(完整word版)2019-2020学年上海市杨浦区民办兰生复旦中学九年级(上)期中化学试卷

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上海市复旦初级中学2019-2020学年九年级12月月考英语试题含详解

上海市复旦初级中学2019-2020学年九年级12月月考英语试题含详解
A. Enough B. too much C. remember D. skills
E. Change F. else G. excited H. makes I. ill
For most students, the time of exams is very stressful(压力大的) and difficult. Some people find exam time sobad that they become_____21_____, because they are afraid of failing; they are afraid of letting their parents and families down. If exams are really making you ill or worried, don’t hide your feelings. Talk to someone about it. If one person doesn’t help you, ask someone_____22_____. How to get through exams? Here are some tips by educational psychologists(教育心理专家):
A. foundB. looked forC. found outD. discovered
st night someone broke ______ our house, but nothing was lost.
A. downB. intoC. upD. out
13.By nine o’clock last night, they ______ 200 pictures from the spaceship.

2019-2020学年上海市杨浦区九年级(上)期末英语试卷

2019-2020学年上海市杨浦区九年级(上)期末英语试卷

2019-2020学年上海市杨浦区九年级(上)期末英语试卷Part 2 Phonetics, Grammar and Vocabulary(第二部分语音、语法和词汇)Ⅱ. Choose the best answer (选择最恰当的答案)1. Which of the following underlined parts in different in pronunciation?()A.r e centB.m e thodC.resp e ctD.e nemy2. My mother was angry_________me because I opened her computer but couldn't put it back.()A.forB.toC.withD.by3. My cat is great, she plays much_________than my big, noisy dog.()A.quietlyB.more quietlyC.most quietlyD.the most quietly4. _________sick need to be looked after, so money must be spent on hospitals.()A.AB.AnC.TheD./5. I've got two cars. _________of them is in very good condition, I'm afraid.()A.BothB.NeitherC.AllD.None6. My neighbor doesn't seem very_________. She always ignores my greeting and avoids me.()A.politelyB.happilyC.patientlyD.friendly7. Our room is always a mess. My roommate never puts_________away.()A.somethingB.anythingC.everythingD.nothing8. We usually go for a walk after supper if there_________a good program on TV.()A.isn'tB.wasn'tC.won't beD.wasn't going to be9. The first computers_________the 1940s were bigger than cars.()A.inB.onC.atD.since10. While his friends_________for him, they heard a scream.()A.waitB.are waitingC.waitedD.were waiting11. ﹣﹣_________I do the exam this year or next year? What do you think?﹣﹣ Next year would be better.()A.CanB.MustC.NeedD.Should12. We heard a strange noise from upstairs, _________I phoned the police.()A.butB.orC.soD.for13. Judy_________ going for a walk, but no one else wanted to.()A.agreedB.offeredC.promisedD.suggested14. My purchases were three weeks late. I had expected the goods_________on time.()A.arriveB.arrivingC.arrivedD.to arrive15. Every year billions of tons rubbish_________in Europe. Are you shocked by this?()A.producesB.are producedC.be producedD.has produced16. The chemicals in cola will be bad for your teeth_________you clean them carefully.()A.unlessB.so thatC.whenD.although17. Wow, _________amazing vase! The artist must have spent a lot of time on it.()A.whatB.what aC.what anD.how18. ﹣﹣_________did the Greeks enter Troy?﹣﹣ They hid in a wooden horse.()A.HowB.WhereC.WhenD.Why19. ﹣﹣ Sorry about the mess I made in your kitchen. I'll clean it up.﹣﹣_________The cleaning lady will do that later.()A.You're welcome.B.That's too bad!C.Oh, it doesn't matter.D.Be careful, please.20. ﹣_________﹣ Yes, please. That's very kind of you.()A.What about seeing a play tonight?B.Shall I help you with the dishes?C.Would you mind passing me the salt?D.May I use your mobile phone?Ⅲ. Co mplete the following passage with the words in the box. Each can only be used once(将下列单词填入空格.每空格限填一词,每词只能填一次)Dogs are known to help humans to protect farm animals, find illegal drugs, and rescue people during natural disasters. But did you know that dogs can also be (1)_______ to help protect plants and animals?Due to their (2)_______ sense of small and their high energy level, dogs can be taught to look for alien species(物种)that endanger local plants and animals.For example, in the United States,over (3)_______million ash trees have been killed by a particular type of beetle from Asia. The females of these insects(4)_______their eggs in the tree's bark. Their young feed on the wood, killing the tree. Since these insect are almost impossible for humans to (5)_______ ,dogs have been brought in to find them out so that humans can take action.Similarly, dogs protect animals that are endangered by illegal hunting. From shark's fin to tiger fur,animal products are sold and traded globally. But dogs' smelling for these products at boarders can make those(1)_______harder to sell, which means fewer animals are likely to be killed.Of all the ways dogs protect endangered animals, smelling for their waste is probably the most (2)_______. Scientists can actually learn a lot from animal waste samples. They can then use these pieces of information to help protect the animal. But (3)_______,depending on the size and location of the animal waste, finding it can be difficult.For protecting endangered species in so many ways, dogs certainly(4)_______ our thanks.They've become not just man's best friend, but the environment's best friend,too.Ⅳ. Complete the sentences with the given words in their proper forms(用括号中所给单词的适当形式完成下列句子,每空格限填一词)Laurent,9,will become one of the________people to graduate from auniversity.(young)We need to concentrate on our________now, while we have the energy.(goal)I forgot to take my textbook. The girl next to me shared________with me.(she)The show was so boring that we fell________after an hour.(sleep)A number of animals have escaped from the zoo________, including a rare white tiger.(recent)The death of the boy's parents left him feeling extremely________and alone.(help)He can correctly________a pack of cards in just 31,16 seconds.(memory)﹣﹣ That's not a very good time of year to travel.﹣﹣ Perhaps not. It was justa________." (think)Ⅴ. Complete the following sentences as required (根据所给要求完成下列句子,每题空格限填一词)The young man denied that he had stolen the necklace.(改为一般疑问句)________the young man________that the had stolen the necklace?My uncle is ________.(对划线部分提问)________does your uncle look________?Follow these instructions and you won't make a mistake.(保持句意基本不变)Follow these instructions and you won't________."Please turn off your mobile phones," said Mr.Brown.(保持句意基本不变)Mr.Brwon told us________off our mobile phones.They have found the lost child alive and well in a London park.(改为被动语态)The lost child________found alive and well in a London park.Could you tell me where to park the car?(保持句意基本不变)Could you tell me where________park the car?find, new skills, you, it, do, to learn, difficult(连词成句)________.Part 3 Reading and Writing (第三部分读写)Ⅵ. Reading comprehension (阅读理解)A. Choose the best answer (根据以下内容,选择最恰当的答案)Crystal was a reporter for her school's newspaper, The Real Story. She'd done several different kinds of stories before, but never anything like the story she was about to do. During an interview, the school nurse, Mrs. Jones, had told Crystal that the school had a secret basement. Crystal was eager to find out what was in the basement. It could be just about anything! So Crystal asked the newspaper's advisor, Mr.Adair, if she could do a feature on the school's basement. He agreed with her that basement might be a fascinating story, especially since a lot of people weren't even aware that it existed.Crystal got permission from the headmaster to go down into the basement as long as some teachers went with her. So Mr.Adair and Mrs.Jones accompanied Crystal when she went to explore the basement.The guard unlocked the basement door, and Crystal and the two teachers slowly ________the narrow staircase. When they reached the bottom of the stairs, Mrs. Jones turned on the light, and she, Mr.Adair, and Crystal began to look around.Crystal soon discovered a large set of bookshelves with a collection of old dusty books on them. "What are those? " she wondered aloud. Neither of the two teachers knew, so Crystal made her way over to look more closely at the books. They turned out to be yearbooks from every year since the school opened. For almost anhour, Crystal, Mr. Adair, and Mrs. Jones paged slowly through the yearbooks. They joked about the old﹣fashioned clothes and hair in the oldest books, and they commented on some of the people they recognized. All of a sudden, Crystal knew the next feature she wanted to write. These books were the school's story just waiting to be told.(1)How did Crystal know about the basement?________A.From Mr.Adair.B. From Mrs.Jones.C. From the headmaster.D. From the newspaper..(2)The underlined word "descended" is closest in meaning to "________".A.touchedB. checkedC. went downD. fell on.(3)Which of the following about the basement might be TRUE?________A.It is no longer in use these days.B. It used to be a school clinic.C. Some clubs hold meeting there.D. Few people dare to visit it..(4)How did Crystal feel about her experience?________A.Frightened.B. Embarrassed.C. Disappointed.D. Excited..(5)Crystal's next featured topic is most likely to be "________".A.the school nurseB. the old yearbooksC. the lost fashionD. an underground library.(6)What's the purpose of the text?________A.To amuse readers with a personal story.B. To warn children not to take adventures.C. To describe how a reporter got a story.D. To explain how a mystery should be solved.B. Choose the best answer and complete the passage (选择最恰当的选项完成短文)Have you ever tried playing the kind of video games that your parents played? The Museum of Science in Manchester, in the UK, has held an exhibition for the last few years, which invites visitors to do(1)_______ that. It offers them the chance to play games from the last 40 years, in various sessions throughout the day.These video game sessions have now become one of the main attractions of the museum.They are full of people every day, playing a wide range of games.However, visitors often choose the ones they're(2)_______. For parents, for example, these are usually the games they used to play in their childhood.There's also a(n)(3)_______purpose to the games. For instance, some old typesof computer, dating back 40 years, are also available in the sessions. They were originally used in classrooms to teach pupils to write their own computer programs. And at the time, it helped lots of young people to do that. Now, the museum is holding workshops that encourage children to learn similar skills ﹣﹣ and they're still very popular.The exhibition also shows how much (4)_______ technology has made over the last 40 years.Parents can often remember playing very simple games.But the games that are played today are more complex.They have better story﹣lines and animation, too. And the players also have to use much more complicated techniques.(5)_______, one serious side of the exhibition is that organisers also want to show that video gaming is an important industry, employing many skilled people.So they hope the exhibition wil(6)_______this message. That way, people who enjoy gaming will also understand all the hard work, talent, imagination that goes into creating these amazing games.(1)A.especiallyB.exactlyC.directlyD.mainly(2)A.ready forB.poor atC.connected toD.familiar with(3)A.dramaticB.officialC.historicalcational(4)A.responseB.progressC.knowledgeD.impression(5)A.HoweverB.ThereforeC.In factD.In conclusion(6)A.linkB.checkC.shareD.recordC. Fill in the blanks with proper words (在短文的空格内填入适当的词,使其内容通顺,每空格限填一词,首字母已给)A rare experienceImagine the situation. You're walking down the crowded high street and s(1)________a complete stranger stops you and says, "Hi! You were on the beach in the south of Spain six years ago. How are you doing?" This stranger isn't necessarily m(2)________. He or she might be a "super ﹣ recogniser". These are people who have the unusual ability to recognise people they have seen only once ﹣﹣ a long time ago, maybe in a crowd.Whatever the differences in looksIt doesn't matter what the person looks like now. People change, get different hair styles,dye their hair or go grey. Wrinkles, new glasses and makeups give them new a (3)________, but the "super ﹣ recognisers" can still recognise them.An inborn skillAlthough scientists have known for a long time that about 2% of people suffer from face ﹣blindness, which means that they have huge problems recognising faces, they are only new realising that some people are the exact o (4)________.Tests have shown that a "super ﹣ recogniser" can identify people that they only saw for a brief moment ﹣﹣ andthis is not an ability that we can d(5)________, it's something we are born with.A great h(6)________The police are starting to use "super ﹣ recognisers" to spot criminal faces in videos of crowds. They look for people with a specific build and facial features like beards and moustaches but they can even recognise quite o (7)________ people, with no noticeable features at all. As well as surprising our holidaymaker in Spain six years later,this ability can be used for a very practical purpose indeed.D. Answer the questions (根据以下内容回答问题)The once was a king who wanted to find the answers to what he considered the three most important questions in life. The three questions are : when is the right time for every action, who are the most important people , and what is the most important thing to do?The king consulted learned men, promising he would richly reward anyone who could answer the questions, but their answers didn't satisfy him.In the end,he decided tovisit a wise hermit who lived in the woods.The hermit would speak only with common people, so the king put on simple clothes.When he got close to the hermit's hut, he ordered his bodyguards to wait behind. He approached the place and found the hermit digging. He asked the wise man his three questions,but the hermit didn't respond. Seeing how old and weak the hermit was, the king began helping him dig.The king dug for hours,but the hermit never answered his questions. Just as the king was about to leave, a wounded man came running to the hut and fainted before them. Then the king carried the wounded man to Hermit's hut and took care of him. The king was exhausted and fell asleep as well. When he awoke, he talked with the man, who said that he had wanted to kill the king but was wounded by his bodyguards. He asked for forgiveness and promised to serve the king faithfully.Afterwards, the king went outside and repeated his questions to the hermit once more. The hermit replied, "If you had not helped me dig earlier, you would have leftand been killed by the man. So, the most important man back then was me, and the most important time was when you were digging. The most important thing to do was tobe kind to me. When the wounded man arrived, he was the most important person,and saving him was the most important thing. By doing so, you made peace with an enemy."The king finally had his three answers. The most important time is the present,as it's the only time we have the power to act. The most important person is the one you are with,and the most important thing is always to (5)______.(1)Was the king satisfied with the answers given by learned men?________(2)Where did the hermit live?________(3)Why did the king help the hermit dig?He thought the hermit________.(4)Who was the wounded man?________(5)What best fits the blank in the last sentence?________(6)What were the king's rewards after he helped the hermit and the wounded man?He________and________.Ⅶ. Writing (作文)In 60﹣100 words, make up a two﹣turn dialogue according to the given situation(根据所给情境编一篇 60﹣100词的两个轮次的对话.标点符号不占格,对话的开头和结尾已给出.)Situation: Tome Brown, a 40﹣year﹣old P. E. teacher, saved a boy from being knocked down by a car. Crystal school reporter wanted to write an article about his good deed(事迹). She interviewed Tom Brown.The interview begins like this:Crystal: Mr. Brown, May I ask you a few questions?Tom Brown: Sure. Please go ahead.Crystal: ________________________________________________.Tom Brown: ______________________________________________________.Crystal: Thank you for you time! It's nice talking to you.(注意:文中不得出现考生的姓名、校名及其他相关信息,否则不予评分.)参考答案与试题解析2019-2020学年上海市杨浦区九年级(上)期末英语试卷Part 2 Phonetics, Grammar and Vocabulary(第二部分语音、语法和词汇)Ⅱ. Choose the best answer (选择最恰当的答案)1.【答案】A【解析】哪个的划线部分的发音不同?2.【答案】C【解析】我妈妈对我很生气,因为我打开了她的电脑却放不回去了.3.【答案】B【解析】我的猫很好,她比我那只吵闹的大狗玩得安静多了.4.【答案】C【解析】病人需要照顾,所以钱必须花在医院里.5.【答案】B【解析】我有两辆汽车,恐怕没有一辆是状况良好的.6.【答案】D【解析】我的邻居看起来不太友好.她总是不理我的招呼,避开我.7.【答案】B【解析】我们的房间总是乱糟糟的.我的室友从不收拾东西.8.【答案】A【解析】如果电视上没有好的节目,我们通常晚饭后去散步.9.【答案】A【解析】上世纪40年代第一台电脑比汽车还大.10.【答案】D【解析】当他的朋友们在等他时,他们听到一声尖叫.11.【答案】D【解析】﹣我应该今年考试还是明年考试?你怎么想?﹣明年会更好.12.【答案】C【解析】我们听到楼上传来一个奇怪的噪音,因此我给警察打电话了.13.【答案】D【解析】朱蒂建议去散步,但没有人想去.14.【答案】D【解析】我买的东西晚了三个星期.我原以为货物会准时到达.15.【答案】B【解析】欧洲每年产生数十亿吨垃圾,你感到震惊了吗?16.【答案】A【解析】可乐里的化学物质对你的牙齿有害,除非你仔细清洗.17.【答案】C【解析】喔,多么好看的花瓶啊!这个艺术家肯定花费了很多时间.18.【答案】A【解析】﹣希腊人如何进入特洛伊城的?﹣他们藏在木马里.19.【答案】C【解析】﹣对不起,我把你的厨房弄得一团糟.我来打扫.﹣哦,没关系.清洁女工稍后会打扫.20.【答案】B【解析】﹣要我帮你洗盘子吗?﹣是的,请.你真的太好了.Ⅲ. Complete the following passage with the words in the box. Each can only be used once(将下列单词填入空格.每空格限填一词,每词只能填一次)【答案】D,A,B,E,C【解析】主要讲述狗不但可以帮助人们保护农场动物、寻找违法药品以及拯救人们,而且他们可以被训练来帮助保护植物和野生动物.【答案】B,C,D,A【解析】本文讲述的是狗通过很多方式能保护濒临危险的物种.Ⅳ. Complete the sentences with the given words in their proper forms(用括号中所给单词的适当形式完成下列句子,每空格限填一词)【答案】youngest【解析】9岁的劳伦特将成为最年轻的大学毕业生之一.【答案】goals【解析】我们现在需要集中精力在我们的目标上,当我们有精力的时候.【答案】hers【解析】我忘了带课本.我旁边的女孩和我分享了她的课本.【答案】asleep【解析】演出太无聊了,一小时后我们就睡着了.【答案】recently【解析】最近一些动物从动物园逃走了,其中包括一只稀有的白虎.【答案】helpless【解析】孩子父母的去世使他感到非常无助和孤独.【答案】memorize【解析】他能在31.16秒内正确地记住一副牌.【答案】thought【解析】﹣﹣那不是一年中旅行的好时机.﹣也许不是.那只是一个想法.Ⅴ. Complete the following sentences as required (根据所给要求完成下列句子,每题空格限填一词)【答案】Did,deny【解析】那个年轻人否认偷了项链.那个年轻人否认偷了项链吗?【答案】tall and wear glasses,What,like【解析】我叔叔很高,戴着眼镜.你叔叔看起来是什么样子的?【答案】gowrong【解析】照着这些指示,你不会犯错的.照着这些指示,你不会出错的.【答案】to turn【解析】"请关掉你们的手机,"Mr.Brown说.Mr.Brown叫我们关掉我们的手机.【答案】has been【解析】他们在伦敦的一个公园里找到了那个走失的孩子.那个走失的孩子在伦敦的一个公园里被找到了.【答案】【解析】你能告诉我在哪里停车吗?你能告诉我在哪里我能停车吗?【答案】Doyoufinditdifficulttolearnnewskills?【解析】Do you find it difficult to learn new skills?你觉得学习新技能很难吗?Part 3 Reading and Writing (第三部分读写)Ⅵ. Reading comprehension (阅读理解)A. Choose the best answer (根据以下内容,选择最恰当的答案)【答案】BCADBC【解析】这篇文章讲述了作为校园记者的Crystal探索学校地下室的故事.B. Choose the best answer and complete the passage (选择最恰当的选项完成短文)【答案】BDDBAC【解析】主要介绍英国的曼彻斯特博物馆过去几年展览过去的游戏并且非常受欢迎.并且讲述它的三个意义:一、具有教育的意义;二、科技在过去的四十年已经取得了重大的进步;三、向人们展示游戏也是一个重要的产业.C. Fill in the blanks with proper words (在短文的空格内填入适当的词,使其内容通顺,每空格限填一词,首字母已给)【答案】uddenly,ad,ppearances,pposite,evelop,elp,rdinary【解析】短文主要介绍了世界上有一种人,能在人群中轻易地辨认出很多年前见过的人,这种拥有超级辨认能力的人对警察有很大的帮助.D. Answer the questions (根据以下内容回答问题)【答案】No, he wasn't.,He lived in a hut/ the woods.,was old and weak,He was the one who wanted to kill the king./ He wasn't the king's enemy.,to be kind to that person/do goodto that person/ cherish the person,(6)He escaped death/made peace with his enemy,got his answers/ learned the wisdom about life本文主要讲述一个国王考虑三个人生问题,即"什么时候是行动的最佳时机,谁是最重要的人,什么是最重要的事".后来国王去拜访一个隐士,在那里帮隐士挖坑、救了一个受伤的人并悉心照顾他,最后国王知道了这三个问题的答案.Ⅶ. Writing (作文)【答案】Crystal: Mr. Brown, May I ask you a few questions?Tom Brown: Sure. Please go ahead.Crystal:When and where the accident happened?(询问事故发生的时间和地点)Tom Brown: At about 8:30, I was walking on Ocean Road when a boy on a bike passed me quickly.【高分句型一】The road was empty at that time so the boy kept riding in the middle.Crystal: How did the accident happened?(询问事故发生的过程)Tom Brown: The boy was listening to music with earplugs and he didn't hear the warning. Suddenly the car speeded up and passed the boy. Unfortunately, it hit the boy. The boy fell off his bike onto the road, but the car didn't stop.Crystal: What did you do then?(询问处理的结果)Tom Brown: The number plate of the car was so dirty that I couldn't see the number clearly.【高分句型二】 The boy was badly hurt so I sent him to the hospital. Then I called the police.Crystal: Thank you for your time! It's nice talking to you.【解析】【高分句型一】At about 8:30, I was walking on Ocean Road when a boy on a bike passed me quickly.在八点半左右,我在大洋路上散步时,这时一个骑自行车的男孩很快从我身边经过.本句使用了be doing…when…结构;when引导时间状语从句,主句用了过去进行时,从句用一般过去时.【高分句型二】The number plate of the car was so dirty that I couldn't see the number clearly.这辆车的车牌太脏了,我看不清楚号码.so…that…如此……以至于……,引导结果状语从句.。

上海民办兰生复旦中学语文九年级上册文言文试卷

上海民办兰生复旦中学语文九年级上册文言文试卷

上海民办兰生复旦中学语文九年级上册文言文试卷一、文言文1.阅读下面的文段,然后回答问题。

范仲淹二岁而孤,家贫无依。

少有大志,每以天下为已任,发愤苦读,或夜昏怠,辄以水沃面;食不给,啖粥而读,既仕,每慷慨论天下事,奋不顾身。

乃至被谗受贬,由参知政事谪守邓州。

仲淹刻苦自励,食不重肉,妻子衣食仅自足而已。

常自诵日:“士当先天下之忧而忧,后天下之乐而乐也。

”(选自《宋名臣言行录》)(1)请解释下列加下划线词的意思。

①辄以水沃面________ ②或夜昏怠________(2)宋濂和范仲淹的求学经历有何相似之处?从中你获得怎样的启迪?2.阅读下文,回答问题。

岳阳楼记①庆历四年春,滕子京谪守巴陵郡。

越明年,政通人和,百废具兴。

乃重修岳阳楼,增其旧制;刻唐贤今人诗赋于其上。

属予作文以记之。

②予观夫巴陵胜状,在洞庭一湖。

衔远山,吞长江,浩浩汤汤,横无际涯:,朝晖夕阴,气象万千。

此则岳阳楼之大观也。

前人之述备矣。

然则北通巫峡,南极潇湘,迁客骚人,多会于此,览物之情,得无异乎?③若夫淫雨霏霏,连月不开,阴风怒号,浊浪排空:日星隐曜,山岳潜形;则有去国怀乡,忧谗畏;商旅不行,樯倾楫摧:薄暮冥冥,登斯楼也,则有去国怀乡,忧谗畏讥,满目萧然,感极而悲者矣。

④至若春和景明,波澜不惊,上下天光,,一碧万顷;沙鸥翔集,锦鳞游泳:岸芷汀兰,郁郁青青。

而或长烟一空,皓月千里,浮光跃金,静影沉璧,渔歌互答,此乐何极!登斯楼也,则有心旷神怡,宠辱偕忘,把酒临风,其喜洋洋者矣。

⑤嗟夫!予尝求古仁人之心,或异二者之为,何哉?不以物喜,不以己悲;居庙堂之高则忧其民:处江湖之远则忧其君。

是进亦忧退亦忧。

然则何时而乐耶?其必曰“先天下之忧而忧,后天下之乐而乐”乎。

噫!微斯人,吾谁与归?(1)解释加下划线词语。

①谪守________②连月不开________③锦鳞游泳________(2)翻译下列句子。

①日星隐曜,山岳潜形。

②居庙堂之高则忧其民,处江湖之远则忧其君。

2020-2021上海民办兰生复旦中学九年级数学上期中模拟试题(含答案)

2020-2021上海民办兰生复旦中学九年级数学上期中模拟试题(含答案)

2020-2021上海民办兰生复旦中学九年级数学上期中模拟试题(含答案)一、选择题1.二次函数y=ax2+bx+c(a≠0)的图象如图所示,那么下列说法正确的是()A.a>0,b>0,c>0 B.a<0,b>0,c>0 C.a<0,b>0,c<0D.a<0,b<0,c>0 2.如图,抛物线y=ax2+bx+c经过点(-1,0),对称轴为直线l.则下列结论:①abc>0;②a-b+c=0;③2a+c<0;④a+b<0.其中所有正确的结论是()A.①③B.②③C.②④D.②③④3.已知抛物线y=x2-2mx-4(m>0)的顶点M关于坐标原点O的对称点为M′,若点M′在这条抛物线上,则点M的坐标为()A.(1,-5)B.(3,-13)C.(2,-8)D.(4,-20)4.下列事件中,属于必然事件的是()A.三角形的外心到三边的距离相等B.某射击运动员射击一次,命中靶心C.任意画一个三角形,其内角和是 180°D.抛一枚硬币,落地后正面朝上5.下列交通标志是中心对称图形的为()A.B.C.D.6.如图,AD、BC是⊙O的两条互相垂直的直径,点P从点O出发,沿O→C→D→O的路线匀速运动.设∠APB=y(单位:度),那么y与点P运动的时间x(单位:秒)的关系图是()A .AB .BC .CD .D7.如图,从一张腰长为90cm ,顶角为120︒的等腰三角形铁皮OAB 中剪出一个最大的扇形OCD ,用此剪下的扇形铁皮围成一个圆锥的侧面(不计损耗),则该圆锥的底面半径为( )A .15cmB .12cmC .10cmD .20cm 8.如图所示,⊙O 是正方形ABCD 的外接圆,P 是⊙O 上不与A 、B 重合的任意一点,则∠APB 等于( )A .45°B .60°C .45° 或135°D .60° 或120° 9.100个大小相同的球,用1至100编号,任意摸出一个球,则摸出的编号是质数的概率是 ( )A .120B .19100C .14D .以上都不对10.如图,圆锥的底面半径r 为6cm ,高h 为8cm ,则圆锥的侧面积为( )A .30πcm 2B .48πcm 2C .60πcm 2D .80πcm 211.有两个一元二次方程2:0M ax bx c ++=,2:0N cx bx a ++=,其中,0ac ≠,a c ≠,下列四个结论中错误的是( )A .如果方程M 有两个不相等的实数根,那么方程N 也有两个不相等的实数B .如果4是方程M 的一个根,那么14是方程N 的另一个根 C .如果方程M 有两根符号相同,那么方程N 的两符号也相同D .如果方程M 和方程N 有一个相同的根,那么这个根必是1x =12.用配方法解方程2890x x ++=,变形后的结果正确的是( )A .()249x +=-B .()247x +=-C .()2425x +=D .()247x += 二、填空题13.二次函数y =ax 2+bx +c 的图象如图11所示,且P =|2a +b|+|3b -2c|,Q =|2a -b|-|3b +2c|,则P ,Q 的大小关系是______.14.新园小区计划在一块长为20米,宽12米的矩形场地上修建三条互相垂直的长方形甬路(一条橫向、两条纵向,且横向、纵向的宽度比为3:2),其余部分种花草.若要使种花草的面积达到144米2.则横向的甬路宽为_____米.15.如图,AB 为⊙O 的直径,点C 、D 在⊙O 上,若∠D =20°,则∠CBA 的度数是__.16.如图,直线l 经过⊙O 的圆心O ,与⊙O 交于A 、B 两点,点C 在⊙O 上,∠AOC =30°,点P 是直线l 上的一个动点(与圆心O 不重合),直线CP 与⊙O 相交于点Q ,且PQ =OQ ,则满足条件的∠OCP 的大小为_______.17.已知x 1,x 2是方程x 2﹣x ﹣3=0的两根,则1211+x x =_____. 18.女生小琳所在班级共有40名学生,其中女生占60%.现学校组织部分女生去市三女中参观,需要从小琳所在班级的女生当中随机抽取一名女生参加,那么小琳被抽到的概率是 .19.若关于 x 的一元二次方程2x 2-x+m=0 有两个相等的实数根,则 m 的值为__________.20.若抛物线的顶点坐标为(2,9),且它在x 轴截得的线段长为6,则该抛物线的表达式为________.三、解答题21.某商场销售某种型号防护面罩,进货价为40元/个.经市场销售发现:售价为50元/个时,每周可以售出100个,若每涨价1元,就会少售出5个.供货厂家规定市场售价不得低于50元/个,且商场每周销售数量不得少于80个.(1)确定商场每周销售这种型号防护面罩所得的利润w (元)与售价x (元/个)之间的函数关系式.(2)当售价x (元/个)定为多少时,商场每周销售这种防护面罩所得的利润w (元)最大?最大利润是多少?22.如图,AB 是O e 的直径,点C D 、在O e 上,且四边形AOCD 是平行四边形,过点D 作O e 的切线,分别交OA 的延长线与OC 的延长线于点E F 、,连接BF 。

2019-2020学年上海市杨浦区初三上期末考试数学试卷(含答案)

2019-2020学年上海市杨浦区初三上期末考试数学试卷(含答案)

杨浦区度第一学期期末质量调研初 三 数 学 试 卷(测试时间:100分钟,满分:150分)考生注意:1.本试卷含三个大题,共25题.答题时,考生务必按答题要求在答题纸规定的位置上作答,在草稿纸、本试卷上答题一律无效.2.除第一、二大题外,其余各题如无特别说明,都必须在答题纸的相应位置上写出证明或计算的主要步骤.一、选择题:(本大题共6题,每题4分,满分24分) 1.如果5x =6y ,那么下列结论正确的是 (A ):6:5x y =; (B ):5:6x y =;(C )5,6x y ==;(D )6,5x y ==.2.下列条件中,一定能判断两个等腰三角形相似的是(A )都含有一个40°的内角; (B )都含有一个50°的内角; (C )都含有一个60°的内角; (D )都含有一个70°的内角.3.如果△ABC ∽△DEF ,A 、B 分别对应D 、E ,且AB ∶DE =1∶2,那么下列等式一定成立的是 (A )BC ∶DE =1∶2; (B ) △ABC 的面积∶△DEF 的面积=1∶2; (C )∠A 的度数∶∠D 的度数=1∶2;(D )△ABC 的周长∶△DEF 的周长=1∶2.4.如果2a b =(,a b 均为非零向量),那么下列结论错误的是(A )//a b ;(B )20a b -=; (C )12b a =; (D )2a b =. 5.如果二次函数2y ax bx c =++(0a ≠)的图像如图所示,那么下列不等式成立的是 (A )0a >; (B )0b <;(C )0ac <;(D )0bc <.6.如图,在△ABC 中,点D 、E 、F 分别在边AB 、AC 、BC 上,且∠AED =∠B ,再将下列四个选项中的一个作为条件,不一定能使得△ADE ∽△BDF 的是(A )EA EDBD BF =; (B )EA EDBF BD =;(C )AD AEBD BF=; (D )BD BABF BC=.二、填空题:(本大题共12题,每题4分,满分48分) 7.抛物线23y x =-的顶点坐标是 .8.化简:112()3()22a b a b --+= . 9.点A (-1,m )和点B (-2,n )都在抛物线2(3)2y x =-+上,则m 与n 的大小关系为m n (填“<”或“>”).10.请写出一个开口向下,且与y 轴的交点坐标为(0,4)的抛物线的表达式 . 11.如图,DE //FG //BC ,AD ∶DF ∶FB =2∶3∶4,如果EG =4,那么AC = .12.如图,在□ABCD 中,AC 、BD 相交于点O ,点E 是OA 的中点,联结BE 并延长交AD 于点F ,如果△AEF的面积是4,那么△BCE 的面积是 .13.Rt △ABC 中,∠C =90°,如果AC =9,cos A =13,那么AB = . 14.如果某人滑雪时沿着一斜坡下滑了130米的同时,在铅垂方向上下降了50米,那么该斜坡的坡度是1∶ .15.如图,Rt △ABC 中,∠C =90°,M 是AB 中点,MH ⊥BC ,垂足为点H ,CM 与AH 交于点O ,如果AB =12,那么CO = .16.已知抛物线22y ax ax c =++,那么点P (-3,4)关于该抛物线的对称轴对称的点的坐标是 . 17.在平面直角坐标系中,将点(-b ,-a )称为点(a ,b )的“关联点”(例如点(-2,-1)是点(1,2)的“关联点”).如果一个点和它的“关联点”在同一象限内,那么这一点在第 象限. 18.如图,在△ABC 中,AB =AC ,将△ABC 绕点A 旋转,当点B 与点C 重合时,点C 落在点D 处,如果sin B =23,BC =6,那么BC 的中点M 和CD 的中点N 的距离是 .C(第18题图)(第11题图)(第12题图)(第15题图)B三、解答题:(本大题共7题,满分78分) 19.(本题满分10分)计算:cos 45tan 45sin 60cot 60cot 452sin 30︒⋅︒-︒⋅︒︒+︒20.(本题满分10分,第(1)、(2)小题各5分) 已知:如图,Rt △ABC 中,∠ACB =90°,sin B =35,点D 、E 分别在边AB 、BC上,且AD ∶DB =2∶3,DE ⊥BC . (1)求∠DCE 的正切值;(2)如果设AB a =,CD b =,试用a 、b 表示AC .21.(本题满分10分)甲、乙两人分别站在相距6米的A 、B 两点练习打羽毛球,已知羽毛球飞行的路线为抛物线的一部分,甲在离地面1米的C 处发出一球,乙在离地面1.5米的D 处成功击球,球飞行过程中的最高点H 与甲的水平距离AE 为4米,现以A 为原点,直线AB 为x 轴,建立平面直角坐标系(如图所示).求羽毛球飞行的路线所在的抛物线的表达式及飞行的最高高度.22.(本题满分10分)如图是某路灯在铅垂面内的示意图,灯柱BC 的高为10米,灯柱BC 与灯杆AB 的夹角为120°.路灯采用锥形灯罩,在地面上的照射区域别为α和45°,且DE 的长为13.3米,从D 、E 两处测得路灯A 的仰角分(第20题图)xtan α=6. 求灯杆AB 的长度.23.(本题满分12分,第(1)小题5分,第(2)小题7分)已知:梯形ABCD 中,AD //BC ,AD =AB ,对角线AC 、BD 交于点E ,点F 在边BC 上,且∠BEF =∠BAC . (1)求证:△AED ∽△CFE ; (2)当EF //DC 时,求证:AE =DE .24.(本题满分12分,第(1)小题3分,第(2)小题5分,第(3)小题4分)在平面直角坐标系xOy 中,抛物线2221y x mx m m =-+--+交 y 轴于点为A ,顶点为D ,对称轴与x轴交于点H .(1)求顶点D 的坐标(用含m 的代数式表示); (2)当抛物线过点(1,-2),且不经过第一象限时,平移此抛物线到抛物线22y x x =-+的位置,求平移的方向和距离; (3)当抛物线顶点D 在第二象限时,如果∠ADH =∠AHO ,求m 的值.25.(本题满分14分,第(1)、(2)小题各6分,第(3)小题2分)已知:矩形ABCD 中,AB =4,BC =3,点M 、N 分别在边AB 、CD 上,直线MN 交矩形对角线AC 于点E ,将△AME 沿直线MN 翻折,点A 落在点P 处,且点P 在射线CB 上. (1)如图1,当EP ⊥BC 时,求CN 的长; (2)如图2,当EP ⊥AC 时,求AM 的长;(第23题图)(3)请写出线段CP 的长的取值范围,及当CP 的长最大时MN 的长.杨浦区初三数学期末试卷参考答案及评分建议一、 选择题:(本大题共6题,每题4分,满分24分) 1、A ; 2、C ; 3、D ; 4、B ; 5、C ; 6、C 二、 填空题:(本大题共12题,每题4分,满分48分)7、(0,-3); 8、142a b -rr ; 9、<;10、24y x =-+等; 11、12; 12、36; 13、27; 14、2.4; 15、4; 16、(1,4); 17、二、四; 18、4 三、解答题:(本大题共7题,满分78分) 19.(本题满分10分)解:原式=12231122⋅+⨯--------------------------------------------------(6分)=1222-----------------------------------------------------------------(2分) (备用图)(图1)ABC D NPME(图2) A BCD N P ME(第25题图)AB CD. --------------------------------------------------------------(2分) 20.(本题满分10分,第(1)、(2)小题各5分) 解:(1)∵∠ACB =90°,sin B =35,∴35AC AB =. -------------------------(1分)∴设AC =3a ,AB =5a . 则BC =4a . ∵AD :DB =2:3,∴AD =2a ,DB =3a . ∵∠ACB =90°即AC ⊥BC ,又DE ⊥BC , ∴AC//DE. ∴DE BD AC AB =, CE ADCB AB=. ∴335DE a a a =, 245CE a a a =. ∴95DE a =,85CE a =.----------(2分) ∵DE ⊥BC ,∴9tan 8DE DCE CE ∠==.-----------------------------(2分) (2)∵AD :DB =2:3,∴AD :AB =2:5. ------------------------------------------------(1分) ∵AB a =,CD b =,∴25AD a =. DC b =-.--------------------(2分) ∵AC AD DC =+,∴25AC a b =-.-----------------------------------(2分)21.(本题满分10分)解:由题意得:C (0,1),D (6,1.5),抛物线的对称轴为直线x =4.----(3分)设抛物线的表达式为()210y ax bx a =++≠-------------------------------------(1分)则据题意得:421.53661ba ab ⎧-=⎪⎨⎪=++⎩. ----------------------------------------------(2分)解得:12413a b ⎧=-⎪⎪⎨⎪=⎪⎩. -------------------------------------------------------------------(2分)∴羽毛球飞行的路线所在的抛物线的表达式为2111243y x x =-++. ------(1分)∵()2154243y x =--+,∴飞行的最高高度为53米. ------------------------(1分) 22.(本题满分10分)解:由题意得∠ADE =α,∠E =45°.----------------------------------------------(2分) 过点A 作AF ⊥CE ,交CE 于点F ,过点B 作BG ⊥AF ,交AF 于点G ,则FG =BC =10. 设AF =x .∵∠E =45°,∴EF =AF =x . 在Rt △ADF 中,∵tan ∠ADF =AFDF,-----------------(1分) ∴DF =tan tan 6AF x xADF α==∠. --------------------------(1分)∵DE =13.3,∴6xx +=13.3. ---------------------------(1分) ∴x =11.4. ---------------------------------------------(1分)∴AG =AF ﹣GF =11.4﹣10=1.4. ------------------------------------------------------------(1分) ∵∠ABC =120°,∴∠ABG =∠ABC ﹣∠CBG =120°﹣90°=30°.-------------------(1分)∴AB =2AG =2.8 ----------------------------------------------------------------------- (1分)答:灯杆AB 的长度为2.8米.------------------------------------------------------------(1分) 23.(本题满分12分,第(1)小题5分,第(2)小题7分) 证明:(1)∵∠BEC =∠BAC+∠ABD , ∠BEC =∠BEF+∠FEC ,又∵∠BEF =∠BAC ,∴∠ABD=∠FEC.------------------------------------ (1分) ∵AD =AB ,∴∠ABD=∠ADB.------------------------------------------------- (1分) ∴∠FEC=∠ADB. -------------------------------------------------------- (1分) ∵AD //BC ,∴∠DAE=∠ECF.--------------------------------------------------- (1分) ∴△AED ∽△CFE. --------------------------------------------------------- (1分) (2)∵EF //D C ,∴∠FEC=∠ECD. --------------------------------------------------- (1分)∵∠ABD=∠FEC ,∴∠ABD=∠ECD.--------------------------------------------- (1分) ∵∠AEB=∠DEC. ∴△AEB ∽△DEC. ----------------------------------------------- (1分)A BC D FG∴AE BEDE CE=.------------------------------------------------------------------------------(1分)∵AD //BC ,∴AE DECE BE=.----------------------------------------------------------------(1分)∴AE AE BE DEDE CE CE BE⋅=⋅.即22AE DE =.-------------------------------------------(1分) ∴ AE =DE . ----------------------------------------------------------------------------- (1分)24.(本题满分12分,第(1)小题3分,第(2)小题5分,第(3)小题4分)解:(1)∵22221()1y x mx m m x m m =-+--+=---+.------------------------(1分) ∴顶点D (m , 1-m ).------------------------------------------------------------------(2分)(2)∵抛物线2221y x mx m m =-+--+过点(1,-2),∴22121m m m -=-+--+.即220m m --=. ---------------------------(1分)∴2m =或1m =-(舍去). ------------------------------------------------------(2分) ∴抛物线的顶点是(2,-1). ∵抛物线22y x x =-+的顶点是(1,1),∴向左平移了1个单位,向上平移了2个单位. -------------------------(2分) (3)∵顶点D 在第二象限,∴0m <.情况1,点A 在y 轴的正半轴上,如图(1).作AG ⊥DH 于点G , ∵A (0,21m m --+),D (m ,-m +1),∴H (,0m ),G (2,1m m m --+)∵∠ADH =∠AHO ,∴tan ∠ADH = tan ∠AHO ,∴AG AODG HO=. ∴2211(1)m m m m m m m ---+=----+-.整理得:20m m +=. ∴1m =-或0m =(舍). --------------(2分)情况2,点A 在y 轴的负半轴上,如图(2).作AG ⊥DH 于点G ,x∵A (0,21m m --+),D (m ,-m +1),∴H (,0m ),G (2,1m m m --+)∵∠ADH =∠AHO ,∴tan ∠ADH = tan ∠AHO ,∴AG AODG HO=. ∴2211(1)m m m m m m m -+-=----+-.整理得:220m m +-=. ∴2m =-或1m =(舍). ---------(2分) ∴1m =-或2m =-.25.(本题满分14分,第(1)、(2)小题各6分,第(3)小题2分) 解:(1)∵△AME 沿直线MN 翻折,点A 落在点P 处, ∴△AME ≌△PME . ∴∠AEM =∠PEM ,AE=PE . ∵ABCD 是矩形,∴AB ⊥BC . ∵EP ⊥BC ,∴AB // EP .∴∠AME =∠PEM . ∴∠AEM =∠AME . ∴AM =AE . ---------------------(2分) ∵ABCD 是矩形,∴AB // DC . ∴AM AECN CE=. ∴CN =CE . ------------------(1分) 设CN = CE =x .∵ABCD 是矩形,AB =4,BC =3,∴AC =5. ∴PE= AE=5- x . ∵EP ⊥BC ,∴4sin 5EP ACB CE =∠=. ∴545x x -=. ---------------------(1分) ∴259x =,即259CN =. ------------------------------------------------------(2分) (2)∵△AME 沿直线MN 翻折,点A 落在点P 处, ∴△AME ≌△PME . ∴AE=PE ,AM=PM . ∵EP ⊥AC ,∴4tan 3EP ACB CE =∠=. ∴43AE CE =. ∵AC =5,∴207AE =,157CE =.∴207PE =. ---------------------(2分)∵EP ⊥AC ,∴257PC ===. ∴254377PB PC BC =-=-=. --------------------------------------(2分) 在Rt △PMB 中,∵222PM PB MB =+,AM=PM .∴2224()(4)7AM AM =+-. ∴10049AM =. --------------------------------------(2分)(3)05CP ≤≤,当CP 最大时MN .--------------------------------------------------(2分)。

上海民办兰生复旦中学语文中考九年级文言文试卷

上海民办兰生复旦中学语文中考九年级文言文试卷

上海民办兰生复旦中学语文中考九年级文言文试卷一、文言文1.阅读下面的文字,回答问题。

【甲】侍中、侍郎郭攸之、费祎、董允等,此皆良实,志虑忠纯,是以先帝简拔以遗陛下。

愚以为宫中之事,事无大小,悉以咨之,然后施行,必能裨补阙漏,有所广益。

将军向宠,性行淑均,晓畅军事,试用于昔日,先帝称之曰能,是以众议举宠为督。

愚以为营中之事,悉以咨之,必能使行阵和睦,优劣得所。

(节选自诸葛亮《出师表》)【乙】高祖①曰:“吾所以有天下者何?项氏所以失天下者何?”高起②等对曰:“陛下使人攻城略③地,所降下者因以予之,与天下④同利也。

项羽妒贤嫉能,战胜而不予人功,得地而不予人利,此所以失天下也。

”高祖曰:“运筹帷幄之中,决胜千里之外,吾不如子房⑤;镇国家,抚百姓,吾不如萧何;连⑥百万之军,战必胜,吾不如韩信。

此三杰,吾能用之。

项羽有一范增而不能用,此其所以为我擒也。

”(节选自《史记·高祖本纪》,有删改)【注】①高祖:指汉高祖刘邦。

②高起:高祖臣子。

③略:攻占。

④天下:这里指刘邦的部属。

⑤子房:西汉谋士张良。

⑥连:率领。

(1)解释下列划线词在文中的意思。

①事无大小,悉以咨之悉:________②性行淑均,晓畅军事晓:________③战胜而不予人功予:________(2)把下面的句子翻译成现代汉语。

①先帝称之曰能,是以众议举宠为督。

①项羽有一范增而不能用,此其所以为我擒也。

(3)下面对【甲】【乙】两个文段理解不正确的一项是( )A.【甲】文中诸葛亮两次提到“先帝”是希望刘禅谨记先帝遗志,谨遵先帝安排,言辞恳切,拳拳之心,溢于言表。

B.【乙】文中,司马迁通过对话描写,将高祖取胜与项羽失败进行对比,刻画了高祖睿智英明的形象。

C.【甲】文中诸葛亮向刘禅举荐郭攸之、费祎、董允等管理“营中”之事,向宠管理“宫中”之事,安排得十分细致周到。

D.【乙】文中高祖非常有自知之明,他能够很清晰地认识到张良、萧何、韩信各自的长处,并让他们各得其所。

上海民办兰生复旦中学九年级上册期末数学试题(word版,含解析)

上海民办兰生复旦中学九年级上册期末数学试题(word版,含解析)

上海民办兰生复旦中学九年级上册期末数学试题(word 版,含解析)一、选择题1.关于x 的一元一次方程122a x m -+=的解为1x =,则a m -的值为( )A .5B .4C .3D .22.某大学生创业团队有研发、管理和操作三个小组,各组的日工资和人数如下表所示.现从管理组分别抽调1人到研发组和操作组,调整后与调整前相比,下列说法中不正确的是( )A .团队平均日工资不变B .团队日工资的方差不变C .团队日工资的中位数不变D .团队日工资的极差不变3.要得到函数y =2(x -1)2+3的图像,可以将函数y =2x 2的图像( ) A .向左平移1个单位长度,再向上平移3个单位长度 B .向左平移1个单位长度,再向下平移3个单位长度 C .向右平移1个单位长度,再向上平移3个单位长度 D .向右平移1个单位长度,再向下平移3个单位长度4.在平面直角坐标系中,点A(0,2)、B(a ,a +2)、C(b ,0)(a >0,b >0),若AB=42且∠ACB 最大时,b 的值为( ) A .226+B .226-+C .242+D .2425.一枚质地匀均的骰子,其六个面上分别标有数字:1,2,3,4,5,6,投掷一次,朝上面的数字大于4的概率是( ) A .12B .13C .23D .166.已知a 是方程x 2+3x ﹣1=0的根,则代数式a 2+3a+2019的值是( ) A .2020 B .﹣2020 C .2021 D .﹣2021 7.下列方程是一元二次方程的是( )A .2321x x =+B .3230x x --C .221x y -=D .20x y +=8.用配方法解方程2890x x ++=,变形后的结果正确的是( ) A .()249x +=-B .()247x +=-C .()2425x +=D .()247x +=9.如图,在□ABCD 中,E 、F 分别是边BC 、CD 的中点,AE 、AF 分别交BD 于点G 、H ,则图中阴影部分图形的面积与□ABCD 的面积之比为( )A .7 : 12B .7 : 24C .13 : 36D .13 : 7210.某同学在解关于x 的方程ax 2+bx +c =0时,只抄对了a =1,b =﹣8,解出其中一个根是x =﹣1.他核对时发现所抄的c 是原方程的c 的相反数,则原方程的根的情况是( )A .有两个不相等的实数根B .有两个相等的实数根C .有一个根是x =1D .不存在实数根11.下列方程中,关于x 的一元二次方程是( ) A .2x ﹣3=xB .2x +3y =5C .2x ﹣x 2=1D .17x x+= 12.已知在△ABC 中,∠ACB =90°,AC =6cm ,BC =8cm ,CM 是它的中线,以C 为圆心,5cm 为半径作⊙C ,则点M 与⊙C 的位置关系为( ) A .点M 在⊙C 上B .点M 在⊙C 内 C .点M 在⊙C 外D .点M 不在⊙C 内13.已知函数2y x bx c =-++的部分图像如图所示,若0y >,则的取值范围是( )A .41x -<<B .21x -<<C .31x -<<D .31x x <->或14.如图,△ABC 中AB 两个顶点在x 轴的上方,点C 的坐标是(﹣1,0),以点C 为位似中心,在x 轴的下方作△ABC 的位似图形△A′B′C′,且△A′B′C′与△ABC 的位似比为2:1.设点B 的对应点B′的横坐标是a ,则点B 的横坐标是( )A .12a -B .1(1)2a -+ C .1(1)2a -- D .1(3)2a -+ 15.二次函数y=ax 2+bx+c (a≠0)的图象如图,给出下列四个结论:①4ac ﹣b 2<0;②4a+c <2b ;③3b+2c <0;④m (am+b )+b <a (m≠﹣1),其中正确结论的个数是( )A .4个B .3个C .2个D .1个二、填空题16.如图,将△ABC 绕点C 顺时针旋转90°得到△EDC ,若点A 、D 、E 在同一条直线上,∠ACD =70°,则∠EDC 的度数是_____.17.如图,△ABC 周长为20cm ,BC=6cm,圆O 是△ABC 的内切圆,圆O 的切线MN 与AB 、CA 相交于点M 、N ,则△AMN 的周长为________cm.18.若记[]x 表示任意实数的整数部分,例如:[]4.24=,21⎡⎤=⎣⎦,…,则123420192020⎡⎤⎡⎤⎡⎤⎡⎤⎡⎤⎡⎤-+-+⋅⋅⋅⋅⋅⋅+-⎣⎦⎣⎦⎣⎦⎣⎦⎣⎦⎣⎦(其中“+”“-”依次相间)的值为______.19.已知点P 是线段AB 的黄金分割点,PA >PB ,AB =4 cm ,则PA =____cm . 20.某企业2017年全年收入720万元,2019年全年收入845万元,若设该企业全年收入的年平均增长率为x ,则可列方程____. 21.抛物线y =3(x+2)2+5的顶点坐标是_____.22.如图,在平面直角坐标系中,直线l :28y x =+与坐标轴分别交于A ,B 两点,点C 在x 正半轴上,且OC =O B .点P 为线段AB (不含端点)上一动点,将线段OP 绕点O 顺时针旋转90°得线段OQ ,连接CQ ,则线段CQ 的最小值为___________.23.抛物线21(5)33y x =--+的顶点坐标是_______. 24.如图,D 、E 分别是△ABC 的边AB ,AC 上的点,AD AB =AEAC,AE =2,EC =6,AB =12,则AD 的长为_____.25.如图(1),在矩形ABCD 中,将矩形折叠,使点B 落在边AD 上,这时折痕与边AD 和BC 分别交于点E 、点F .然后再展开铺平,以B 、E 、F 为顶点的△BEF 称为矩形ABCD 的“折痕三角形”.如图(2),在矩形ABCD 中,AB=2,BC=4,当“折痕△BEF”面积最大时,点E 的坐标为_________________________.26.已知关于x 的一元二次方程2230x x k -+=有两个不相等的实数根,则k 的取值范围是________.27.在Rt △ABC 中,两直角边的长分别为6和8,则这个三角形的外接圆半径长为_____. 28.如图,在△ABC 中,AC :BC :AB =3:4:5,⊙O 沿着△ABC 的内部边缘滚动一圈,若⊙O 的半径为1,且圆心O 运动的路径长为18,则△ABC 的周长为_____.29.若关于x 的一元二次方程22(1)0k x x k -+-=的一个根为1,则k 的值为__________. 30.如图,四边形ABCD 中,∠A =∠B =90°,AB =5cm ,AD =3cm ,BC =2cm ,P 是AB 上一点,若以P 、A 、D 为顶点的三角形与△PBC 相似,则PA =_____cm .三、解答题31.某商店专门销售某种品牌的玩具,成本为30元/件,每天的销售量y (件)与销售单价x (元)之间存在着如图所示的一次函数关系.(1)求y与x之间的函数关系式;(2)当销售单价为多少元时,每天获取的利润最大,最大利润是多少?(3)为了保证每天的利润不低于3640元,试确定该玩具销售单价的范围.32.如图,AB为⊙O的直径,AC、DC为弦,∠ACD=60°,P为AB延长线上的点,∠APD=30°.(1)求证:DP是⊙O的切线;(2)若⊙O的半径为3cm,求图中阴影部分的面积.33.如图,AB为O的直径,PD切O于点C,交AB的延长线于点D,且D A∠=∠.2∠的度数.(1)求D(2)若O的半径为2,求BD的长.34.超市销售某种儿童玩具,如果每件利润为40元(市场管理部门规定,该种玩具每件利润不能超过60元),每天可售出50件.根据市场调查发现,销售单价每增加2元,每天销售量会减少1件.设销售单价增加x元,每天售出y件.(1)请写出y与x之间的函数表达式;(2)当x为多少时,超市每天销售这种玩具可获利润2250元?(3)设超市每天销售这种玩具可获利w元,当x为多少时w最大,最大值是多少?⊥交AB于点P,过点B的直线交OP的延长线于35.如图,AB是⊙O的弦,OP OA点C,且BC是⊙O的切线.(1)判断CBP ∆的形状,并说明理由; (2)若6,2OA OP ==,求CB 的长;(3)设AOP ∆的面积是1,S BCP ∆的面积是2S ,且1225S S =.若⊙O 的半径为6,45BP =,求tan APO ∠.四、压轴题36.如图,在Rt △ABC 中,∠A=90°,0是BC 边上一点,以O 为圆心的半圆与AB 边相切于点D ,与BC 边交于点E 、F ,连接OD ,已知BD=3,tan ∠BOD=34,CF=83.(1)求⊙O 的半径OD ; (2)求证:AC 是⊙O 的切线; (3)求图中两阴影部分面积的和.37.如图1,ABC ∆是⊙O 的内接等腰三角形,点D 是弧AC 上异于,A C 的一个动点,射线AD 交底边BC 所在的直线于点E ,连结BD 交AC 于点F . (1)求证:ADB CDE ∠=∠;(2)若7BD =,3CD =,①求AD DE •的值;②如图2,若AC BD ⊥,求tan ACB ∠;(3)若5tan 2CDE ∠=,记AD x =,ABC ∆面积和DBC ∆面积的差为y ,直接写出y 关于x 的函数关系式.38.在平面直角坐标系xOy 中,对于任意三点A ,B ,C ,给出如下定义:如果矩形的任何一条边均与某条坐标轴平行,且A ,B ,C 三点都在矩形的内部或边界上,则称该矩形为点A,B,C的覆盖矩形.点A,B,C的所有覆盖矩形中,面积最小的矩形称为点A,B,C的最优覆盖矩形.例如,下图中的矩形A1B1C1D1,A2B2C2D2,AB3C3D3都是点A,B,C的覆盖矩形,其中矩形AB3C3D3是点A,B,C的最优覆盖矩形.(1)已知A(﹣2,3),B(5,0),C(t,﹣2).①当t=2时,点A,B,C的最优覆盖矩形的面积为;②若点A,B,C的最优覆盖矩形的面积为40,求直线AC的表达式;(2)已知点D(1,1).E(m,n)是函数y=4x(x>0)的图象上一点,⊙P是点O,D,E的一个面积最小的最优覆盖矩形的外接圆,求出⊙P的半径r的取值范围.39.如图,在⊙O中,弦AB、CD相交于点E,AC=BD,点D在AB上,连接CO,并延长CO交线段AB于点F,连接OA、OB,且OA=5,tan∠OBA=12.(1)求证:∠OBA=∠OCD;(2)当△AOF是直角三角形时,求EF的长;(3)是否存在点F,使得S△CEF=4S△BOF,若存在,请求EF的长,若不存在,请说明理由.40.如图,PA切⊙O于点A,射线PC交⊙O于C、B两点,半径OD⊥BC于E,连接BD、DC和OA,DA交BP于点F;(1)求证:∠ADC+∠CBD=12∠AOD;(2)在不添加任何辅助线的情况下,请直接写出图中相等的线段.【参考答案】***试卷处理标记,请不要删除一、选择题 1.D 解析:D 【解析】 【分析】满足题意的有两点,一是此方程为一元一次方程,即未知数x 的次数为1;二是方程的解为x=1,即1使等式成立,根据两点列式求解. 【详解】 解:根据题意得, a-1=1,2+m=2, 解得,a=2,m=0, ∴a-m=2. 故选:D. 【点睛】本题考查一元一次方程的定义及方程解的定义,对定义的理解是解答此题的关键.2.B解析:B 【解析】 【分析】根据平均数、方差、中位数和众数的定义分别对每一项进行分析,即可得出答案. 【详解】解:调整前的平均数是:26042804300443⨯+⨯+⨯⨯=280;调整后的平均数是:260528023005525⨯+⨯+⨯++=280; 故A 正确;调整前的方差是:()()()222142602804280280430028012⎡⎤-+-+-⎣⎦=8003; 调整后的方差是:()()()222152602802280280530028012⎡⎤-+-+-⎣⎦=10003; 故B 错误;调整前:把这些数从小到大排列为:260,260,260,260,280,280,280,280,300,300,300,300;最中间两个数的平均数是:280,则中位数是280,调整后:把这些数从小到大排列为:260,260,260,260,260,280,280,300,300,300,300,300;最中间两个数的平均数是:280,则中位数是280, 故C 正确;调整前的极差是40,调整后的极差也是40,则极差不变, 故D 正确. 故选B. 【点睛】此题考查了平均数、方差、中位数和极差的概念,掌握各个数据的计算方法是关键.3.C解析:C 【解析】 【分析】找到两个抛物线的顶点,根据抛物线的顶点即可判断是如何平移得到. 【详解】解:∵y =2(x -1)2+3的顶点坐标为(1,3),y=2x 2的顶点坐标为(0,0),∴将抛物线y=2x 2向右平移1个单位,再向上平移3个单位,可得到抛物线y =2(x -1)2+3 故选:C . 【点睛】本题考查了二次函数图象与几何变换,解答时注意抓住点的平移规律和求出关键点顶点坐标.4.B解析:B 【解析】 【分析】根据圆周角大于对应的圆外角可得当ABC ∆的外接圆与x 轴相切时,ACB ∠有最大值,此时圆心F 的横坐标与C 点的横坐标相同,并且在经过AB 中点且与直线AB 垂直的直线上,根据FB=FC 列出关于b 的方程求解即可. 【详解】解:∵AB=A(0,2)、B(a ,a +2)= 解得a =4或a =-4(因为a >0,舍去) ∴B(4,6),设直线AB 的解析式为y=kx+2, 将B(4,6)代入可得k =1,所以y=x+2,利用圆周角大于对应的圆外角得当ABC ∆的外接圆与x 轴相切时,ACB ∠有最大值. 如下图,G 为AB 中点,()2,4G ,设过点G 且垂直于AB 的直线:l y x m =-+, 将()2,4G 代入可得6m =,所以6y x =-+.设圆心(),6F b b -+,由FC FB =,可知()()()2226466b b b -+=-+-+-,解得262b =(已舍去负值).故选:B. 【点睛】本题考查圆的综合题,一次函数的应用和已知两点坐标,用勾股定理求两点距离.能结合圆的切线和圆周角定理构建图形找到C 点的位置是解决此题的关键.5.B解析:B 【解析】 【分析】直接得出朝上面的数字大于4的个数,再利用概率公式求出答案. 【详解】∵一枚质地均匀的骰子,其六个面上分别标有数字1,2,3,4,5,6,投掷一次, ∴共有6种情况,其中朝上面的数字大于4的情况有2种, ∴朝上一面的数字是朝上面的数字大于4的概率为:2163=, 故选:B . 【点睛】本题考查简单的概率求法,概率=所求情况数与总情况数的比;熟练掌握概率公式是解题关键.解析:A【解析】【分析】根据一元二次方程的解的定义,将a 代入已知方程,即可求得a 2+3a 的值,然后再代入求值即可.【详解】解:根据题意,得a 2+3a ﹣1=0,解得:a 2+3a =1,所以a 2+3a+2019=1+2019=2020.故选:A.【点睛】此题考查的是一元二次方程的解,掌握一元二次方程解的定义是解决此题的关键7.A解析:A【解析】【分析】根据一元二次方程的定义逐一判断即可.【详解】解:A . 2321x x =+是一元二次方程,故本选项符合题意;B . 3230x x --是一元三次方程,故本选项不符合题意;C . 221x y -=是二元二次方程,故本选项不符合题意;D . 20x y +=是二元一次方程,故本选项不符合题意;故选A .【点睛】此题考查的是一元二次方程的判断,掌握一元二次方程的定义是解决此题的关键.8.D解析:D【解析】【分析】先将常数项移到右侧,然后两边同时加上一次项系数一半的平方,配方后进行判断即可.【详解】2890x x ++=,289x x +=-,2228494x x ++=-+,所以()247x +=,【点睛】本题考查了配方法解一元二次方程,熟练掌握配方法的一般步骤以及注意事项是解题的关键.9.B解析:B【解析】【分析】根据已知条件想办法证明BG=GH=DH ,即可解决问题;【详解】解:∵四边形ABCD 是平行四边形,∴AB ∥CD ,AD ∥BC ,AB=CD ,AD=BC ,∵DF=CF ,BE=CE , ∴12DH DF HB AB ==,12BG BE DG AD ==, ∴13DH BG BD BD ==, ∴BG=GH=DH ,∴S △ABG =S △AGH =S △ADH ,∴S 平行四边形ABCD =6 S △AGH ,∴S △AGH :ABCD S 平行四边形=1:6,∵E 、F 分别是边BC 、CD 的中点, ∴12EF BD =, ∴14EFC BCDD S S =, ∴18EFC ABCD SS =四边形, ∴1176824AGH EFC ABCD S S S +=+=四边形=7∶24, 故选B.【点睛】本题考查了平行四边形的性质、平行线分线段成比例定理、等底同高的三角形面积性质,题目的综合性很强,难度中等.10.A解析:A【解析】【分析】直接把已知数据代入进而得出c 的值,再解方程根据根的判别式分析即可.【详解】∵x =﹣1为方程x 2﹣8x ﹣c =0的根,1+8﹣c =0,解得c =9,∴原方程为x 2-8x +9=0,∵24b ac ∆=-=(﹣8)2-4×9>0,∴方程有两个不相等的实数根.故选:A .【点睛】本题考查一元二次方程的解、一元二次方程根的判别式,解题的关键是掌握一元二次方程根的判别式,对于一元二次方程()200++=≠ax bx c a ,根的情况由24b ac ∆=-来判别,当24b ac ->0时,方程有两个不相等的实数根,当24b ac -=0时,方程有两个相等的实数根,当24b ac -<0时,方程没有实数根.11.C解析:C【解析】【分析】利用一元二次方程的定义判断即可.【详解】A 、方程2x ﹣3=x 为一元一次方程,不符合题意;B 、方程2x +3y =5是二元一次方程,不符合题意;C 、方程2x ﹣x 2=1是一元二次方程,符合题意;D 、方程x +1x=7是分式方程,不符合题意, 故选:C .【点睛】本题考查了一元一次方程的问题,掌握一元一次方程的定义是解题的关键.12.A解析:A【解析】【分析】根据题意可求得CM 的长,再根据点和圆的位置关系判断即可.【详解】如图,∵由勾股定理得2268 ,∵CM 是AB 的中线,∴CM=5cm ,∴d=r ,所以点M 在⊙C 上,故选A .【点睛】本题考查了点和圆的位置关系,解决的根据是点在圆上⇔圆心到点的距离=圆的半径.13.C解析:C【解析】【分析】根据抛物线的对称性确定抛物线与x 轴的另一个交点为(−3,0),然后观察函数图象,找出抛物线在x 轴上方的部分所对应的自变量的范围即可.【详解】∵y =ax 2+bx +c 的对称轴为直线x =−1,与x 轴的一个交点为(1,0),∴抛物线与x 轴的另一个交点为(−3,0),∴当−3<x <1时,y >0.故选:C .【点睛】此题主要考查二次函数的图像与性质,解题的关键是根据函数对称轴找到抛物线与x 轴的交点.14.D解析:D【解析】【分析】设点B 的横坐标为x ,然后表示出BC 、B′C 的横坐标的距离,再根据位似变换的概念列式计算.【详解】设点B 的横坐标为x ,则B 、C 间的横坐标的长度为﹣1﹣x ,B′、C 间的横坐标的长度为a+1,∵△ABC 放大到原来的2倍得到△A′B′C ,∴2(﹣1﹣x )=a+1,解得x=﹣12(a+3),故选:D.【点睛】本题考查了位似变换,坐标与图形的性质,根据位似变换的定义,利用两点间的横坐标的距离等于对应边的比列出方程是解题的关键.15.B解析:B【解析】【分析】【详解】解:∵抛物线和x轴有两个交点,∴b2﹣4ac>0,∴4ac﹣b2<0,∴①正确;∵对称轴是直线x﹣1,和x轴的一个交点在点(0,0)和点(1,0)之间,∴抛物线和x轴的另一个交点在(﹣3,0)和(﹣2,0)之间,∴把(﹣2,0)代入抛物线得:y=4a﹣2b+c>0,∴4a+c>2b,∴②错误;∵把(1,0)代入抛物线得:y=a+b+c<0,∴2a+2b+2c<0,∵b=2a,∴3b,2c<0,∴③正确;∵抛物线的对称轴是直线x=﹣1,∴y=a﹣b+c的值最大,即把(m,0)(m≠0)代入得:y=am2+bm+c<a﹣b+c,∴am2+bm+b<a,即m(am+b)+b<a,∴④正确;即正确的有3个,故选B.考点:二次函数图象与系数的关系二、填空题16.115°【解析】【分析】根据∠EDC=180°﹣∠E﹣∠DCE,想办法求出∠E,∠DCE即可.【详解】由题意可知:CA=CE,∠ACE=90°,∴∠E=∠CAE=45°,∵∠ACD=7解析:115°【解析】【分析】根据∠EDC=180°﹣∠E﹣∠DCE,想办法求出∠E,∠DCE即可.【详解】由题意可知:CA=CE,∠ACE=90°,∴∠E=∠CAE=45°,∵∠ACD=70°,∴∠DCE=20°,∴∠EDC=180°﹣∠E﹣∠DCE=180°﹣45°﹣20°=115°,故答案为115°.【点睛】本题考查了旋转的性质,等腰直角三角形的性质,三角形的内角和定理等知识,解题的关键是灵活运用所学知识,问题,属于中考常考题型.17.8【解析】【分析】先作出辅助线,连接切点,利用内切圆的性质得到BE=BF,CE=CG,ME=MH,NG=NH,再利用等量代换即可解题.【详解】解:∵圆O是△ABC的内切圆,MN是圆O的切线解析:8【解析】【分析】先作出辅助线,连接切点,利用内切圆的性质得到BE=BF,CE=CG,ME=MH,NG=NH,再利用等量代换即可解题.【详解】解:∵圆O是△ABC的内切圆,MN是圆O的切线,如下图,连接各切点,有切线长定理易得,BE=BF,CE=CG,ME=MH,NG=NH,∵△ABC周长为20cm, BC=6cm,∴BC=CE+BE=CG+BF=6cm,∴△AMN 的周长=AM+AN+MN=AM+AN+FM+GN=AF+AG,又∵AF+AG=AB+AC-(BF+CG)=20-6-6=8cm故答案是8【点睛】本题考查了三角形内接圆的性质,切线长定理的应用,中等难度,熟练掌握等量代换的方法是解题关键.18.-22【解析】【分析】先确定的整数部分的规律,根据题意确定算式的运算规律,再进行实数运算.【详解】解:观察数据12=1,22=4,32=9,42=16,52=25,62=36的特征,得出数 解析:-22【解析】【分析】 1,2,32020的整数部分的规律,根据题意确定算式123420192020⎡⎡⎡⎤⎡-+-+⋅⋅⋅⋅⋅⋅+-⎣⎣⎣⎦⎣的运算规律,再进行实数运算. 【详解】解:观察数据12=1,22=4,32=9,42=16,52=25,62=36的特征,得出数据1,2,3,4……2020中,算术平方根是1的有3个,算术平方根是2的有5个,算数平方根是3的有7个,算数平方根是4的有9个,…其中432=1849,442=1936,452=2025,所以在1⎡⎤⎣⎦、22020⎡⋅⋅⋅⋅⋅⋅⎣中,算术平方根依次为1,2,3……43的个数分别为3,5,7,9……个,均为奇数个,最大算数平方根为44的有85个,所以123420192020⎡⎡⎡⎤⎡-+-+⋅⋅⋅⋅⋅⋅+-⎣⎣⎣⎦⎣=1-2+3-4+…+43-44= -22 【点睛】本题考查自定义运算,通过正整数的算术平方根的整数部分出现的规律,找到算式中相同加数的个数及符号的规律,方能进行运算.19.2-2【解析】【分析】根据黄金分割点的定义,知AP 是较长线段;则AP=AB ,代入运算即可.【详解】解:由于P 为线段AB=4的黄金分割点,且AP 是较长线段;则AP=4×=cm ,故答案为解析:2【解析】【分析】根据黄金分割点的定义,知AP 是较长线段;则AB ,代入运算即可. 【详解】解:由于P 为线段AB=4的黄金分割点,且AP 是较长线段;则AP=4×12=)21cm ,故答案为:(2)cm.【点睛】此题考查了黄金分割的定义,应该识记黄金分割的公式:较短的线段=,难度一般. 20.720(1+x )2=845.【解析】【分析】增长率问题,一般用增长后的量=增长前的量×(1+增长率),参照本题,如果该企业全年收入的年平均增长率为x ,根据2017年全年收入720万元,2019 解析:720(1+x )2=845.【解析】【分析】增长率问题,一般用增长后的量=增长前的量×(1+增长率),参照本题,如果该企业全年收入的年平均增长率为x ,根据2017年全年收入720万元,2019年全年收入845万元,即可得出方程.【详解】解:设该企业全年收入的年平均增长率为x ,则2018的全年收入为:720×(1+x )2019的全年收入为:720×(1+x)2.那么可得方程:720(1+x)2=845.故答案为:720(1+x)2=845.【点睛】本题考查了一元二次方程的运用,解此类题的关键是掌握等量关系式:增长后的量=增长前的量×(1+增长率).21.(﹣2,5)【解析】【分析】已知抛物线的顶点式,可直接写出顶点坐标.【详解】解:由y=3(x+2)2+5,根据顶点式的坐标特点可知,顶点坐标为(﹣2,5).故答案为:(﹣2,5).【点解析:(﹣2,5)【解析】【分析】已知抛物线的顶点式,可直接写出顶点坐标.【详解】解:由y=3(x+2)2+5,根据顶点式的坐标特点可知,顶点坐标为(﹣2,5).故答案为:(﹣2,5).【点睛】本题考查二次函数的性质,熟知二次函数的顶点式是解题的关键,即在y=a(x-h)2+k中,顶点坐标为(h,k),对称轴为x=h.22.【解析】【分析】在OA上取使,得,则,根据点到直线的距离垂线段最短可知当⊥AB时,CP最小,由相似求出的最小值即可.【详解】解:如图,在OA上取使,∵,∴,在△和△QOC中,,解析:455【解析】【分析】在OA上取'C使'OC OC=,得'OPC OQC≅,则CQ=C'P,根据点到直线的距离垂线段最短可知当'PC⊥AB时,CP最小,由相似求出C'P的最小值即可.【详解】解:如图,在OA上取'C使'OC OC=,∵90AOC POQ∠=∠=︒,∴'POC QOC∠=∠,在△'POC和△QOC中,''OP OQPOC QOCOC OC=⎧⎪∠=∠⎨⎪=⎩,∴△'POC≌△QOC(SAS),∴'PC QC=∴当'PC最小时,QC最小,过'C点作''C P⊥AB,∵直线l:28y x=+与坐标轴分别交于A,B两点,∴A坐标为:(0,8);B点(-4,0),∵'4OC OC OB===,∴22228445AB OA OB++=''4AC OA OC=-=.∵'''OB C Psin BAOAB AC∠==,''445C P=,∴4''55C P=∴线段CQ455【点睛】 本题主要考查了一次函数图像与坐标轴的交点及三角形全等的判定和性质、垂线段最短等知识,解题的关键是正确寻找全等三角形解决问题,学会利用垂线段最短解决最值问题,属于中考压轴题.23.(5,3)【解析】【分析】根据二次函数顶点式的性质直接求解.【详解】解:抛物线的顶点坐标是(5,3)故答案为:(5,3).【点睛】本题考查二次函数性质其顶点坐标为(h ,k ),题目比较解析:(5,3)【解析】【分析】根据二次函数顶点式2()y a x h k =-+的性质直接求解.【详解】 解:抛物线21(5)33y x =--+的顶点坐标是(5,3)故答案为:(5,3).【点睛】本题考查二次函数性质2()y a x h k =-+其顶点坐标为(h ,k ),题目比较简单. 24.3【解析】【分析】把AE =2,EC =6,AB =12代入已知比例式,即可求出答案.【详解】解:∵=,AE =2,EC =6,AB =12,∴=,解得:AD =3,故答案为:3.【点睛】本题解析:3 【解析】【分析】把AE =2,EC =6,AB =12代入已知比例式,即可求出答案.【详解】解:∵AD AB =AE AC,AE =2,EC =6,AB =12, ∴12AD =226, 解得:AD =3,故答案为:3.【点睛】 本题考查了成比例线段,灵活的将已知线段的长度代入比例式是解题的关键.25.(,2).【解析】【分析】【详解】解:如图,当点B 与点D 重合时,△BEF 面积最大,设BE=DE=x ,则AE=4-x ,在RT △ABE 中,∵EA2+AB2=BE2,∴(4-x )2+22=解析:(32,2). 【解析】【分析】【详解】解:如图,当点B 与点D 重合时,△BEF 面积最大,设BE=DE=x ,则AE=4-x ,在RT △ABE 中,∵EA 2+AB 2=BE 2,∴(4-x )2+22=x 2,∴x=52,∴BE=ED=52,AE=AD-ED=32,∴点E坐标(32,2).故答案为:(32,2).【点睛】本题考查翻折变换(折叠问题),利用数形结合思想解题是关键.26.【解析】【分析】根据一元二次方程的根的判别式,建立关于k的不等式,求出k的取值范围.【详解】根据一元二次方程的根的判别式,建立关于k的不等式,求出k的取值范围. ,,方程有两个不相等的实数解析:3k<【解析】【分析】根据一元二次方程的根的判别式,建立关于k的不等式,求出k的取值范围.【详解】根据一元二次方程的根的判别式,建立关于k的不等式,求出k的取值范围.1a,b=-,c k=方程有两个不相等的实数根,241240b ac k∴∆=-=->,3k∴<.故答案为:3k<.【点睛】本题考查了根的判别式.总结:一元二次方程根的情况与判别式△的关系:(1)△>0⇔方程有两个不相等的实数根;(2)△=0⇔方程有两个相等的实数根;(3)△<0⇔方程没有实数根.27.5【解析】【分析】根据直角三角形外接圆的直径是斜边的长进行求解即可.【详解】由勾股定理得:AB==10,∵∠ACB=90°,∴AB是⊙O的直径,∴这个三角形的外接圆直径是10;∴这解析:5【解析】【分析】根据直角三角形外接圆的直径是斜边的长进行求解即可.【详解】由勾股定理得:AB=22=10,68∵∠ACB=90°,∴AB是⊙O的直径,∴这个三角形的外接圆直径是10;∴这个三角形的外接圆半径长为5,故答案为5.【点睛】本题考查了90度的圆周角所对的弦是直径,熟练掌握是解题的关键.28.30【解析】【分析】如图,首先利用勾股定理判定△ABC是直角三角形,由题意得圆心O所能达到的区域是△DEG,且与△ABC三边相切,设切点分别为G、H、P、Q、M、N,连接DH、DG、EP、EQ解析:30【解析】【分析】如图,首先利用勾股定理判定△ABC是直角三角形,由题意得圆心O所能达到的区域是△DEG,且与△ABC三边相切,设切点分别为G、H、P、Q、M、N,连接DH、DG、EP、EQ、FM、FN,根据切线性质可得:AG=AH,PC=CQ,BN=BM,DG、EP分别垂直于AC,EQ、FN分别垂直于BC,FM、DH分别垂直于AB,继而则有矩形DEPG、矩形EQNF、矩形DFMH,从而可知DE=GP,EF=QN,DF=HM,DE∥GP,DF∥HM,EF∥QN,∠PEF =90°,根据题意可知四边形CPEQ是边长为1的正方形,根据相似三角形的判定可得△DEF ∽△ACB ,根据相似三角形的性质可知:DE ∶EF ∶FD =AC ∶CB ∶BA =3∶4∶5,进而根据圆心O 运动的路径长列出方程,求解算出DE 、EF 、FD 的长,根据矩形的性质可得:GP 、QN 、MH 的长,根据切线长定理可设:AG =AH =x ,BN =BM =y ,根据线段的和差表示出AC 、BC 、AB 的长,进而根据AC ∶CB ∶BA =3∶4∶5列出比例式,继而求出x 、y 的值,进而即可求解△ABC 的周长.【详解】∵AC ∶CB ∶BA =3∶4∶5,设AC =3a ,CB =4a ,BA =5a (a >0)∴()()()222222=345AC CB a a a BA ++==∴△ABC 是直角三角形,设⊙O 沿着△ABC 的内部边缘滚动一圈,如图所示,连接DE 、EF 、DF ,设切点分别为G 、H 、P 、Q 、M 、N ,连接DH 、DG 、EP 、EQ 、FM 、FN ,根据切线性质可得:AG =AH ,PC =CQ ,BN =BMDG 、EP 分别垂直于AC ,EQ 、FN 分别垂直于BC ,FM 、DH 分别垂直于AB ,∴DG ∥EP ,EQ ∥FN ,FM ∥DH ,∵⊙O 的半径为1∴DG =DH =PE =QE =FN =FM =1,则有矩形DEPG 、矩形EQNF 、矩形DFMH ,∴DE =GP ,EF =QN ,DF =HM ,DE ∥GP ,DF ∥HM ,EF ∥QN,∠PEF =90°又∵∠CPE =∠CQE =90°, PE =QE =1∴四边形CPEQ 是正方形,∴PC =PE =EQ =CQ =1,∵⊙O 的半径为1,且圆心O 运动的路径长为18,∴DE +EF +DF =18,∵DE ∥AC ,DF ∥AB ,EF ∥BC ,∴∠DEF =∠ACB ,∠DFE =∠ABC ,∴△DEF ∽△ABC ,∴DE :EF :DF =AC :BC :AB =3:4:5,设DE =3k (k >0),则EF =4k ,DF =5k ,∵DE +EF +DF =18,∴3k +4k +5k =18,解得k =32, ∴DE =3k =92,EF =4k =6,DF =5k =152, 根据切线长定理,设AG =AH =x ,BN =BM =y ,则AC =AG +GP +CP =x +92+1=x +5.5, BC =CQ +QN +BN =1+6+y =y +7,AB =AH +HM +BM =x +152+y =x +y +7.5, ∵AC :BC :AB =3:4:5, ∴(x +5.5):(y +7):(x +y +7.5)=3:4:5,解得x =2,y =3,∴AC =7.5,BC =10,AB =12.5,∴AC +BC +AB =30.所以△ABC 的周长为30.故答案为30.【点睛】本题是一道动图形问题,考查切线的性质定理、相似三角形的判定与性质、矩形的判定与性质、解直角三角形等知识点,解题的关键是确定圆心O 的轨迹,学会作辅助线构造相似三角形,综合运用上述知识点.29.0【解析】把x =1代入方程得,,即,解得.此方程为一元二次方程,,即,故答案为0.解析:0【解析】把x =1代入方程得,2110k k -+-=,即20k k -=,解得120,1k k ==.此方程为一元二次方程,10k ∴-≠,即1k ≠,0.k ∴=故答案为0.30.2或3【解析】【分析】根据相似三角形的判定与性质,当若点A ,P ,D 分别与点B ,C ,P 对应,与若点A ,P ,D 分别与点B ,P ,C 对应,分别分析得出AP 的长度即可.【详解】解:设AP =xcm .则解析:2或3【解析】【分析】根据相似三角形的判定与性质,当若点A ,P ,D 分别与点B ,C ,P 对应,与若点A ,P ,D 分别与点B ,P ,C 对应,分别分析得出AP 的长度即可.【详解】解:设AP =xcm .则BP =AB ﹣AP =(5﹣x )cm以A ,D ,P 为顶点的三角形与以B ,C ,P 为顶点的三角形相似,①当AD :PB =PA :BC 时,352x x =-, 解得x =2或3.②当AD :BC =PA +PB 时,3=25x x-,解得x =3, ∴当A ,D ,P 为顶点的三角形与以B ,C ,P 为顶点的三角形相似,AP 的值为2或3. 故答案为2或3.【点睛】本题考查了相似三角形的问题,掌握相似三角形的性质以及判定定理是解题的关键.三、解答题31.(1)10700y x =-+;(2)销售单价为50元时,每天获取的利润最大,最大利润是4000元;(3)44≤x ≤56【解析】【分析】(1)直接利用待定系数法求出一次函数解析式即可;(2)利用w=销量乘以每件利润进而得出关系式求出答案;(3)利用w=3640,进而解方程,再利用二次函数增减性得出答案.【详解】解:(1)y 与x 之间的函数关系式为:y kx b =+把(35,350),(55,150)代入得:由题意得:3503515055k b k b =+⎧⎨=+⎩解得:10700k b =-⎧⎨=⎩∴y 与x 之间的函数关系式为:10700y x =-+.(2)设销售利润为W 元则W=(x ﹣30)•y =(x ﹣30)(﹣10x +700),W =﹣10x 2+1000x ﹣21000W =﹣10(x ﹣50)2+4000∴当销售单价为50元时,每天获取的利润最大,最大利润是4000元.(3)令W =3640∴﹣10(x ﹣50)2+4000=3640∴x 1=44,x 2=56如图所示,由图象得:当44≤x ≤56时,每天利润不低于3640元.【点睛】此题主要考查了二次函数的应用以及待定系数法求一次函数解析式,正确掌握二次函数的性质是解题关键.32.(1)证明见解析;(22933()22cm . 【解析】【分析】(1)连接OD ,求出∠AOD ,求出∠DOB ,求出∠ODP ,根据切线判定推出即可. (2)求出OP 、DP 长,分别求出扇形DOB 和△ODP 面积,即可求出答案.【详解】解:(1)证明:连接OD ,。

兰生复旦中学2019年自主招生测试题

兰生复旦中学2019年自主招生测试题

兰生复旦中学2019届九年级自主招生数学模拟试题3班级 座号 姓名 成绩一、填空、选择题(每题5分共50分)1、从1-,1,2这三个数中,任取两个不同的数作为一次函数y ax b =+的系数,a b ,则一次函数y ax b =+的图象不经过第三象限的概率是 .2、 定义一种运算*“”:当a b ≥时,22a b a b *=+;当a b <时,22a b a b *=-,则方程212x *=的解是3、 方程2(2000)1999200110x x +⨯-=较小的一个根是________.4、 已知:如图,⊙O 是△ABC 的外接圆,AD 是BC 边上的高,BD =8cm ,CD =3cm ,AD =6cm ,则直径AM =________cm .5、科学家研究表明,当人的下肢长与身高之比为0.618时,看起来最美,某成年女士身高为153 cm ,下肢长为92 cm ,该女士穿的高跟鞋鞋跟的最佳高度约为______________ cm.(精确到0.1 cm)6、使不等式2x x <成立的x 的取值范围是( )A .1x >B .1x <-C .11x -<<D . 10x -<<或01x <<7、按下列图示的程序计算,若开始输入的值为x =3,则最后输出的结果是A .6B .21C .156D .2318、如图,P (x ,y )是以坐标原点为圆心,5为半径的圆周上的点,若x ,y 都是整数,则这样的点共有 ( )(A )4个 (B )8个 (C )12个 (D )16个9、如图,在矩形ABCD 中,AB=3,AD=4,P 是AD 上动点,PE ⊥AC 于E ,PF ⊥BD 于F ,则PE+PF 的值为( )A.512B.2C.25D.513 10、 如图,在梯形ABCD 中,AD //BC ,对角线AC ⊥BD ,且AC =12,BD =9,则此梯形的 中位线长是 ( )A .10B .212 C .152 D .12 第9题 第10题A第4题 B CD M · O输入x 计算(1)2x x +的值>100输出结果否 是 P y -5-555.................... A D B二、解答题11、(8分)京津城际铁路将于2008年8月1日开通运营,预计高速列车在北京、天津间单程直达运行时间为半小时.某次试车时,试验列车由北京到天津的行驶时间比预计时间多用了6分钟,由天津返回北京的行驶时间与预计时间相同.如果这次试车时,由天津返回北京比去天津时平均每小时多行驶40千米,那么这次试车时由北京到天津的平均速度是每小时多少千米?12、(10分)用剪刀将形状如图1所示的矩形纸片ABCD 沿着直线CM 剪成两部分,其中M为AD 的中点.用这两部分纸片可以拼成一些新图形,例如图2中的Rt △BCE 就是拼成的一个图形.(1)用这两部分纸片除了可以拼成图2中的Rt △BCE 外,还可以拼成一些四边形.请你试一试,把拼好的四边形分别画在图3、图4的虚框内.(4分)(2)若利用这两部分纸片拼成的Rt △BCE 是等腰直角三角形,设原矩形纸片中的边AB 和BC 的长分别为a 厘米、b 厘米,且a 、b 恰好是关于x 的方程01)1(2=++--m x m x 的两个实数根,试求出原矩形纸片的面积.(6分)EB AC B A MC D M 图3 图4 图1 图213、如图Rt △ABC 的两条直角边4BC 3AC ==、, 点P 是边BC 上的一动点(P 不与B 重合)以P 为圆心作⊙P 与BA 相切于点M 。

上海市2019-2020学年九年级上学期期中英语试卷精选汇编:词汇运用专题

上海市2019-2020学年九年级上学期期中英语试卷精选汇编:词汇运用专题

词汇运用专题2020届九年级上学期英语期中试卷IV. Complete the sentences with the given words in their proper forms: (共8分)54. Tom became famous in her (forty).55. Tu Yoyo is the first Chinese to win a Nobel Prize in science.(city)56. We must be responsible for for others and for society. (we)57. The boy is quite . He always tries to finish the tasks by himself.(depend)58. Cormorant fishing is a .Chinese skill as well as paper cutting (tradition)59. I'm sure you will in working out the problem unless you give up.(success)60.Walk along the street and you will see the market in five minutes.(straight)61. The fans all felt very sad about the of the famous actor.(dead)54.【答案】forties.【解析】in one’s forties 固定用法:在某人四十岁时。

55.【答案】citizen【解析】公民:citizen56.【答案】ourselves【解析】介词+宾语for ourselves57.【答案】independent【解析】be 动词+adj58.【答案】traditional【解析】be 动词+adj59.【答案】succeed【解析】be succeed in doing sth 固定用法:成功做某事60.【答案】independent【解析】be 动词+adj61.【答案】straightly【解析】副词修辞动词上海市上外民办劲松中学2020届九年级上学期期中测试III. Complete the sentences with the given words in their proper forms(用括号中所给单词的适当形式完成下列句子。

2019-2020学年上海市杨浦区民办兰生复旦中学九年级(上)期中化学试卷

2019-2020学年上海市杨浦区民办兰生复旦中学九年级(上)期中化学试卷

2019-2020学年上海市杨浦区民办兰生复旦中学九年级(上)期中化学试卷、选择题(共计24分)1. (1分)下列变化属于化学变化的是()A .树根“变”根雕B .玉石“变”印章C •水果“变”果汁D •葡萄“变”美酒2. (1分)常州博物馆启用了“真空充氮杀虫灭菌消毒机”来处理和保护文物。

即将文物置于该机器内,三天后氮气浓度可达99.99%;再密闭三天左右,好氧菌、厌氧菌和丝状霉菌都被杀灭。

下列有关氮气说法错误的是()A •氮气还能用于灯泡填充气B •通常情况下氮气的化学性质很活泼C •氮气不能供给呼吸D •高浓度氮气可抑制菌类的生长3. (1分)上海市垃圾分类实行的“四分类”标准,废荧光灯管属于()4. (1分)以下变化中,氧元素由化合态全部变为游离态都是()A .分离液态空气制取氧气B .铁丝燃烧C .电解水D .双氧水制取氧气5. (1分)根据分析证明:健康人的头发每克约含铁130mg、锌167〜172mg、铝5mg、硼7mg等。

这里的铁、锌、铝、硼是指()A •分子B •原子C.元素 D •单质C .胆矶:C U SO4?5H2OD .水银:Hg6. (1分)物质的俗名、化学式一致的是()A .水:H2O2B .熟石灰:CaOA B C DX金属溶液纯净物化合反应Y单质乳浊液化合物氧化反应8 (1分)下列有关2molH2O2的解释正确的是()A .含有2molO2B .与质量为36克水含有相同数量的氢原子C .该物质的摩尔质量为68g/molD •共含有约1.204 X 1024个氧原子9. (1分)下列物质放入水中能形成无色溶液的是()A •高锰酸钾B •硝酸钾C.碳酸钙 D •胆矶10. (1分)计算一定质量的纯净物所含的微粒个数,下列量没有用处的是()A •微粒大小B •微粒的质量C •阿伏伽德罗常数D •摩尔质量11. (1分)氮化硅(Si3N4)是一种高温陶瓷材料,硬度大、熔点高、化学性质稳定。

2019-2020学年上海市杨浦区民办兰生复旦中学九年级(上)期中化学试卷

2019-2020学年上海市杨浦区民办兰生复旦中学九年级(上)期中化学试卷

2019-2020学年上海市杨浦区民办兰生复旦中学九年级(上)期中化学试卷一、选择题(共计24分)1.(1分)下列变化属于化学变化的是()A.树根“变”根雕B.玉石“变”印章C.水果“变”果汁D.葡萄“变”美酒2.(1分)常州博物馆启用了“真空充氮杀虫灭菌消毒机”来处理和保护文物。

即将文物置于该机器内,三天后氮气浓度可达99.99%;再密闭三天左右,好氧菌、厌氧菌和丝状霉菌都被杀灭。

下列有关氮气说法错误的是()A.氮气还能用于灯泡填充气B.通常情况下氮气的化学性质很活泼C.氮气不能供给呼吸D.高浓度氮气可抑制菌类的生长3.(1分)上海市垃圾分类实行的“四分类”标准,废荧光灯管属于()A.B.C.D.4.(1分)以下变化中,氧元素由化合态全部变为游离态都是()A.分离液态空气制取氧气B.铁丝燃烧C.电解水D.双氧水制取氧气5.(1分)根据分析证明:健康人的头发每克约含铁130mg、锌167~172mg、铝5mg、硼7mg等。

这里的铁、锌、铝、硼是指()A.分子B.原子C.元素D.单质6.(1分)物质的俗名、化学式一致的是()A.水:H2O2B.熟石灰:CaOC.胆矾:CuSO4•5H2O D.水银:Hg7.(1分)下列选项符合图示从属关系的是()A.A B.B C.C D.D8.(1分)下列有关2molH2O2的解释正确的是()A.含有2molO2B.与质量为36克水含有相同数量的氢原子C.该物质的摩尔质量为68g/molD.共含有约1.204×1024个氧原子9.(1分)下列物质放入水中能形成无色溶液的是()A.高锰酸钾B.硝酸钾C.碳酸钙D.胆矾10.(1分)计算一定质量的纯净物所含的微粒个数,下列量没有用处的是()A.微粒大小B.微粒的质量C.阿伏伽德罗常数D.摩尔质量11.(1分)氮化硅(Si3N4)是一种高温陶瓷材料,硬度大、熔点高、化学性质稳定。

有电子、军事和核工业等方面有着广泛的应用,若Si3N4中Si显+4价,则下列物质中的N的化合价与氮化硅中N的化合价相同的是()A.NH3B.N2C.N2O3D.HNO312.(1分)氮化硅(Si3N4)是一种高温陶瓷材料,硬度大、熔点高、化学性质稳定。

上海市2019-2020学年九年级上学期期中英语试卷精选汇编:选词填空专题

上海市2019-2020学年九年级上学期期中英语试卷精选汇编:选词填空专题

选词填空专题上海市嘉定区江桥中学2020届九年级上学期英语期中试卷III. Complete the following passage with the words or phrases in the box. Each can only be used once: (共8分)Sydney is a young city. Its (46)_________ goes back just over 200 years. But in Australia, it is the oldest city. It is also the country's largest city and the most populous (人口稠密的) city of Australia. The climate of Sydney is very (47) __________ It's not too cold during the winter and not too hot during the summer. The sky is blue, the air is fresh, and birds sing in the garden. PeopleW ho live in Sydney (48) __________ to have an easy life style. They will tell you, “Don’t worry."Many people think that Sydney is one of the most attractive cities in the world. It hasmany tall and modern buildings. Among them, Centre point Tower,is the tallest. Standing on the 305-meter tower, you will have a great (49) __________of the city.[答案]:B D A E[解析]:(46)B 通过句子结构判断,主语位置缺少基础的名词或代词。

上海市2019-2020学年九年级上学期期中英语试卷精选汇编:单项选择专题

上海市2019-2020学年九年级上学期期中英语试卷精选汇编:单项选择专题

单项选择专题上海市嘉定区江桥中学2020届九年级上学期英语期中试卷26. Which of the following words is pronounced /'ekspə:t/?A. expectB. exceptC. expertD. express【答案】C27. Which of the following underlined parts is different in pronunciation from the others?A.What's the meaning of this word?B.Her boyfriend's death made her very sad.C. You can take this seat because there is no one else here.D.My dream job is to be an astronaut.【答案】B28.The doctor asked me to take the medicine before meals, three at time.A. anB. aC. /D. the【答案】B29. All the girls took part in the activity Ellie because of her broken leg.A. besideB. besidesC. exceptD. except for【答案】C30. Our decision depends his behavior.A. onB.inC. withD. for【答案】A31. Excuse me, the coat is big wear. Could you give me another one?A. so; thatB. too; toC. enough; thatD. such; that【答案】B32. Although Sally tried her best, she didn't manage the record.A. breakB. breakingC. to breakingD. to break【答案】D33. He said in the maths contest was his next goal.A. winB. winningC. wonD. have won【答案】B34. 一Great changes in Shanghai in the past few years.A. have taken placeB. took placeC. take placeD. will take place【答案】A35.___ exciting travelling abroad with my family will be!A. WhatB. What aC. HowD. What an【答案】C36. Please keep quiet. Lucy is busy studying for the final exam.A. inB. /C. onD. with【答案】B37. By 2018, my parents_ around ten countries in Europe.A. have travelledB. had travelledC. will travelD. are travelling【答案】B38. After retiring, my teacher Mr. Dai moved to the village he was born sixty years ago.A. whenB. WhichC. WhereD. What【答案】C39.Some people have no choice but_ their dogs in small spaces.A. keepB. keepingC. to keepD. kept【答案】C40. Jenny's parents gave her huge amounts of money, which enabled her a small flat.A. BuyB. buysC. to buyD. buying【答案】C41. Peter John is keen on watching movies They often go 10 the cinema together.A. both;andB. not only; but alsoC. neither, norD. either, or【答案】B42. Food___ in the refrigerator in summer or it will go bad.A. must keepB. mustn’t keepC. must be keptD. mustn't be kept【答案】C43.-I can not tell whether Mary is fat or thin at present, I have not seen her for a long time.B. andB. butC.soD. for【答案】D44. Would you mind my turning on the recorder?________Please go ahead.A. No. I don'tB. Never mindC. Of course notD. You'd better not【答案】C45.I don't think young children should do part-time jobs.By doing that, they can not only make money but also make new friends.A. Neither do IB.I disagreeC. I think soD. I don't like it【答案】B上海市上外民办劲松中学2020届九年级上学期期中测试I. Choose the best answer(选择最恰当的答案)(共35分)( ) 1. Which of the following words matches the sound/ˈwɔ:tə /?A) want B) water C) waste D) waiter( ) 2. Joanna wants to be _______ journalist when she finishes university.A) a B) an C) the D) /( ) 3. You can find many _______ in the Science and Technology Museum on Sundays.A) information B) fun C) children D) story( ) 4. Lin Lin often practices English _______ chatting with her American friend.A) to B) by C) for D) with( ) 5. It’s known that France is famous _______ its wine and beautiful scenery.A) on B) in C) at D) for( ) 6. Jason will invite you and _______ to his birthday party this coming Sunday.A) I B) me C) my D) myself( ) 7. --- How heavily it rained this early morning!--- Yes. But _______ of the students in our class was late for school.A)some B)all C)none D)both( ) 8. I’ve got two tickets for tonight’s football match. One is for me, ________ is for you.A) other B) the other C) others D) another( ) 9. --- Chinese athletes won 11 gold medals in the 2010 Winter Olympics.--- _______ exciting news it is!A)What B) What an C) What a D) How( ) 10. Kitty’s hometown is developing ________ this year than ever before.A) quickly B) much quickly C) more quickly D) the most quickly ( ) 11. --- _______ do you water the plants?--- Twice a day.A) How long B) How many times C) How often D) How soon ( ) 12.Why not ask your teacher for help if you can’t finish _______ the report by yourself?A) write B) writes C) writing D) to write( ) 13.--- What do you think of the biscuit?--- I like it very much. It tastes _______.A) good B) terrible C) well D) terribly ( ) 14. Sue told me that she _______ shopping with her sister the next day.A)will go B) would go C) goes D) has gone ( ) 15. --- Where is Peter?--- He _______ volleyball with his friends in the school gym.A) plays B) played C) was playing D) is playing ( ) 16. --- Hasn't Jim come yet?--- No, and I _______ for him for nearly 2 hours.A) wait B) waited C) have waited D) had waited ( ) 17. I _______ dinner at my friend’s house when you called me yesterday evening.A) had B) have had C) was having D) are having ( ) 18. Liu Qian _______ to perform magic by Hunan TV last week.A) will be invited B) is invited C) are invited D) was invited ( ) 19. Lily, please be quiet! The others ______ hear clearly.A) mustn’t B) can’t C) needn’t D) shouldn’t ( ) 20. We often hear the girl _______ happily in her room.A) sing B) to sing C) sang D) sings( ) 21. The thief refused _______ the policeman’s questions at first.A) answer B) answering C) to answer D) answered ( ) 22. “_______ exercise every day, my child. It’s good for your health,” Father said.A)Taking B) Took C) Take D) Takes( ) 23. Mum, my classmates are waiting outside, _______ I must go now.A) or B) so C) but D) though( ) 24. _______ he got up early, he didn’t catch the bus.A) Although B) Because C) After D) If( ) 25. Jenny often helps others and her parents are very _______ what she does.A) ready for B) strict with C) afraid of D) satisfied with ( ) 26. You did very well in your homework. You have made _______ mistakes in it.A) few B) a few C) little D) a little( ) 27. Try to guess its meaning when you meet with a new word. Don’t _______ your dictionary all the time.A) work on B) try on C) keep on D) depend on ( ) 28. _______ human beings ________ animals can live without air.A) Both, and B) Either, or C) Neither, nor D) Not only, but also ( ) 29. If I _______ free time tomorrow, I will go fishing with you.A)will have B) have C) had D) has( ) 30. Jenny has been a member of the Dance Club _______ last month.A) since B) for C) as soon as D) until( ) 31. He bought the newest type of camera in this shop, _______?A) d idn’t he B) did he C) doesn’t he D) does he ( ) 32. --- Thank you very much for coming to see me off.--- _______A) I’m fine. B) Nice to see you off. C) My duty. D) My pleasure. ( ) 33. --- Do you mind my smoking here?--- _______ Look at the sign. It says, “No smoking”.A) Of course not. B) No, I don’t. C) You’d better not. D) Sorry, I won’t. ( ) 34. --- Would you like some more soup?--- ________ It is delicious, but I’ve had enough.A)Yes, please. B) No, thank you. C) Nothing more. D) I’d like some. ( ) 35. --- I couldn’t get any tickets for the basketball game.--- ________A)Good idea! B) What a pity! C) OK. Anytime. D) Yes, I’d love to.1-5:BACBD 6-10: BCBAC 11-15: CCABD 16-20: CCDBA 21-25: CABAD 26-30: ADCBA 31-35: ADCBB上海市杨浦区兰生复旦2019-2020学年上学期九年级英语期中考试题Ⅱ. Choose the best answer (选择最恰当的答案)22. Which of the following underlined part is different in pronunciation ?A. reachB. deathC. leafD. beaten23. For ________ time being, this house suits all of our family’s needs.A. aB. anC. theD. /24. He broke the car windows ________ a stone to reach the driver and get him out.A. inB. withC. byD. on25. We can have a barbecue. We ________ one since last summer.A. didn’t haveB. haven’t hadC. don’t haveD. won’t have26. We are wondering whether technology has brought us ________ problems as it has solved.A. as manyB. as muchC. moreD. most27. He ________ said to Corey that the school was going to host a graduation party.A. lovelyB. friendlyC. lonelyD. softly28. The museum has some ________ which isn’t needed any more.A. itemB. to showC. showingD. to showing29. True friendship requires us ________ trust and support.A. showB. to showC. showingD. to showing30. Imagine ________ up in a house that is not the one you went to sleep in. How strange!A. wakeB. wokeC. wakingD. to wake31. Some people say we can see the Great Wall from the moon, ________ it’s not true.A. butB. orC. soD. and32. Nowadays, this type of shark ________ only in North America.A. findsB. foundC. be foundD. is found33. Children ________ visit the museum only if they’re with an adult.A. canB. mustC. needD. should34. -- Are you parents in? --- No, they’ve gone to visit a friends of ________.A. theyB. themC. theirsD. themselves35. Players train regularly to stay in shape ________ they can be better prepared for the games.A. althoughB. sinceC. unlessD. so that36. -- ________.-- No, thanks. But actually, Mrs Lin needs a student to take the boxes to her office.A. Can you pass me the books over there?B. Would you like to go to the library?C. Would you like me to take these books?D. Why don’t you study in the library?22. B 23. C 24. B 25. B 26. A27. D 28. D 29. B 30. C 31. A32. D 33. A 34. C 35. D 36. C上海市世界外国语中学2019-2020学年第一学期期中考试初三英语试卷31. We make a recorder play music by __________ the PLAY button.A. pressB. pressingC. to pressD. pressed32. Who knows the answer __________ question No. 8?A. toB. forC. ofD. about33. Today is April Fools’ Day. Let’s __________ on Mr. Li.A. celebrateB. make sureC. make jokesD. play a trick34. Are men better than women __________ new ideas?A. in createB. at createC. in creatingD. at creating35. Accountants no longer use __________ calculators a lot. Instead, they use computers very often.A. electricB. electricalC. electronicD. electricity36. Many teachers believe __________ textbooks, children also learn a lot __________ life.A. except, ofB. besides, fromC. except, fromD. besides, about37. Computers can’t take the place of human beings __________.A. at the presentB. for the first timeC. for the time beingD. at a time38. How much money we can make __________ how hard we work.A. dependsB. depends onC. dependingD. depending on39. I only drank a little coke, and __________ is kept in the fridge.A. the otherB. othersC. the othersD. the rest40. Helen __________ from the chair and __________ a question to the teacher.A. raised, raisedB. rose, roseC. rose, raisedD. raised, rose41. I __________ my key in her office. What was worse, I __________ her address and number.A. forgot, forgotB. left, leftC. left, forgotD. forgot, left42. Tom __________ any help because he can finish the work on time himself.A. needn’tB. doesn’t need toC. doesn’t needD. needn’t to43. The firemen put out the serious fire before any damage was caused, __________?A. did theyB. didn’t theyC. was itD. wasn’t it44. -- How long have you worked here?-- __________.A. Two years agoB. Since two yearsC. For 1999D. Since 2 years ago45. -- Where is Betty?-- She __________ for a few minutes.A. has goneB. has leftC. has returnedD. has been away46. __________ you work today, successful you will be in the future.A. More hardly, moreB. The harder, the moreC. More harder, moreD. The more hardly, the more47. Actions __________ to protect the environment since last week.A. have takenB. were takenC. have been takenD. will be taken48. -- __________ can you be ready, Andy?-- In ten minutes.A. How muchB. How oftenC. How longD. How soon49. With the development of Information Technology, __________ emails __________ becoming more and more convenient.A. Send, isB. Sending, isC. Sending, areD. To send, are50. People __________ clothes by hand. Nowadays washing machines __________ it.A. used to wash, are used to doB. are used to washing, are used to doingC. used to washing, are used to doD. are used to wash, used to do51. __________ young people like to play computer games on the Internet and __________ like to do online shopping.A. One, othersB. One, the otherC. Some, othersD. Some, the others52. __________ great fun the children are having in the Century Park)A. HowB. WhatC. What anD. What a53. He asked me __________ I was busy with those days.A. thatB. ifC. whyD. what54. The teacher asked __________.A. which subject I like bestB. who would be the winnerC. if we understood or notD. that I would return the next week55. -- Would you mind my smoking here?-- __________ Look at the sign. It says, "No smoking."A. Never mind.B. Of course not.C. No, I don't.D. You'd better not.56. -- Would you like me to buy the ticket for you?-- __________ I can book it by phone.A. Yes, please.B. Yes, I'd love to.C. No, thanks.D. No, you wouldn't.上海市浦东新区南片联合体2019-2020学年九年级(上)期中英语试卷II. Choose the best answer. (选出最恰当的答案)(共20分)12.(1分)Which of the following underlined parts is different in pronunciation?()A.bread B.cheap C.reach D.meat【分析】找出一个划线部分读音与其他三个不同的选项?bread面包;cheap便宜的;reach到达;meat 肉.【解答】bread的音标是[bred];cheap的音标是[tʃiːp];reach的音标是[riːtʃ];meat的音标是[miːt].因此可知bread中的划线部分与其他几项不一样.故选:A.【点评】首先要掌握这几个词的发音,然后根据具体的题目,就可以做正确选择.13.(1分)We should give a hand to ________ poor when they need help in life.()A.a B.an C.the D./【分析】当穷人在生活中需要帮助的时候,我们应当帮助他们.【解答】考察冠词the 的用法;a 、an与数词one 同源,是"一个"的意思,a用于辅音音素前,而an则用于元音音素前.the 和某些形容词连用,使形容词名词化,代表一类人或物,the poor意为"穷苦人"这一类人.本题We should give a hand to ________ poor,我们应当给穷苦人帮助,应当指的是一类人,不是个别的穷苦人,故选:C.【点评】the的用法:特指双方熟悉,上文已经提及;世上独一无二,方位名词、乐器;某些专有名词,外加复数姓氏;序数词和最高级,习惯用语要特记.14.(1分)The China International Import Expo was held ________ November 5,2018.()A.on B.at C.in D.for【分析】中国国际进口博览会于2018年11月5日举行.【解答】on表示在具体的某一天;at表示在具体的时刻前,in在一天的上午、下午、晚上等前;for其后跟一段时间.此处表示2018年11月5日,故用介词on.故选:A.【点评】掌握时间介词的用法,是解答本题的关键.15.(1分)The children enjoyed at the Christmas party.()A.they B.their C.themselves D.them【分析】孩子们在圣诞派对上玩得很开心.【解答】考查反身代词的用法;反身代词:表示我自己myself,你自己yourself,他自己himself,我们自己ourselves,你们自己yourselves,他们自己themselves等的词叫做反身代词.第一,二人称反身代词构成是由形容词性物主代词加"﹣self"(复数加﹣selves )构成.第三人称反身代词是由人称代词宾格形式加﹣self (复数加﹣selves )构成.本句中The children enjoyed at the Christmas party.孩子们在圣诞派对上玩得和开心,enjoy oneself 玩得开心、过得愉快,本句的主语the children为复数形式,对应的反身代词应为themselves,故选:C.【点评】反身代词和形容词性物主代词、名词性物主代词的用法区别,需要重点加以辨析,出错几率很大.16.(1分)The child walked as ________ as possible to keep up with his father.()A.quickly B.more quicklyC.most quickly D.the most quickly【分析】这孩子走得尽可能快,以跟上他父亲的步伐.【解答】根据as…as…这一结构的用法可知,它中间应该是形容词或副词的原级形式,意思是"和…一样…",表示两者在程度上相同.这里修饰动词用副词形式.故选:A.【点评】本题考查副词的原级比较,基础题,掌握该知识点是解题的关键,再根据题干即可作出选择.17.(1分)________ of my classmates has been to the new Ocean Park.()A.None B.Both C.All D.Some【分析】我的同学都没有去过新的海洋公园.【解答】none没有,both两者都,all所有,some一些,根据句意"我的同学都没有去过新的海洋公园"可知使用none.故选:A.【点评】本题考查代词的辨析,先区分所给代词的用法,再结合语境确定正确答案.18.(1分)A:Mum,must I wash the dishes now?B:No,you ________.You can do it after you watch the TV news.()A.can't B.shouldn't C.mustn't D.needn't【分析】﹣妈妈,我现在必须洗碗吗?﹣不,你不必要.你可以看完电视新闻后做.【解答】can't不可能,不能,shouldn't不应该,mustn't禁止,needn't不必;根据句意"妈妈,我现在必须洗碗吗?不,你不必要.你可以看完电视新闻后做"可知,要用"不必",其它选项语意不通.故选:D.【点评】考查情态动词,牢记情态动词的含义和用法,进行对比,排除错误的答案,从而做出正确的答案.19.(1分)I have got _____ much work to do _______ I don't have time to play with my friends.()A.so;that B.such;that C.too;to D.enough;to【分析】我有如此多的工作要做,以至于我没有时间和我的朋友们一起玩.【解答】so…that…如此…以致于…,so修饰形容词或副词;such…that…,如此…以致于…,such修饰名词,但是名词前面有much时,用so,排除B;too…to…太…而不能…,enough to足够做…,to后面要跟动词原形,排除CD;根据句意"我有如此多的工作要做,以至于我没有时间和我的朋友们一起玩"可知,要用固定句式so…that…故选:A.【点评】考查固定句式,要熟练掌握固定句式的构成及用法.20.(1分)________ wonderful film we have seen!()A.What B.What a C.What an D.How【分析】我们看了多精彩的电影啊!【解答】分析句子结构可知考查感叹句,在感叹句中,what修饰名词,how修饰形容词/副词,根据wonderful film we have seen可知句型结构为:What +a/an+形容词+单数名词+主语+谓语动词!film 是可数名词单数形式.故选:B.【点评】解答此类试题时,务必根据题目的要求,在准确理解句子意思的前提下,结合感叹句的构成准确作答.21.(1分)The restaurant is popular ________ the food and service are good.()A.unless B.because C.until D.although【分析】这个饭店很受欢迎,原因是食品和服务都很好.【解答】unless意思是"如果不"引导条件状语从句;because意思是"因为"引导原因状语从句;until意思是"直到…"引导时间状语从句;although意思是"尽管"引导让步状语从句.通过分析句子结构可知,这句话的意思是"这个饭店很受欢迎,原因是食品和服务都很好."故选:B.【点评】首先要掌握这几个词的意思以及用法,然后根据具体的题目,就可以做正确选择.22.(1分)The pianist usually practises ________ the piano for six hours every day.()A.play B.plays C.playing D.to play【分析】这位钢琴家通常每天练习弹六个小时的钢琴.【解答】practise doing sth.练习做某事,固定搭配,所以空处动词需用其动名词形式playing.故选:C.【点评】此题需要熟悉选项意思,并结合语境确定答案23.(1分)The little girl looks ________ and everyone likes her.()A.wonderfully B.gentlyC.beautifully D.lovely【分析】这个小女孩看起来很可爱,每个人都喜欢她.【解答】wonderfully非常,很好地,副词;gently温柔地,温和地,副词;beautifully美丽地,副词;lovely 可爱的,美丽的,形容词.look看起来,系动词,后需接形容词作表语,lovely可爱的,美丽的,形容词,符合题意.故选:D.【点评】此题需要熟悉选项意思,并结合语境确定答案.24.(1分)China ________ a lot since the year 1978.()A.changes B.will changeC.has changed D.changed【分析】自1978年以来,中国发生了很大的变化.【解答】根据句意"自1978年以来,中国发生了很大的变化"和since the year 1978可知,要用现在完成时,其构成为have/ has done,其它选项不符合语法.故选:C.【点评】对时态的考查,要求牢记各种时态的构成形式,结合时间状语和上下文的关系,找出正确答案.25.(1分)Look! The students ________ activities in the playground.()A.are doing B.will do C.do D.did【分析】看!学生们正在操场上做活动.【解答】根据句意"看!学生们正在操场上做活动"和look可知,要用现在进行时,其构成为am/ is/ are doing,其它选项不符合语法.故选:A.【点评】对时态的考查,要求牢记各种时态的构成形式,结合时间状语和上下文的关系,找出正确答案.26.(1分)The boy ________ for her in the rain for half an hour before she came.()A.waits B.waited C.has waited D.had waited【分析】在她来之前,那个男孩在雨中等了她半个小时.【解答】根据句意"在她来之前,那个男孩在雨中等了她半个小时"和before she came可知,要用过去完成时,其构成为had done,其它选项不符合语法.故选:D.【点评】对时态的考查,要求牢记各种时态的构成形式,结合时间状语和上下文的关系,找出正确答案.27.(1分)It's quite hot outside.You had better _________ off your jacket.()A.taking B.to take C.taken D.take【分析】外面相当地热.你最好脱掉你的夹克.【解答】had better 后面要跟动词原形.是固定搭配,意思是"最好".A是现在分词;B是不定式;C是过去分词;D是动词原形.故选:D.【点评】考查动词的固定搭配,要牢记动词固定搭配的含义及用法,进行比较分析,选择正确答案.28.(1分)﹣________ do you go to Beijing on business?﹣About three times a month.()A.How long B.How soon C.How far D.How often【分析】你多久去一次北京出差?﹣大约一个月三次.【解答】考查频度.How long和How soon多久,问的是时间段;How far多远,问的是距离;How often 多久一次,问的是频度.根据About three times a month.可知问的是频度,用how often.故选:D.【点评】仔细分析句子的结构,根据About three times a month.结合选项作答.29.(1分)﹣What do you want for breakfast?﹣Either noodles or bread ________ OK.()A.are B.is C.be D.am【分析】你早餐想吃什么?面条或面包都可以.【解答】either…or连接两个并列成分作主语,谓语动词的人称和数用就近原则,bread是不可数名词,后面用is.故选:B.【点评】考查系动词,牢记系动词的含义和用法,进行对比,排除错误的答案,从而做出正确的答案.30.(1分)A:I'm sorry I'm late,Tony.B:______,Arthur.I've just got here myself,so I wasn't waiting for long.()A.It's my pleasure B.That's all right.C.Enjoy your time D.Good idea【分析】A:对不起,我迟到了,托尼.B:没关系,亚瑟.我也刚到这儿,所以没等多久.【解答】考查情境对话.A.It's my pleasure我的荣幸;B.That's all right.没关系.C.Enjoy your time 过的愉快;D.Good idea好主意.根据I've just got here myself,so I wasn't waiting for long.可知应说没关系.故选:B.【点评】首先迅速地浏览一遍对话,根据对话的情境,I've just got here myself,so I wasn't waiting for long,结合选项作答.31.(1分)﹣Let's clean our neighborhood this weekend,shall we?﹣____ ()A.Yes,please.B.Never mind.C.Well done.D.That's a good idea.【分析】﹣我们这个周末打扫一下我们的社区好吗?﹣是个好主意.【解答】A.请B.不介意C.干的好D.好主意,根据﹣Let's clean our neighborhood this weekend,shall we?,可知是一种建议,所以选项D符合题意,句意:﹣我们这个周末打扫一下我们的社区好吗?﹣是个好主意.故选:D.。

2019-2020学年上海市杨浦区兰生复旦九上英语期中考试卷

2019-2020学年上海市杨浦区兰生复旦九上英语期中考试卷

Part 2 Phonetics, Grammar and Vocabulary(第二部分语音、语法和词汇)Ⅱ. Choose the best answer (选择最恰当的答案)22. Which of the following underlined part is different in pronunciation ?A. reachB. deathC. leafD. beaten23. For ________ time being, this house suits all of our family’s needs.A. aB. anC. theD. /24. He broke the car windows ________ a stone to reach the driver and get him out.A. inB. withC. byD. on25. We can have a barbecue. We ________ one since last summer.A. didn’t haveB. haven’t hadC. don’t haveD. won’t have26. We are wondering whether technology has brought us ________ problems as it has solved.A. as manyB. as muchC. moreD. most27. He ________ said to Corey that the school was going to host a graduation party.A. lovelyB. friendlyC. lonelyD. softly28. The museum has some ________ which isn’t needed any more.A. itemB. to showC. showingD. to showing29. True friendship requires us ________ trust and support.A. showB. to showC. showingD. to showing30. Imagine ________ up in a house that is not the one you went to sleep in. How strange!A. wakeB. wokeC. wakingD. to wake31. Some people say we can see the Great Wall from the moon, ________ it’s not true.A. butB. orC. soD. and32. Nowadays, this type of shark ________ only in North America.A. findsB. foundC. be foundD. is found33. Children ________ visit the museum only if they’re with an adul t.A. canB. mustC. needD. should34. -- Are you parents in? --- No, they’ve gone to visit a friends of ________.A. theyB. themC. theirsD. themselves35. Players train regularly to stay in shape ________ they can be better prepared for the games.A. althoughB. sinceC. unlessD. so that36. -- ________.-- No, thanks. But actually, Mrs Lin needs a student to take the boxes to her office.A. Can you pass me the books over there?B. Would you like to go to the library?C. Would you like me to take these books?D. Why don’t you study in the library?Ⅲ. complete the following passage with the words in the box. Each can only be used once(将下列单词填入空格。

【精品初中语文试卷】2019-2020学年上海市杨浦区九年级(上)期末语文试卷+答案

【精品初中语文试卷】2019-2020学年上海市杨浦区九年级(上)期末语文试卷+答案

2019-2020学年上海市杨浦区九年级(上)期末语文试卷一、文言文阅读(40分)1.(15分)默写(1)几处早莺争暖树,。

(《钱塘湖春行》)(2)欲为圣朝除弊事,。

(《左迁至蓝关示侄孙湘)》)(3),带月荷锄归。

(《归园田居》)(4),盖以诱敌。

(《狼》)(5)《陋室铭》中表现主人公交往之雅的句子是,。

2.(12分)阅读下面古诗文,完成下列各题。

【甲】行路难(其一)金樽清酒斗十千,玉盘珍羞直万钱。

停杯投箸不能食,拔剑四顾心茫然。

欲渡黄河冰塞川,将登太行雪满山。

闲来垂钓碧溪上,忽复乘舟梦日边。

行路难,行路难,多歧路,今安在?长风破浪会有时,直挂云帆济沧海。

【乙】醉翁亭记(节选)若夫日出而林霏开,云归而岩穴暝,晦明变化者,山间之朝暮也野芳发而幽香,佳木秀而繁阴,风霜高洁,水落而石出者,山间之四时也。

朝而往,暮而归,四时之景不同,而乐亦无穷也。

(1)甲诗作者为代诗人;乙文作者为文学家。

(2)用现代汉语翻译下面的句子。

日出而林霏开,云归而岩穴暝。

(3)下面选项恰当的一项是A.甲诗起首两句以夸张手法叙写欢乐豪放的宴饮,体现作者与友人相处时的愉悦心情。

B.乙文作者细致描绘山间一日及四时的自然风光,体现其寄情山水乐无穷的怡然自得。

C.甲诗具体描绘渡黄河登泰山时艰难困顿的遭遇形象体现了诗歌“行路难”的主题。

D.乙文具体描绘琅琊山朝暮与四季景色之美,意在表作者对醉翁亭秀丽环境的欣赏。

(4)教材将甲诗和乙文编入同一单元,意在告诉我们,人生中难免种种,对此,我们要、地面对。

3.(12分)阅读下文,完成下列各题。

刘宣苦读成才景泰间,吉安刘公宣代戍于京师龙驤卫,为卫使畜马,昼夜读书厩中,使初不知也。

公偶与塾师论《春秋》,师惊异之,以语使,使乃优遇之。

未几,发解及第取解时,刘文恭公铉主试,讶其文,谓必山林老儒之作,及启封乃公也,人始识公,而文恭知人之益著。

(焦竤《玉堂丛语》)【注释】①刘公宣:即刘宣。

②卫使:人名。

③发解及第(刘宣参加科举考试)发榜考中解元。

2020-2021学年上海市杨浦区兰生复旦中学九年级上学期期中数学仿真试卷 (Word版 含解析)

2020-2021学年上海市杨浦区兰生复旦中学九年级上学期期中数学仿真试卷 (Word版 含解析)

2020-2021学年上海市杨浦区兰生复旦中学九年级(上)期中数学仿真试卷一、选择题(共6小题).1.已知线段a、b、c,求作第四比例线段x,则以下正确的作图是()A.B.C.D.2.如图,在梯形ABCD中,AB∥CD,过O的直线MN∥CD,则=()A.B.C.D.3.如图,在△ABC中,D、E分别在AB、AC上,DE∥BC,EF∥CD交AB于F,那么下列比例式中正确的是()A.B.C.D.4.已知点E、F分别在△ABC的AB、AC边上,则下列判断正确的是()A.若△AEF与△ABC相似,则EF∥BCB.若AE×BE=AF×FC,则△AEF与△ABC相似C.若,则△AEF与△ABC相似D.若AF•BE=AE•FC,则△AEF与△ABC相似5.下列正确的是()A.B.为单位向量,则C.平面内向量、,总存在实数m使得向量D.若,,,则、就是在、方向上的分向量6.如图,在直角梯形ABCD中,DC∥AB,∠DAB=90°,AC⊥BC,AC=BC,∠ABC的平分线分别交AD、AC于点E,F,则的值是()A.B.C.D.二.填空题(共12小题).7.若,那么的值为.8.计算:tan15°•tan45°•tan75°=.9.若是与非零向量反向的单位向量,那么=.10.如图,在△ABC中,BC=6,G是△ABC的重心,过G作边BC的平行线交AC于点H,则GH的长为.11.二次函数y=ax2﹣3x+a2﹣a的图象经过原点,则a=.12.若过⊙O内一点M的最长弦为10,最短弦为6,则OM的长为.13.已知⊙O的半径为13,弦AB=24,CD=10,且AB∥CD,则弦AB与CD之间的距离为.14.如图,一桥拱呈抛物线状,桥的最大高度是16米,跨度是40米,在线段AB上离中心M处5米的地方,桥的高度是m(π取3.14).15.小亮的身高为1.8米,他在路灯下的影子长为2米;小亮距路灯杆底部为3米,则路灯灯泡距离地面的高度为米.16.如图,△ABC中,BC=5,AC=3,△ABC绕着C点旋转到△A′B′C的位置,那么△BB′C与△AA′C的面积之比为.17.如图,在Rt△ABC中,∠BAC=90°,AD⊥BC于点D,O为AC边中点,=2,连接BO交AD于F,作OE⊥OB交BC边于点E,则的值=.18.将一个无盖正方体纸盒展开(如图①),沿虚线剪开,用得到的5张纸片(其中4张是全等的直角三角形纸片)拼成一个正方形(如图②).则所剪得的直角三角形较短的与较长的直角边的比是.三.解答题(本大题共7小题,19-22题每题10分,23-24题每题12分,25题14分,共78分)19.计算:3tan30°+cos60°﹣+2sin245°20.已知在直角坐标系中,点A的坐标是(﹣3,1),将线段OA绕着点O顺时针旋转90°得到OB.(1)求点B的坐标;(2)求过A、B、O三点的抛物线的解析式;(3)设点B关于抛物线的对称轴L的对称点为C,求△ABC的面积.21.如图,在平行四边形ABCD中,过点B作BE⊥CD,垂足为E,连接AE,F为AE上一点,且∠BFE=∠C.(1)求证:△ABF∽△EAD;(2)若AD=3,∠BAE=30°,求BF的长.(计算结果保留根号)22.已知:如图,△ABC中,点E在中线AD上,∠DEB=∠ABC.求证:(1)DB2=DE•DA;(2)∠DCE=∠DAC.23.如图,△ABC中,D为BC边上的一点,E在AD上,过点E作直线l分别和AB、AC 两边交于点P和点Q,且EP=EQ.(1)当点P和点B重合的时候,求证:;(2)当P、Q不与A、B、C三点重合时,求证:.24.如图,在平面直角坐标系xOy中,正方形OABC的边长为2cm,点A、C分别在y轴和负半轴和x轴的正半轴上,抛物线y=ax2+bx+c(a≠0)经过的A、B,且12a+5c=0.(1)求抛物线的解析式;(2)若点P由点A开始边以2cm/s的速度向点B移动,同时点Q由点B开始沿BC边以1cm/s的速度向点C移动.当一点到达终点时,另一点也停止运动.①当移动开始后第t秒时,设S=PQ2(cm),试写出s与t之间的函数关系式,并写出t的取值范围.②当t取何值时,S取得最小值?此时在抛物线上是否存在点R,使得以P、B、Q、R为顶点的四边形是平行四边形?若存在,直接写出点R的坐标,若不存在,请说明理由.25.已知:在Rt△ABC中,∠C=90°,AC=4,∠A=60°,CD是边AB上的中线,直线BM∥AC,E是边CA延长线上一点,ED交直线BM于点F,将△EDC沿CD翻折得△E′DC,射线DE′交直线BM于点G.(1)如图1,当CD⊥EF时,求BF的值;(2)如图2,当点G在点F的右侧时;①求证:△BDF∽△BGD;②设AE=x,△DFG的面积为y,求y关于x的函数解析式,并写出x的取值范围;(3)如果△DFG的面积为,求AE的长.参考答案一.选择题(本大题共有6题,每题4分,共24分)1.已知线段a、b、c,求作第四比例线段x,则以下正确的作图是()A.B.C.D.【分析】根据第四比例线段的定义列出比例式,再根据平行线分线段成比例定理对各选项图形列出比例式即可得解.解:∵线段x为线段a、b、c的第四比例线段,∴=,∴正确的作图是B;故选:B.2.如图,在梯形ABCD中,AB∥CD,过O的直线MN∥CD,则=()A.B.C.D.【分析】先得到MN∥AB,利用平行线分线段成比例定理得到=,=,=,则可判断ON=OM,再证明△AON∽△ACD得到=①,证明△COM∽△CAB得到=②,把两式相加后利用等式的性质可得到+=.解:∵AB∥CD,MN∥CD,∴MN∥AB,∵ON∥AB,OM∥AB,∴=,=,∵=,∴=,∴ON=OM,∵ON∥CD,∴△AON∽△ACD,∴=①,∵OM∥AB,∴△COM∽△CAB,∴=②,①+②得+=1,即+=1,∴+=.故选:B.3.如图,在△ABC中,D、E分别在AB、AC上,DE∥BC,EF∥CD交AB于F,那么下列比例式中正确的是()A.B.C.D.【分析】根据平行线分线段成比例定理和相似三角形的性质找准线段的对应关系,对各选项分析判断后利用排除法求解.解:A、∵EF∥CD,DE∥BC,∴,,∵CE≠AC,∴.故本答案错误;B、∵DE∥BC,EF∥CD,∴,,∴,∵AD≠DF,∴,故本答案错误;C、∵EF∥CD,DE∥BC,∴,,∴.∵AD≠DF,∴,故本答案错误;D、∵DE∥BC,EF∥CD,∴,,∴,故本答案正确.故选:D.4.已知点E、F分别在△ABC的AB、AC边上,则下列判断正确的是()A.若△AEF与△ABC相似,则EF∥BCB.若AE×BE=AF×FC,则△AEF与△ABC相似C.若,则△AEF与△ABC相似D.若AF•BE=AE•FC,则△AEF与△ABC相似【分析】根据三角形相似的判定定理一一判断即可.解:选项A错误,∵△AEF与△ABC相似,可能是∠AEF=∠C,推不出EF∥BC.选项B错误,由AE×BE=AF×FC,推不出△AEF与△ABC相似.选项C错误,由,推不出△AEF与△ABC相似.选项D正确.理由:∵AF•BE=AE•FC,∴=,∴EF∥BC,∴△AEF∽△ABC.故选:D.5.下列正确的是()A.B.为单位向量,则C.平面内向量、,总存在实数m使得向量D.若,,,则、就是在、方向上的分向量【分析】根据平面向量的性质一一判断即可.解:A、|k|=k•|,正确.B、为单位向量,则=||•,错误,应该是=±||•.C、平面内向量、,总存在实数m使得向量=m,错误,因为与不一定是平行向量.D、若,,,则、就是在、方向上的分向量,错误,也可能是在、反方向上的分向量.故选:A.6.如图,在直角梯形ABCD中,DC∥AB,∠DAB=90°,AC⊥BC,AC=BC,∠ABC的平分线分别交AD、AC于点E,F,则的值是()A.B.C.D.【分析】作FG⊥AB于点G,由AE∥FG,得出=,求出Rt△BGF≌Rt△BCF,再由AB=BC求解.解:作FG⊥AB于点G,∵∠DAB=90°,∴AE∥FG,∴=,∵AC⊥BC,∴∠ACB=90°,又∵BE是∠ABC的平分线,∴FG=FC,在Rt△BGF和Rt△BCF中,∴Rt△BGF≌Rt△BCF(HL),∴CB=GB,∵AC=BC,∴∠CBA=45°,∴AB=BC,∴====+1.故选:C.二.填空题(本大题共有12题,每题4分,共48分)7.若,那么的值为.【分析】根据已知得出b=a,再代入要求的式子进行计算即可得出答案.解:∵,∴b=a,∴==;故答案为:.8.计算:tan15°•tan45°•tan75°=1.【分析】直接利用锐角三角函数关系以及特殊角的三角函数值代入得出答案.解:原式=tan15°•tan75°•tan45°=1×1=1.故答案为:1.9.若是与非零向量反向的单位向量,那么=﹣||.【分析】根据向量的几何意义填空即可.解:若是与非零向量反向的单位向量,那么=﹣|•,故答案为﹣||.10.如图,在△ABC中,BC=6,G是△ABC的重心,过G作边BC的平行线交AC于点H,则GH的长为2.【分析】连接AG,并延长AG交BC于D;根据重心的性质知:D是BC中点,且AG:AD=2:3;可根据平行线分线段成比例定理得出的线段比例关系式及CD的长求出GH 的值.解:如图,连接AG,并延长AG交BC于D;∵G是△ABC的重心,∴AG:GD=2:3,且D是BC的中点;∵GH∥BC,∴==;∵CD=BC=3,∴GH=2.11.二次函数y=ax2﹣3x+a2﹣a的图象经过原点,则a=1.【分析】将(0,0)代入二次函数的解析式即可求出a的值.解:将(0,0)代入y=ax2﹣3x+a2﹣a,∴0=a2﹣a,∴a=0(舍去)或a=1,故答案为:1.12.若过⊙O内一点M的最长弦为10,最短弦为6,则OM的长为4.【分析】根据垂径定理及勾股定理即可求出.解:由已知可知,最长的弦是过M的直径AB,最短的是垂直平分直径的弦CD,已知AB=10,CD=6,则OD=5,MD=3,由勾股定理得OM=4.故答案为:4.13.已知⊙O的半径为13,弦AB=24,CD=10,且AB∥CD,则弦AB与CD之间的距离为7或17.【分析】分两种情况进行讨论:①弦AB和CD在圆心同侧;②弦AB和CD在圆心异侧;作出半径和弦心距,利用勾股定理和垂径定理求解即可.解:①当弦AB和CD在圆心同侧时,如图1,∵AB=24,CD=10,∴AE=12,CF=5,∵OA=OC=13,∴EO=5,OF=12,∴EF=12﹣5=7;②当弦AB和CD在圆心异侧时,如图2,∵AB=24,CD=10,∴AE=12,CF=5,∵OA=OC=13,∴EO=5,OF=12,∴EF=OF+OE=17.∴AB与CD之间的距离为7或17.故答案为7或17.14.如图,一桥拱呈抛物线状,桥的最大高度是16米,跨度是40米,在线段AB上离中心M处5米的地方,桥的高度是15m(π取3.14).【分析】根据题意假设解析式为y=ax2+bx+c,用待定系数法求出解析式.然后把自变量的值代入求解对应函数值即可.解:设抛物线的方程为y=ax2+bx+c已知抛物线经过(0,16),(﹣20,0),(20,0),故可得,可得a=﹣,b=0,c=16,故解析式为y=﹣x2+16,当x=5时,y=15m.15.小亮的身高为1.8米,他在路灯下的影子长为2米;小亮距路灯杆底部为3米,则路灯灯泡距离地面的高度为 4.5米.【分析】根据已知得出图形,进而利用相似三角形的判定与性质求出即可.解:结合题意画出图形得:∴△ADC∽△AEB,∴=,∵小亮的身高为1.8米,他在路灯下的影子长为2米;小亮距路灯杆底部为3米,∴AC=2,BC=3,CD=1.8,∴=,解得:BE=4.5,故答案为:4.5.16.如图,△ABC中,BC=5,AC=3,△ABC绕着C点旋转到△A′B′C的位置,那么△BB′C与△AA′C的面积之比为.【分析】由旋转的性质可得AC=CA',BC=CB',∠BCB'=∠ACA',可证△ACA'∽△BCB',由相似三角形的面积比等于相似比的平方可求解.解:∵△ABC绕着C点旋转到△A′B′C的位置,∴AC=CA',BC=CB',∠BCB'=∠ACA',∴,∴△ACA'∽△BCB',∴=()2=,故答案为:.17.如图,在Rt△ABC中,∠BAC=90°,AD⊥BC于点D,O为AC边中点,=2,连接BO交AD于F,作OE⊥OB交BC边于点E,则的值=2.【分析】先证明∠BAF=∠C,∠ABF=∠COE,作OH⊥AC,交BC于H,易证:△OEH 和△OFA相似,可得,由三角形中位线定理可得OH=AB,OA=OC=AC,即可求解.解:∵AD⊥BC,∴∠DAC+∠C=90°.∵∠BAC=90°,∴∠BAF=∠C.∵OE⊥OB,∴∠BOA+∠COE=90°,∵∠BOA+∠ABF=90°,∴∠ABF=∠COE.过O作AC的垂线交BC于H,则OH∥AB,∵∠ABF=∠COE,∠BAF=∠C.∴∠AFB=∠OEC,∴∠AFO=∠HEO,而∠BAF=∠C,∴∠FAO=∠EHO,∴△OEH∽△OFA,∴,又∵O为AC的中点,OH∥AB.∴OH为△ABC的中位线,∴OH=AB,OA=OC=AC,而,∴,即,故答案为:2.18.将一个无盖正方体纸盒展开(如图①),沿虚线剪开,用得到的5张纸片(其中4张是全等的直角三角形纸片)拼成一个正方形(如图②).则所剪得的直角三角形较短的与较长的直角边的比是1:2.【分析】本题考查了拼摆的问题,仔细观察图形的特点作答.解:由图可得,所剪得的直角三角形较短的边是原正方体棱长的一半,而较长的直角边正好是原正方体的棱长,所以所剪得的直角三角形较短的与较长的直角边的比是1:2.三.解答题(本大题共7小题,19-22题每题10分,23-24题每题12分,25题14分,共78分)19.计算:3tan30°+cos60°﹣+2sin245°【分析】直接利用特殊角的三角函数值和二次根式的性质分别化简得出答案.解:原式=3×+﹣+2×()2=+﹣+1=.20.已知在直角坐标系中,点A的坐标是(﹣3,1),将线段OA绕着点O顺时针旋转90°得到OB.(1)求点B的坐标;(2)求过A、B、O三点的抛物线的解析式;(3)设点B关于抛物线的对称轴L的对称点为C,求△ABC的面积.【分析】(1)本题可通过构建全等三角形来求解.过点A作AH⊥x轴,过点B作BM ⊥y轴,根据旋转的性质可知:OA=OB,而∠MOB与∠AOH都是∠AOM的余角,因此两角相等,因此这两个直角三角形就全等,那么OH=OM,AH=BM,由此可得出B点坐标.(2)根据求出的B点坐标以及已知的A、O的坐标即可用待定系数法求抛物线的解析式.(3)先根据抛物线的解析式求出抛物线的对称轴及C点坐标,即可得出BC的长,求三角形ABC的面积时,可以BC为底,以A、B纵坐标差的绝对值为高来求解.解:(1)过点A作AH⊥x轴,过点B作BM⊥y轴,由题意得OA=OB,∠AOH=∠BOM,∴△AOH≌△BOM∵A的坐标是(﹣3,1),∴AH=BM=1,OH=OM=3∴B点坐标为(1,3)(2)设抛物线的解析式为y=ax2+bx+c则.得∴抛物线的解析式为y=x2+x(3)对称轴为x=﹣∴C的坐标为(﹣,3)∴S△ABC=BC•h BC=×(1+)×2=.21.如图,在平行四边形ABCD中,过点B作BE⊥CD,垂足为E,连接AE,F为AE上一点,且∠BFE=∠C.(1)求证:△ABF∽△EAD;(2)若AD=3,∠BAE=30°,求BF的长.(计算结果保留根号)【分析】(1)可通过证明∠BAF=∠AED,∠AFB=∠D,证得△ABF∽△EAD;(2)根据平行线的性质得到BE⊥AB,根据三角函数的定义得到tan∠BAE=,根据相似三角形的性质即可得到结论.【解答】(1)证明:在平行四边形ABCD中,∵∠D+∠C=180°,AB∥CD,∴∠BAF=∠AED.∵∠AFB+∠BFE=180°,∠D+∠C=180°,∠BFE=∠C,∴∠AFB=∠D,∴△ABF∽△EAD;(2)解:∵BE⊥CD,AB∥CD,∴BE⊥AB.∴∠ABE=90°.在Rt△ABE中,∠BAE=30°,∴tan∠BAE=,∵由(1)知,△ABF∽△EAD,∴,∵AD=3,∴BF=.22.已知:如图,△ABC中,点E在中线AD上,∠DEB=∠ABC.求证:(1)DB2=DE•DA;(2)∠DCE=∠DAC.【分析】(1)根据已知可证△BDE∽△DAB,得到,即证BD2=AD•DE.(2)在(1)的基础上,因为CD=BD,可证,即可证△DEC∽△DCA,得到∠DCE=∠DAC.【解答】证明:(1)在△BDE和△DAB中∵∠DEB=∠ABC,∠BDE=∠ADB,(1分)∴△BDE∽△ADB,(1分)∴,(1分)∴BD2=AD•DE.(1分)(2)∵AD是中线,∴CD=BD,∴CD2=AD•DE,∴,(1分)又∠ADC=∠CDE,(1分)∴△DEC∽△DCA,(1分)∴∠DCE=∠DAC.(1分)23.如图,△ABC中,D为BC边上的一点,E在AD上,过点E作直线l分别和AB、AC两边交于点P和点Q,且EP=EQ.(1)当点P和点B重合的时候,求证:;(2)当P、Q不与A、B、C三点重合时,求证:.【分析】(1)过点Q作QF∥BC交AD于F,由相似三角形的性质可得=,可得BD=FQ,EF=DE,通过证明△AFQ∽△ADC,可得,即可得结论;(2)过点Q作QF∥BC交AD于F,过点P作PH∥BC交AD于H,由相似三角形的性质可得,可得PH=FQ,EF=HE,由相似三角形的性质可得,,即可得结论.【解答】证明:(1)如图,过点Q作QF∥BC交AD于F,∴△FQE∽△DPE,∴=,又∵QE=EP,∴BD=FQ,EF=DE,∵QF∥CD,∴△AFQ∽△ADC,∴,∴,∴;(2)如图,过点Q作QF∥BC交AD于F,过点P作PH∥BC交AD于H,∴QF∥PH,∴△FQE∽△HPE,∴,又∵QE=EP,∴PH=FQ,EF=HE,∵FQ∥BC,∴△AQF∽△ACD,∴,∵PH∥BC,∴△APH∽△ABD,∴,∴===.24.如图,在平面直角坐标系xOy中,正方形OABC的边长为2cm,点A、C分别在y轴和负半轴和x轴的正半轴上,抛物线y=ax2+bx+c(a≠0)经过的A、B,且12a+5c=0.(1)求抛物线的解析式;(2)若点P由点A开始边以2cm/s的速度向点B移动,同时点Q由点B开始沿BC边以1cm/s的速度向点C移动.当一点到达终点时,另一点也停止运动.①当移动开始后第t秒时,设S=PQ2(cm),试写出s与t之间的函数关系式,并写出t的取值范围.②当t取何值时,S取得最小值?此时在抛物线上是否存在点R,使得以P、B、Q、R为顶点的四边形是平行四边形?若存在,直接写出点R的坐标,若不存在,请说明理由.【分析】(1)根据已知条件,结合正方形的性质求出A、B点的坐标,利用待定系数法可求解;(2)①用t表示出PB、BQ的长,利用勾股定理建立起它们之间的关系;②利用①中关系式,根据二次函数的性质求出S取最小值时的t的取值,计算出PB、BQ的长,然后分三种情况讨论利用平行四边形的性质可求解.解:(1)据题意知:A(0,﹣2),B(2,﹣2),∵A点在抛物线上,∴c=﹣2,∵12a+5c=0,∴a=,由AB=2知抛物线的对称轴为:x=1,即:﹣=1,∴b=﹣,∴抛物线的解析式为:y=x2﹣x﹣2;(2)①由图象知:PB=2﹣2t,BQ=t,∴S=PQ2=PB2+BQ2=(2﹣2t)2+t2,即S=5t2﹣8t+4(0≤t≤1);②假设存在点R,可构成以P、B、R、Q为顶点的平行四边形,∵S=5t2﹣8t+4(0≤t≤1),∴S=5(t﹣)2+(0≤t≤1),∴当t=时,S取得最小值;这时PB=2﹣=0.4,BQ=0.8,P(1.6,﹣2),Q(2,﹣1.2),分情况讨论:若PB与PQ为边,这时QR=PB=0.4,QR∥PB,则:R的坐标为(2.4,﹣1.2),代入y=x2﹣x﹣2,左右两边相等,∴这时存在R(2.4,﹣1.2)满足题意;若PB与QB为边,这时PR=QB,PR=QB=0.8,则:R的坐标为(1.6,﹣1.2),代入y=x2﹣x﹣2,左右两边不相等,R不在抛物线上;若PQ与QB为边,这时PR=QB,PR∥QB,则:R的坐标为(1.6,﹣2.8),代入y=x2﹣x﹣2,左右不相等,R不在抛物线上.综上所述,存在一点R(2.4,﹣1.2)满足题意.25.已知:在Rt△ABC中,∠C=90°,AC=4,∠A=60°,CD是边AB上的中线,直线BM∥AC,E是边CA延长线上一点,ED交直线BM于点F,将△EDC沿CD翻折得△E′DC,射线DE′交直线BM于点G.(1)如图1,当CD⊥EF时,求BF的值;(2)如图2,当点G在点F的右侧时;①求证:△BDF∽△BGD;②设AE=x,△DFG的面积为y,求y关于x的函数解析式,并写出x的取值范围;(3)如果△DFG的面积为,求AE的长.【分析】(1)由∠ACB=90°,AD=BD,利用斜边上的中线等于斜边的一半得到CD =AD=BD,再由∠BAC=60°,得到三角形ADC为等边三角形,由AC的长求出AD与BD的长,同时求出∠ABC=30°,由BM与AC平行,利用两直线平行内错角相等得到∠MBC=∠ACB=90°,再由CD垂直于EF,得到∠CDE和∠CDF都为直角,在直角三角形EDC中,求出∠DEC为30°,利用两直线平行内错角相等可得出∠BFD也为30°,而由∠CDE﹣∠CDA求出∠EDA为30°,利用对顶角相等得到∠BDF为30°,即∠BFD =∠BDF,利用等角对等边可得出BD=BF,由BD的长即可求出BF的长;(2)当点G在点F的右侧时,如图2所示,①由翻折,得∠E′CD=∠ACD=60°,得到一对内错角相等,利用内错角相等两直线平行,得到CE′∥AB,再由两直线平行得到一对内错角相等,利用等量代换得到∠BDG=∠BFD,再由一对公共角,利用两对应角相等的两三角形相似可得出△BDF∽△BGD;②由△BDF∽△BGD得比例,将各自的值代入即可列出y与x的函数关系式,求出x的范围即可;(3)分两种情况考虑:(i)当点G在点F的右侧时,在y与x的关系式中,令y=6列出关于x的方程,求出方程的解得到x的值,即为AE的长;(ii)当点G在点F的左侧时,如图3所示,列出此时y与x的关系式,令y=6列出关于x的方程,求出方程的解得到x的值,即为AE的长,综上,得到所有满足题意的AE的长.解:(1)∵∠ACB=90°,AD=BD,∴CD=AD=BD,∵∠BAC=60°,∴∠ADC=∠ACD=60°,∠ABC=30°,AD=BD=AC,∵AC=4,∴AD=BD=AC=4,∵BM∥AC,∴∠MBC=∠ACB=90°,又∵CD⊥EF,∴∠CDF=90°,∴∠BDF=30°,∴∠BFD=30°,∴∠BDF=∠BFD,∴BF=BD=4;(2)①证明:由翻折,得∠E′CD=∠ACD=60°,∴∠ADC=∠E′CD,∴CE′∥AB,∴∠CE′D=∠BDG,∵BM∥AC,∴∠CED=∠BFD,又∵∠CE′D=∠CED,∴∠BDG=∠BFD,∵∠DBF=∠GBD,∴△BDF∽△BGD;②由△BDF∽△BGD,得=,∵D为AB的中点,∴BD=AD,又∵BM∥AC,∴∠DBF=∠DAE,∠BFD=∠DEA,在△BFD和△AED中,∵,∴△BFD≌△AED(AAS),∴BF=AE=x,∴=,∴BG=,在Rt△ABC中,AB=8,AC=4,根据勾股定理得:BC==4,∵点D到直线BM的距离d=BC=2,∴S△DFG=FG•d=(BG﹣BF)•d,即y=×(﹣x)×2=﹣x(0<x<4);(3)(i)当点G在点F的右侧时,由题意,得6=﹣x,整理,得x2+6x﹣16=0,解得x1=2,x2=﹣8(不合题意,舍去);(ii)当点G在点F的左侧时,如图3所示:同理得到S△DFG=FG•d=(BF﹣BG)•d,即y=x﹣(x>4),由题意,得6=x﹣,整理,得x2﹣6x﹣16=0,解得x3=8,x4=﹣2(不合题意,舍去),综上所述,AE的值为2或8.。

上海民办兰生复旦中学初中数学九年级下期中复习题(专题培优)

上海民办兰生复旦中学初中数学九年级下期中复习题(专题培优)

一、选择题1.(0分)[ID:11121]如图,以O为圆心,任意长为半径画弧,与射线OM交于点A,再以A为圆心,AO长为半径画弧,两弧交于点B,画射线OB.则cos∠AOB的值等于()A.√33B.12C.√22D.√322.(0分)[ID:11118]已知线段a、b,求作线段x,使22bxa,正确的作法是()A.B.C.D.3.(0分)[ID:11107]如图,平面直角坐标系中,点A是x轴上任意一点,BC平行于x轴,分别交y=3x(x>0)、y=kx(x<0)的图象于B、C两点,若△ABC的面积为2,则k值为()A .﹣1B .1C .12-D .12 4.(0分)[ID :11101]下列判断中,不正确的有( )A .三边对应成比例的两个三角形相似B .两边对应成比例,且有一个角相等的两个三角形相似C .斜边与一条直角边对应成比例的两个直角三角形相似D .有一个角是100°的两个等腰三角形相似5.(0分)[ID :11100]若37a b =,则b a a -等于( ) A .34 B .43 C .73 D .376.(0分)[ID :11091]已知两个相似三角形的面积比为 4:9,则周长的比为 ( ) A .2:3B .4:9C .3:2D .2:37.(0分)[ID :11088]如图,在正方形ABCD 中,N 为边AD 上一点,连接BN .过点A 作AP ⊥BN 于点P ,连接CP ,M 为边AB 上一点,连接PM ,∠PMA =∠PCB ,连接CM ,有以下结论:①△PAM ∽△PBC ;②PM ⊥PC ;③M 、P 、C 、B 四点共圆;④AN =AM .其中正确的个数为( )A .4B .3C .2D .18.(0分)[ID :11085]如图,过反比例函数的图像上一点A 作AB ⊥轴于点B ,连接AO ,若S △AOB =2,则的值为( )A .2B .3C .4D .59.(0分)[ID :11069]如图所示,在平行四边形ABCD 中,AC 与BD 相交于点O ,E 为OD 的中点,连接AE 并延长交DC 于点F ,则DF :FC=( )A.1:3 B.1:4 C.2:3 D.1:210.(0分)[ID:11065]已知线段a、b、c、d满足ab=cd,把它改写成比例式,错误的是()A.a:d=c:b B.a:b=c:d C.c:a=d:b D.b:c=a:d 11.(0分)[ID:11064]如图,△ABC中AB两个顶点在x轴的上方,点C的坐标是(﹣1,0),以点C为位似中心,在x轴的下方作△ABC的位似图形△A′B′C′,且△A′B′C′与△ABC的位似比为2:1.设点B的对应点B′的横坐标是a,则点B的横坐标是()A.12a-B.1(1)2a-+C.1(1)2a--D.1(3)2a-+12.(0分)[ID:11060]在平面直角坐标系中,将点(2,l)向右平移3个单位长度,则所得的点的坐标是()A.(0,5)B.(5,1)C.(2,4)D.(4,2)13.(0分)[ID:11055]若反比例函数2yx=-的图象上有两个不同的点关于y轴的对称点都在一次函数y=-x+m的图象上,则m的取值范围是()A.22m>B.-22m<C.22-22m m>或<D.-2222m<<14.(0分)[ID:11033]给出下列函数:①y=﹣3x+2;②y=3x;③y=2x2;④y=3x,上述函数中符合条作“当x>1时,函数值y随自变量x增大而增大“的是()A.①③B.③④C.②④D.②③15.(0分)[ID:11059]如图,在△ABC中,AC=8,∠ABC=60°,∠C=45°,AD⊥BC,垂足为D,∠ABC的平分线交AD于点E,则AE的长为A .423B .22C .823D .32二、填空题16.(0分)[ID :11230]如图,在△ABC 中,CD 、BE 分别是△ABC 的边AB 、AC 上的中线,则DF EF BF CF++=________。

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2019-2020学年上海市杨浦区民办兰生复旦中学九年级(上)期中化学试卷一、选择题(共计24分)1.(1分)下列变化属于化学变化的是()A.树根“变”根雕B.玉石“变”印章C.水果“变”果汁D.葡萄“变”美酒2.(1分)常州博物馆启用了“真空充氮杀虫灭菌消毒机”来处理和保护文物。

即将文物置于该机器内,三天后氮气浓度可达99.99%;再密闭三天左右,好氧菌、厌氧菌和丝状霉菌都被杀灭。

下列有关氮气说法错误的是()A.氮气还能用于灯泡填充气B.通常情况下氮气的化学性质很活泼C.氮气不能供给呼吸D.高浓度氮气可抑制菌类的生长3.(1分)上海市垃圾分类实行的“四分类”标准,废荧光灯管属于()A.B.C.D.4.(1分)以下变化中,氧元素由化合态全部变为游离态都是()A.分离液态空气制取氧气B.铁丝燃烧C.电解水D.双氧水制取氧气5.(1分)根据分析证明:健康人的头发每克约含铁130mg、锌167~172mg、铝5mg、硼7mg等。

这里的铁、锌、铝、硼是指()A.分子B.原子C.元素D.单质6.(1分)物质的俗名、化学式一致的是()A.水:H2O2B.熟石灰:CaOC.胆矾:CuSO4•5H2O D.水银:Hg7.(1分)下列选项符合图示从属关系的是()A B C DX金属溶液纯净物化合反应Y单质乳浊液化合物氧化反应A.A B.B C.C D.D8.(1分)下列有关2molH2O2的解释正确的是()A.含有2molO2B.与质量为36克水含有相同数量的氢原子C.该物质的摩尔质量为68g/molD.共含有约1.204×1024个氧原子9.(1分)下列物质放入水中能形成无色溶液的是()A.高锰酸钾B.硝酸钾C.碳酸钙D.胆矾10.(1分)计算一定质量的纯净物所含的微粒个数,下列量没有用处的是()A.微粒大小B.微粒的质量C.阿伏伽德罗常数D.摩尔质量11.(1分)氮化硅(Si3N4)是一种高温陶瓷材料,硬度大、熔点高、化学性质稳定。

有电子、军事和核工业等方面有着广泛的应用,若Si3N4中Si显+4价,则下列物质中的N的化合价与氮化硅中N的化合价相同的是()A.NH3B.N2C.N2O3D.HNO312.(1分)氮化硅(Si3N4)是一种高温陶瓷材料,硬度大、熔点高、化学性质稳定。

有电子、军事和核工业等方面有着广泛的应用,下列关于氮化硅(Si3N4)的说法正确的是()A.氯化硅是由三个硅原子与四个氧原子构成B.1mol氮化硅的质量与10mol氮气质量相等C.氮化硅中硅元素与氮元素的质量比为3:2D.70克氮化硅中含有硅元素30克13.(1分)氮化硅(Si3N4)是一种高温陶瓷材料,硬度大、熔点高、化学性质稳定。

有电子、军事和核工业等方面有着广泛的应用,某生成氮化硅(Si 3N 4)反应的化学方程为:3SiCl 4+4NH 3═Si 3N 4+12HCl ,下列说法不正确的是( )A .该反应既不是化合反应,也不是分解反应B .该反应中NH 3与HCl 物质的量之比为1:3C .3克四氯化硅与4克氨气可以恰好完全反应D .反应后生成物分子总个数比反应前增加了14.(1分)推理是化学学习中常用的思维方法,下列推理正确的是( ) A .纯净物中往往只含有一种分子,则由同种分子构成的物质是纯净物B .混合物中至少含有两种物质,则混合物中至少含有两种元素C .化学变化伴随有能量变化,则有能量变化的变化一定是化学变化D .均一稳定的混合物是溶液,水均一稳定,则水属于溶液 15.(1分)实验设计不合理的是( )A .检查装置气密性B .证明MnO 2的催化作用C .探究同种物质在不同溶剂中的溶解性D .探究空气中氧气的体积分数A .AB .BC .CD .D16.(1分)溶解度曲线是溶解度表示方法之一。

曲线上任意一点表示的是( ) A .溶液达到饱和时溶解的溶质的质量B .一定温度和一定质量的溶剂里溶解的溶质的质量C .该温度时,100 g 溶剂里溶解的溶质的质量D .该温度时,溶液处于饱和状态17.(1分)加热氯酸钾和高锰酸钾混合物一段时间后,试管内剩余固体种类不可能是( ) A .5种B .4种C .3种D .2种18.(1分)下列实验现象的记录,正确的是()A.过氧化氢溶液和二氧化锰制氧气时产生大量气体,黑色固体逐渐消失B.碳酸钙和稀盐酸反应时产生大量气泡,白色固体逐渐消失C.向硫酸铜溶液中逐渐滴加氢氧化钠溶液,白色沉淀逐渐增加D.向一定量的生石灰中加水,放出热量,产生大量白烟19.(1分)将8.0g铜与碳的混合物在氧气中充分灼烧后,冷却,称量,发现反应后的固体质量仍为8.0g,则原混合物中铜的物质的量为()A.0.01mol B.0.02mol C.0.05mol D.0.1mol20.(1分)下列实验基本操作中正确的是()A.吸取液体B.倾倒液体C.过滤D.实验室制氧气21.(1分)打开一瓶盐汽水,有大量二氧化碳气体逸出,相关分析正确的是()A.盐汽水中只有二氧化碳一种溶质,逸出后剩余汽水中不含二氧化碳B.打开汽水瓶,因为温度升高,二氧化碳溶解度变小导致气体逸出C.打开汽水瓶盖,因为瓶内气压减小,导致气体逸出,形成当时条件下二氧化碳的不饱和溶液D.该盐汽水打开瞬间,是当时条件下二氧化碳的饱和溶液,氯化钠的不饱和溶液22.(1分)在密闭容器中有甲、乙、丙、丁四种物质,在一定条作下反应,测得反应前及反应过程中的两个时刻各物质的质量分数如表所示。

表中a、b、c、d分别表示相应物质的质量分数,下列数据正确的是()甲乙丙丁反应前70%14%6%10%反应中a7%18%b反应后54%c30%dA.a=56%B.b=10%C.c=0D.d=10%23.(1分)40℃时,甲、乙物质饱和溶液降温至20℃时,对此过程判断一定正确的是()A.现象:有析出晶体,甲溶液析出固体质量>乙溶液析出固体质量B.溶解度:溶解度都变小,20℃时S甲=S乙C.溶液状态:都为饱和溶液,甲溶质的质量分数>乙溶质的质量分数D.溶剂变化:溶剂的质量不变,甲溶液中溶剂质量>乙溶液中溶剂质量24.(1分)常温下,往盛放适量M物质的烧杯中逐渐加入N物质并充分搅拌。

如图横坐标x表示N物质的质量,纵坐标y表示烧杯中的某物理量(见表)。

下列实验与图象对应关系合理的是()M N(x)yA水氧化钙溶质的质量B水氢氧化钠溶液的温度C饱和石灰水氧化钙溶液中溶质的质量D饱和硫酸铜溶液胆矾溶液的质量A.A B.B C.C D.D二、填空题(47分)25.(5分)利用所学化学知识完成下列填空。

写出下列操作所用的主要仪器或用品的名称:①取少量二氧化锰倒入试管中,②夹持点燃的镁条,③量取5.6mL水,④盛放稀硫酸的试剂瓶,⑤粗盐提纯实验中每一步都要用到的玻璃仪器。

26.(4分)用化学用语表示下列物质或微粒:①60个碳原子,②n个氮分子,③硝酸,④碳酸氢铵中氢元素显正一价。

27.(4分)在学过的反应中选择合适的,写出符合下列条件的化学方程式:①银白色固体反应后变成白色固体:。

②银白色固体反应后变成黑色固体:。

③液体颜色由紫色变成红色:。

④液体颜色由红色变为无色:。

28.(3分)用A代替“大于”,B代替“小于”,C代替“等于”,D代替“无法判断”,请分别选用“A”、“B”、“C”或“D”填空。

(1)沸点:在同温同压下,氧气氮气;(2)微粒直径大小:分子原子;(3)50mL酒精与150mL水充分混合:溶液的体积200mL;(4)碳元素的质量分数:葡萄糖(C6H12O6)醋酸(C2H4O2)。

29.(12分)如图所示为电解水的实验装置图,请你根据图回答下列问题:(1)图1中与电源负极相连的b管处产生的现象是:。

检验a管处产生的气体方法是。

该实验得出结论:水是由组成的。

(2)根据图2,可以了解有关该变化过程的一个微观信息是。

(3)理论上其中一个电极产生5mL气体时,另一个电极产生的气体体积可能是,该实验所得正极气体与负极气体的质量比是,说说你求出两种气体质量比的依据是。

(4)由于氧气在水中的溶解性比氢气略大,所以电解刚开始时,正极收集到气体的体积与负极气体之比略1:2(填大于、小于、等于),但是随着反应进行,收集到的正极气体与负极气体的体积比越来越接近1:2,可能是原因是和。

(5)水是重要的自然资源,与生活生产关系密切,生活中常用的直饮水,生产中采用“活性炭+超滤膜+紫外线”组合工艺,其中活性炭的作用是,紫外线相当于自来水生产中氯气或二氧化氯的作用,其作用是。

(6)下列物质中,即不是溶液,也不是乳浊液是的。

(填序号)A.医用酒精B.油水混合物C.石灰乳D.纯净水30.(12分)如图是a、b、c三种物质(均不含结晶水)的溶解度曲线,请回答下列问题。

(1)图P点的含义为。

(2)t2℃时,23克c的饱和溶液中含有3克c,则此时c的溶解度是。

(3)t2℃时,将等质量的a、b、c三种物质的饱和溶液分别降温到t1℃时,所得溶液中各项关系符合b>a>c的是。

(填序号)A.溶解度B.溶剂质量C.溶液质量D.溶质的质量分数(4)三种物质的饱和溶液的溶质质量分数关系为b>a≥c时的温度为t,则t的取值范围是。

(5)t1℃时,a、b、c溶液的溶质质量分数相等均为x%,其中一定是不饱和溶液的是,x%的取值范围是。

(6)t2℃时,向两只分别盛有20克a和20克c固体的烧杯中,各加入100g水,充分溶解后,能形成不饱和溶液的物质是,其不饱和溶液的质量为克。

(7)采用一种操作方法,将上述(6)中某个烧杯内的剩余固体全部溶解,变为不饱和溶液,下列说法正确的是。

(填写序号)A.可降低温度到t1℃B.可以恒温加溶剂50克C.以恒温蒸发10克水D.不断升高温度(8)若需由一份150克a物质与12克c物质形成的混合物中提纯a,应采用结晶法,具体操作是向混合物中加入克t3℃的热水,使样品完全溶解后,再蒸发去克水后冷却到t1℃时过滤,此时得到最多的纯净物a物质。

31.(1分)向如图所示的烧杯中逐滴加水,各种量的变化所加水的质量变化正确的是。

(填序号)32.(2分)向装有一定量饱和石灰水的烧杯中加入少量生石灰,请在如图绘制溶液质量与溶液中溶质质量分数随时间变化的曲线:、,33.(1分)如图是A物质(不含结晶水)的溶解度曲线,N点表示t3℃时A物质的(填“饱和”或“不饱和”)溶液。

34.(3分)如图甲、乙、丙、丁是M、N两点代表的溶液相互转化的途径路线图。

(用→表示转化方向)。

其中,采用“先将N点溶液降温至M点对应的温度后,然后再加入”,这一措施来实现甲图中N→M转化的途径,请在乙、丙、丁中任寻一个,写出实现M→N转化的措施。

选择序号:,对应措施。

三、实验题(本题共29分)35.(10分)某同学用如图1装置对质量守恒定律进行验证,反应前称出装置连药品总质量,然后将气球内大量大理石倒入锥形瓶中与稀盐酸充分反应后,再称量:(1)观察到的现象有,天平示数无明显改变,粗略验证了质量守恒定律。

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