x的函数解析式是y=(2)解:由题意可得,当0≤x≤9时,1﹣0.1x>0.5,可得,x<5,则当x≤x<5且x为正整数时,选择B品牌的共享单车;当0≤x≤9时,1﹣0.1x=0.5,得x=5,则x=5时,选择A或B品牌的共享单车消费一样;当0≤x≤9时,1﹣0.1x<0.5,得x>5,则x>5且x为正整数,选择A品牌的共享单车;当x≥10且x为正整数时,0.1<0.5,故答案为:项A品牌的共享单车.【考点】二元一次方程组的应用,一次函数的应用【解析】【分析】(1)可分为0≤x≤9且x为正整数或x≥10且x为正整数两种情况列出y与x的函数关系式;(2)分为0≤x≤9;0≤x≤9;0≤x≤9;当x≥10四种情况列出关于x的方程或不等式,然后再进行求解即可.24.【答案】(1)解:∵∠M=∠N=∠MBC=90°,∴四边形MNCB是矩形,∵MB=MN=2,∴矩形MNCB是正方形,∴NC=CB=2,由折叠得:AN=AC= NC=1,Rt△ACB中,由勾股定理得:AB= = ,∴AD=AB= ,∴CD=AD﹣AC= ﹣1;(2)解:四边形ABQD是菱形,理由是:由折叠得:AB=AD,∠BAQ=∠QAD,∵BQ∥AD,∴∠BQA=∠QAD,∴∠BAQ=∠BQA,∴AB=BQ,∴BQ=AD,BQ∥AD,∴四边形ABQD是平行四边形,∵AB=AD,∴四边形ABQD是菱形.【考点】正方形的判定与性质【解析】【分析】(1)首先证明四边形MNCB为正方形,然后再依据折叠的性质得到:CA=1,AB=AD,最后再依据CD=AD-AC求解即可;(2)根据平行线的性质和折叠的性质可得到∠BAQ=∠BQA,然后依据等角对等边的性质得到AB=BQ,接下来,依据一组对边平行且相等的四边形为平行四边形可证明四边形ABQD是平行四边形,再由AB=AD,可得四边形ABQD是菱形.25.【答案】(1)解:如图1中,延长BP交DE于M.∵四边形ABCD是正方形,∴CB=CD,∠BCP=∠DCE=90°,∵CP=CE,∴△BCP≌△DCE,∴∠BCP=∠CDE,∵∠CBP+∠CPB=90°,∠CPB=∠DPM,∴∠CDE+∠DPM=90°,∴∠DMP=90°,∴BP⊥DE.(2)解:由题意S1﹣S2= (4+x)•x﹣•(4﹣x)•x=x2(0<x<4).(3)解:①如图2中,当∠PBF=30°时,∵∠CPE=∠CEP=∠DPF=45°,∠FDP=90°,∴∠PFD=∠DPF=45°,∴DF=DP,∵AD=CD,∴AF=PC,∵AB=BC,∠A=∠BCP=90°,∴△BAF≌△BCP,∴∠ABF=∠CBP=30°,∴x=PC=BC•tan30°= ,∴S1﹣S2=x2= .②如图3中,当∠PBF=45°时,在CB上截取CN=CP,理解PN.由①可知△ABF≌△BCP,∴∠ABF=∠CBP,∵∠PBF=45°,∴∠CBP=22.5°,∵∠CNP=∠NBP+∠NPB=45°,∴∠NBP=∠NPB=22.5°,∴BN=PN= x,∴x+x=4,∴x=4 ﹣4,∴S1﹣S2=(4 ﹣4)2=48﹣32 .【考点】正方形的性质【解析】【分析】(1)首先延长BP交DE于M.然后依据SAS可证明△BCP≌△DCE,依据全等三角形的性质可得到∠BCP=∠CDE,由∠CBP+∠CPB=90°,∠CPB=∠DPM,即可推出∠CDE+∠DPM=90°;(2)根据题意可得到S1-S2=S△PBE-S△PDE,然后依据三角形的面积公式列出函数关系式即可;(3)分当∠PBF=30°和∠PBF=45°两种情形分别求出PC的长,最后再利用(2)中结论进行计算即可.。

广东省广州市越秀区2016-2017学年八年级下册数学期末考试试卷(解析版)

广东省广州市越秀区2016-2017学年八年级下册数学期末考试试卷(解析版)

广东省广州市越秀区2016-2017学年八年级下册数学期末考试试卷(解析版)一、选择题1.下列式子没有意义的是()A. B.C.D.2.下列计算中,正确的是()A.÷ =B.(4 )2=8 C.=2 D. 2 ×2 =23.刻画一组数据波动大小的统计量是()A.平均数B.方差 C.众数 D.中位数4.在暑假到来之前,某机构向八年级学生推荐了A,B,C三条游学线路,现对全级学生喜欢哪一条游学线路作调查,以决定最终的游学线路,下面的统计量中最值得关注的是()A.方差B.平均数 C.中位数 D.众数5.关于正比例函数y=﹣2x,下列结论中正确的是()A.函数图象经过点(﹣2,1) B.y随x的增大而减小C.函数图象经过第一、三象限 D.不论x取何值,总有y <06.以下列各组数据中的三个数作为三角形的边长,其中能构成直角三角形的是()A.2,3,4B.,,C.1,,2 D.7,8,97.若一个直角三角形的一条直角边长是5cm,另一条直角边比斜边短1cm,则斜边长为()cm.A.10B.11 C.12 D.13 8.如图,在菱形ABCD中,对角线AC,BD相交于点O,AB=5,AC=6,则菱形ABCD的面积是()A.24B.26 C.30 D.489.在下列命题中,是假命题的是()A.有一个角是直角的平行四边形是矩形B.一组邻边相等的矩形是正方形C.一组对边平行且相等的四边形是平行四边形D.有两组邻边相等的四边形是菱形10.已知平面上四点A(0,0),B(10,0),C(12,6),D(2,6),直线y=mx﹣3m+6将四边形ABCD 分成面积相等的两部分,则m的值为()A. B.﹣1 C.2D.二、填空题11.已知a= +2,b= ﹣2,则ab=________.12.一次函数y=kx+b(k≠0)中,x与y的部分对应值如下表:x ﹣2 ﹣1 0 1 2y ﹣6 ﹣4 ﹣2 0 2那么,一元一次方程kx+b=0的解是x=________.13.如图是一次函数y=mx+n的图象,则关于x的不等式mx+n>2的解集是________.14.一组数据:2017、2017、2017、2017、2017,它的方差是________.15.考古学家们发现了几块大约完成于公元前2000年左右的古巴比伦的泥版书,据专家们考证,其中一块上面刻有如下问题:“一根长度为30个单位的棍子直立在墙上,当其上端垂直滑下6个单位时,请问其下端离开墙角有多远?”,这个问题的答案是:其下端离开墙角________个单位.16.如图所示,在Rt△ABC中,∠A=90°,DE∥BC,F,G,H,I分别是DE,BE,BC,CD的中点,连接FG,GH,HI,IF,FH,GI.对于下列结论:①∠GFI=90°;②GH=GI;③GI= (BC﹣DE);④四边形FGHI是正方形.其中正确的是________(请写出所有正确结论的序号).三、解答题17.计算:(+ ﹣)× .18.如图,在△ABC中,AD⊥BC,AB=5,BD=4,CD= .(1)求AD的长.(2)求△ABC的周长.19.如图在平行四边形ABCD中,AC交BD于点O,AE⊥BD,CF⊥BD,垂足分别为E、F,求证:四边形AECF 为平行四边形.20.下表是某校八年级(1)班43名学生右眼视力的检查结果.视力4.0 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 5.0人数1 2 5 4 3 5 1 1 5 10 6(1)该班学生右眼视力的平均数是________(结果保留1位小数).(2)该班学生右眼视力的中位数是________.(3)该班小鸣同学右眼视力是 4.5,能不能说小鸣同学的右眼视力处于全班同学的中上水平?试说明理由.21.如图,正方形ABCD的对角线相交于点O,BC=6,延长BC至点E,使得CE=8,点F是DE的中点,连接CF、OF.(1)求OF的长.(2)求CF的长.22.如图,在平面直角坐标系中,直线y=kx+b经过点A(﹣30,0)和点B(0,15),直线y=x+5与直线y=kx+b相交于点P,与y轴交于点C.(1)求直线y=kx+b的解析式.(2)求△PBC的面积.23.2016年下半年开始,不同品牌的共享单车出现在城市的大街小巷.现已知A品牌共享单车计费方式为:初始骑行单价为1元/半小时,不足半小时按半小时计算.内设邀请机制,每邀请一位好友注册认证并充值押金成功,双方骑行单价均降价0.1元/半小时,骑行单价最低可降至0.1元/半小时(比如,某用户邀请了3位好友,则骑行单价为0.7元/半小时).B品牌共享单车计费方式为:0.5元/半小时,不足半小时按半小时计算.(1)某用户准备选择A品牌共享单车使用,设该用户邀请好友x名(x为整数,x≥0),该用户的骑行单价为y元/半小时.请写出y关于x的函数解析式.(2)若有A,B两种品牌的共享单车各一辆供某用户一人选择使用,请你根据该用户已邀请好友的人数,给出经济实惠的选择建议.24.下面我们做一次折叠活动:第一步,在一张宽为2的矩形纸片的一端,利用图(1)的方法折出一个正方形,然后把纸片展平,折痕为MC;第二步,如图(2),把这个正方形折成两个相等的矩形,再把纸片展平,折痕为FA;第三步,折出内侧矩形FACB的对角线AB,并将AB折到图(3)中所示的AD处,折痕为AQ.根据以上的操作过程,完成下列问题:(1)求CD的长.(2)请判断四边形ABQD的形状,并说明你的理由.25.如图,正方形ABCD中,AB=4,P是CD边上的动点(P点不与C、D重合),过点P作直线与BC的延长线交于点E,与AD交于点F,且CP=CE,连接DE、BP、BF,设CP═x,△PBF的面积为S1,△PDE的面积为S2.(1)求证:BP⊥DE.(2)求S1﹣S2关于x的函数解析式,并写出x的取值范围.(3)分别求当∠PBF=30°和∠PBF=45°时,S1﹣S2的值.答案解析部分一、<b >选择题</b>1.【答案】B【考点】二次根式有意义的条件【解析】【解答】A、有意义,A不合题意;B、没有意义,B符合题意;C、有意义,C不合题意;D、有意义,D不合题意;故答案为:B.【分析】依据二次根式被开放数为非负数求解即可.2.【答案】C【考点】二次根式的性质与化简,二次根式的乘除法【解析】【解答】解:A、原式= = =3,A不符合题意;B、原式=32,B不符合题意;C、原式=|﹣2|=2,C符合题意;D、原式=4 ,D不符合题意;故答案为:C.【分析】依据二次根式的除法法则可对A作出判断;依据二次根式的性质可对B、C作出判断,依据二次根式的乘法法则可对D作出判断.3.【答案】B【考点】统计量的选择【解析】【解答】由于方差反映数据的波动情况,衡量一组数据波动大小的统计量是方差.故答案为:B.【分析】方差是反应一组数据波动大小的量.4.【答案】D【考点】统计量的选择【解析】【解答】由于众数是数据中出现次数最多的数,故全级学生喜欢的游学线路最值得关注的应该是统计调查数据的众数.故答案为:D.【分析】决定最终的线路应改由多数人员的意见决定,故此可得到问题的答案.5.【答案】B【考点】正比例函数的图象和性质【解析】【解答】解:A、当x=﹣2时,y=﹣2×(﹣2)=4,即图象经过点(﹣2,4),不经过点(﹣2,1),故本选项错误;B、由于k=﹣2<0,所以y随x的增大而减小,故本选项正确;C、由于k=﹣2<0,所以图象经过二、四象限,故本选项错误;D、∵x>0时,y<0,x<0时,y>0,∴不论x为何值,总有y<0错误,故本选项错误.故答案为:B.【分析】依据正比例函数的图像和性质可对B、C、D作出判断,将x=-2代入函数解析式可求得y的值,从而可对A作出判断.6.【答案】C【考点】勾股定理的逆定理【解析】【解答】A、22+32≠42,故不是直角三角形,A不符合题意;B、()2+()2≠()2,故不是直角三角形,B不符合题意;C、12+()2=22,故是直角三角形,C符合题意;D、72+82≠92,故不是直角三角形,D不符合题意;故答案为:C.【分析】依据勾股定理的逆定理进行判断即可.7.【答案】D【考点】勾股定理【解析】【解答】设斜边长为xcm,则另一条直角边为(x﹣1)cm,由勾股定理得,x2=52+(x﹣1)2,解得,x=13,则斜边长为13cm,故答案为:D.【分析】设斜边长为xcm,则另一条直角边为(x-1)cm,然后依据勾股定理列方程求解即可.8.【答案】A【考点】菱形的性质【解析】【解答】∵四边形ABCD是菱形,∴OA=OC=3,OB=OD,AC⊥BD,在Rt△AOB中,∠AOB=90°,根据勾股定理,得:OB= ,= ,=4,∴BD=2OB=8,∴S菱形ABCD=×AC×BD= ×6×8=24.故答案为:A.【分析】根据菱形的对角线互相垂直且互相平分可得到AC⊥BD,且AO=OC=3,然后依据勾股定理可求得BO 的长,从而可得到BD的长,最后依据菱形的面积等于对角线乘积的一半求解即可.9.【答案】D【考点】命题与定理【解析】【解答】A、有一个角是直角的平行四边形是矩形,正确,A不符合题意;B、一组邻边相等的矩形是正方形,正确,B不符合题意;;C、一组对边平行且相等的四边形是平行四边形,正确,C不符合题意;D、有两组邻边相等且平行的四边形是菱形,错误,D不符合题意.故答案为:D.【分析】首先依据矩形的定义、正方形的判定定理、平行四边形的判定定理、菱形的判定定理判定命题的对错,从而可做出判断.10.【答案】 B【考点】待定系数法求一次函数解析式【解析】【解答】解:如图,∵A(0,0),B(10,0),C(12,6),D(2,6),∴AB=10﹣0=10,CD=12﹣2=10,又点C、D的纵坐标相同,∴AB∥CD且AB=CD,∴四边形ABCD是平行四边形,∵12÷2=6,6÷2=3,∴对角线交点P的坐标是(6,3),∵直线y=mx﹣3m+6将四边形ABCD分成面积相等的两部分,∴直线y=mx﹣3m+6经过点P,∴6m﹣3m+6=3,解得m=﹣1.故答案为:B.【分析】首先依据各点的坐标可确定出四边形ABCD为平行四边形,然后可求得两对角线交点的坐标,然后由直线平分线四边形的面积可知直线经过点(6,3),最后将点(6,3)代入直线解析式求解即可.二、<b >填空题</b>11.【答案】 1【考点】分母有理化【解析】【解答】解:∵a= +2,b= ﹣2,∴ab=(+2)(﹣2)=5﹣4=1,故答案为: 1【分析】依据平方差公式和二次根式的性质进行计算即可.12.【答案】 1【考点】一次函数与一元一次方程【解析】【解答】解:根据上表中的数据值,当y=0时,x=1,即一元一次方程kx+b=0的解是x=1.故答案是:1.【分析】依据表格找出当y=0时,对应的x的取值即可.13.【答案】x>0【考点】一次函数与一元一次不等式【解析】【解答】解:由题意,可知一次函数y=mx+n的图象经过点(0,2),且y随x的增大而增大,所以关于x的不等式mx+n>2的解集是x>0.故答案为:x>0.【分析】不等式的解集为当y>2时,函数自变量的取值范围.14.【答案】0【考点】方差【解析】【解答】解:该组数据一样,没有波动,方差为0,故答案为:0.【分析】方差的意义或利用方差公式进行解答即可.15.【答案】18【考点】勾股定理的应用【解析】【解答】解:∵PC=AB=30,PA=6,∴AC=24,∴BC= = =18,∴下端离开墙角18个单位.故答案为:18.【分析】根据题意可得到PC=AB=30,AC=24,然后在Rt△ABC中利用勾股定理求出CB的长即可.16.【答案】①③【考点】中点四边形【解析】【解答】解:延长IF交AB于K,∵DF=EF,BG=GE,∴FG= BD,GF∥AB,同理IF∥AC,HI= BD,HI∥BD,∴∠BKI=∠A=90°,∴∠GFI=∠BKI=90°,∴GF⊥FI,故①正确,∴FG=HI,FG∥HI,∴四边形FGHI是平行四边形,∵∠GFI=90°,∴四边形FGHI是矩形,故②④错误,延长EI交BC于N,则△DEI≌△CNI,∴DE=CN,EJ=JN,∵EG=GB,EI=IN,∴GI= BHN= (BC﹣DE),故③正确,故答案为①③.【分析】对于①,延长IF交AB于K,然后根据两直线平行同位角相等进行解答即可;对于②和④.只要证明四边形FGHI是矩形即可判断;对于③,先延长EI交BC于N,然后再证明△DEI≌△CNI,依据全等三角形的性质可得到DE=CN,EJ=JN,然后再结合中点的定义可推出GI=HN=(BC-DE).三、<b >解答题</b>17.【答案】解:原式=(6 + ﹣3 )× = × =7.【考点】二次根式的混合运算【解析】【分析】先将各二次根式化简为最简二次根式,然后再合并同类二次根式,最后,在依据二次根式的乘法法则进行计算即可.18.【答案】(1)解:在Rt△ABD中,AD= =3(2)解:在Rt△ACD中,AC= =2 ,则△ABC的周长=AB+AC+BC=5+4+ +2 =9+3【考点】勾股定理【解析】【分析】(1)在Rt△ABD中,依据勾股定理可求得AD的长;(2)在Rt△ACD中,依据勾股定理可求得AC的长,然后再依据三角形的周长等于三边长度之和求解即可.19.【答案】证明:∵四边形ABCD是平行四边形,∴AB=CD,AB∥CD,∴∠ABE=∠CDF,∵AE⊥BD,CF⊥BD,∴AE∥CF,∠AEB=∠CFD=90°,在△AEB和△CFD中,∵,∴△AEB≌△CFD(AAS),∴AE=CF,∴四边形AECF是平行四边形.【考点】平行四边形的判定与性质【解析】【分析】首先依据四边形的性质可得AB=CD,AB∥CD,然后再证明AE∥CF,接下来,利用AAS证得△AEB≌△CFD,依据全等三角形的性质可得到AE=CF,最后依据一组对边相等且平行的四边形是平行四边形进行证明即可.20.【答案】(1)4.6(2)4.7(3)解:不能,∵小鸣同学右眼视力是 4.5,小于中位数 4.7,∴不能说小鸣同学的右眼视力处于全班同学的中上水平.【考点】中位数、众数【解析】【解答】解:(1)该班学生右眼视力的平均数是×(4.0+4.1×2+4.2×5+4.3×4+4.4×3+4.5×5+4.6+4.7+4.8×5+4.9×10+5.0×6)≈4.6,故答案为: 4.6;(2)由于共有43个数据,其中位数为第22个数据,即中位数为 4.7,(3)不能,∵小鸣同学右眼视力是 4.5,小于中位数 4.7,∴不能说小鸣同学的右眼视力处于全班同学的中上水平.故答案为:(1)4.6;(2)4.7;(3)不能.【分析】(1)根据加权平均数公式求解即可;(2)首先将这组数据按照从小到大的顺序排列,中位数为第22个数据;(3)根据小鸣同学右眼视力是 4.5,小于中位数 4.7,故此可得到问题的答案.21.【答案】(1)解:∵四边形ABCD是正方形,∴BC=CD=6,∠BCD=∠ECD=90°,OB=OD,∵CE=8,∴BE=14,∵OB=OD,DF=FE,∴OF= BE=7.(2)解:在Rt△DCE中,DE= = =10,∵DF=FE,∴CF= DE=5.【考点】正方形的性质【解析】【分析】(1)由正方形的性质可知O为BD的中点,故此OF是△DBE的中位线,然后依据三角形中位线的性质解答即可;(2)在Rt△DCE中,利用勾股定理求出DE,再利用直角三角形斜边上中线等于斜边的一半求解即可.22.【答案】(1)解:将点A(﹣30,0)、B(0,15)代入y=kx+b,,解得:,∴直线y=kx+b的解析式为y= x+15.(2)解:联立两直线解析式成方程组,,解得:,∴点P的坐标为(20,25).当x=0时,y=x+5=5,∴点C的坐标为(0,5),∴BC=15﹣5=10,∴S△PBC= BC?x P= ×10×20=100.【考点】两条直线相交或平行问题【解析】【分析】(1)将点A和点B的坐标代入直线的解析式得到关于k、b的方程组,从而可求得k、b 的值,于是可得到直线AB的解析式;(2)联立两直线解析式成方程组,通过解方程组可得出点P的坐标,由一次函数图象上点的坐标特征可求出点C的坐标,进而可得出线段BC的长度,最后利用三角形的面积公式求解即可.23.【答案】(1)解:由题意可得,当0≤x≤9且x为正整数时,y=1﹣0.1x,当x≥10且x为正整数时,y=0.1,即y关于x的函数解析式是y=(2)解:由题意可得,当0≤x≤9时,1﹣0.1x>0.5,可得,x<5,则当x≤x<5且x为正整数时,选择B品牌的共享单车;当0≤x≤9时,1﹣0.1x=0.5,得x=5,则x=5时,选择A或B品牌的共享单车消费一样;当0≤x≤9时,1﹣0.1x<0.5,得x>5,则x>5且x为正整数,选择A品牌的共享单车;当x≥10且x为正整数时,0.1<0.5,故答案为:项A品牌的共享单车.【考点】二元一次方程组的应用,一次函数的应用【解析】【分析】(1)可分为0≤x≤9且x为正整数或x≥10且x为正整数两种情况列出y与x的函数关系式;(2)分为0≤x≤9;0≤x≤9;0≤x≤9;当x≥10四种情况列出关于x的方程或不等式,然后再进行求解即可.24.【答案】(1)解:∵∠M=∠N=∠MBC=90°,∴四边形MNCB是矩形,∵MB=MN=2,∴矩形MNCB是正方形,∴NC=CB=2,由折叠得:AN=AC= NC=1,Rt△ACB中,由勾股定理得:AB= = ,∴AD=AB= ,∴CD=AD﹣AC= ﹣1;(2)解:四边形ABQD是菱形,理由是:由折叠得:AB=AD,∠BAQ=∠QAD,∵BQ∥AD,∴∠BQA=∠QAD,∴∠BAQ=∠BQA,∴AB=BQ,∴BQ=AD,BQ∥AD,∴四边形ABQD是平行四边形,∵AB=AD,∴四边形ABQD是菱形.【考点】正方形的判定与性质【解析】【分析】(1)首先证明四边形MNCB为正方形,然后再依据折叠的性质得到:CA=1,AB=AD,最后再依据CD=AD-AC求解即可;(2)根据平行线的性质和折叠的性质可得到∠BAQ=∠BQA,然后依据等角对等边的性质得到AB=BQ,接下来,依据一组对边平行且相等的四边形为平行四边形可证明四边形ABQD是平行四边形,再由AB=AD,可得四边形ABQD是菱形.25.【答案】(1)解:如图1中,延长BP交DE于M.∵四边形ABCD是正方形,∴CB=CD,∠BCP=∠DCE=90°,∵CP=CE,∴△BCP≌△DCE,∴∠BCP=∠CDE,∵∠CBP+∠CPB=90°,∠CPB=∠DPM,∴∠CDE+∠DPM=90°,∴∠DMP=90°,∴BP⊥DE.(2)解:由题意S1﹣S2= (4+x)?x﹣?(4﹣x)?x=x2(0<x<4).(3)解:①如图2中,当∠PBF=30°时,∵∠CPE=∠CEP=∠DPF=45°,∠FDP=90°,∴∠PFD=∠DPF=45°,∴DF=DP,∵AD=CD,∴AF=PC,∵AB=BC,∠A=∠BCP=90°,∴△BAF≌△BCP,∴∠ABF=∠CBP=30°,∴x=PC=BC?tan30°= ,∴S1﹣S2=x2= .②如图3中,当∠PBF=45°时,在CB上截取CN=CP,理解PN.由①可知△ABF≌△BCP,∴∠ABF=∠CBP,∵∠PBF=45°,∴∠CBP=22.5°,∵∠CNP=∠NBP+∠NPB=45°,∴∠NBP=∠NPB=22.5°,∴BN=PN= x,∴x+x=4,∴x=4 ﹣4,∴S1﹣S2=(4 ﹣4)2=48﹣32 .【考点】正方形的性质【解析】【分析】(1)首先延长BP交DE于M.然后依据SAS可证明△BCP≌△DCE,依据全等三角形的性质可得到∠BCP=∠CDE,由∠CBP+∠CPB=90°,∠CPB=∠DPM,即可推出∠CDE+∠DPM=90°;(2)根据题意可得到S1-S2=S△PBE-S△PDE,然后依据三角形的面积公式列出函数关系式即可;(3)分当∠PBF=30°和∠PBF=45°两种情形分别求出PC的长,最后再利用(2)中结论进行计算即可.。

广州大学附属中学2016-2017学年第二学期期中考试 初二年级 数学 试卷及参考答案

广州大学附属中学2016-2017学年第二学期期中考试 初二年级 数学 试卷及参考答案
一、选择题(本大题共10小题,每小题3分,共30分)
1. 【答案】D 2. 【答案】C 3. 【答案】B 4. 【答案】D 5. 【答案】A 6. 【答案】D 7. 【答案】D 8. 【答案】B 9. 【答案】D 10. 【答案】D
二、填空题(本大题共6小题,每小题3分,共18分)
11. 【答案】 12. 【答案】 13. 【答案】 或
米.则 与 的函数关系式为
.(不需要写出自变量的取值范围.)
后两车相距
16. 如图,四边形 ,则
是边长为 的正方形纸片,将其沿
的长为

折叠,使点 落在 边上 点处,点 的对应点为 ,若
三、解答题(本大题共7小题,共72分)
17. 计算:
(1)

(2)

18. 如图, 表示一骑自行车者, 表示一骑摩托车者沿相同的路线由甲地到乙地行驶过程的函数图像,两地相距 千米,请根 据图像解决下列问题:
20. 如图,平行四边形
的对角线 与 相交于 ,


,求证:平行四边形
是菱形.
21. 如图,矩形 四边形.
中, 为 中点, 过点 分别交 、 于点 、 ,连接 和 ,求证:四边形
是平行
22. 如图,以
的三边为边分别作等边



(1) 求证:

(2) 求证:四边形
(3) 解答下列问题:
1
满足
2
满足
. 是平行四边形.
(1) 骑
者出发早,早
小时;骑
者早到达目的地,早
小时;他们在离甲地

里处相遇.
(2) 求两个人在途中行驶的平均速度分别是多少?

广州市越秀区广州大学附属中学2016-2017学年第二学期期末考试 初二年级 物理 试卷及参考答案

广州市越秀区广州大学附属中学2016-2017学年第二学期期末考试 初二年级 物理 试卷及参考答案

B.木块两次的动能一样大D.木块两次受到的拉力和摩擦力均相等活塞式抽水机把水抽到高处是利用( )B.连通器D.液体压强与流速的关系浮力变大压强变小千克,则这种液体的密度是(,静止放在如图所示的斜面上,请作出物体所受重力的示意图,与重力平衡的力(即重力的平衡力),则剩余部,则它们的密度之比是 ,质量之比是 .的速度沿水平方向匀速移动了的瓶子(瓶子厚度不计),放置在面积为的水平桌面中央,当瓶中倒入甲 乙根据图象求物体作匀速直线运动的速度.若该物体在水平拉力作用下沿同一水平面以的速度向右作匀速直线运动,求:浸入水中称时,弹簧测力计的示数是多少.在探究“压力的作用效果跟什么因素有关”的实验中,某同学做了如图所示的实验,比较实验的甲、乙两图可在探究“滑动摩擦力的大小与什么因素有关”时,小明利用图所示的实验装置进行实验,他通过比较实验的 两图可得:在接触面粗糙程度一定时,滑动摩擦力的大小与压力大小有关.小明利用实验探究浮力大小和哪些因素有关.他把金属块挂在弹簧测力计上,将它分别浸入水和酒精中的不同位置,如图所示.实验中,小明先后在水平桌面铺上粗糙程度不同的物体(毛巾、棉布、木板),做了三次实验,每次都让小车从同一斜面的 位置由静止开始滑下,是为了使小车在滑到斜面底端时具有相同的速度.比较小车在不同表面滑行的最大距离,可以得出:在初速度相同的条件下,水平面越光滑,小车受到的摩擦越小,小车运动的距离就越 .进一步推理可知,若水平面绝对光滑,则小车会在水平面上做 运动.物块浸没在水中时所受到的浮力.溢出水杯的水没装满,导致本实验测量的.(选填“”或“”)浮的盐水中,漂浮时密度计受到的浮力的 (选填“上方”或“下方”).小刚将自制好的密度计放入另一种液体中,沉到容器底,不好测量这种液体密度,如图,可采取的办【答案】【答案】【答案】(1)水对瓶底的压力为,压强是(2)瓶子对桌面的压力为,压强为【答案】(1)(2)1。

2016-2017学年广东省广州大学附中八年级(下)期末数学试卷(解析版)

2016-2017学年广东省广州大学附中八年级(下)期末数学试卷(解析版)

2016-2017学年广东省广州大学附中八年级(下)期末数学试卷一、选择题(共10小题,每小题3分,共30分)1.(3分)下列各组数中,不能作为直角三角形的三边长的是()A.3,4,5B.6,8,10C.5,5,6D.5,12,13 2.(3分)在下列条件中,能够判定四边形是菱形的是()A.两条对角线相等B.两条对角线相等且互相垂直C.两条对角线互相垂直D.两条对角线互相垂直平分3.(3分)a是任意实数,下列各式中:①;②;③;④;⑤,一定是二次根式的个数是()A.1B.2C.3D.44.(3分)若直线y=2x+3与y=3x﹣2b相交于x轴上,则b的值是()A.b=﹣3B.b=﹣C.b=﹣D.b=65.(3分)如图,平行四边形ABCD中,对角线AC,BD相交于O,AC=8,BD=10,则边AB的取值范围是()A.8<AB<10B.1<AB<9C.4<AB<5D.2<AB<18 6.(3分)已知一次函数y=kx+b,y随着x的增大而减小,且kb<0,则在直角坐标系内它的大致图象是()A.B.C.D.7.(3分)某中学足球队的19名队员的年龄如表所示:这19名队员年龄的众数和中位数分别是()A.13岁,14岁B.14岁,14岁C.14岁,13岁D.14岁,15岁8.(3分)如图,四边形ABCD中,AC=8,BD=6,且AC⊥BD,连接四边形ABCD各边中点得到四边形EFGH,下列说法错误的是()A.四边形EFGH是矩形B.四边形EFGH的周长是14C.四边形EFGH的面积是12D.四边形ABCD的面积是489.(3分)已知1<x≤2,则|x﹣3|+的值为()A.2x﹣5B.﹣2C.5﹣2x D.210.(3分)如图,长方体的长为15,宽为10,高为20,点B离点C的距离为5,一只蚂蚁如果要沿着长方体的表面从点A爬到点B,需要爬行的最短距离是()A.5B.25C.10+5D.35二、填空题(共6小题,每小题3分,共18分)11.(3分)当a时,有意义.12.(3分)某水库的水位在5小时内持续上涨,初始的水位高度为6米,水位以每小时0.3米的速度匀速上升,则水库的水位高度y米与时间x小时(0≤x≤5)的函数关系式为.13.(3分)设甲组数据:6,6,6的方差为s甲2,乙组数据:1,1,2的方差为s乙2,则s甲2与s乙2的大小关系是.14.(3分)已知一次函数y=ax+b的图象经过第二、三、四象限,与x轴的交点为(﹣2,0),则不等式ax+b<0 的解集是.15.(3分)如图是“赵爽弦图”,△ABH、△BCG、△CDF和△DAE是四个全等的直角三角形,四边形ABCD和EFGH都是正方形.如果AB=10,EF=2,那么AH等于.16.(3分)如图,在菱形ABCD中,AC与BD于点O,AE⊥CD,且AE=OD,若AO+OD+AD =3+,则菱形ABCD的面积是.三、解答题(共7大题,总计72分)17.(4分)计算:2b﹣(4a+)(a>0,b>0).18.(6分)已知直线y=2x+4与x轴的交点为A,与y轴的交点为B,点C(a,0)是x轴正半轴上一动点.(1)求△ABC的面积S关于a的函数解析式(不写自变量a的取值范围).(2)如M(1,3)是线段BC上一点,求△ABM的面积.19.(8分)某初中学校欲向高一级学校推荐一名学生,根据规定的推荐程序:首先由本年级200名学生民主投票,每人只能推荐一人(不设弃权票),选出了票数最多的甲、乙、丙三人.投票结果统计如图一:其次,对三名候选人进行了笔试和面试两项测试.各项成绩如下表所示:图二是某同学根据上表绘制的一个不完全的条形图.请你根据以上信息解答下列问题:(1)补全图一和图二;(2)请计算每名候选人的得票数;(3)若每名候选人得一票记1分,投票、笔试、面试三项得分按照2:5:3的比确定,计算三名候选人的平均成绩,成绩高的将被录取,应该录取谁?20.(8分)如图,正方形ABCD中,AB=6,点E在边CD上,且CD=3DE.将△ADE沿AE对折至△AFE,延长EF交边BC于点G,连接AG、CF.(1)证明△ABG≌△AFG;(2)求BG的长;(3)求△FGC的面积.21.(8分)如图,已知△ABC是等腰三角形,顶角∠BAC=α(α<60°),D是BC边上的一点,连接AD,线段AD绕点A顺时针旋转α到AE,过点E作BC的平行线,交AB于点F,连接DE,BE,DF.(1)求证:BE=CD;(2)若AD⊥BC,试判断四边形BDFE的形状,并给出证明.22.(8分)甲、乙两车分别从A、B两地同时出发,甲车匀速前往B地,到达B地立即以另一速度按原路匀速返回到A地;乙车匀速前往A地,设甲、乙两车距A地的路程为y (千米),甲车行驶的时间为x(时),y与x之间的函数图象如图所示.(1)求甲车从A地到达B地的行驶时间;(2)求甲车返回时y与x之间的函数关系式,并写出自变量x的取值范围;(3)求乙车到达A地时甲车距A地的路程.23.(9分)广安某水果店计划购进甲、乙两种新出产的水果共140千克,这两种水果的进价、售价如表所示:(1)若该水果店预计进货款为1000元,则这两种水果各购进多少千克?(2)若该水果店决定乙种水果的进货量不超过甲种水果的进货量的3倍,应怎样安排进货才能使水果店在销售完这批水果时获利最多?此时利润为多少元?24.(10分)如图,四边形ABCD中,∠A=∠ABC=90°,AD=1,BC=3,E是边CD的中点,连接BE并延长与AD的延长线相交于点F.(1)求证:四边形BDFC是平行四边形;(2)若△BCD是等腰三角形,求四边形BDFC的面积.25.(11分)如图(1),在平面直角坐标系中,点O是坐标原点,四边形ABCO是菱形,点A的坐标为(﹣3,4),点C在x轴的正半轴上,直线AC交y轴于点M,AB边交y轴于点H.(1)求直线AC的解析式.(2)连接BM,如图(2),动点P从点A出发,沿折线ABC方向以2个单位/秒的速度向终点C匀速运动,设△PMB的面积为S(S≠0),点P的运动时间为t秒,求S与t之间的函数关系式(要求写出自变量t的取值范围).(3)在(2)的条件下,当t为何值时,∠MPB与∠BCO互为余角,并求此时直线OP的解析式.2016-2017学年广东省广州大学附中八年级(下)期末数学试卷参考答案与试题解析一、选择题(共10小题,每小题3分,共30分)1.【解答】解:A、32+42=52,符合勾股定理的逆定理,故错误;B、62+82=102,符合勾股定理的逆定理,故错误;C、52+52≠62,不符合勾股定理的逆定理,故正确;D、52+122=132,符合勾股定理的逆定理,故错误.故选:C.2.【解答】解:菱形的判定方法有三种:①定义:一组邻边相等的平行四边形是菱形;②四边相等;③对角线互相垂直平分的四边形是菱形,故选D.3.【解答】解:二次根式有②③④,共3个,故选:C.4.【解答】解:∵直线y=2x+3与直线y=3x﹣2b相交于x轴上,∴2x+3=0,x=,∴两直线的交点坐标为(,0),把此点坐标代入直线y=3x﹣2b得,×3﹣2b=0,∴b=﹣.故选:C.5.【解答】解:∵四边形ABCD是平行四边形,AC=8,BD=10,∴OA=OC=4,OB=OD=5,在△AOB中,由三角形三边关系定理得:5﹣4<AB<5+4,即1<AB<9,故选:B.6.【解答】解:∵一次函数y=kx+b,y随着x的增大而减小∴k<0又∵kb<0∴b>0∴此一次函数图象过第一,二,四象限.故选:A.7.【解答】解:由表可知14岁出现次数最多,有6次,所以众数为14岁;这组数据的中位数为第10个数据,即中位数为14岁,故选:B.8.【解答】解:∵点E、F、G、H分别是边AB、BC、CD、DA的中点,∴EF=AC,GH=AC,∴EF=GH,同理EH=FG∴四边形EFGH是平行四边形;又∵对角线AC、BD互相垂直,∴EF与FG垂直.∴四边形EFGH是矩形,故选项A正确,不符合题意;∵AC=8,BD=6,且AC⊥BD,∴四边形ABCD的面积=AC•BD=24,故选项D错误,符合题意;∵四边形EFGH是矩形,且HG=AC=4,HE=BD=3∴四边形EFGH的面积=3×4=12,故选项C正确,不符合题意;∵EF=AC=4,HE=BD=3,∴四边形EFGH的周长=2(3+4)=14,所以选项B正确,不符合题意,故选:D.9.【解答】解:∵1<x≤2,∴x﹣3<0,x﹣2≤0,∴原式=3﹣x+(2﹣x)=5﹣2x.故选:C.10.【解答】解:将长方体展开,连接A、B,根据两点之间线段最短,(1)如图,BD=10+5=15,AD=20,由勾股定理得:AB====25.(2)如图,BC=5,AC=20+10=30,由勾股定理得,AB====5.(3)只要把长方体的右侧表面剪开与上面这个侧面所在的平面形成一个长方形,如图:∵长方体的宽为10,高为20,点B离点C的距离是5,∴BD=CD+BC=20+5=25,AD=10,在直角三角形ABD中,根据勾股定理得:∴AB===5;由于25<5<5,故选:B.二、填空题(共6小题,每小题3分,共18分)11.【解答】解:当1﹣2a≥0,即a≤时,二次根式有意义,故答案为:≤.12.【解答】解:因为初始的水位高度为6米,水位以每小时0.3米的速度匀速上升,所以k=0.3,b=6,根据题意可得:y=6+0.3x(0≤x≤5),故答案为:y=6+0.3x.13.【解答】解:因为甲组的数据都相等,没有波动,而乙组数有波动,所以s甲2<s乙2.故答案为:s甲2<s乙2.14.【解答】解:∵一次函数的图象经过第二、三、四象限,且与x轴的交点为(﹣2,0),∴a<0,b=2a,∴ax+b=ax+2a=a(x+2)∴ax+b<0的解集即为a(x+2)<0的解集,∴x>﹣2故答案为:x>﹣215.【解答】解:∵AB=10,EF=2,∴大正方形的面积是100,小正方形的面积是4,∴四个直角三角形面积和为100﹣4=96,设AE为a,DE为b,即4×ab=96,∴2ab=96,a2+b2=100,∴(a+b)2=a2+b2+2ab=100+96=196,∴a+b=14,∵a﹣b=2,解得:a=8,b=6,∴AE=8,DE=6,∴AH=8﹣2=6.故答案为:6.16.【解答】解:∵四边形ABCD是菱形,∴AD=DC,AC⊥BD,∵AE⊥CD,∴∠DOA=∠AED=90°,在Rt△AOD和Rt△DEA中,,∴Rt△AOD≌Rt△DEA(HL),∴∠DAO=∠ADE,∵AD=DC,∴∠DAC=∠DCA,∴∠DAC=∠DCA=∠ADC,∴△ADC是等边三角形,∴∠ADC=60°,∴∠ADO=∠ADC=30°,∴AD=2AO,OD=AO,∵AO+OD+AD=3+,∴AO+AO+2AO=3+,∴AO=1,OD=,∴AC=2AO=2,BD=2OD=2,∴菱形ABCD的面积是:AC•BD=×2×2=2.故答案为:2.三、解答题(共7大题,总计72分)17.【解答】解:原式=2b×﹣4a×﹣3=2﹣4﹣3=﹣5.18.【解答】解:(1)∵直线y=2x+4与x轴的交点为A,与y轴的交点为B,∴A(﹣2,0),B(0,4),又∵点C(a,0)在x轴正半轴上,∴S=•(a+2)•4=2a+4;(2)设直线BC的解析式为y=kx+b,把B(0,4)、C(1,3)代入得,解得,∴直线BC的解析式为y=﹣x+4,∴C点坐标为(4,0),又∵M(1,3)是线段BC上一点,∴△ABM的面积=△ABC的面积﹣△MAC的面积=×4×6﹣×3×6=3.19.【解答】解:(1)(2)甲的票数是:200×34%=68(票),乙的票数是:200×30%=60(票),丙的票数是:200×28%=56(票);(3)甲的平均成绩:,乙的平均成绩:,丙的平均成绩:,∵乙的平均成绩最高,∴应该录取乙.20.【解答】解:(1)在正方形ABCD中,AD=AB=BC=CD,∠D=∠B=∠BCD=90°,∵将△ADE沿AE对折至△AFE,∴AD=AF,DE=EF,∠D=∠AFE=90°,∴AB=AF,∠B=∠AFG=90°,又∵AG=AG,在Rt△ABG和Rt△AFG中,∵,∴△ABG≌△AFG(HL);(2)∵CD=3DE∴DE=2,CE=4,设BG=x,则CG=6﹣x,GE=x+2∵GE2=CG2+CE2∴(x+2)2=(6﹣x)2+42,解得x=3∴BG=3;(3)过C作CM⊥GF于M,∵BG=GF=3,∴CG=3,EC=6﹣2=4,∴GE==5,CM•GE=GC•EC,∴CM×5=3×4,∴CM=2.4,∴.21.【解答】证明:(1)∵△ABC是等腰三角形,顶角∠BAC=α(α<60°),线段AD绕点A顺时针旋转α到AE,∴AB=AC,∵∠EAD=∠BAC,∴∠BAE=∠CAD,在△ACD和△ABE中,,∴△ACD≌△ABE(SAS),∴BE=CD;(2)∵AD⊥BC,∴BD=CD,∴BE=BD=CD,∠BAD=∠CAD,∴∠BAE=∠BAD,在△ABD和△ABE中,,∴△ABD≌△ABE(SAS),∴∠EBF=∠DBF,∵EF∥BC,∴∠DBF=∠EFB,∴∠EBF=∠EFB,∴EB=EF=BD,∴四边形EFDB是平行四边形,∵EF=EB,∴四边形BDFE为菱形.22.【解答】解:(1)300÷(180÷1.5)=2.5(小时),答:甲车从A地到达B地的行驶时间是2.5小时;(2)设甲车返回时y与x之间的函数关系式为y=kx+b,∴,解得:,∴甲车返回时y与x之间的函数关系式是y=﹣100x+550(2.5≤x≤5.5);(3)300÷[(300﹣180)÷1.5]=3.75小时,当x=3.75时,y=175千米,答:乙车到达A地时甲车距A地的路程是175千米.23.【解答】解:(1)设购进甲种水果x千克,则购进乙种水果(140﹣x)千克,根据题意可得:5x+9(140﹣x)=1000,解得:x=65,∴140﹣x=75(千克),答:购进甲种水果65千克,乙种水果75千克;(2)由图表可得:甲种水果每千克利润为:3元,乙种水果每千克利润为:4元,设总利润为W,由题意可得出:W=3x+4(140﹣x)=﹣x+560,故W随x的增大而减小,则x越小W越大,因为该水果店决定乙种水果的进货量不超过甲种水果的进货量的3倍,∴140﹣x≤3x,解得:x≥35,∴当x=35时,W最大=﹣35+560=525(元),故140﹣35=105(kg).答:当甲购进35千克,乙种水果105千克时,此时利润最大为525元.24.【解答】(1)证明:∵∠A=∠ABC=90°,∴BC∥AD,∴∠CBE=∠DFE,在△BEC与△FED中,,∴△BEC≌△FED,∴BE=FE,又∵E是边CD的中点,∴CE=DE,∴四边形BDFC是平行四边形;(2)①BC=BD=3时,由勾股定理得,AB===2,所以,四边形BDFC的面积=3×2=6;②BC=CD=3时,过点C作CG⊥AF于G,则四边形AGCB是矩形,所以,AG=BC=3,所以,DG=AG﹣AD=3﹣1=2,由勾股定理得,CG===,所以,四边形BDFC的面积=3×=3;③BD=CD时,BC边上的中线应该与BC垂直,从而得到BC=2AD=2,矛盾,此时不成立;综上所述,四边形BDFC的面积是6或3.25.【解答】解:(1)过点A作AE⊥x轴垂足为E,如图(1)∵A(﹣3,4),∴AE=4 OE=3,∴OA==5,∵四边形ABCO为菱形,∴OC=CB=BA=0A=5,∴C(5,0)设直线AC的解析式为:y=kx+b,∵,∴,∴直线AC的解析式为y=﹣x+.(2)由(1)得M点坐标为(0,),∴OM=,如图(1),当P点在AB边上运动时由题意得OH=4,∴HM=OH﹣OM=4﹣=,∴s=BP•MH=(5﹣2t)•,∴s=﹣t+(0≤t<),当P点在BC边上运动时,记为P1,∵∠OCM=∠BCM,CO=CB,CM=CM,∴△OMC≌△BMC,∴OM=BM=,∠MOC=∠MBC=90°,∴S=P1B•BM=(2t﹣5),∴S=t﹣(<t≤5),(3)设OP与AC相交于点Q连接OB交AC于点K,∵∠AOC=∠ABC,∴∠AOM=∠ABM,∵∠MPB+∠BCO=90°,∠BAO=∠BCO,∠BAO+∠AOH=90°,∴∠MPB=∠AOH,∴∠MPB=∠MBH.当P点在AB边上运动时,如图1,∵∠MPB=∠MBH,∴PM=BM,∵MH⊥PB,∴PH=HB=2,∴P A=AH﹣PH=1,∴P(﹣2,4),∴直线OP的解析式为y=﹣2x,∵A(﹣3,4),∴AP=1,∴t=1÷1=,当P点在BC边上运动时,如图2,∵∠BHM=∠PBM=90°,∠MPB=∠MBH,∴tan∠MPB=tan∠MBH,∴=,即=,∴BP=,∵B(2,4),C(5,0),∴直线BC解析式为y=﹣x+,设P(m,﹣m+),∴BP2=(m﹣2)2+(﹣m+﹣4)2=,∴m=0(舍)或m=4,∴P(4,),∴直线OP的解析式为y=x,∵B(2,4),∴BP=,∴t=(2+3+)÷2=.。

2016-2017学年广东省广州市越秀区八年级(下)期末数学试卷(解析版)

2016-2017学年广东省广州市越秀区八年级(下)期末数学试卷(解析版)

2016-2017学年广东省广州市越秀区八年级(下)期末数学试卷一、选择题(共10小题,每小题3分,满分30分)1.(3分)下列式子没有意义的是()A.B.C.D.2.(3分)下列计算中,正确的是()A.÷=B.(4)2=8C.=2D.2×2=2 3.(3分)刻画一组数据波动大小的统计量是()A.平均数B.方差C.众数D.中位数4.(3分)在暑假到来之前,某机构向八年级学生推荐了A,B,C三条游学线路,现对全级学生喜欢哪一条游学线路作调查,以决定最终的游学线路,下面的统计量中最值得关注的是()A.方差B.平均数C.中位数D.众数5.(3分)关于正比例函数y=﹣2x,下列结论中正确的是()A.函数图象经过点(﹣2,1)B.y随x的增大而减小C.函数图象经过第一、三象限D.不论x取何值,总有y<06.(3分)以下列各组数据中的三个数作为三角形的边长,其中能构成直角三角形的是()A.2,3,4B.,,C.1,,2D.7,8,97.(3分)若一个直角三角形的一条直角边长是5cm,另一条直角边比斜边短1cm,则斜边长为()cm.A.10B.11C.12D.138.(3分)如图,在菱形ABCD中,对角线AC,BD相交于点O,AB=5,AC=6,则菱形ABCD的面积是()A.24B.26C.30D.489.(3分)在下列命题中,是假命题的是()A.有一个角是直角的平行四边形是矩形B.一组邻边相等的矩形是正方形C.一组对边平行且相等的四边形是平行四边形D.有两组邻边相等的四边形是菱形10.(3分)已知平面上四点A(0,0),B(10,0),C(12,6),D(2,6),直线y=mx ﹣3m+6将四边形ABCD分成面积相等的两部分,则m的值为()A.B.﹣1C.2D.二、填空题:本大题共6小题,每小题3分,满分18分.11.(3分)已知a=+2,b=﹣2,则ab=.12.(3分)一次函数y=kx+b(k≠0)中,x与y的部分对应值如下表:那么,一元一次方程kx+b=0的解是x=.13.(3分)如图是一次函数y=mx+n的图象,则关于x的不等式mx+n>2的解集是.14.(3分)一组数据:2017、2017、2017、2017、2017,它的方差是.15.(3分)考古学家们发现了几块大约完成于公元前2000年左右的古巴比伦的泥版书,据专家们考证,其中一块上面刻有如下问题:“一根长度为30个单位的棍子直立在墙上,当其上端垂直滑下6个单位时,请问其下端离开墙角有多远?”,这个问题的答案是:其下端离开墙角个单位.16.(3分)如图所示,在Rt△ABC中,∠A=90°,DE∥BC,F,G,H,I分别是DE,BE,BC,CD的中点,连接FG,GH,HI,IF,FH,GI.对于下列结论:①∠GFI=90°;②GH=GI;③GI=(BC﹣DE);④四边形FGHI是正方形.其中正确的是(请写出所有正确结论的序号).三、解答题:本大题共9小题,满分72分.解答须写出文字说明、推理过程和演算步骤.17.(6分)计算:(+﹣)×.18.(8分)如图,在△ABC中,AD⊥BC,AB=5,BD=4,CD=.(1)求AD的长.(2)求△ABC的周长.19.(8分)如图在平行四边形ABCD中,AC交BD于点O,AE⊥BD,CF⊥BD,垂足分别为E、F,求证:四边形AECF为平行四边形.20.(8分)下表是某校八年级(1)班43名学生右眼视力的检查结果.(1)该班学生右眼视力的平均数是(结果保留1位小数).(2)该班学生右眼视力的中位数是.(3)该班小鸣同学右眼视力是4.5,能不能说小鸣同学的右眼视力处于全班同学的中上水平?试说明理由.21.(8分)如图,正方形ABCD的对角线相交于点O,BC=6,延长BC至点E,使得CE =8,点F是DE的中点,连接CF、OF.(1)求OF的长.(2)求CF的长.22.(8分)如图,在平面直角坐标系中,直线y=kx+b经过点A(﹣30,0)和点B(0,15),直线y=x+5与直线y=kx+b相交于点P,与y轴交于点C.(1)求直线y=kx+b的解析式.(2)求△PBC的面积.23.(8分)2016年下半年开始,不同品牌的共享单车出现在城市的大街小巷.现已知A品牌共享单车计费方式为:初始骑行单价为1元/半小时,不足半小时按半小时计算.内设邀请机制,每邀请一位好友注册认证并充值押金成功,双方骑行单价均降价0.1元/半小时,骑行单价最低可降至0.1元/半小时(比如,某用户邀请了3位好友,则骑行单价为0.7元/半小时).B品牌共享单车计费方式为:0.5元/半小时,不足半小时按半小时计算.(1)某用户准备选择A品牌共享单车使用,设该用户邀请好友x名(x为整数,x≥0),该用户的骑行单价为y元/半小时.请写出y关于x的函数解析式.(2)若有A,B两种品牌的共享单车各一辆供某用户一人选择使用,请你根据该用户已邀请好友的人数,给出经济实惠的选择建议.24.(8分)下面我们做一次折叠活动:第一步,在一张宽为2的矩形纸片的一端,利用图(1)的方法折出一个正方形,然后把纸片展平,折痕为MC;第二步,如图(2),把这个正方形折成两个相等的矩形,再把纸片展平,折痕为F A;第三步,折出内侧矩形F ACB的对角线AB,并将AB折到图(3)中所示的AD处,折痕为AQ.根据以上的操作过程,完成下列问题:(1)求CD的长.(2)请判断四边形ABQD的形状,并说明你的理由.25.(10分)如图,正方形ABCD中,AB=4,P是CD边上的动点(P点不与C、D重合),过点P作直线与BC的延长线交于点E,与AD交于点F,且CP=CE,连接DE、BP、BF,设CP═x,△PBF的面积为S1,△PDE的面积为S2.(1)求证:BP⊥DE.(2)求S1﹣S2关于x的函数解析式,并写出x的取值范围.(3)分别求当∠PBF=30°和∠PBF=45°时,S1﹣S2的值.2016-2017学年广东省广州市越秀区八年级(下)期末数学试卷参考答案与试题解析一、选择题(共10小题,每小题3分,满分30分)1.【解答】解:A、有意义,故此选项不合题意;B、没有意义,故此选项符合题意;C、有意义,故此选项不合题意;D、有意义,故此选项不合题意;故选:B.2.【解答】解:A、原式===3,不符合题意;B、原式=32,不符合题意;C、原式=|﹣2|=2,符合题意;D、原式=4,不符合题意;故选:C.3.【解答】解:由于方差反映数据的波动情况,衡量一组数据波动大小的统计量是方差.故选:B.4.【解答】解:由于众数是数据中出现次数最多的数,故全级学生喜欢的游学线路最值得关注的应该是统计调查数据的众数.故选:D.5.【解答】解:A、当x=﹣2时,y=﹣2×(﹣2)=4,即图象经过点(﹣2,4),不经过点(﹣2,1),故本选项错误;B、由于k=﹣2<0,所以y随x的增大而减小,故本选项正确;C、由于k=﹣2<0,所以图象经过二、四象限,故本选项错误;D、∵x>0时,y<0,x<0时,y>0,∴不论x为何值,总有y<0错误,故本选项错误.故选:B.6.【解答】解:A、22+32≠42,故不是直角三角形;B、()2+()2≠()2,故不是直角三角形;C、12+()2=22,故是直角三角形;D、72+82≠92,故不是直角三角形;故选:C.7.【解答】解:设斜边长为xcm,则另一条直角边为(x﹣1)cm,由勾股定理得,x2=52+(x﹣1)2,解得,x=13,则斜边长为13cm,故选:D.8.【解答】解:∵四边形ABCD是菱形,∴OA=OC=3,OB=OD,AC⊥BD,在Rt△AOB中,∠AOB=90°,根据勾股定理,得:OB=,=,=4,∴BD=2OB=8,∴S菱形ABCD=×AC×BD=×6×8=24.故选:A.9.【解答】解:A、有一个角是直角的平行四边形是矩形,正确;B、一组邻边相等的矩形是正方形,正确;C、一组对边平行且相等的四边形是平行四边形,正确;D、有两组邻边相等且平行的四边形是菱形,错误;故选:D.10.【解答】解:如图,∵A(0,0),B(10,0),C(12,6),D(2,6),∴AB=10﹣0=10,CD=12﹣2=10,又点C、D的纵坐标相同,∴AB∥CD且AB=CD,∴四边形ABCD是平行四边形,∵12÷2=6,6÷2=3,∴对角线交点P的坐标是(6,3),∵直线y=mx﹣3m+6将四边形ABCD分成面积相等的两部分,∴直线y=mx﹣3m+6经过点P,∴6m﹣3m+6=3,解得m=﹣1.故选:B.二、填空题:本大题共6小题,每小题3分,满分18分.11.【解答】解:∵a=+2,b=﹣2,∴ab=(+2)(﹣2)=5﹣4=1,故答案为:112.【解答】解:根据上表中的数据值,当y=0时,x=1,即一元一次方程kx+b=0的解是x=1.故答案是:1.13.【解答】解:由题意,可知一次函数y=mx+n的图象经过点(0,2),且y随x的增大而增大,所以关于x的不等式mx+n>2的解集是x>0.故答案为x>0.14.【解答】解:该组数据一样,没有波动,方差为0,故答案为0.15.【解答】解:∵PC=AB=30,P A=6,∴AC=24,∴BC===18,∴下端离开墙角18个单位.故答案为18.16.【解答】解:延长IF交AB于K,∵DF=EF,BG=GE,∴FG=BD,GF∥AB,同理IF∥AC,HI=BD,HI∥BD,∴∠BKI=∠A=90°,∴∠GFI=∠BKI=90°,∴GF⊥FI,故①正确,∴FG=HI,FG∥HI,∴四边形FGHI是平行四边形,∵∠GFI=90°,∴四边形FGHI是矩形,故②④错误,延长EI交BC于N,则△DEI≌△CNI,∴DE=CN,EJ=JN,∵EG=GB,EI=IN,∴GI=BHN=(BC﹣DE),故③正确,故答案为①③.三、解答题:本大题共9小题,满分72分.解答须写出文字说明、推理过程和演算步骤.17.【解答】解:原式=(6+﹣﹣3)×=×=7.18.【解答】解:(1)在Rt△ABD中,AD==3;(2)在Rt△ACD中,AC==2,则△ABC的周长=AB+AC+BC=5+4++2=9+3.19.【解答】证明:∵四边形ABCD是平行四边形,∴AB=CD,AB∥CD,∴∠ABE=∠CDF,∵AE⊥BD,CF⊥BD,∴AE∥CF,∠AEB=∠CFD=90°,在△AEB和△CFD中,∵,∴△AEB≌△CFD(AAS),∴AE=CF,∴四边形AECF是平行四边形.20.【解答】解:(1)该班学生右眼视力的平均数是×(4.0+4.1×2+4.2×5+4.3×4+4.4×3+4.5×5+4.6+4.7+4.8×5+4.9×10+5.0×6)≈4.6,故答案为:4.6;(2)由于共有43个数据,其中位数为第22个数据,即中位数为4.7,故答案为:4.7;(3)不能,∵小鸣同学右眼视力是4.5,小于中位数4.7,∴不能说小鸣同学的右眼视力处于全班同学的中上水平.21.【解答】解:(1)∵四边形ABCD是正方形,∴BC=CD=6,∠BCD=∠ECD=90°,OB=OD,∵CE=8,∴BE=14,∵OB=OD,DF=FE,∴OF=BE=7.(2)在Rt△DCE中,DE===10,∵DF=FE,∴CF=DE=5.22.【解答】解:(1)将点A(﹣30,0)、B(0,15)代入y=kx+b,,解得:,∴直线y=kx+b的解析式为y=x+15.(2)联立两直线解析式成方程组,,解得:,∴点P的坐标为(20,25).当x=0时,y=x+5=5,∴点C的坐标为(0,5),∴BC=15﹣5=10,∴S△PBC=BC•x P=×10×20=100.23.【解答】解:(1)由题意可得,当0≤x≤9且x为正整数时,y=1﹣0.1x,当x≥10且x为正整数时,y=0.1,即y关于x的函数解析式是y=;(2)由题意可得,当0≤x≤9时,1﹣0.1x>0.5,可得,x<5,则当x≤x<5且x为正整数时,选择B品牌的共享单车;当0≤x≤9时,1﹣0.1x=0.5,得x=5,则x=5时,选择A或B品牌的共享单车消费一样;当0≤x≤9时,1﹣0.1x<0.5,得x>5,则x>5且x为正整数,选择A品牌的共享单车;当x≥10且x为正整数时,0.1<0.5,故选项A品牌的共享单车.24.【解答】解:(1)∵∠M=∠N=∠MBC=90°,∴四边形MNCB是矩形,∵MB=MN=2,∴矩形MNCB是正方形,∴NC=CB=2,由折叠得:AN=AC=NC=1,Rt△ACB中,由勾股定理得:AB==,∴AD=AB=,∴CD=AD﹣AC=﹣1;(2)四边形ABQD是菱形,理由是:由折叠得:AB=AD,∠BAQ=∠QAD,∵BQ∥AD,∴∠BQA=∠QAD,∴∠BAQ=∠BQA,∴AB=BQ,∴BQ=AD,BQ∥AD,∴四边形ABQD是平行四边形,∵AB=AD,∴四边形ABQD是菱形.25.【解答】解:(1)如图1中,延长BP交DE于M.∵四边形ABCD是正方形,∴CB=CD,∠BCP=∠DCE=90°,∵CP=CE,∴△BCP≌△DCE,∴∠BCP=∠CDE,∵∠CBP+∠CPB=90°,∠CPB=∠DPM,∴∠CDE+∠DPM=90°,∴∠DMP=90°,∴BP⊥DE.(2)由题意S1﹣S2=[16﹣2x﹣2x﹣(4﹣x)2]﹣•(4﹣x)•x=8﹣2x(0<x<4).(3)①如图2中,当∠PBF=30°时,∵∠CPE=∠CEP=∠DPF=45°,∠FDP=90°,∴∠PFD=∠DPF=45°,∴DF=DP,∵AD=CD,∴AF=PC,∵AB=BC,∠A=∠BCP=90°,∴△BAF≌△BCP,∴∠ABF=∠CBP=30°,∴x=PC=BC•tan30°=,∴S1﹣S2=8﹣2x=8﹣.②如图3中,当∠PBF=45°时,在CB上截取CN=CP,理解PN.由①可知△ABF≌△BCP,∴∠ABF=∠CBP,∵∠PBF=45°,∴∠CBP=22.5°,∵∠CNP=∠NBP+∠NPB=45°,∴∠NBP=∠NPB=22.5°,∴BN=PN=x,∴x+x=4,∴x=4﹣4,∴S1﹣S2=8﹣2x=16﹣8.。

最新广东省广州市广州大学附属中学-2017学年下期八年级期末考试数学试题(无答案)

最新广东省广州市广州大学附属中学-2017学年下期八年级期末考试数学试题(无答案)

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2016-2017年广东省广州大学附中八年级(下)期中数学试卷(解析版)

2016-2017年广东省广州大学附中八年级(下)期中数学试卷(解析版)

2016-2017学年广东省广州大学附中八年级(下)期中数学试卷一、选择题(本大题共10小题,每小题3分,共30分)1.(3分)下列式子中,属于最简二次根式的是()A.B.C.D.2.(3分)下列计算正确的是()A.﹣=B.=±3C.÷=D.=﹣3 3.(3分)在一个直角三角形中,已知两直角边分别为6cm,8cm,则下列结论不正确的是()A.斜边长为10cm B.周长为25cmC.面积为24cm2D.斜边上的中线长为5cm4.(3分)已知平行四边形ABCD中,∠A=110°,则∠B的度数为()A.110°B.100°C.80°D.70°5.(3分)下列不能判断四边形ABCD是平行四边形的是()A.AB=CD,AD=BC B.AB∥CD,AD=BCC.AB∥CD,AD∥BC D.∠A=∠C,∠B=∠D6.(3分)如图下列各曲线中表示y是x的函数的是()A.B.C.D.7.(3分)下列命题的逆命题是真命题的是()A.若两个实数相等,则这两个实数的平方相等B.若两个角是直角,则这两个角相等C.若AB=5,BC=4,CA=3,则△ABC是直角三角形D.若一个四边形的对角线互相垂直且平分,则这个四边形是菱形8.(3分)矩形、菱形、正方形都具有的性质是()A.对角线相等B.对角线互相平分C.对角线互相垂直D.对角线平分对角9.(3分)顺次连接四边形ABCD四边中点得到新的四边形为菱形,那么原四边形ABCD为()A.矩形B.菱形C.对角线相等的四边形D.对角线垂直的四边形10.(3分)如图,点O(0,0),A(0,1)是正方形OAA1B的两个顶点,以OA1对角线为边作正方形OA1A2B1,再以正方形的对角线OA2作正方形OA2A3B3,…,依此规律,则点A8的坐标是()A.(﹣8,2)B.(0,8)C.(0,8)D.(0,16)二、填空题(本大题共6小题,每小题3分,共18分)11.(3分)若二次根式有意义,则x的取值范围是.12.(3分)如图由于台风的影响,一棵树在离地面6m处折断,树顶落在离树干底部8m处,则这棵树在折断前(不包括树根)长度是m.13.(3分)一直角三角形的两边长分别为5和12,则第三边的长是.14.(3分)如图,在菱形ABCD中,AC、BD相交于点O,E为AB的中点,若OE=2,则菱形ABCD的周长是.15.(3分)甲、乙两车沿直线同向行驶,车速分别为15米/秒和25米/秒.先甲车在乙车前500米处,设x秒(0≤x≤50)后两车相距y米.则y与x的函数关系式为(不需要写出自变量取值范围).16.(3分)如图,四边形ABCD是边长为9的正方形纸片,将其沿MN折叠,使点A落在BC边上A′点处,点D的对应点为点D′,若A′B=3,则DM 的长为.三、解答题(本大题共7小题,满分72分.解答应写出文字说明、证明过程或演算步骤)17.(8分)(1)×﹣4×(2)+﹣.18.(8分)如图,l1表示一骑自行车者、l2表示一骑摩托车者沿相同的路线由甲地到乙地行驶过程的函数图象,两地相距80千米,请根据图象解决下列问题:(1)骑者出发早,早小时;骑者早到达目的地,早小时;他们在离甲地公里处相遇.(2)求两个人在途中行驶的平均速度分别是多少.19.(10分)如图,正方形网格中,每个小正方形的边长均为1,每个小正方形的顶点叫格点,以格点为顶点按下列要求画图:(1)在图 中画一条线段MN,使MN=;(2)在图 中画一个三边长均为无理数,且各边都不相等的直角△DEF.20.(10分)如图,平行四边形ABCD的对角线AC与BD相交于O,AB=5,CO=4,OD=3,求证:平行四边形ABCD是菱形.21.(10分)如图,矩形ABCD中,O为BD中点,PQ过点P分别交AD、BC 于点P、Q,连接BP和DQ,求证:四边形PBQD是平行四边形.22.(12分)如图,以△ABC的三边为边分别作等边△ACD、△ABE、△BCF (1)求证:△EBF≌△DFC;(2)求证:四边形AEFD是平行四边形;(3)①△ABC满足时,四边形AEFD是菱形.(无需证明)②△ABC满足时,四边形AEFD是矩形.(无需证明)③△ABC满足时,四边形AEFD是正方形.(无需证明)23.(14分)如图,在等腰△ACE中,已知CA=CE=2,AE=2c,点B、D、M分别是边AC、CE、AE的中点,以BC、CD为边长分别作正方形BCGF和CDHN,连结FM、FH、MH.(1)求△ACE的面积;(2)试探究△FMH是否是等腰直角三角形?并对结论给予证明;(3)当∠GCN=30°时,求△FMH的面积.2016-2017学年广东省广州大学附中八年级(下)期中数学试卷参考答案与试题解析一、选择题(本大题共10小题,每小题3分,共30分)1.(3分)下列式子中,属于最简二次根式的是()A.B.C.D.【考点】74:最简二次根式.【解答】解:∵=2,=,=,而中被开方数30不含能开得尽方的因数,∴属于最简二次根式的是,故选:D.2.(3分)下列计算正确的是()A.﹣=B.=±3C.÷=D.=﹣3【考点】79:二次根式的混合运算.【解答】解:A、错误,不是同类二次根式不能合并;B、错误=3;C、正确.D、错误.=3;故选:C.3.(3分)在一个直角三角形中,已知两直角边分别为6cm,8cm,则下列结论不正确的是()A.斜边长为10cm B.周长为25cmC.面积为24cm2D.斜边上的中线长为5cm【考点】KP:直角三角形斜边上的中线;KQ:勾股定理.【解答】解:∵在一个直角三角形中,已知两直角边分别为6cm,8cm,∴直角三角形的面积=×6×8=24cm2,故选项C不符合题意;∴斜边==10cm,故选项A不符合题意;∴斜边上的中线长为5cm,故选项D不符合题意;∵三边长分别为6cm,8cm,10cm,∴三角形的周长=24cm,故选项B符合题意,故选:B.4.(3分)已知平行四边形ABCD中,∠A=110°,则∠B的度数为()A.110°B.100°C.80°D.70°【考点】L5:平行四边形的性质.【解答】解:∵四边形ABCD是平行四边形,∴AD∥BC,∴∠A+∠B=180°,∵∠A=110°,∴∠B=70°,故选:D.5.(3分)下列不能判断四边形ABCD是平行四边形的是()A.AB=CD,AD=BC B.AB∥CD,AD=BCC.AB∥CD,AD∥BC D.∠A=∠C,∠B=∠D【考点】L6:平行四边形的判定.【解答】解:平行四边形的定义:两组对边分别平行的四边形叫做平行四边形.∴C能判断,平行四边形判定定理1,两组对角分别相等的四边形是平行四边形;∴D能判断;平行四边形判定定理2,两组对边分别相等的四边形是平行四边形;∴A能判定;平行四边形判定定理3,对角线互相平分的四边形是平行四边形;平行四边形判定定理4,一组对边平行相等的四边形是平行四边形;故选:B.6.(3分)如图下列各曲线中表示y是x的函数的是()A.B.C.D.【考点】E2:函数的概念.【解答】解:A、图象满足对于x的每一个取值,y都有唯一确定的值与之对应关系,故A符合题意;B、图象不满足对于x的每一个取值,y都有唯一确定的值与之对应关系,故B不符合题意;C、图象不满足对于x的每一个取值,y都有唯一确定的值与之对应关系,故C不符合题意;D、图象不满足对于x的每一个取值,y都有唯一确定的值与之对应关系,故D不符合题意;故选:A.7.(3分)下列命题的逆命题是真命题的是()A.若两个实数相等,则这两个实数的平方相等B.若两个角是直角,则这两个角相等C.若AB=5,BC=4,CA=3,则△ABC是直角三角形D.若一个四边形的对角线互相垂直且平分,则这个四边形是菱形【考点】O1:命题与定理.【解答】解:A、逆命题为:若两个实数的平方相等,则这两个数相等,此逆命题为假命题;B、逆命题为:若两个角相等,则这两个角都是直角,此逆命题为假命题;C、逆命题为:若△ABC是直角三角形,则AB=5,BC=4,CA=3,此逆命题为假命题;D、逆命题为:菱形的对角线互相垂直且平分,此逆命题为真命题.故选:D.8.(3分)矩形、菱形、正方形都具有的性质是()A.对角线相等B.对角线互相平分C.对角线互相垂直D.对角线平分对角【考点】L1:多边形.【解答】解:矩形、菱形、正方形都具有的性质是对角线互相平分.故选:B.9.(3分)顺次连接四边形ABCD四边中点得到新的四边形为菱形,那么原四边形ABCD为()A.矩形B.菱形C.对角线相等的四边形D.对角线垂直的四边形【考点】LN:中点四边形.【解答】解:∵E,F,G,H分别是边AD,DC,CB,AB的中点,∴EF=AC,EH∥AC,FG=AC,FG∥AC,EF=BD,∴EH∥FG,EF=FG,∴四边形EFGH是平行四边形,∵一组邻边相等的四边形是菱形,∴若AC=BD,则四边形是菱形.故选:C.10.(3分)如图,点O(0,0),A(0,1)是正方形OAA1B的两个顶点,以OA1对角线为边作正方形OA1A2B1,再以正方形的对角线OA2作正方形OA2A3B3,…,依此规律,则点A8的坐标是()A.(﹣8,2)B.(0,8)C.(0,8)D.(0,16)【考点】D2:规律型:点的坐标.【解答】解:根据题意和图形可看出每经过一次变化,都顺时针旋转45°,边长都乘以,∵从A到A3经过了3次变化,∵45°×3=135°,1×()3=2.∴点A3所在的正方形的边长为2,点A3位置在第四象限.∴点A3的坐标是(2,﹣2);可得出:A1点坐标为(1,1),A2点坐标为(2,0),A3点坐标为(2,﹣2),A4点坐标为(0,﹣4),A5点坐标为(﹣4,﹣4),A6(﹣8,0),A7(﹣8,8),A8(0,16),故选:D.二、填空题(本大题共6小题,每小题3分,共18分)11.(3分)若二次根式有意义,则x的取值范围是x≥2.【考点】72:二次根式有意义的条件.【解答】解:根据题意,使二次根式有意义,即x﹣2≥0,解得x≥2;故答案为:x≥2.12.(3分)如图由于台风的影响,一棵树在离地面6m处折断,树顶落在离树干底部8m处,则这棵树在折断前(不包括树根)长度是16m.【考点】KU:勾股定理的应用.【解答】解:由题意得BC=8m,AC=6m,在直角三角形ABC中,根据勾股定理得:AB==10(米).所以大树的高度是10+6=16(米).故答案为:16.13.(3分)一直角三角形的两边长分别为5和12,则第三边的长是13或.【考点】KQ:勾股定理.【解答】解:设第三边为x,(1)若12是直角边,则第三边x是斜边,由勾股定理得:52+122=x2,∴x=13;(2)若12是斜边,则第三边x为直角边,由勾股定理得:52+x2=122,∴x=;∴第三边的长为13或.故答案为:13或.14.(3分)如图,在菱形ABCD中,AC、BD相交于点O,E为AB的中点,若OE=2,则菱形ABCD的周长是16.【考点】KP:直角三角形斜边上的中线;L8:菱形的性质.【解答】解:∵在菱形ABCD中,AC、BD相交于点O,E为AB的中点,∴EO是△ABC的中位线,∵OE=2,∴BC=4,则菱形ABCD的周长是:4×4=16.故答案为:16.15.(3分)甲、乙两车沿直线同向行驶,车速分别为15米/秒和25米/秒.先甲车在乙车前500米处,设x秒(0≤x≤50)后两车相距y米.则y与x的函数关系式为y=﹣10x+500(不需要写出自变量取值范围).【考点】E3:函数关系式;E4:函数自变量的取值范围.【解答】解:由题意可得:y=500﹣(25﹣15)x=﹣10x+500.故答案为:y=﹣10x+500.16.(3分)如图,四边形ABCD是边长为9的正方形纸片,将其沿MN折叠,使点A落在BC边上A′点处,点D的对应点为点D′,若A′B=3,则DM 的长为2.【考点】LE:正方形的性质;PB:翻折变换(折叠问题).【解答】解:如图所示:连结AM、A′M.由翻折的性质可知:DM=D′M,AM=A′M.设MD=x,则MC=9﹣x.∵A′B=3,BC=9,∴A′C=6.在Rt△MCA′中,MA′2=A′C2+MC2=36+(9﹣x)2,在Rt△ADM中,AM2=AD2+DM2=81+x2.∴36+(9﹣x)2=81+x2,解得x=2,即DM=2.故答案为:2三、解答题(本大题共7小题,满分72分.解答应写出文字说明、证明过程或演算步骤)17.(8分)(1)×﹣4×(2)+﹣.【考点】79:二次根式的混合运算.【解答】解:(1)原式=﹣=2﹣=;(2)原式=3+5﹣3b=8﹣3b.18.(8分)如图,l1表示一骑自行车者、l2表示一骑摩托车者沿相同的路线由甲地到乙地行驶过程的函数图象,两地相距80千米,请根据图象解决下列问题:(1)骑自行车者出发早,早3小时;骑摩托车者早到达目的地,早3小时;他们在离甲地40公里处相遇.(2)求两个人在途中行驶的平均速度分别是多少.【考点】FH:一次函数的应用.【解答】解:(1)观察函数图象可知:骑自行车者出发早,早3小时;骑摩托车者早到达目的地,早3小时;他们在离甲地40公里处相遇.故答案为:自行车;3;摩托车;3;40.(2)骑自行车者的平均速度为80÷8=10(千米/小时);骑摩托车者的平均速度为80÷(5﹣3)=40(千米/小时).答:骑自行车者在途中行驶的平均速度为10千米/小时,骑摩托车者在途中行驶的平均速度为40千米/小时.19.(10分)如图,正方形网格中,每个小正方形的边长均为1,每个小正方形的顶点叫格点,以格点为顶点按下列要求画图:(1)在图 中画一条线段MN,使MN=;(2)在图 中画一个三边长均为无理数,且各边都不相等的直角△DEF.【考点】KQ:勾股定理.【解答】解:如图所示:20.(10分)如图,平行四边形ABCD的对角线AC与BD相交于O,AB=5,CO=4,OD=3,求证:平行四边形ABCD是菱形.【考点】L5:平行四边形的性质;L9:菱形的判定.【解答】证明:∵四边形ABCD为平行四边形,∴CD=AB=5,∵CO=4,OD=3,∴OC2+OD2=AB2,∴△COD为直角三角形,∴AC⊥BD,∴四边形ABCD为菱形.21.(10分)如图,矩形ABCD中,O为BD中点,PQ过点P分别交AD、BC 于点P、Q,连接BP和DQ,求证:四边形PBQD是平行四边形.【考点】L6:平行四边形的判定;LB:矩形的性质.【解答】证明:∵四边形ABCD是矩形,∴AD∥BC,∴∠PDO=∠QBO,在△POD和△QOB中,,∴△POD≌△QOB(ASA),∴OP=OQ;又∵O为BD的中点,∴OB=OD,∴四边形PBQD为平行四边形;22.(12分)如图,以△ABC的三边为边分别作等边△ACD、△ABE、△BCF (1)求证:△EBF≌△DFC;(2)求证:四边形AEFD是平行四边形;(3)①△ABC满足AB=AC时,四边形AEFD是菱形.(无需证明)②△ABC满足∠BAC=150°时,四边形AEFD是矩形.(无需证明)③△ABC满足AB=AC,∠BAC=150°时,四边形AEFD是正方形.(无需证明)【考点】KD:全等三角形的判定与性质;KK:等边三角形的性质;L7:平行四边形的判定与性质;LA:菱形的判定与性质;LD:矩形的判定与性质;LF:正方形的判定.【解答】解:(1)∵△ABE、△BCF为等边三角形,∴AB=BE=AE,BC=CF=FB,∠ABE=∠CBF=60°,∴∠ABE﹣∠ABF=∠FBC﹣∠ABF,即∠CBA=∠FBE,在△ABC和△EBF中,,∴△ABC≌△EBF(SAS),∴EF=AC,又∵△ADC为等边三角形,∴CD=AD=AC,∴EF=AD=DC,同理可得△ABC≌△DFC,∴DF=AB=AE=DF,∴四边形AEFD是平行四边形;∴∠FEA=∠ADF,∴∠FEA+∠AEB=∠ADF+∠ADC,即∠FEB=∠CDF,在△FEB和△CDF中,.∴△EBF≌△DFC(SAS),(2)∵△EBF≌△DFC,∴EB=DF,EF=DC.∵△ACD和△ABE为等边三角形,∴AD=DC,AE=BE,∴AD=EF,AE=DF∴四边形AEFD是平行四边形;(3)①若AB=AC,则平行四边形AEFD是菱形;此时AE=AB=AC=AD,即△ABC是等腰三角形;故△ABC满足AB=AC时,四边形AEFD是菱形;②若∠BAC=150°,则平行四边形AEFD是矩形;由(1)知四边形AEFD是平行四边形,则∠EAD=90°时,可得平行四边形AEFD 是矩形,∴∠BAC=360°﹣60°﹣60°﹣90°=150°,即△ABC满足∠BAC=150°时,四边形AEFD是矩形;③综合①②的结论知:当△ABC是顶角∠BAC是150°的等腰三角形时,四边形AEFD是正方形.故答案是:①AB=AC;②∠BAC=150°;③AB=AC,∠BAC=150°.23.(14分)如图,在等腰△ACE中,已知CA=CE=2,AE=2c,点B、D、M 分别是边AC、CE、AE的中点,以BC、CD为边长分别作正方形BCGF和CDHN,连结FM、FH、MH.(1)求△ACE的面积;(2)试探究△FMH是否是等腰直角三角形?并对结论给予证明;(3)当∠GCN=30°时,求△FMH的面积.【考点】KU:勾股定理的应用.【解答】解:(1)连结CM,∵CA=CE=2,M分别是边AE的中点,∴CM⊥AE.…(1分)在RT△ACM中,,由勾股定理得,.∴S=AE•CM=c.…(2分)△ACE(2)△FMH是等腰直角三角形.…(3分)证明:连结BM,DM.∵CA=CE=2,点B、D、M分别是边AC、CE、AE的中点,∴BC=CD=BM=DM=1.…(4分)∴四边形BCDM是边长为1的菱形,∴∠CBM=∠CDM.∴∠CBM+∠FBC=∠CDM+∠HDC,即∠FBM=∠HDM,∴△FBM≌△MDH.…(4分)∴FM=MH,且∠FMB=∠HMD(设大小为θ).又设∠A=α,则∠BMA=∠DME=∠E=∠A=α,∠MDC=2α.在△MDH中,DM=DH=1,∴∠DHM=∠DMH=θ,由三角形内角和定理可有:∴∠DHM+∠DMH+∠MDH=180°,得:θ+θ+2α+90°=180°,∴α+θ=45°.…(5分)∴∠FMH=180°﹣∠AMH﹣∠CMH=180°﹣2(α+θ)=90°.∴△FMH是等腰直角三角形.…(6分)(3)在等腰△ACE中,∠ACE=180°﹣2α,又当∠GCN=30°时,∠ACE=360°﹣∠GCN=180°﹣30°=150°从而有:180°﹣2α=150°,又α+θ=45°,得θ=30°,α=15°.…(7分)如图,作△HMD的边MD上的高HQ,则由勾股定理有:,,…(8分)∴△FMH的面积.…(9分)。

2017年广州市中大附中八年级下学期期末英语试卷及答案

2017年广州市中大附中八年级下学期期末英语试卷及答案

2017年广州市中大附中八年级下学期期末英语试卷一、语法选择(共15小题;每小题1分,满分15分)Growing up is not always easy.When facing difficulties,courage and a spirit of independence can be more useful than crying for help.That's what’sthe story I’ve heard recently tells me..The young man was born in1982in a____1____family in a small town.When he was only 11,his father became mentally ill and one day came back with an abandoned baby girl.A year____3____hard life later,his____2____left home with his younger brother because she couldn’tand pressure from her mad husband.The12-year-old boy began to____4____his whole family:he took care of not only his sick father,but also the little girl.____5_____,he didn't give up ter the went to college.To care for the little girl,he rented a room near his college and____6____her education by doing part-time jobs.After the young man’ss tory went public,people were deeply impressed by his behavior.____7____his life was hard,he didn’tabandon his father and his sister because____8____ needed his help very much.And he____9____donations from others.He said he felt____10____ because of kind offers,but he could depend on his own work.With his hard-won money,he____11____helped other students in trouble.Today when many young people worry more about their own happiness,this young man’sbehavior reminds us of what we usually____12____:love and care for others.Nobody could continue to live____13____these.Through his hard life,the young man is regarded as a____14____.So when facing difficulties,don't____15____bad luck.Consider what more you could do for your family and society and you'll find the world smiling back.1.A.rich B.happy C.poor D.sad2.A.father B.mother C.sister D.brother3.A.understand B.stand C.believe D.think4.A.support anize C.hate D.like5.A.At times B.Sometimes C.At the sametime D.For the first time6.A.planted B.afforded C.ended D.went7.A.Since B.Unless C.If D.Although8.A.they B.he C.she D.you9.A.received B.expected C.refused D.got10.A.sorry B.worried C.encouraged D.sad11.A.even B.hardly C.still D.too12.A.get usedto B.worry about C.pay attention to D.look forward to13.A.with B.without C.by D.for14.A.fool B.volunteer C.hero D.boy15.A.knock into B.dream about C.talk about plain about二、完形填空(共10小题;每小题 1.5分,满分15分)Different people have different ideas about time.People in the USA think that it is ____16____to know the time.In cities in America,there are____17____in stations,factories and other buildings.Radio announcers give you the correct time during the day.Most Americans also have watches with them____18____they go.They want to do certain things____19____certain time.They don't like to be late.They think everyone is supposedto do____20____on time.But time is not____21____important to everybody in the world.When you visit a country in South America,you will____22____that people there don't like to rush.If you had an appointment with somebody,he could probably be late____23____he may not want to arrive on time.In South America,even the radio programs may not begin on time.The men on the radio may not think it isimportant to tell the exact____24____.People in South America think that clocks or watches are just machines.They think that you will____25____a clock or a watch control your life if you do everything on time.16.A.quiet eless C.important D.easy17.A.clocks B.cars C.books D.watches18.A.whenever B.however C.whatever D.wherever19.A.at B.on C.for D.in20.A.nothing B.everything C.something D.some things21.A.such B.as C.so D.only22.A.understand B.find C.forget D.remember23.A.while B.but C.though D.because24.A.time B.place C.weather D.news25.A.put B.want C.let D.expect三、阅读理解第一节:阅读理解(共20小题;每小题2分,满分40分)(A)Day falls on November11th.No one is quite sure exactly who first thought it up,and Singles’the most widely-accepted story is emerged from the dorms of Nanjing University in1993when four single male students got together to discuss how to break free of loneliness of single life.One suggested that because of the ones in the date,November11would be a good day to organizeDay,started by a small group of friends,gradually grew into activities for singles.Singles’single young people could use as an excuseDay,a day China’ssomething like the anti-Valentine’sto get together and have fun together.Day is about shopping—Originally about being with friends and having fun,now Singles’a long story,but the short answer is Alibaba. mostly online shopping.How did this happen?It’sBy the late2000s,most of China’sDay.ThereInternet users were familiar with Singles’might have been some shops online and offline suffering sales on that day earlier,but no majorDay online sale in2009.company until Alibaba launched its first Singles’In that first year,Alibaba was only major ecommerce company to offer a sale,and it featured just27brands offering discounts via its T-mall marketplace.The sale was definitely successful, enough to redefine the holiday on its own.In the following year,Alibaba went bigger, but it wasn’toffering more brands and deeper discounts.But other companies had noticed the potential of the 2009sales bonanza and decided to follow suit.E-commerce platforms like JD had their first majorDay went from a T-mall sale to something that Singles’Day sales in2010,and overnight Singles’was beginning to look like Cyber Monday in some western countries.Over the next few years,Alibaba,JD,and other Chinese e-commerce players all expandedDay in2012, their one-day discounts,and sales grew exponentially(以指数方式).On Singles’marketplaces,Taobao and T-mall,did about$3billion in sales.In2013that number Alibaba’snearly doubled,and Chinese shoppers had obliterated America’sCyber Monday spending recordsonly getting bigger.in just the first few hours of the sale.And it’sDay was first set________.26.We can know that Singles’A.by a large e-commerce company called Alibaba.B.for the young people who want to have fun togetherC.to organize activities for singles to get rid of their lonelinessDayD.to express something against Valentine’sDay,single young people were most likely to________.27.Before2009on Singles’A.just spend the day aloneB.have fun with friendsC.do some online shoppingD.go back home to visit parents28.Which of the following statements is TRUE according to the passage?A.By2009,Singles’Day have already became a famous online-shopping day.Day.B.Cyber Monday in some western countries is to celebrate Singles’marketplaces did about nearly$4billion in sales.C.In2013,Alibaba’sD.Alibaba was the first major e-commerce company to offer a sale on Singles’Day.29.According to the last two paragraphs,we can know that________.DayA.Alibaba played an important part in the new definition of Singles’B.Other Chinese e-commerce company like JD followed Cyber MondyC.Alibaba offered more brands and deeper discount to make it biggerD.Alibaba,Taobao,T-mall and JD are different e-commerce dompanies30.What is the passagemainly about?Day. B.The life of modern young single people.A.The origin of Singles’C.The e-commerce company Alibaba.D.How Singles’Day become a shopping day.(B)Dear Lan Lan,I’mnow writing to you at Hartsop,a village in the Lake District,a place to have most beautiful scenery in English couple here to spend Christmas with them.This was planned for overseas students to know British way of life by living with British families.We came two days before Christmas and during our stay,we have had everything we expected.Christmas turkey,Christmas cake,Christmas puddings,Christmas parties and Christmasgifts under the Christmas tree.All are exciting and amusing,but above all these,we are deeply touched by the hospitality of the family,Roger and Anne Marie.Anne Marie was a nurse and all thesedays,she had been busy cooking meals,washing dishes and showing us around.Roger,who was a doctor,knows a lot about China and still wants to know more.He plays us Chinese music and it seems to me he does better in that than we do.In the evenings,we all sit around the fireplace with Tim,a lovely dog,by our side.Like long-time-no-see friends,we talk about all the things that have happenedor we hope to happen in our lives.There is always so much to tell and to know.After three months away from home,we are again feeling how sweet a home can be.Roger and Anne Marrie are not like most of the other people we have met who always make us feel we are foreigners in a foreign country.They show such kindness to us that they bring us a person-to-person feeling,instead of a British-to-Chinese feeling.They make us believe that though there are differences of languages and cultures,one may always expect to find in every corner of the world the feeling of love and being loved.How I wish Mum and Dad could meet them!How I wish you were here with me!How strange it is that the more I feel at home here,the more I miss my real home and all of you.We always talk about“when w e go back home next year…”a nd soon we will turn this into“when w eDay is coming.Miss you.go back home THIS year…”f or the New Year’sLoveCao Wen31.The writer is________now in England.A.a Chinese visitor to an English familyB.a Chinese student studyingC.a Chinese teacher workingD.a Chinese student living32.In Hartsop,Cao Wen felt________.A.quite at homeB.the English home better than her homeC.she was a foreigner in a foreign countryD.she was one member of the family33.It seemsthat Roger________.A.can speak Chinese very wellB.knows something about Chinese for a long timeC.once lived in China for a long timeD.can do everything better than the Chinese34.When you are away from home,it is good for you to have the feeling________.A.of being aloneB.of going home soonC.of love and being lovedD.as a foreigneropinion,the best home is________.35.In the writer’sA.the home they stayed in EnglandB.the home that you left behindown homeC.Chinese homeD.one’s(C)Super Typhoon Megi crashed into the Philippines on Monday,leaving a trail of destruction in its path.Thousands of residents fled to safety as heavy rains and strong winds caused flooding, collapsed trees and cut off power.It was the strongest storm to hit the island nation in four years.At least seven people died in the storm,and at least six others were injured by falling trees, broken rooftops and glass,officials said.More than4,150residents were forced to take shelter in schools,town halls,churches and relatives'homes.A typhoon is a tropical cyclone,or hurricane,that occurs in the western Pacific or Indianwhen winds exceed150mph.Megi crashed ashore Oceans.The storm is named a“super typhoon”in the Isabela province of the northern Philippines with whipping winds that reached162mph. The storm created big waves in Palanan Bay.All ships and fishing boats were advised to stay in ports.Several airline flights were canceled.Luckily,the winds blew high from the ground,sparing many rice fields from damage,says Alvaro Antonio,the governor of Cagayan province.The region also took a hard hit from the storm.Thousands of military reserve officers and volunteers stood by to help with relief efforts. Evacuations and emergency preparations for the storm were made days in advance.Retired army Major General Benito Ramos heads the Philippines'disaster-preparedness agency.He said readying for the typhoon was like“preparing f or war.”Notes:①tropical cyclone热带气旋②evacuation n.疏散③collapsed adj.倒塌的36.How many people were homeless in this natural disaster?A.At least six.B.At least seven.C.About13.D.More than4,150.37.Which of the following description about typhoon is RIGHT?A.It is a tropical cyclone occurs in the eastern Pacific.B.It is a hurricane occurs in the Atlantic Ocean.C.When winds exceed100mph,the storm is named a“super typhoon”.D.It can bring a great destruction.38.What does the underlined word mean?A.Avoiding.B.Loosing.C.Taking. D.Protecting.39.Which of the following didn't get involved in this disaster?A.Military reserve officers.B.Volunteers.C.Army D.Disaster-preparednessagency.40.What can we infer from the major General's words?A.The preparing work is all done by soldiers.B.The preparing work is a hard and responsible task.C.The army should get into the relief efforts.D.They did the same thing with soldiers.(D)No one is sure how the ancient Egyptians built the pyramids near Cairo.But a new study.Long-ago builders could have attached wooden pole s to the suggeststhey used a little rock‘n’rollstones and rolled then across the sand,the scientists say.proposing is possible,”physicist Daniel Bonn said.“Technically,I think what they’reno obvious People have long puzzled over how the Egyptians moved such huge rocks.And there’sanswer.On average,each of the two million big stones weighed about as much as a large pickup truck.The Egyptians somehow moved the stone blocks to the pyramid site from about one kilometer away.The most popular view is that Egyptian workers slid the blocks along smooth paths.Many scientists suspect workers first would have put the blocks on sleds(滑板).Then they would have dragged them along paths.To make the work easier,workers may have lubricated the paths either with wet clay or with the fat from cattle.Bonn has now tested this idea by building small sleds and dragging heavy objects over sand.Evidence from the sand supports this idea.Researchersfound small amounts of fat,as well as a large amount of stone and the remains of paths.However,physicist JosephWest think there might have been a simpler way,led the newstudy.West said,“Iwas inspired while watching a television program showing how sleds mightthey just try rolling the have helped with pyramid construction.I thought,‘Why don’tthings?’”Asquare could be turned into a rough sort of wheel by attaching wooden poles to its.sides,he realized.That,he notes,should make a block of stone“alot easierto roll than a square”So he tried it.He and his students tied some poles to each of four sides of a30-kilogram stone block.That action turned the block into somewhat a wheel.Then they placed the block on the ground.They wrapped one end of a rope around the block and pulled.The researchers found they could easily roll the block along different kinds of paths.They calculated that rolling the block required about as much force as moving it along a slippery(滑的)path.tested his idea on larger blocks,but he thinks rolling has clear advantages over West hasn’thave needed to carry cattle fat or water to smooth the paths.sliding.At least,workers wouldn’twidely believed that the stone blocks were moved to the pyramid site by________.41.It’sA.rolling them on roadsB.pushing them over the sandC.sliding them on smooth pathsD.dragging them on some polesParagraph4means________.42.The underlined part“lubricatedthe paths”inA.made the path wetB.made the path hardC.made the path wideD.made the path slipperyParagraph7refer to?43.What does the underlined word“it”inA.Rolling the blocks with poles attached.B.Rolling the blocks on wooden wheels.C.Rolling poles to move the blocks.D.rolling the blocks with fat.44.Why is rolling better than sliding according to West?A.Becausemore force is needed for sliding.B.Becauserolling work can be done by fewer cattle.C.Becausesliding on smooth road is more dangerous.D.Becauseless preparation on path is needed for rolling.45.What is the text mainly aboutA.An experiment on ways of moving blocks to the pyramid site.B.An application of the method of moving blocks to the pyramid site.C.An argument about different methods of moving blocks to the pyramid site.D.An introduction to a possible new way of moving blocks to the pyramid site.第二节阅读填空Taking good notes is a time-saving skill that will help you to become a better student in several ways.____46____Second,your notes are excellent materials to refer to when you are studying for a test.Third,note-taking offers variety to your study time and helps you to hold your interest.You will want to take notes during classroom discussions and while reading textbook or doing research for a report.____47____Whenever or however you take to tell and to know notes,keep in mind that note-taking is a selective process.____48____The following methods may work best for you.●Read the text quickly to find the main facts and ideas in it.●Carefully read the text and watch for words that can show main points and supporting facts.●Write your notes in your own words.●____49____●Note any questions or ideas you may have about what was said or written.As you take notes,you may want to use your own shorthand(速记).When you do,be sure that you understand your symbols and that you use them all the time.____50____e words,not complete sentences.B.Otherwise,you may not be able to read your notes later.C.You will also want to develop your own method for taking notes.D.That means you must first decide what is important enough to include in your notes.E.First,the simple act of writing something down makes it easier for you to understand and remember it.五、写作(共三节,满分35分)第一节单词拼写51.When meeting something difficult or dangerous,he showed great c________.s________our hands or raise your hand.52.When she came in,she didn’t53.The police asked the shopkeeper to d________the thief so that they could know more about the theft.great!We have s________another goal!54.That’s55.Some of the stones w________twenty-six tons.56.I prefer the quiet countryside to the n________cities.第二节完成句子57.妈妈正在为圣诞晚餐做准备。

2017越秀区八年级下学期期末考试数学试题

2017越秀区八年级下学期期末考试数学试题

2017-2018学年第二学期教学质量检测试卷本试卷共三大题,满分120分,考试时间90分钟,不能使用计算器。

一、选择题(本题共有10小题,每小题2分,共20分)注意:每小题有四个选项,其中有且仅有一项符合题意,选错、不选、多选或涂改不清的均不给分.1.在下列四个函数中,是一次函数的是( ).A .2y x =B .21y x =+C .21y x =+D .16y x=+2.某班要从9名百米跑成绩各不相同的同学中选4名参加4×100米接力赛,而这9名同学只知道自己的成绩,要想让他们知道自己是否入选,老师只需公布他们成绩的( ).A .平均数B .中位数C .众数D .方差3.如图,每个小正方形的边长为1,A,B,C 是小正方形的顶点,则∠ABC 的度数为( ).A. 90°B. 60° C 45° D. 30°第3题 第10题4.若直角三角形一条直角边长为6,斜边长为10,则斜边上的高是( ).A .125B .245C .5D .105.教练要从甲、乙两名射击运动员中选一名成绩较稳定的运动员参加比赛。

两人在相同条件下各打了5发子弹,命中环数如下:甲:9、 8 、 7、 7、 9 ; 乙: 10、 8 、 9 、 7、 6;那么应该选( ).A .甲B .乙C .甲、乙都可以D .无法确定6.下列给出的条件中,能判定四边形ABCD 为平行四边形的是( ).A .AB=CD=DAB .AB ∥CD,AD=BCC .AB ∥CD,∠A=∠CD .∠A=∠B , ∠C=∠D7.为了了解某社区居民的用水情况,随机抽取20户居民进行了调查,下表是所抽查各户居民2018年5月份用水量的调查结果:( ).A .中位数是10(吨)B .众数是8(吨)C .平均数是10(吨)D .样本容量是208.如果12a ≤≤,则2441a a a -++-的值是( ).A .1B .1-C .23a -D .32a -9.函数y mx b =+与y bx =的大致图像是( ).10.如图,矩形ABCD 的面积为210cm ,它的两条对角线交于点1O ,以AB 、1AO 为两邻边作平行四边形11ABC O ,平行四边形11ABC O 的对角线交于点2O ,同样以AB,2AO 为两邻边作平行四边形22ABC O ,……,依次类推,则平行四边形n n ABC O 的面积为( ).A .2210cm nB .21102n cm +C .212n cmD .2102n cm二、填空题(本题共有6小题,每小题3分,共18分)11.如果一组数据:8,7,5,x ,9,4的平均数为6,那么x 的值是__________.12.若二次根式12x-在实数范围内有意义,则实数x 的取值范围是________. 13.在平行四边形ABCD 中,对角线AC 、BD 相交于O ,AC+BD=10,BC=3,则△AOD 的周长为_________.14.已知直线23y x m =+-的图像经过x 轴的正半轴,则m 的取值范围为__________.15.已知一个直角三角形的两边分别为6cm 和8cm ,则第三边为__________cm .16.某汽车在某一直线道路上行驶,该车离出发地的距离s (千米)和行驶时间t (小时)之间的函数关系如图所示(折线ABCDE )。

广州市越秀区二中2016-2017学年第二学期期中考试 初二年级 数学 试卷及参考答案

广州市越秀区二中2016-2017学年第二学期期中考试 初二年级 数学 试卷及参考答案

24. 在矩形
中,将对角线 绕着点 逆时针旋转 度得到
于 ,连接 、 .
(1) 若
,依题意以尺规作图方式补全图 .
,连接 ,作
的平分线 ,交
(2) 在( )的条件下,若

,求 的长.
(3) 当旋转角度 为多少度时,线段 、 、 满足

2016~2017学年广东广州越秀二中初二下期中试卷
一、选择题(本大题共10小题,每小题3分,共30分)
1. 【答案】B 2. 【答案】C 3. 【答案】B 4. 【答案】D 5. 【答案】C 6. 【答案】C 7. 【答案】A 8. 【答案】C 9. 【答案】C 10. 【答案】D
二、填空题(本大题共6小题,每小题3分,共18分)
11. 【答案】 12. 【答案】 13. 【答案】
14. 【答案】
(2) 若线段 是一条水渠,且点 在边 上,
,请直接写出 的长度.
22. 有两张相同的矩形纸片

(1) 如图,若将其中一张矩形纸片
,其中


沿着 折叠,点 落在点 处,设 与 相交于点 ,求 的长.
(2) 如图,若将这两张矩形纸片交叉叠放,试判断四边形
的形状,并证明.
23. 定义:如果一个正整数 不是完全平方数,且到离它最近的完全平方数 的距离是一个完全平方数,那么我们把 叫做近似 方数,把面积为近似方数 且有一条边长度为 的矩形叫做近似方形.(注:完全平方数是整数的平方). (1) 是不是近似方数?请说明理由. (2) 在一个面积为 的近似方形中能否按图所示的方式剪出两个正方形,面积分别为 和 ?请通过计算说明.
2016~2017学年广东广州越秀二中初二下期中试卷
一、选择题(本大题共10小题,每小题3分,共30分)
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( )如图乙,将薄纸片置于下唇底部,沿着纸片的上表面吹气,纸片升起,可知吹气过程空气
对纸片上表面的压力
下表面的压力(选填“大于”、“等于”、“小于”),证明气体压强与流
速的关系是

三、计算题
19 2016~2017学年广东广州越秀区广州大学附属中学初二下学期期末第20题 如图是山地自行车比赛的用车,其相关数据见下表
力的作用,该力的方向是

(2)“玉兔”月球车的轮子做得宽大,能
才不至于陷入松软的月壤中,而不使用充气轮胎
是因为月球表面大气压较
(选填“大”或“小”),充气轮胎容易爆胎.
(3)宇航员向前抛出一个小球,小球离开手后仍能继续向前运动,这是由于小球具有

13 2016~2017学年广东广州越秀区广州大学附属中学初二下学期期末第14题
满水的溢水杯中,它漂浮在水面上,从杯中溢出的水重( )
A. 大于
B. 等于
C. 小于
D. 无法判断
二、填空题
12 2016~2017学年广东广州越秀区广州大学附属中学初二下学期期末第13题
(1)宇航员王亚平在“天宫一号”上展示了一个近乎正球形的大水球,而地球上我们看到的水珠一
般都是扁球形的,这是由于地球上的水珠受到了
A. 管内液体上升是由于瓶内气压小于瓶外气压 B. 由图可知此时瓶内气压等于瓶外气压
C. 拿着该装置由一楼到二十楼管内液面会不断 D. 拿着该装置由一楼到二十楼管内液面会保持
上升
不变
11 2016~2017学年广东广州越秀区广州大学附属中学初二下学期期末第12题3分
把一小球放入盛满酒精的溢水杯中,它沉入容器底部,从杯中溢出 酒精;若将该小球放入盛
(1) 机械人工作时,主机内部的气压 使吸盘吸附在玻璃上.
(2)
(选填“大于”、“小于”或“等于”)外界大气压,而
工作过程发现窗户擦的不干净,于是用真空泵向外多抽几次气,擦窗机器人在竖直玻璃板
上水平移动时受到的摩擦力跟抽气前相比
(选填“变大”、“变小”或“不变”).
(3) 吸盘与玻璃的接触面积为
,若吸盘在此面积上对玻璃的压强为
在救灾中,通往灾区的道路严重受损,救援部队果断地开辟了水路,利用快艇转运受困群众,下
表是某种快艇部分技术参数,关于该快艇以下说法正确的是( 取
)( )
油箱体积
排水量
限速
快艇自身质量
A. 快艇在安全范围内装运的人数越多,船底受到的压强越小 B. 快艇一次最多能运载 人(人均质量为 ) C. 快艇空载时所受浮力为 D. 快艇满载时排开水的体积为
车架材料 车架材料体积/
车架质量/ 整车质量/ 车轮接触面积/
碳纤维
(1) 求碳纤维车架的密度.
(2) 若运动员的质量是 ,求比赛中的山地自行车对地面的压强( 取
).
(3) 比赛运动员在某路段匀速骑行 ,耗时
,若该路段阻力为总重力的 倍,求
他的骑行功率.
20 2016~2017学年广东广州越秀区广州大学附属中学初二下学期期末第21题
22 2016~2017学年广东广州越秀区广州大学附属中学初二下学期期末第23题 小明猜想:滑动摩擦力大小与以下三个因素有关 ①接触面所受的压力大小;②接触面的粗糙程度;③接触面面积大小 (1)
为了探究“滑动摩擦力大小与与接触面面积大小”的关系,小明利用 、 两正方体木块 (大小如右图)、书桌的水平面和弹簧测力计进行实验探究.
1 当小卡片平衡时,小华将小卡片转过一个角度,松手后小卡片
(选
填“能”或“不能”)平衡.设计此实验步骤的目的是为了探究

2 在探究同一问题时,小明将木块放在水平桌面上,设计了如图乙所示的实验,同学
们认为小华的实验优于小明的实验,其主要原因是

A. 减少摩擦力对实验结果的影响
B. 小书线运动
B.
段的速度大于 段的
C.
段受到的摩擦力为
D. 段受到摩擦力大小等于
(2) 在 、 段做功分别为

A.
B.
C.
D.
,功率分别为
、 ,则( )
7 2016~2017学年广东广州越秀区广州大学附属中学初二下学期期末第8题3分 共2个 在水平桌面上,有两个相同的圆柱形容器,肉盛相等质量的盐水,将同一鸡蛋分别放入其中,鸡 蛋静止如图所示.鸡蛋在甲、乙两杯中所受浮力分别为 和 ,盐水对容器底的压强分别为 和 ,则( )
1 第 次实验时示意图如下,存在的错误是

2 重新准确做出第 次实验后,再做了第 次实验,在方框内画出第 次实验时的示意 图.
3 上述两次实验操作均正确,以“·”表示被拉物体,在方框内画出物体在水平方向的受 力示意图.
(2) 其中第 次测力计的示数如图17所示,

第 次实验,测得物体所受的摩擦力为 .
如图所示,鲸静止在海里.图中黑点表示鲸鱼受力的作用点,画出鲸鱼的受力分析.鲸在下潜过
程中所受浮力将
(选填“变大”、“变小”或“不变”).下潜过程中鲸鱼的胸腔受到海水的压
强将会不断
(选填“变大”、“变小”或“不变”).
15 2016~2017学年广东广州越秀区广州大学附属中学初二下学期期末第16题 如右图所示,木箱静止在斜面上,用力的示意图画出木箱所受到的重力和木箱对斜面的压力.
如图所示,空中加油机正在给匀速水平飞行的战斗机加油,加油后战斗机仍以原来的高度和速度 做匀速飞行,则战斗机的( )
A. 动能不变,势能不变,机械能不变 C. 动能增加,势能不变,机械能增加
B. 动能不变,势能减少,机械能减少 D. 动能增加,势能增加,机械能增加
10 2016~2017学年广东广州越秀区广州大学附属中学初二下学期期末第11题3分 如图,取一个两端开口的细玻璃管,在它上面画上刻度,使玻璃管穿过与瓶子密封的橡皮塞插入 水中,从管子上端吹少量的气体,则( )
,则吸盘对玻璃的压力是

18 2016~2017学年广东广州越秀区广州大学附属中学初二下学期期末第19题
( )托里拆利测量大气压强值实验如图甲所示,当时的大气压强等于
高水银柱所产
生的压强.
( )若在实验过程中不小心有气泡进入玻璃管内,则管内外水银液面高度差将
(选填“变
大”、“变小”或“不变”).
通过比较,小明发现:
,由此可以初步认为

23 2016~2017学年广东广州越秀区广州大学附属中学初二下学期期末第24题 共2个 下面是测量鹅卵石体积的两种实验设计,都用到水,水的密度用 表示。
(1) 因鹅卵石较大,放不大下量筒,故采用的方法如下:
①按图所示方法进行测量,鹅卵石的体积是
②从图 到图 的操作去除鹅卵石会带走一些水,鹅卵石的体积测量值比真实
一个西瓜慢慢放入装满水的桶中时,排出 的水,最后漂浮如图所示.( 取



(1) 求西瓜漂浮时排开水的体积. (2) 求西瓜的重力. (3) 从水中缓慢托起西瓜,直到它离开水面的过程中,小明感到:竖直向上用的力越来越大.
请用学过的知识解释这种现象.
四、实验探究题
21 2016~2017学年广东广州越秀区广州大学附属中学初二下学期期末第22题 实验探究题 (1) 如图甲是小华同学探究二力平衡条件时的实验情景.
2016~2017学年广东广州越秀区广州大学附属中 学初二下学期六校联考期末物理试卷
一、选择题
1 2016~2017学年广东广州越秀区广州大学附属中学初二下学期期末第1题3分 如图所示,下列说法正确的是( )
A. 小鸟受到的重力与树枝对小鸟的支持力是一对平衡力 B. 小鸟对树枝的压力与树枝对小鸟的支持力是一对平衡力 C. 小鸟受到的重力与树枝受到的重力是一对平衡力 D. 小鸟受到的重力与小鸟对树枝的压力是一对平衡力

(选填“丙”或“丁”)中的液体是盐水.静止时木块在盐水中受到的浮

在清水中受到的浮力.(选填“大于”、“小于”或“等于”)
17 2017年广东广州番禺区初三下学期初三一模第20题3分 共2个 如图所示是“擦窗机器人”,它的“腹部”有吸盘,当擦窗机器人的真空泵将吸盘内的空气向外抽出 时,它能牢牢地吸在竖直玻璃上.
16 2016~2017学年广东广州越秀区广州大学附属中学初二下学期期末第17题 为了鉴别两个烧杯中哪个装的是清水哪杯是盐水.
(1) 小明将压强计的金属盒先后浸入到两杯液体中,如图甲和乙所示.请指出他实验操作不妥
之处:

(2) 小红用一长方形小木块,先后放入两杯液体中,小木块静止时的情况如图丙和丁.
小明用 的水平推力推重为 的箱子,箱子纹丝不动,这时小明对物体
(选
填“有”或“没有”)做功,判断依据是
.小明再用 的水平推力推着箱子在 内沿水平
地面匀速前进
,请在图甲位置中画出物体在水平方向受到的力,则物体移动的距离为图

段(选填“ ”或“ ”).这个过程中推力对物体做功的功率为

14 2016~2017学年广东广州越秀区广州大学附属中学初二下学期期末第15题
A.
B.
C.
D.
8 2016~2017学年北京东城区北京市第二中学初二下学期期中第13题2分 共8个 如图所示玻璃管两端开口处蒙的橡皮膜绷紧程度相同.将此装置置于水中,下图中的哪幅图能反 映橡皮膜受到水的压强后的凹凸情况( )
A.
B.
C.
D.
9 2016~2017学年安徽阜阳初二下学期期末第11题3分 共5个

(选填“偏小”或“偏大”).
(2) 不用量筒,只是用符合要求的弹簧测力计等如图所示器材,请你设计实验: ①写出实验过程及要测量的量 ②鹅卵石的体积的表达式(用所测出的物理量表示)
4 2016~2017学年广东广州越秀区广州大学附属中学初二下学期期末第4题3分 如图所示是双人花样滑冰运动中男运动员拉着女运动员在冰面做圆锥摆动的精彩场面,当女运动 员在水平面上做匀速圆周运动时,则下列有关女运动员的判断正确的是( )
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