三校 8A(8A U5-7)第一学期第三次月度联考答卷纸

合集下载

第一学期高三期中三校联考试卷

第一学期高三期中三校联考试卷

20XX年中学测试中学试题试卷科目:年级:考点:监考老师:日期:20XX学年第一学期高三期中三校联考试卷考试时间90分钟,卷面满分为100分一、选择题(本题共10小题,每小题4分,共40分。

在每小题给出的四个选项中,有的小题只有一个选项正确,有的小题有多个选项正确。

全部选对的得4分,选的不全得2分,有选错的或不答的得0分。

)1、20XX年我国自行研制的“枭龙”战机在四川某地试飞成功。

假设该战机起飞前从静止开始做匀加速直线运动,达到起飞速度v所需时间为t,则起飞前的运动距离为……………………()A. vt B. vt/2 C. 2vt D. 不能确定2、如图,一个体重650N的人,把重为50N的物体挂在棒的一端,棒放在肩膀上,另一端用手抓住使棒处于平衡状态,这时手向下的拉力是100N。

若不计棒的重力,则这时人对地面的压力为()A.650N B.700NC.550ND.750N3、如图所示,一车西瓜随汽车一起沿水平路面向右做匀加速直线运动,加速度大小为a,车中质量为m的西瓜(该西瓜不与车接触)受到其它西瓜的作用力的合力大小为……………()A、mgB、maC、22gam+D、)(agm+4、已知甲、乙两物体的质量相同,图是甲、乙两运动物体的s-t 图像,由图像可以知道…………………()A.甲做变速运动,乙做匀速运动B.两物体的初速度都为零C.在t1时间内两物体的平均速度相等D. 相遇时,两物体的动量相同5、起重机用钢绳吊起一重物,竖直向上做匀加速直线运动。

若不计空气的阻力,则钢绳的拉力对重物所做的功()A.等于重物增加的机械能 B.等于重物增加的动能C .大于重物增加的机械能D .大于重物增加的动能6、狗拉雪橇沿位于水平面内的圆弧形轨道路匀速行驶,图为四个关于雪橇受到的牵引力F 及摩擦力f 的示意图(O 为圆心),其中正确的是( )7、一艘宇宙飞船在预定轨道上做匀速圆周运动,在该飞船的密封舱内,下列实验能够进行的是…………………………( )8、图为一列横波在传播过程中两个时刻的波形图,实线表示t 1=0时刻的波形,虚线是t 2=1.5s 时刻的波形,且t 2-t 1小于一个周期,则下列说法中正确的是……( ) A .波的振幅是10cm B .波长一定等于40cm C .波一定沿x 轴正方向传播D .波的周期可能是2s ,也可能是6s 。

最新江苏省三校2022-2022年八年级上第三次月考英语试卷含答案2

最新江苏省三校2022-2022年八年级上第三次月考英语试卷含答案2

上学期阶段(jiēduàn)测试初二英语试题第I卷(选择题)一、单项选择(25分)1.This is____ orange and that is____ map.A. a; anB. an; aC. a; aD. an; an2.He had , but he didn’t write .A. enough time;careful enoughB.enough time; carefully enoughC.enough time;enough carefulD.time enough;enough carefully3.—Happy birthday to you, Mary!— .A. Thanks a lotB. The same to youC. You’re welcomeD. It’s kind of you4.-- do the children take art lessons ? --Three or four times a week.A. How longB. How many timesC. How soonD.How often5.There will be less ________.A.trees B.people C.pollution D.cars6.Betty swims ______ than I, But I doesn’t swim ______ Jim.A. badly, as good asB. bader, as well asC. worse, so better asD. worse, as well as7.—Which do you like ____ , Chinese , Math or English?—English.A.bestB. betterC. wellD. Good8.If we keep planting trees, our country will become ______.A. more and more beautifullyB. more or more beautifullyC. more or more beautifulD. more and more beautiful9.―The best time ______ Yangzhou is in spring.―Yes, I think so.A. visitsB. to visitC. visitD. visiting10.__________ important for children __________ by themselves.A. That’s; to learnB. It’s; to learnC. That’s; learningD. It’s; learning11.– Is the math problem ______?-- Yes, I can work it out ______.A. easy; easilyB. easy ; easyC. easily ; easyD. easily ; easily 12.There_______a sports meeting in our school next week.A.is going to beB.is going to haveC.is going to hasD.is going to do 13.----I think Mr Li is one of ______ in our school.----Me too. He’s friendly and helpful.A. the most popular teachersB. the popular teachersC. the most popular teacherD. most popular teachers14.I think there will be _____ people and ______pollution.A. less, fewerB. less, moreC. fewer, less D.fewer, fewer 15.It is ___________ cold today and there is ___________rain.A. too much; too muchB. much too; much tooC. too much; much tooD. much too; too much16.— Can you come to my birthday party?—________.A. I can’t go thereB. Yes, I willC. No, I can’t there.D. Sure, I’d like to 17.-----What should we do to make the salad?-------First ,wash the vegetables and _______.A. cut them upB. cut up themC. cut it upD. cut up it18._____ bananas and_____ yogurt do you need for your banana milk shake. A. How many; how many B. How many; how muchC. How much; how muchD. How much; how many19.___ weather it is! Let’s go for a picnic!A .What fine B. What a bad C. How bad D. How a fine 20.—Shanghai is really a beautiful city and there are many places of interest.—So it is. Why not _______ here for a long time?A. stay B to stay C. staying D. stays21.Li Ming enjoys _____ English every morning.He is good ____English.A.to read; at B.to read; for C.reading; for D.reading; at22.He can play tennis better than ___________ in the class.A.any boys B.any other boy C.any boy D.any other 23.The box is ________ heavy _________ I can’t carry it.A.too; toB. enough; toC. as ; asD. so ; that24.My mother says my friend is similar ______me,but l think she is different ______me.A. to:fromB. as:fromC. to:toD. as:to25.— are you going to do that?— I am going to take acting lessons.A. WhenB. HowC. WhereD. What二、完形填空(10分)Most children like watching TV very much. And 26 children like watching cartoons(卡通片) every day. But watching TV too much is bad 27 them.Why do children like watching TV? 28 they think it is interesting and easyto watch TV. They can hear the voices(声音(shēngyīn)) and 29 the moving pictures at the same time. They can see and learn a lot about 30 countries. They also learn newer and better ways of doing something. They may find the 31 is becoming smaller and smaller.Many children watch TV only 32 Saturday and Sunday evenings. Because they are very 33 with their lessons on weekdays. But some children watch TV every night. So they go to bed too 34 , and they can’t have a good rest. Watching TV too 35 makes them very tired. It is bad for their health and eyes.26.A. both B. all C. no D.neither27.A. for B. at C. With D.in28.A. If B. Because C. So D.but29.A. listen B. taste C. watch D.smell30.A. they B. them C. their D. we31.A. school B. City C. world D.shop32.A. on B. At C. in D.of33.A. happy B. angry C. busy D.free34.A. new B. early C. late D.old35.A. little B. much C. many D.lot三、阅读理解(40分)ATom, Jean, Jack and Lucy are talking about the movie theaters in the city. My name’s Tom. I go to the Movie Palace twice a month. It’s about twokilometers from my home. The service is good. But the price of a ticket I’m Jean. I go to Moon Cinema more often. There are often new movies onshow and the screens are big. Most importantly, the price is cheaper. I’m Jack. I live about 5 kilometers from Moon Cinema and 8 kilometersfrom Movie Palace. I think the seats in Moon Cinema are hard and the I’m Lucy. I prefer to go to Movie Palace. It’s in the center of thecity and is very modern. I like shopping and there are some shopping36.Who go to a cinema once a month?A. Tom and JeanB. Jean and JackC. Jack and LucyD. Lucy and Tom. 37.How much is a full-price ticket in Moon Cinema?A. 50 yuan.B. 100 yuan.C. 30 yuan.D. 60 yuan.38.Why does Tom never go to Moon Cinema?A. The ticket is expensive.B. The seats are hard.C. The screens are big.D. It’s far from his home.39.Jack’s home is ___kilometers farther from Movie Palace than Tom’s.A. 3B. 5C. 6D. 1040.Which of the following is NOT true?A. Tom thinks the service in Movie Palace is good.B. Jean and Jack prefer Movie Palace to Moon Cinema.C. Lucy likes Movie Palace because she can go shopping.D. Movie Palace is in the middle of the city.BI like to get up late, so my ideal(理想(lǐxiǎng)的)school starts at 9: 00 a. m. It finishes at 3: 00 p. m. So we can do lots of sports after school.In my ideal school, there is a big dining hall. We have an hour for lunch. We can listen to music in the hall. We can have maths every day because I think maths is very interesting. The classes are very small. There are 15 students in each class. We can have a big library with a lot of useful books. We can also have a swimming pool.After school, we only have half an hour of homework every day. Every month, we can go on a school trip to a museum(博物馆)or a theater.41. How many hours for lessons are there in my ideal school?A. Three hours.B. Four hours.C. Five hours.D. Six hours.42.The underlined word “useful” means in Chinese.A. 有好处(hǎo chu)的B. 感兴趣的C. 有用(yǒu yònɡ)的D. 有趣(yǒuqù)的43.Why do we have maths every day?A. Because it is interesting.B. Because it is very important.C. Because it is very useful.D. Because I will be a maths teacher.44.How many students are there in my class in my ideal school?A. Sixteen.B. FifteenC. Fourteen.D. Thirteen.45.What’s the best title(标题(biāotí))for the passage?A. My schoolB. My ideal subjectC. My school dayD. My ideal schoolCHappiness is for everyone. You don’t need to care about those who have beautiful houses with large gardens and swimming pools or those who have nice cars and lots of money and so on. Why? Because those who have big houses may often feel lonely and those who have cars may want to walk on the country roads学校_____________ 班级____________ 姓名____________ 考号__________--------------------------------------装---------------------------------------订----------------------------------------线---------------------------------- in their free time.In fact, happiness is always around you ,When you are in trouble at school, your friends will help you; when you study hard at your lessons, your parents are always taking good care of your life and your health; when you get success, your friends will say congratulations(祝贺(zhùhè)) to you; when you do something good to others, you will feel happy, too. All these are your happiness. If you notice them, you can see that happiness is always around you. Happiness is not the same as money. It is a feeling of your heart. When you are poor,you can also say you are very happy, because you have something else that can’t be bought with money. When you meet with difficulties, you can say loudly you are very happy, because you have more chances to challenge yourself. As the saying goes, life is like a revolving(旋转(xuánzhuǎn)) door. When it closes, it also opens. If you take every chance you get, you can be a happy and lucky person. 46.Happiness is for____.A. those who have large and beautiful houses.B. those who have lots of money.C. all people.D. some people47.When you are in trouble at school,________.A. your friends will help you.B. your classmates will laugh at youC. you will be happyD. you’ll cry.48.Which is TRUE according to the passage?A. When you get success, your friends will be very proud of youB. You can get help from others when you make mistakes.C. Happiness is not the same as moneyD. All of the above.49.We say “Happiness is not the same as money.”, because_______A. money always brings happinessB. money doesn’t always bring happiness.C. everything can be bought with money.D. money is very important.50.Which is the best title of the passage?A. HappinessB. Happy and LuckyC. Life and Success.D. Money and happinessDEarly to bed and early to rise makes a man healthy, wealthy and wise.This is an old English saying (谚语(yànyǔ)). It means that we should go to bed early at night and get up early in the morning. If we do, we shall be healthy. We shall also be rich and clever.Is this true? Perhaps it is. The body must have enough sleep. Children of yourage need ten hours’ sleep every night. If you go to bed late, you can’t have enough sleep. The n you can’t think carefully and your homework will be wrong. You will not be wise and you may not become wealthy!Some people go to bed late at night and get up late in the morning. This is not good for them. We should sleep at night when it is dark. The dark helps us sleep well. When the daylight comes, we should get up. This is the time for exercise. If the body is not used, it will become weak. Exercise keeps it strong.Exercise helps the blood (血液(xuèyè)) to move around inside the body. Blood takes nutrition (营养) to all parts of our bodies. The brains (大脑) in our heads also need blood. We think with our brains. If we keep our bodies healthy, and take exercise, we can think better!Our bodies also need air to breathe (呼吸(hūxī)). Without air we will die. G et up early in the morning and we can have plenty of clean, fresh air. That will keep us healthy and happy.根据(gēnjù)短文内容,选择正确答案。

高三第一学期第三次月考英语考试(2020-2021学年度)

高三第一学期第三次月考英语考试(2020-2021学年度)

高三第一学期第三次月考英语考试(2020-2021学年度)第I卷(三部分,共105分)第一部分:听力理解(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。

每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1.How did Charles travel through Australia?A.By bus. B.By car. C.By train.2.Where is the man speaker now?A.In a hotel. B.In his home. C.In a restaurant.3.What do we know about the man?A.He wants to get a new position.B.He is asking the woman for help.C.He enjoys letter writing.4.Who is probably the man speaker?A.A lawyer. B.A driver. C.A policeman.5.What was Mary probably doing when the conversation took place?A.Having supper out with her classmate.B.Doing homework with her classmate.C.Attending a party at a classmate’s home.第二节(共15小题;每题1.5分,共22.5分)听下面5段对话或独白。

每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。

2021年高三级第一学期第三次月考答案

2021年高三级第一学期第三次月考答案

第一部分阅读理解(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)1-----15 BADA BADC ACD DCDA第二节(共5小题;每小题2分,满分10分)16. AD 17. D 18. B 19. A 20.AC第二部分:语言知识运用(共两节,满分75分)第一节完形填空(共40小题;每小题1.5分,满分60分)21----40 CADBC ADBCC ABDCA BDCAB41-----60 ABADC DBACD ABCCB DCADA第二节(共10小题;每小题1.5分,满分15分)61.operation 62.better 63.the 64.have herard 65.which 66.with 67.one 68.How69.chatting 70.so that/in order that第三部分写作(共两节,满分35分)第一节短文改错:(共10小题;每小题1分,满分10分)1. invite改inviting2. your 改my3.of去掉4. but前加and5. subject改subjects6. had改have7. Poorly改poor8.chance后加to 9. (in order) to改that 10. that改which第二节书面表达: (满分25分)Dear Tom,I’m Li Hua, a student of a middle school. I learned quite by cha nce thatyou needed a book to improve your Chinese. I happen to have one, which I thinkmight help you.The book is just intended for beginners at your level. Not only does it include the basic conversations in our daily life, it is also a window through which you can get to know Chinese culture and customs. Edited by three language experts, it is widely used by many foreign learners.If it is convenient to you, let’s meet at the entrance to Jiangnan Park at 3 p.m. this Sunday. If not, try to find another time that is suitable for both of us.Yours,Li Hua30391 76B7 皷37619 92F3 鋳25926 6546 敆K]21014 5216 刖o'Le36642 8F22 輢36726 8F76 轶524055 5DF7 巷21099 526B 剫。

七年级英语第三次月度联考试卷及答案

七年级英语第三次月度联考试卷及答案

三校2014-2015学年度第一学期第三次月度联考七年级英语试题(考试时间:120分钟满分:150分)第一部分选择题(90分)一、听力(共25小题; 每小题l分,计25分)第一部分听对话,回答问题。

本部分共有10道小题, 每小题你将听到一段对话, 每段对话听两遍。

在听每段对话前, 你将有5秒钟的时间阅读题目;听完后, 你还有5秒钟的时间选出你认为最合适的备选答案。

1. What does the boy usually do after class?A. B. C.2.Which does the girl like best?A. B. C.3.Where does the boy often go?A. B. C.4. Which festival are they talking about?A. B. C.B)听对话,选择最佳答案,听两遍5. What's the girl's favourite food?A. EggsB. CakesC. Hamburgers6. How often does the girl play football?A. OftenB. NeverC. Seldom7. Where are they talking ?A. At schoolB. In a bookshopC. In a clothes shop8. How much does the girl spend on the CDs?A. 5 yuanB. 4 yuanC. 20 yuan9.Who calls 110 ?A. The manB. The womanC. The crying girl10.Who is the letter from?A. It's from America.B. It's from Millie.C. It's from Li Lei.第二部分听对话或短文,回答问题。

【八年级】2021 2021学年八年级语文上册第三次联考试卷(新人教带答案)

【八年级】2021 2021学年八年级语文上册第三次联考试卷(新人教带答案)

【八年级】2021 2021学年八年级语文上册第三次联考试卷(新人教带答案)【八年级】2021-2021学年八年级语文上册第三次联考试卷(新人教带答案)2022-2022学年八年级中文第一卷第三卷联考试卷(新人带答案授课)一、书写(5分包含卷面分)请抄写下面的句子。

书写要正确、规范、整洁。

智者不惑,仁者不忧。

二、积累和应用(25分)1.请根据拼音写出相应的汉字(5分)纯胚Gō瓶身牡丹就像你的第一套衣服/兰然檀香透过窗户为我担心李ǎ或án()/写在宣纸上,然后把它放在一半/釉色XuáNRǎn(n)女士身材魅力被隐藏/而你yān的微笑正在萌芽/你的美丽分散/去我不能去的地方(方文山青花瓷)2.古诗文名句填空(9分)(1)一只鹤在晴朗的天空,在云端。

秋词(2)________________,千山高复低。

《鲁山山行》(3)挥手吧。

送朋友(4)陶渊明在《桃花源诗》中有嬴氏乱天纪,贤者避其世的诗句,《桃花源记》中的_______________,_____________一句对此作了印证。

(5) ___________________________。

(6)《春望》中移情于物的句子是:_______________,________________。

3.根据上下文和所给单词的意思,在下列句子的横线处选择并填写汉字。

其中一个错误是()(3分)a.中国烹饪,无比神秘,难以复制。

从深山到闹市,厨艺的传授,仍然(尊、遵)循口耳相传、心领神会的传统方式。

“尊”有“敬重、推崇”的意思,“遵”有“沿用、依照”的意思,横线处应填“遵”。

b、贬谪永州的柳宗元写了著名的《永州八记》,以抒发自己的忧郁症,在闲暇(闲暇与遐想)中游山玩水,在风景中抒发感情。

“休闲”意味着“休闲”,“遐想”意味着“长期”。

水平线应该充满“幻想”。

c.随着我国经济发展,国力强盛,不少流失海外的珍贵文物有机会完(璧、壁)归赵,重新回到祖国的怀抱。

八校高三3月试题含解析试题

八校高三3月试题含解析试题

卜人入州八九几市潮王学校师范大学附属、高级、高新一中、铁一、西工大附中等八校2021届高三语文3月联考试题〔含解析〕本套试卷分第一卷(阅读题)和第二卷(表达题两局部)一共150分。

考试时间是是150分钟。

本卷须知:2.选择题答案使需要用2B铅笔填涂,如需改动,用橡皮擦干净后再选涂其它答案标号;非选择题用0.5毫米的黑色中性(签字)笔或者碳素笔书写,字体工整,笔迹清楚。

3.请按照题号在各题的答题区域(黑色线框)内答题,超出答题区域书写之答案无效。

4.保持纸面清洁,不折叠,不破损。

5.做选考题时,考生按照题目要求答题并需要用2B铅笔在答题纸上把所选题目对应的题号涂黑。

第一卷阅读题一、现代文阅读(36分)(一)阐述类文本阅读(此题一共3小题,9分)阅读下面的文字,完成各题。

历史上潮到1200多年前,中国唐朝诗人杜甫的一首很著名的诗句“HY亦有限,列国自有疆。

苟能制侵陵,岂在多杀伤〞深化反映了中国人的HY文化,中国HY事的防御思想正是这种HY文化的详细表现。

中国人为什么会有这样的HY文化呢首先,源于中国农耕民族强烈的中土意识。

历史上,中国是一个典型的农业社会,农业文化天然具有“保守性〞。

眷恋故土、安土重迁成为古代中国人的普遍心态。

此外,自给自足的消费方式,使中国人在物质生活上无须外求。

这些反映在HY事上就形成了固土自守,以德怀远的HY防御思想,对外侵略战争在古代中国不具备其原始驱动力。

矗立千年、横亘于中国北疆的万里长城,既是抵御北方游牧民族入侵的HY事屏障,也是中国传统HY防御思想的物化和缩影。

与此相对,游牧民族和海洋民族以放牧和贸易为生,大范围的迁徙和流动成为其生存所必需的主要方式,战争成为其获取生活必需品和争夺海上贸易份额的主要手段,侵略和征服在其文化传承中被视为荣耀之举。

不同的消费和生活方式,产生了不同的HY文化。

概括来说,中国的HY事理论是内向型、防御性的。

更注重HY事谋略的运用:从HY层面上讲,谋略主要是强调防患于未然,避难于无形。

八年级数学上册第三次月考试卷(12月份)含答案A3版

八年级数学上册第三次月考试卷(12月份)含答案A3版

2019-2020学年八年级(上)月考数学试卷(12月份)副标题得分1.下列几何图形一定是轴对称图形的是()A. 三角形B. 梯形C. 等腰三角形D. 直角三角形2.下列长度的三根小木棒能构成三角形的是()A. 2cm,3cm,5cmB. 7cm,4cm,2cmC. 3cm,4cm,8cmD. 3cm,3cm,4cm3.下列运算正确的是()A. 2a+a=3a2B. (−2a)3=−8a3C. (a2)3÷a5=1D. 3a3⋅2a2=6a64.下列各式计算正确的是()A. (x+y)2=x 2+y2B. (x−5)(x+6)=x2−30C. (−x+1)(−x−1)=x2−1 D. (x−12y)2=x 2−xy+12y25.如图,小明从O点出发,前进6米后向右转20°,再前进6米后又向右转20°,…,这样一直走下去,他第一次回到出发点O时一共走了()A. 72米B. 108米C. 144米D. 120米6.等腰三角形一腰上的高与另一腰的夹角为45°,则等腰三角形的底角为()A. 67°B. 67.5°C. 22.5°D. 67.5°或22.5°7.下列命题中正确的有()①已知任意一边和一个锐角对应相等的两个直角三角形全等.②任意两角和一边对应相等的两个三角形全等.③已知任意两边和一角对应相等的两个三角形全等.④已知腰和顶角对应相等的两个等腰三角形全等.⑤如果两个三角形有两条边及其中一边上的中线分别相等,那么这两个三角形全等.A. 1个B. 2个C. 3个D. 4个8.如图,P是等边三角形ABC内一点,∠APB,∠BPC,∠CPA的大小之比为5:6:7,则以PA,PB,PC为边的三角形三内角大小之比(从小到大)是()A. 2:3:4B. 3:4:5C. 4:5:6D.以上结果都不对9.如图是由8个全等的长方形组成的大正方形,线段AB的端点都在小矩形的顶点上,如果点P是某个小矩形的顶点,连接PA、PB,那么使△ABP为等腰三角形的点P的个数是()A. 3个B. 4个C. 5个D. 6个10.如图,DB=DC,∠BAC=∠BDC=120°,DM⊥AC,E为BA延长线上的点,∠BAC的角平分线交BC于N,∠ABC的外角平分线交CA的延长线于点P,连接PN交AB于K,连接CK,则下列结论正确的是()①∠ABD=∠ACD;②DA平分∠EAC;③当点A在DB左侧运动时,AC+ABAM为定值;④∠CKN=30°A. ①③④B. ②③④C. ①②④D. ①②③11.若一个n边形的外角和与它的内角和之和为1800°,则边数n=______.12.若x2−2(m−3)x+16是完全平方式,则m的值是______.13.已知(x2+px+8)(x2−3x+q)展开后不含x2与x3的项,则q p=______.14.如图,△ABC中,AB=AC,点D为BC上一点,以AD为腰作等腰△ADE,且AD=AE,∠BAC=∠DAE=30°,连接CE,若BD=2,S△DCE=√3,则CD的长为______.15.如图,△ABC中,∠BAC=90°,AB=AC,D是△ABC内一点,∠DAC=∠DCA=15°,则∠BDA=______.16.△ABC中,∠B=80°,∠BAC=40°,D为BC上一点,若DA平分∠BAC,BD=2,BC=5,则AB=______.17.(1)计算:(−2a4)2⋅(−a3)÷(−a2)3(2)若x2+x−2019=0,求(2x+3)(2x−3)−x(5x+4)−(x−1)218.如图,点E、F在BC上,BE=CF,EG=GF,∠B=∠C,AF与DE交于点G,求证:AB=DC.19.阅读下列文字:我们知道对于一个图形,通过不同的方法计算图形的面积,可以得到一个数学等式,例如由图1可以得到(a+2b)(a+b)=a2+3ab+2b2.请解答下列问题:(1)写出图2中所表示的数学等式______;(2)利用(1)中所得到的结论,解决下面的问题:已知a+b+c=9,ab+bc+ac=29,求a2+b2+c2的值;(3)小明同学打算用x张边长为a和y张边长为b的小正方形,z张相邻两边长分别为a、b的长方形纸片拼出了一个面积为(3a+5b)(4a+7b)的长方形,那么他总共需要多少张纸片?20.如图,在△ABC中,AB=2,BC=4,其两条外角平分线AD、CD交于点D,且∠ADC=45°,连接BD交AC于点P,过点P作PE⊥AC交BC于点F,交AB的延长线于点E.(1)求证:∠ABC=90°.(2)求S△PFC:S△PBF的值.21.如图,在直角坐标系中,△ABC三个顶点的坐标分别是A(1,1),B(4,2),C(3,4).(1)请画出△ABC关于y轴对称的△A1B1C1;(2)△A1B1C1的面积为______.(3)在x轴上求作一点P,使△PAB周长最小,请画出△PAB,并直接写出点P的坐标.22.已知:BF为△ABC的外角∠ABE的平分线,D为BF上一点,且AD=CD.(1)如图1,过点D作DH⊥CE于点H,若AB=8,BC=6,求BH的长.(2)如图2,若∠ABC=24°,∠ABD=78°,∠BAD=60°,求∠BAC的度数.23.(1)如图1,等腰Rt△ABC中,∠CAB=90°,点H在BC边上,连AH,作等腰Rt△HFA,∠HFA=90°.求证:AF=CF.(2)如图2,等腰Rt△ABC中,∠CAB=90°,D在BC上,AD⊥AE,AD=AE,G为CD中点,求证:AG⊥BE.(3)如图3,等腰Rt△ABC中,∠BAC=90°,过C作CD//AB,CD=8,连AD,在AD上取一点E使AE=AB,连BE交AC于F,若AF=9,则AD=______.24.已知,在直角坐标系中,A(−a,0),B(b,0),C(0,c),且满足b=√a −c+√c−a+2(1)如图1,过B作BD⊥AC,交y轴于M,垂足为D,求M点的坐标.(2)如图2,若a=3√2,AC=6,点P为线段AC上一点,D为x轴负半轴上一点,且PD=PO,∠DPO=45°,求点D的坐标.(3)如图3,M在OC上,E在AC上,满足∠CME=∠OMA,EF⊥AM交AO于G,垂足为F,试猜想线段OG,OM,CM三者之间的数量关系,并给出证明.答案和解析1.【答案】C【解析】解:A、不一定是轴对称图形,故此选项错误;B、不一定是轴对称图形,故此选项错误;C、一定是轴对称图形,故此选项正确;D、不一定是轴对称图形,故此选项错误.故选:C.如果一个图形沿一条直线折叠,直线两旁的部分能够互相重合,这个图形叫做轴对称图形,根据轴对称图形的概念求解.本题考查了轴对称图形的概念,轴对称图形的关键是寻找对称轴,图形两部分沿对称轴折叠后可重合.2.【答案】D【解析】解:A、因为2+3=5,所以不能构成三角形,故A错误;B、因为2+4<7,所以不能构成三角形,故B错误;C、因为3+4<8,所以不能构成三角形,故C错误;D、因为3+3>4,所以能构成三角形,故D正确.故选:D.依据三角形任意两边之和大于第三边求解即可.本题主要考查的是三角形的三边关系,掌握三角形的三边关系是解题的关键.3.【答案】B【解析】解:A、2a+a=3a,故本选项不符合题意.B、(−2a)3=−8a3,故本选项符合题意.C、(a2)3÷a5=a,故本选项不符合题意.D、3a3⋅2a2=6a5,故本选项不符合题意,故选:B.根据合并同类项法则、幂的乘方和积的乘方,完全平方公式,同底数幂的乘法求出每个式子的值,再判断即可.本题考查了合并同类项法则、幂的乘方和积的乘方,完全平方公式,同底数幂的乘法等知识点,能求出每个式子的值是解此题的关键.4.【答案】C【解析】解:A、(x+y)2=x 2+2xy+y2,故此选项错误;B、(x−5)(x+6)=x2+x−30,故此选项错误;C、(−x+1)(−x−1)=x2−1,正确;D、(x−12y)2=x 2−xy+14y2,故此选项错误;故选:C.直接利用乘法公式进而分别计算得出答案.此题主要考查了整式的混合运算,正确运用乘法公式是解题关键.5.【答案】B【解析】解:依题意可知,小陈所走路径为正多边形,设这个正多边形的边数为n,则20n=360,解得n=18,∴他第一次回到出发点O时一共走了:6×18=108(米),故选:B.利用多边形外角和等于360度即可求出答案.本题考查了多边形的外角和,正多边形的判定与性质.关键是根据每一个外角判断多边形的边数.6.【答案】D【解析】解:有两种情况;(1)如图当△ABC是锐角三角形时,BD⊥AC于D,则∠ADB=90°,已知∠ABD=45°,∴∠A=90°−45°=45°,∵AB=AC,∴∠ABC=∠C=12×(180°−45°)=67.5°;(2)如图,当△EFG是钝角三角形时,FH⊥EG于H,则∠FHE=90°,已知∠HFE=45°,∴∠HEF=90°−45°=45°,∴∠FEG=180°−45°=135°,∵EF=EG,∴∠EFG=∠G =12×(180°−135°)=22.5°,综合(1)(2)得:等腰三角形的底角是67.5°或22.5°.故选:D.先知三角形有两种情况(1)(2),求出每种情况的顶角的度数,再利用等边对等角的性质(两底角相等)和三角形的内角和定理,即可求出底角的度数.本题考查了三角形有关高问题有两种情况的理解和掌握,能否利用三角形的内角和定理和等腰三角形的性质,知三角形的一个角能否求其它两角.7.【答案】D【解析】解:已知任意一边和一个锐角对应相等的两个直角三角形全等,所以①正确;任意两角和一边对应相等的两个三角形全等,所以②正确;已知任意两边和它们的夹角对应相等的两个三角形全等,所以③错误;已知腰和顶角对应相等的两个等腰三角形全等,所以④正确;如果两个三角形有两条边及其中一边上的中线分别相等,那么这两个三角形全等,所以⑤正确.故选:D.利用三角形全等的判定方法对①②③进行判断;利用等腰三角形的性质和三角形全等的判定方法对④进行判断;利用三角形中线的定义和三角形全等的判定方法对⑤进行判断.本题考查了命题与定理:命题的“真”“假”是就命题的内容而言.任何一个命题非真即假.要说明一个命题的正确性,一般需要推理、论证,而判断一个命题是假命题,只需举出一个反例即可.也考查了全等三角形的判定.8.【答案】A【解析】解:如图,将△APB绕A点逆时针旋转60°得△AP′C,显然有△AP′C≌△APB,连PP′,∵AP′=AP,∠P′AP=60°,∴△AP′P是等边三角形,∴PP′=AP,∵P′C=PB,∴△P′CP的三边长分别为PA,PB,PC,∵∠APB+∠BPC+∠CPA=360°,∠APB:∠BPC:∠CPA=5:6:7,∴∠APB=100°,∠BPC=120°,∠CPA=140°,∴∠PP′C=∠AP′C−∠AP′P=∠APB−∠AP′P=100°−60°=40°,∠P′PC=∠APC−∠APP′=140°−60°=80°,∠PCP′=180°−(40°+80°)=60°,∴∠PP′C:∠PCP′:∠P′PC=2:3:4.故选A.将△APB绕A点逆时针旋转60°得△AP′C,显然有△AP′C≌△APB,连PP′,则AP′=AP,∠P′AP=60°,得到△AP′P是等边三角形,PP′=AP,所以△P′CP的三边长分别为PA,PB,PC;再由∠APB+∠BPC+∠CPA= 360°,∠APB:∠BPC:∠CPA=5:6:7,得到∠APB=100°,∠BPC=120°,∠CPA=140°,这样可分别求出∠PP′C=∠AP′C−∠AP′P=∠APB−∠AP′P=100°−60°=40°,∠P′PC=∠APC−∠APP′=140°−60°=80°,∠PCP′=180°−(40°+80°)=60°,即可得到答案.本题考查了旋转的性质:旋转前后的两个图形全等,对应点与旋转中心的连线段的夹角等于旋转角,对应点到旋转中心的距离相等.也考查了等边三角形的性质.9.【答案】D【解析】解:如图所示,使△ABP为等腰直角三角形的点P的个数是6,故选:D.根据等腰直角三角形的判定即可得到结论.本题考查了等腰直角三角形的判定,正确的找出符合条件的点P是解题的关键.10.【答案】C【解析】解:如图,∵∠BAC=∠BDC=120°,∴A,B,C,D四点共圆,DB=DC,作四边形ABCD的外接圆⊙O,∴∠ABD=∠ACD,故①正确,作DN⊥AE于N.∵DM⊥AC,∴∠DMC=∠DNB=90°,∵∠DCM=∠DBN,DC=DB,∴△DMC≌△DNB(AAS),∴DM=DN,BN=CM,∵DN⊥AE,DM⊥AC,∴DA平分∠EAC,故②正确,∵∠DNA=∠DMA=90°,AD=AD,DN=DM,∴△ADN≌△ADM(HL),∴AN=AM,∴AC+AB=BN−AN+AM+CM=2CM,∴AC+ABAM =2CMAM≠定值,故③错误,作KG⊥AP于G,KH⊥AN于H,延长AN,在AN上取一点J,使得KJ=KC.∵∠BAC=120°,AN平分∠BAC,∴∠PAB=∠BAN=60°,∴KG=KH,∵∠KGC=∠KHJ=90°,KJ=KC,KH=KG,∴Rt△KHJ≌Rt△KGC(HL),∴∠HKJ=∠GKC,∴∠CKJ=∠KGH=∠AKG+∠AHK=30°+30°=60°,∵KJ=KC,∴△KJC是等边三角形,∴∠KCJ=∠KJC=∠CKJ=60°,作PT⊥JA交JA的延长线于T,PR⊥CB于R,PW⊥AB于W,KL⊥BC于L.∵BP平分∠ABR,PA平分∠TAB,∴PE=PW,PW=PT,∴PR=PT,∵PR⊥NR,PT⊥NT,∴PN平分∠RNT,∵KH⊥NT,KL⊥NR,∴KL=KH,∵KH=KG,∴KL=KG,∵KL⊥CL,KG⊥CG,∴∠KCG=∠KCL=∠NJK,∵∠KCJ=∠KJC,∴∠NCJ=∠NJC,∴NC=NJ,∵KN=KN,AC=KJ,∴△KNC≌△KNJ(SSS),∴∠NKC=∠NKJ=30°,故④正确.故选:C.①正确.利用圆周角定理证明即可.②正确,构造全等三角形解决问题即可.③错误,作DN⊥AE于N.证明△ADN≌△ADM(HL),推出AN=AM,推出AC+AB=BN−AN+AM+ CM=2CM,推出AC+ABAM=2CMAM≠定值.④正确.作KG⊥AP于G,KH⊥AN于H,延长AN,在AN上取一点J,使得KJ=KC.作PT⊥JA交JA的延长线于T,PR⊥CB于R,PW⊥AB于W,KL⊥BC于L.想办法证明△KCJ是等边三角形,证明△KNC≌△KNJ(SSS)即可解决问题.本题属于三角形综合题,考查了圆周角定理,角平分线的性质定理,全等三角形的判定和性质,等边三角形的判定和性质等知识,解题的关键是学会添加辅助线,构造全等三角形解决问题,属于中考选择题中的压轴题.11.【答案】10【解析】解:由题意得(n−2)⋅180°+360°=1800°,解得n=10.故答案为:10根据n边形的内角和可以表示成(n−2)⋅180°,外角和为360°,根据题意列方程求解.本题考查了多边形的内角和公式与外角和定理,多边形的外角和与边数无关,任何多边形的外角和都是360°.12.【答案】7或−1【解析】【解答】解:∵x2−2(m−3)x+16是完全平方式,∴−(m−3)=±4,解得:m=7或m=−1,故答案为:7或−1【分析】此题考查了完全平方式,熟练掌握完全平方公式是解本题的关键.利用完全平方公式的结构特征判断即可确定出m的值.13.【答案】1【解析】解:(x2+px+8)(x2−3x+q)=x4−3x3+qx2+px3−3px2+pqx+8x2−24x+8q=x4+(p−3)x3+(−3p+q+8)x2+(pq−24)x+8q,∵(x2+px+8)(x2−3x+q)展开后不含x2与x3的项,∴p−3=0,−3p+q+8=0,解得p=3,q=1,∴q p=13=1.故答案为:1.根据多项式乘多项式的法则把式子展开,找到所有x2与x3的所有系数,令其为0,可求出p,q的值,再代入计算即可求解.本题主要考查了多项式乘多项式的运算,注意当要求多项式中不含有哪一项时,应让这一项的系数为0.14.【答案】2√3【解析】解:∵∠BAC=∠DAE ,∴∠BAC−∠DAC=∠DAE−∠DAC,∴∠BAD=∠EAC,在△ABD和△ACE中,{AB=AC∠BAD=∠EAC AD=AE,∴△ABD≌△ACE(SAS),∴BD=CE=2;过D作DF⊥EC交EC的延长线于F,∵△ABD≌△ACE,∴∠ACE=∠B,∵∠BAC=30°,∴∠B+∠ACB=150°,∴∠BCE=∠ACB+∠ACE=150°,∴∠DCF=30°,∴DF=12CD,∵△DCE的面积为√3,∴12DF⋅CE=12×2DF=√3,∴DF=√3,∴CD=2DF=2√3,故答案为:2√3.根据全等三角形的性质得到BD=CE=2;过D作DF⊥EC交EC的延长线于F,求得∠DCF=30°,根据直角三角形的性质得到DF=12CD,根据三角形的面积求得DF=√3,于是得到结论.本题考查等腰三角形的性质以及三角形全等的判定和性质,灵活运用相关的判定定理和性质定理是解题的关键.15.【答案】75°【解析】解:如图:以AD为边,在△ADB中作等边三角形ADE,连接BE.∵∠BAE=90°−60°−15°=15°,即∠BAE=∠CAD,在△AEB和△ADC中,∵{AE=AD∠BAE=∠CADAB=AC,∴△EAB≌△DAC(SAS),∴∠BEA=∠CDA=180°−15°−15°=150°,∴∠BED=360°−∠BEA−60°=150°,即∠BEA=∠BED;在△AEB和△DEB中∵{AE=ED∠AEB=∠DEBBE=BE∴△BEA≌△BED(SAS),∴BA=BD,∠DBE=∠ABE=15°∴∠ABD=30°∴∠BDA =180°−30°2=75°故答案为:75°.先作辅助线,以AD为边,在△ADB中作等边三角形ADE,连接BE.可证得△EAB≌△DAC,再证得△BEA≌△BED,即可得结论.本题主要考查了三角形全等的判定和性质,涉及到等边三角形的性质、三角形内角和定理、周角的定义等知识点,正确作出辅助线是解题的关键.16.【答案】7【解析】解:延长CB至G,使CG=AC,连接AG,过A作AH⊥BC于H,∵∠ABC=80°,∠BAC=40°,∴∠C=60°,∴△ACG为等边三角形,∴∠CAG=60°,∵AD平分∠BAC,∴∠BAD=∠CAD=20°,∴∠ADB=∠C+∠DAC=60°+20°=80°,∵∠ABC=80°,∴∠ABC=∠ADB,∴AB=AD,∴DH=BH=12BD=1,∵BC=5,∴CH=5−1=4,∵△ACG是等边三角形,AH⊥CG,∴∠CAH=30°,∴AC=2CH=8,AH=4√3,由勾股定理得:AB=√BH2+AH2=√12+(4√3)2=7,故答案为:7.作辅助线,构建等边三角形,先证明△ACG为等边三角形,得∠CAG=60°,根据三角形内角和定理可得∠C= 60°,由角平分线的定理和三角形外角的性质得:∠ABC=∠ADB,所以AB=AD,由等腰三角形三线合一的性质得DH=BH=1,由直角三角形30度角的性质和勾股定理可得结论.本题主要考查了等边三角形的判定与性质,等腰三角形的判定,勾股定理等知识,正确作辅助线是本题的关键.17.【答案】解:(1)(−2a4)2⋅(−a3)÷(−a2)3=4a8⋅(−a3)÷(−a6)=−4a11÷(−a6)=4a5;(2)∵x2+x−2019=0,∴x2+x=2019,(2x+3)(2x−3)−x(5x+4)−(x−1)2=4x2−9−5x2−4x−(x2−2x+1)=4x2−9−5x2−4x−x2+2x−1=−2x2−2x−10=−2(x2+x)−10=−2×2019−10=−4048.【解析】(1)直接利用积的乘方运算法则以及整式的乘除运算法则计算得出答案;(2)直接利用乘法公式化简,再把已知代入求出答案.此题主要考查了整式的混合运算,正确掌握相关运算法则是解题关键.18.【答案】证明:∵BE=CF,∴BE+EF=CF+EF,即BF=CE,∵EG=GF,∴∠DEF=∠AFB,在△ABF和△DCE中,∵{∠B=∠CBF=CE∠AFB=∠DEC,∴△ABF≌△DCE(ASA),∴AB=DC(全等三角形对应边相等).【解析】根据BE=CF推出BF=CE,用EG=GF推出∠AFB=∠DEC,然后利用“角边角”证明△ABF和△DCE全等,根据全等三角形对应边相等即可证明.本题考查了全等三角形的判定与性质,根据BE=CF推出BF=CE,从而得到三角形全等的条件是解题的关键.19.【答案】(a+b+c)2=a2+b2+c2+2ab+2ac+2bc【解析】解:(1)根据阅读材料,观察图2中所表示的数学等式:(a+b+c)2=a2+b2+c2+2ab+2ac+2bc故答案为:(a+b+c)2=a2+b2+c2+2ab+2ac+2bc(2)∵(a+b+c)2=a2+b2+c2+2ab+2ac+2bc∴(a+b+c)2−2(ab+ac+bc)=a2+b2+c2∴a2+b2+c2=81−58=23答:a2+b2+c2的值为23.(3)∵(3a+5b)(4a+7b)=12a2+41ab+35b212+41+35=88答:总共需要88张纸片.(1)根据阅读材料即可写出数学等式;(2)根据(1)中所得到的结论,代入求值即可;(3)根据多项式乘以多项式,再根据(1)的思想,即可得出结论.本题考查了因式分解的应用、完全平方公式的几何背景,解决本题的关键是利用数形结合思想.20.【答案】解:(1)设∠BAC=α,∠BCA=β,∵AD、CD为△ABC外角平分线,∴∠DAC=12(180°−∠BAC)=90°−12α∠DCA=12(180°−∠BCA)=90°−12β∵∠DAC+∠DCA+∠ADC=180°即90°−12α+90°−12β+45°=180°∴α+β=90°∴∠ABC=180°−(α+β)=90°.(2)如图所示:过点D作DN⊥AB于点N,DM⊥BC于点M,DH⊥AC于点H,∵AD平分∠NAC,CD平分∠ACM,∴DN=DH,DH=DM,∴DN=DM,∴BD平分∠ABC,∵∠ABC=90°,∴∠PBC=45°,过点P作PG⊥BD交BC于点G,如图,∴∠PBG=∠PGB=45°,∴PB=PG,∵∠PCG+∠BAC=90°,∠E+∠BAC=90°,∴∠PCG=∠E,∵PE⊥AC,∴∠CPG+∠GPF=90°,∵∠EPB+∠GPF=90°,∴∠CPG=∠EPB,∴△PGC≌△PBE(AAS)∴PE=PC,∵∠PCF=∠E,∠CPF=∠EPA=90°,∴△PCF≌△PEA(ASA),∴CF=AE,设BF=x,则CF=AE=4−x,BE=AE−AB=4−x−2=2−x,∵∠ACB=∠E,∠ABC=∠FBE=90°,∴△ABC∽△FBE,∴BFBE =ABBC=12,即x2−x =12,解得x=23,∴CF=4−23=103∴S△PFCS△PBF=CFBF=23×310=15即S△PFC:S△PBF的值为15.【解析】(1)设∠BAC=α,∠BCA=β,然后分别表示出∠DAC和∠DCA,利用三角形内角和可求出α+β=90°,即可得证;(2)由角平分线的性质可得BD平分∠ABC,过点P作PG⊥BD交BC于点G,证明△PBE≌△PGC,然后证明△PCF≌△PEA,可得CF=AE,设BF=x,则CF=AE=4−x,可得BE=2−x,由三角形相似得BF 与BE的比例关系可解得x,得到BF与FC的比例关系即为面积比.本题考查了三角形全等、角平分线的性质、相似三角,形的判定与性质、三角形的面积,解决本题的关键是熟练应用以上知识.21.【答案】3.5【解析】解:(1)如图所示,△A1B1C1即为所求;(2)△A1B1C1的面积为:3×3−12×1×2−12×1×3−12×2×3=9−1−1.5−3=3.5;故答案为:3.5.(3)如图所示,△PAB即为所求,点P的坐标为(2,0).(1)依据轴对称的性质进行作图,即可得到△A1B1C1;(2)依据割补法进行计算,即可得到△A1B1C1的面积;(3)作点A关于x轴的对称点A′,连接A′B,交x轴于点P,则△PAB周长最小.本题考查了作图−轴对称变换:几何图形都可看做是由点组成,我们在画一个图形的轴对称图形时,也是先从确定一些特殊的对称点开始的;凡是涉及最短距离的问题,一般要考虑线段的性质定理,结合轴对称变换来解决,多数情况要作点关于某直线的对称点.22.【答案】(1)解:作DG⊥AB于G,∵BF为△ABC的外角∠ABE的平分线,DH⊥CE于点H,∴DG=DH,在Rt△BDG和Rt△BDH中{DG=DHBD=BD∴Rt△BDG≌Rt△BDH(HL),∴BG=BH,在Rt△ADG和Rt△CDH中{AD=DCDG=DH∴Rt△ADG≌Rt△CDH(HL),∴AG=CH,∴AB−BH=BC+BH,∵AB=8,BC=6,∴BH=1;(2)解:作DG⊥AB于G,DH⊥CE于点H,同理:Rt△ADG≌Rt△CDH,∴∠BAD=∠BCD=60°,∵∠ADO+∠AOD+∠DAO=180°,∠OBC+∠BOC+∠BCO=180°,∠AOD=∠BOC,∴∠ADO=∠CBO,即∠ADC=∠ABC=24°.∵AD=DC,∴∠DAC=∠ACD=12(180°−∠ADC)=78°,∴∠ACB=∠ACD+∠BCD=78°+60°=138°,∴∠BAC=180°−∠ACB−∠ABC=180°−138°−24°=18°.【解析】(1)作DG⊥AB于G,根据角平分线的性质,得出DG=DH,进而证得Rt△BDG≌Rt△BDH,得到BG=BH,证得Rt△ADG≌Rt△CDH,得到AG=CH,即可得到AB−BH=BC+BH,求得BH=1;(2)根据等腰三角形的性质,三角形内角和定理以及对顶角相等,即可解决问题.本题考查三角形综合题、等腰三角形的判定和性质、全等三角形的判定和性质.角平分线的性质定理等知识,解题的关键是学会添加常用辅助线构造全等三角形解决问题,属于中考压轴题.23.【答案】17【解析】证明:(1)如图1,过点F作FD//AB交AC于点D,交BC于点G,在CD上取ED=FD,∴△EFD为等腰直角三角形,∠FEA=∠HGF=45°,又∵∠HFA+∠DFA=∠DFA+∠FAE=90°,∴∠HFG=∠EAF,又∵∠FEA=∠HGF=45°,FA=FH,∴△FHG≌△FAE(AAS),∴FG=EA,∴FG=DG+FD=CD+ED,EA=ED+DA,∴CD=DA .即FD垂直平分CA,∴FA=FC.(2)如图2,延长AG至T,使AG=GT,∵CG=GD,∠AGC=∠TGD,∴△ACG≌△TDG(SAS),∴∠ACG=∠GDF=45°,∴∠ADT=∠TDC+∠CDA=∠TDC+∠DAB+45°=90°+∠DAB=∠EAB,又∵AD=AE,DT=AC=AB,∴△ADT≌△EAB(SAS),∴∠TAD=∠AEB,∴∠EAT+∠AEB=∠EAB+∠TAD=90°,∴AG⊥BE;(3)解:如图3,作EM⊥AD交AC延长于点M,∵∠EDC=∠EMC,AB=AC=AE,∠AEM=∠ACD,∴△AEM≌△ACD(AAS),∴ME=CD=8,∵∠ABE=∠AEB,∠AFB=∠EFM,∠ABF+∠AFB=90°=∠AEF+∠FEM,∴∠EFM=∠FEM,∴FM=ME,∴AD=AM=AF+FM=AF+EM=AF+CD=8+9=17.故答案为:17.(1)过点F作FD//AB交AC于点D,交BC于点G,在CD上取ED=FD,证得∠HFG=∠EAF,根据AAS 证明△FHG≌△FAE,得出FG=EA,则CD=DA,结论得证;(2)延长AG至T,使AG=GT,根据SAS可证明△ACG≌△TDG,可得∠ACG=∠GDF=45°,证得∠ADT=∠EAB,证明△ADT≌△EAB,可得∠TAD=∠AEB,则∠EAT+∠AEB=∠EAB+∠TAD=90°;(3)作EM⊥AD交AC延长于点M,证明△AEM≌△ACD,可得ME=CD=8,证得∠EFM=∠FEM,则FM= ME,则答案可求出.本题属于三角形综合题,考查了等腰直角三角形的判定和性质,全等三角形的判定和性质,等腰三角形的性质等知识,解题的关键是熟练掌握基本知识,正确作出辅助线.24.【答案】解:(1)∵a−c≥0,c−a≥0,∴a=c,∴△AOC为等腰直角三角形,∠ACO=45°,∵BD⊥AC,∴∠CMD=∠BMO=45°,∴OM=OB=2,∴M(0,2).(2)∵∠APD+∠CPO=180°−∠DPO=135°,∠CPO+∠COP=180°−∠ACO=135°,∴∠APD=∠COP,∵∠PCO=∠PAD=45°,PD=PO,∴△PDA≌△OPC(AAS),∴PA=OC=3√2,DA =PC=6−3√2,∴OD=OA−DA=3√2−(6−3√2)=6√2−6.∴D(6−6√2,0).(3)OM=CM−OG.证明如下:作ON//EG,CH//AO,且CH与GE的延长线交于点H,∵EF⊥AM,∴ON⊥AM,∴∠NCO=∠AOM=90°,∴∠NCO=∠MAO,∵OA=OC,∴△AOM≌△OCN(ASA),∴OM=CN,∠CNO=∠AMO,∵∠ACO=45°,∴∠HCE=45°,∵∠CME=∠OMA,∵ON//GH,∴∠CNO=∠H,∴∠H=∠CME,∵CM=CM,∴△CME≌△CEH(ASA),∴CH=CM,∴OM=CN=CH−HN=CM−OG.【解析】(1)由二次根式的性质得出a=c,根据等腰直角三角形的性质求出∠CMD和∠BMO的度数,则OM= OB=2;(2)根据等腰直角三角形的性质得到∠APD=∠COP,证明△PDA≌△OPC,根据全等三角形的性质得到OC= PA,根据坐标与图形性质可得到答案;(3)作ON//EG,CH//AO,CH与GE的延长线交于点H,可得出∠NCO=∠MAO,根据ASA可证明△AOM≌△OCN,得到OM=CN,证明△CME≌△CEH,得到CH=CM,即可得出结论.本题是三角形综合题,考查了二次根式的性质、全等三角形的判定和性质、等腰直角三角形的性质、坐标与图形性质,掌握全等三角形的判定定理和性质定理是解题的关键.。

三校2021-2022年七年级上第三次月考英语试卷含答案

三校2021-2022年七年级上第三次月考英语试卷含答案

2021--2021学年(xuénián)上学期阶段性测试初一英语试题(shìtí)第一卷选择题(共70分)I. 单项选择(xuǎnzé)(30分,每题1分)1. —What can you see in the box ?—Well, I see ___ID card.A. theB. aC. anD. /2. — This is my little sister. _______ name is Jenny, we all like_______.A. She; sheB. Her; herC. She; herD. Her; she3. —A pencil and a book __on the desk.A. isB. areC. amD. be4. —Is this your pencil? ——Yes, it’s___.A. IB. meC. myD. mine5. —____is he? ——He’s my English teacher.A. HowB. WhatC. WhereD. Who6. These arephotos ____my family.A. atB. ofC. inD. on7. —Where's my pencil box?—I ______ know.A.aren't B.don't C.isn't D.be not8. —______ is my skirt?—It's behind the door.A.Where B.What C.Whose D.Which9. This is my friend, John Black. Black is her_____.A. nameB. first nameC. last nameD. two names10. There are many pictures ______ the wall.A.from B.on C.about D.in11. That’s an______ question.A. interestingB.boringC.difficultD.relaxing12.Do you______ TV every day?A. lookB.seeC.watchD.read13.I have a sister, a brother ________ a cousin.A. orB. forC. forD. and14.—Does Jim like basketball or baseball?—________________________/A. Yes, he doesB. No, He doesn’tC. He likes basketballD. He doesn’t like baseball15. —Let’s ______ computer games. —That sounds ______. I like it.A. play; more goodB. plays; difficultC. play; funD. plays; interesting16. We play sports ______.A. everydayB. every dayC. one dayD. a day17. They have bananas _______ lunch.A. atB. forC. inD. on18. There is some_______on the table.A. tomatoB.chickenC.eggD.bananas19.—What do you and your cousin ______?—I______a baseball and my cousin______two volleyballs.A.have;has;have B.have;have;have C.has;have;has D.have;have;has 20.Mary______ salad.A.doesn't like B.don't like C.doesn't likes D.like21.— _______ is this T-shirt? — It’s seven dollars.A. WhatB. HowC. How muchD. How many22.— ____________? — Yes, please. I want a bag.A. Can you help meB. Can I help youC. How are youD. What do you want to buy23.The students come to see their teacher _______ September 10th.A. atB. inC. onD. of34.—Are those _______ bikes?— Yes, they are.A. Jim and TomB. Jim’s and Tom’sC. Jim and Tom’sD. Jim’s and Tom25.The skirts are very cheap(便宜(biànyí)的). I’ll ________ them.A. sellB. bringC. takeD. eat26.— _______ is his birthday? — It’s on December 19th.A. HowB. WhatC. WhenD. Where27.— ______ your sister _______ a birthday party every year?— Yes, she does.A. Does,hasB. Does,haveC. Do,haveD. Do,has28.We have black and white hats _______ $15.A. forB. atC. inD. on29.You look beautiful _______ this pink dress.A. forB. onC. inD. with30. There are _____ months in a year. The ____ month is December.A. twelve,twelveB. twelfth,twelfthC. twelve,twelfthD. twelfth,twelveII.完形填空。

广东省重点学校2023-2024届学年高三上学期第三次月考英语试卷(含答案)

广东省重点学校2023-2024届学年高三上学期第三次月考英语试卷(含答案)

深圳外国语学校2023—2024学年度第一学期高三年级第三次月考英语试卷本试卷分选择题和非选择题两部分,共9页,满分120分,考试用时120分钟。

注意事项:1.答题前,考生务必用黑色字迹的钢笔或签字笔将自己的姓名、班级、座位号等相关信息填写在答题卷指定区域内。

2.选择题每小题选出答案后,用2B铅笔把答题卷上对应题目的答案标号涂黑;如需改动,用橡皮擦干净后,再选涂其它答案;不能答在试卷上。

3.非选择题必须用黑色字迹的钢笔或签字笔作答,答案必须写在答题卷各题目指定区域内的相应位置上;如需改动,先划掉原来的答案,然后再写上新的答案;不准使用铅笔和涂改液。

不按以上要求作答的答案无效。

4.考生必须保持答题卷的整洁。

第二部分阅读(共两节,满分50分)第一节(共15小题;每小题2. 5分,满分37. 5分)阅读下列短文,从每题所给的A、B、C、D四个选项中选出最佳选项。

ABy our very rough calculations, Reader’s Digest has publish ed 35,000 articles in nearly 1200 issues. These small pages have held some very big names, including U.S. presidents, world leaders,sports legends, and, indeed, the biggest contributors of all, everyday Americans with a story to tell. These are some of our proudest moments.How to Keep Young Mentally by Mary B. MullettThis first article in the first issue highlighted inventor Alexander Graham Bell and his belief in lifelong learning: “The first essential of any real education is to observe. Observe! Remem ber! Compare!” It was an appropriate beginning, reflecting our self-educated founder’s endless curiosity.I’ve Come to Clean Your Shoes by Madge HarrahThe morning of a family funeral, an acquaintance shows up unannounced and says, “I’ve come to clean your shoes.” He spends the day quietly shining every pair in the house. The writer ends with: “Now, whenever I hear of an acquaintance who has lost a loved one, I try to think of one specific task suiting that person’s need. And if the person says, ‘How did yo u know I needed that done ’ I reply, ‘it’s because a man once cleaned my shoes. ’ ”Strange Encounter on Coho Creek by Morris Homer ErwinA miner spends days camping in the Alaskan wilderness, working hard to earn the trust of a mother wolf stuck in a trap before she and her four pups starve to death. Eventually, he is able to free her. Four years later, he encounters a wolf in the same meadow. Yes,the same wolf. When we shared this classic on in 2019, it went viral and it has now been read by many millions online and in print. How Honest Are We by Ralph Kinney BennettOur famous “wallet drop” set up the ultimate test of honesty: If we left wallets in cities around the country, how many would be returned Well over half, it turned out — 67 percent. The most honest city Seattle, whose upstanding residents returned nine out of ten wallets.21. What do we know about the acquaintance in the story I’ve Come to Clean Your ShoesA. He shows concern by cleaning shoes.B. He cleans shoes at the owner’s request.C. He comforts the owner by telling his own story.D. He visits the house where the owner is severely sick.22. Which article gained significant popularity onlineA. How to Keep Young Mentally.B. I’ve Come to Clean Your Shoes.C. Strange Encounter on Coho Creek.D. How Honest Are We.23. Where can the text be foundA. In a history book.B. In a magazine.C. In a novel.D. In a brochure. BI settled into my seat on a plane bound for Cuba feeling frustrated. When I planned the trip, I had assumed that my Cuban partnersand I would go to the field directly to collect samples from rivers. That’ s how I’ d done fieldwork in Namibia and Bolivia. But not in Cuba, it seemed. Five days earlier, a Cuban scientist had emailed to inform me that we would only be meeting to talk about our planned project. Sampling would happen during a later trip.At the airport, one of my partners greeted me. We drove to the research center where he worked, and then toured every lab in the building. I met scientists, technicians, students and even the cook. I was impressed that I was introduced to each person. The lack of hierarchy (等级制度) was unlike anything I had experienced before in academia (学术界).The next day, we met again to brainstorm. Together, we looked at maps to plan how we were going to collect samples. Had it not been for the Cubans, I would have been unaware that the maps I had left out some new reservoirs (水库). Local involvement and knowledge were key —making me wonder what I’ d missed working without such a team in other places.Six months later, I flew back to Cuba and this time, we headed to the field. I was impressed again by the lengths to which my Cuban partners went to ensure that all team members were treated equally. We drove around Cuba in bright yellow minibuses, and each minibus had a mix of members at all seniority levels. In thefield, all members sweated together.On the last night of the trip, we searched for a restaurant that could seat all 14 of us at one table. When a restaurant couldn’t accommodate the team without separating us, my partners insisted that we move on and find a place with a large enough table.In 26 years as a professor, I have never been a fan of academia’s hierarchy. I want everyone working with me to feel as though they are part of a team. But my Cuban partners take teamwork to another level entirely. They make it clear that all team members are valued, that everyone is equal, and that true team work makes for better science.24. Why did the author feel frustrated in paragraph 1A. He was asked to host a meeting in Cuba.B. He was told to change his planned project.C. He couldn’t do his work in his usual way in Cuba.D. He spent a long time waiting for his plane to Cuba.25. What was wrong with the author’s mapA. It was torn up.B. It was outdated.C. It was too old to read.D. It was about another water area.26. What can prove that the Cuban team didn’t have hierarchyA. They drove minibuses to the field.B. They were friendly to the author.C. They worked in different groups.D. They had dinner at one table.27. Which of the following can be a suitable title for the textA. A fruitful trip in CubaB. Impressed by a Cuban teamC. Respect seniors in the teamD. You’ll never know until you tryCThe conventional wisdom about insects has been that they are unthinking, unfeeling creatures whose behavior is entirely hardwired (天生的). But in the 1990s researchers began making surprising discoveries about insect minds. Some species of wasps (黄蜂) recognize their nest mates’ faces and acquire impressive social skills. For example, they can infer the fighting strengths of other wasps relative to their own just by watching other wasps fight among themselves.Given the substantial work on the complexity of insect cognition (认知) , it might seem surprising that it took scientists so long to ask whether, if they are that smart, could also be sentient, capable of feeling. Since we have no direct window into the inner world of an animal that cannot verbally communicate its thoughts and feelings, the question of whether insects are sentient remained academic.15 years ago, I performed an experiment in which we asked whether bumblebees could learn about threat from their naturalenemies. We built a plastic spider model with a mechanism that would briefly trap a bumblebee between two sponges before releasing it. The bumblebees showed a significant change in their behavior after being attacked by the robotic spider. Perhaps unsurprisingly, they learned to avoid flowers with spiders and meticulously scanned every flower before landing. Curiously, however, they sometimes even fled from imaginary threats, scanning and then abandoning a perfectly safe, spider-free flower. Although this incidental observation did not constitute formal evidence of an emotion-like state, it did open the door to the idea that such states might exist in insects.Some research suggested that insects might have positive states of mind. Researchers discovered that bees actively seek out drugs such as nicotine and caffeine when given the choice and even treat themselves with nicotine when sick. Male fruit flies stressed by being robbed of mating opportunities prefer food containing alcohol, and bees even show withdrawal symptoms when removed from an alcohol-rich diet.Why would insects consume mind-alte ring substances if there isn’t a mind to alter But these suggestive hints of negative and positive mind states still fell short of what was needed to demonstrate that insects are sentient.28. What does the example of the wasps indicateA. Insects show signs of intelligence.B. Insects can do complex calculations.C. Insects can socialize in a skillful way.D. Insects live in highly complex societies.29. What does the underlined word “meticulously” in paragraph 3 probably meanA. Hesitantly.B. Casually.C. Eagerly.D. Carefully.30. What was unexpected about the bumblebees’ behavior in the experimentA. They avoided flowers with spiders.B. They settled on flowers despite threats.C. They might abandon spider-free flowers.D. They might get scared away by other insects.31. What does the text mainly discussA. What insects’ various behavior can reveal.B. How insects communicate their thoughts.C. What amazing powers insects possess.D. Whether insects are capable of feeling.DWhile teenagers who are at risk of depression with risky behaviorsdrinking alcohol, smoking cigarettes and cutting classes often alert parents and teachers that serious problems are brewing, a new study finds that there’s another group of adolescents who are in nearly as much danger of experiencing the same mental symptoms. These teens use tons of media, get insufficient sleep and have a sedentary (不爱活动的) lifestyle. Of course, that may sound like a description of every teenager on the planet. But the study warns that it is teenagers who engage in all three of these practices in the extreme that are truly in a dangerous position. Because their behaviors are not usually seen as a red flag, these young people have been called the “invisible risk” group by the study’s authors. The study’s authors surveyed 15,395 students and analyzed nine risk behaviors, including excessive alcohol use, illegal drug use, heavy smoking and high media use. Their aim was to determine the relationship between these risk behaviors and mental health issues in teenagers. The group that scored high on all nine of the risk behaviors was most likely to show symptoms of depression; in all, nearly 15% of this group reported being depressed, compared with just 4% of the low-risk group. But the invisible group wasn’t far behind the high-risk set, with more than 13% of them exhibiting depression.The findings caught Carli off guard. “We didn’t expect that,” hesays. “The high-risk group and low-risk group are obvious, but this third group was not only unexpected. It was so distinct and so larger — nearly one third of our sample — that it became a key finding of the study.”Carli says that one of the most significant things about his study is that it provides new early warning signs for parents, teachers and mental health-care providers. And early identifications, support and treatment for mental health issues, he says, are the best ways to keep them from turning into full-blown disorders.32. Which teenager probably belongs to the “invisible group”A. A teenager who drinks frequently.B. A teenager who exercises regularly.C. A teenager who skips school.D. A teenager who suffers from a lack of sleep.33. What can we know about the new studyA. It was conducted by analyzing and comparing the previous data.B. It was intended to dig into the reasons for depression.C. It revealed an alarming rate of the invisible group suffering depression.D. Its findings were under expectation of the research team.34. What is Carli’s attitude toward the findingsA. Unclear.B. Positive.C. Doubtful.D. Indifferent.35. The author wrote this passage to _____________.A. introduce a new therapy for teens’ mental disorderB. warn about the unobserved signals for teens’ mental problemsC. share a novel psychological experiment with teensD. caution teens against developing unhealthy habits第二节(共5小题;每小题2.5分,满分12.5分)阅读下面短文,从短文后的选项中选出可以填入空白处的最佳选项。

第三次三校联考试题答案

第三次三校联考试题答案

08—09学年度第三次三校联考英语试题参考答案2009年04月16日一、听力〔20小题,每一小题1分, 共20分〕1-5:ACABA6-10:ABACB11-15:AACAC16-20:ABCBC二、单项选择〔15小题,第小题1分,15分〕21-25:DCBCB26-30: AAABD31-35:CABBC三、完形填空〔20小题,每一小题2分,共40分〕36-40 DACBC41-45 CBADD46-50 BBCDA51-55 DBCBA四、阅读理解(20小题,第小题3分,共60分)56-60:ACCDC 61-65:AACBA 66-70:ADDBC 71-75:DACCD五、短文改错: (10小题, 每小1分, 共10分)76. very 前加a 77. and改为but 78. being改为be 79. for改为by 80.√81. card改为cards 82. quarreled改为quarrels 83.fixing改为fixed 84. good改为well 85. 去掉to六、书面表达:〔25分〕一、评分原如此1.此题总分为25分,按5个档次给分。

2.评分时,先根据文章的内容和语言初步确定其所属档次,然后以该档次的要求来衡量,确定或调整档次,最后给分。

3.词数少于80和多于120的,从总分中减去2分。

4.评分时,应注意的主要内容为:内容要点、应用词汇和语法结构的数量和准确性与上下文的连贯性。

5.拼写与标点符号是语言准确性的一个方面,评分时,应视其对交际的影响程度予以考虑英、美拼写与词汇用法均可承受。

6.如书写较差,以至影响交际,将分数降低一个档次。

What We High School Students Learn fromthe Global Financial CrisisThe global financial crisis broke out in 2008, which caused serious economic slide and increased job loss.What can we learn from it?First , we should put all our heart into study to broaden our knowledge in various fields. Second, it is required that we should develop our abilities in many ways and improve our personal qualities to achieve our goals in this highly competitive society. Next, learning to cooperate with others is another thing we must do, because sense of teamwork contributes to success more quickly. Last but not least, as young generation in China, w e’d better set our long-term plans to meet what society needs.At present I think only by working hard can our ability be fully embodied. I’ll spar e no effort to study hard to achieve my goal.。

2024-2025学年广东省三校高二数学上学期第一次月考试卷及答案解析

2024-2025学年广东省三校高二数学上学期第一次月考试卷及答案解析

2024-2025学年广东省三校高二数学上学期第一次月考试卷一、单选题(本大题共8小题,每小题5分,共40分.每小题给出的四个选项中,只有一个选项是正确的.请把正确的选项填涂在答题卡相应的位置上.)1.如图所示,在三棱台A B C ABC '''-中,截去三棱锥A ABC '-,则剩余部分是()A.三棱锥B.四棱锥C.三棱柱D.组合体【答案】B 【解析】【分析】根据图形和棱锥的定义及结构特征,即可得出结论.【详解】三棱台A B C ABC '''-中,沿平面A BC '截去三棱锥A ABC '-,剩余的部分是以A '为顶点,四边形BCC B ''为底面的四棱锥A BCC B '''-.故选:B2.棱长为1的正四面体的表面积为()A.B. C. D.【答案】A 【解析】【分析】利用三角形的面积公式可得出正四面体的表面积.【详解】棱长为1的正四面体的表面积为2211341sin 6041222S =⨯⨯⨯=⨯⨯⨯= .故选:A.3.如图,在正四棱台1111ABCD A B C D -中,,,,E F G H 分别为棱1111,,,A D B C BC AD 的中点,则()A.直线HE 与直线GF 是异面直线B.直线HE 与直线1BB 是异面直线C.直线HE 与直线1CC 共面D.直线HE 与直线BF 共面【答案】C 【解析】【分析】由正四棱台的结构特征,侧棱的延长线交于同一点,,HE GF 的延长线必过此点,可判断选项中的线线位置关系.【详解】延长1111,,,AA BB CC DD ,由正四棱台的性质可得侧棱1111,,,AA BB CC DD 的延长线交于同一点,设该交点为P .,,,E F G H 分别为棱1111,,,A D B C BC AD 的中点,延长,HE GF ,则,HE GF 的延长线必过点P ,则直线HE 与直线GF 相交于点P ;与直线1BB 相交于点P ;与直线1CC 相交于点P ;与直线BF 是异面直线.故选:C.4.底面积是π,侧面积是3π的圆锥的体积是()A. B.C.2π3D.π3【答案】D 【解析】【分析】先利用圆锥的侧面积公式求出母线长,进而求出高,再利用圆锥的体积公式求解.【详解】设圆锥的母线长为l ,高为h ,半径为r ,则2ππS r ==底且=π3πS r l ⨯⨯=侧,故1,3r l ==h ∴===∴圆锥的体积为21π1π33⨯⨯⨯=.故选:D .5.已知正方体1111ABCD A B C D -中,E 为11B C 中点,则异面直线1BA 与C 所成角的余弦值为()A.10B.5C.10D.5【答案】D 【解析】【分析】连接1CD ,1D E ,根据异面直线所成角的定义,转化为求1D CE ∠(或其补角),然后在1D CE 中用余弦定理即可解得.【详解】连接1CD ,1D E,如图:因为1111ABCD A B C D -为正方体可得11//CD BA ,所以1D CE ∠(或其补角)是异面直线1BA 与C 所成角,设正方体的棱长为a,1CD ===,155,22CE D E a ===,在1D CE 中,2221111cos 2CD CE D E D CE CD CE +-∠=⋅⋅22255244a a a +-=5=,所以异面直线1BA 与C 所成角的余弦值是5.故选:D.6.如图,在正四棱台1111ABCD A B C D -中,1114,2,3AB A B AA ===,则该正四棱台的体积为()A.1129B.1409C.1123D.1403【答案】A 【解析】【分析】作出截面,过点1A 作1A E AC ⊥,结合等腰梯形的性质得到高,再计算体积即可.【详解】过11,AC A C 作出截面如图所示,过点1A 作1A E AC ⊥,垂足为E ,易知1A E 为正四棱台1111ABCD A B C D -的高,因为1124,AB A B ==,所以由勾股定理得11AC A C ==,又113CC AA ==,则在等腰梯形11ACC A 中,AE =,所以143A E ===,所以所求体积为11111114112((1643339ABCD A B C D V S S A E =⨯++⋅=⨯++⨯=.故选:A .7.我国古代数学专著《九章算术》中有这样一个问题:“今有木长二丈,围之三尺.葛生其下,缠木七周,上与木齐.问葛长几何?”其意思为:“圆木长2丈,圆周长为3尺,葛藤从圆木的底部开始向上生长,绕圆木7周,顶部刚好与圆木平齐,问葛藤长为多少?"若1丈10=尺,则葛藤最少长()A.21尺 B.25尺C.29尺D.33尺【答案】C 【解析】【分析】根据题意知,圆柱的侧面展开图是矩形,且矩形的长为21(尺),高为20尺,则葛藤的最少长度为矩形的对角线长,利用勾股定理可求得结果.【详解】根据题意知,圆柱的侧面展开图是矩形,如下图所示,矩形的高(即圆木长)为20尺,矩形的底边长为7321⨯=(尺),29=(尺).故选:C.8.如图所示,在正方体1111ABCD A B C D -中,E ,F 分别为1AA ,AB 上的中点,且E F P 点是正方形11ABB A 内的动点,若1C P ∥平面1CD EF ,则P 点的轨迹长度为()A. B.3πC.D.π【答案】C 【解析】【分析】取11A B 的中点H ,1B B 的中点为G ,连接11,,,,GH C H C G EG HF ,可得四边形11EGC D 是平行四边形,可得1C G ∥1D E ,同理可得1C H ∥CF .可得面面平行,进而得出P 点的轨迹.【详解】如图所示,取11A B 的中点H ,1B B 的中点为G ,连接11,,,,GH C H C G EG HF ,则11A B ∥EG ,11A B EG =,且11A B ∥11C D ,1111A B C D =,可得EG ∥11C D ,且11EG C D =,可知四边形11EGC D 是平行四边形,则1C G ∥1D E ,且1C G ⊄平面1CD EF ,1D E ⊄平面1CD EF ,可得1C G ∥平面1CD EF ,同理可得:1C H ∥平面1CD EF ,且111C H C G C = ,11,C H C G ⊂平面1C GH ,可知平面1C GH ∥平面1CD EF ,又因为P 点是正方形11ABB A 内的动点,1C P ∥平面1CD EF ,所以点M 在线段GH 上,由题意可知:1111,22GH A B EF A B ==,可得GH EF ==,所以P 故选:C.二、多选题(本大题共3小题,每小题6分,共计18分.每小题给出的四个选项中,有多项符合题目要求.全部选对得6分,选对但不全的部分分,有选错的得0分.)9.已知α,β是两个不同的平面,l ,m 是两条不同的直线,有如下四个命题,其中正确的是()A.若αβ⊥,l β⊥,则l α∥B.若m β⊥,l m ∥,l α⊂,则αβ⊥C.若αβ∥,m α⊥,l β⊂,则l m ⊥D.若m αβ= ,l α∥,则l m∥【答案】BC 【解析】【分析】根据空间中垂直关系的转化可判断ABC 的正误,根据线面平行定义可判断D 的正误.【详解】对于A ,若αβ⊥,l β⊥,则l α∥或l α⊂,故A 错误;对于B ,若m β⊥,l m ∥,则l β⊥,而l α⊂,故αβ⊥,故B 正确;对于C ,若αβ∥,m α⊥,则m β⊥,而l β⊂,故l m ⊥,故C 正确;对于D ,若m αβ= ,l α∥,则l m ∥或,l m 异面,故D 错误,故选:BC10.在实践课上,小华将透明塑料制成了一个长方体容器1111ABCD A B C D -,如图(1),2AB BC ==,15A A =,在容器内灌进一些水()14D H DH =,现固定容器底面一边BC 于地面上,再将容器倾斜,如图(2),则()A.有水的部分始终呈三棱柱或四棱柱B.棱11A D 与水面所在平面平行C.水面EFGH 所在四边形的面积为定值D.当容器倾斜成如图(3)所示时,EF 的最小值为【答案】ABD 【解析】【分析】由棱柱的概述判断A ;由线面平行判定定理判断B ;计算EFGH S 可判断C ;利用基本不等式可判断D.【详解】由棱柱的定义知,选项A 正确;对于选项B ,由于11A D BC ∥,BC FG ∥,所以11A D FG ∥,且11A D 不在水面所在平面内,所以棱11A D 与水面所在平面平行,选项B 正确;对于选项C ,在图(1)中,4EFGH S FG EF BC AB =⋅=⋅=,在图(2)中,4EFGH S FG EF AB BC =⋅>⋅=,选项C 错误;对于选项D ,12212V BE BF BC =⨯⨯=⋅⋅⋅水,所以4BE BF ⋅=.22228EF BE BF BE BF =+≥⋅=,当且仅当2BE BF ==时,等号成立,所以EF 的最小值为,选项D 正确.故选:ABD .11.半正多面体(semiregular solid )亦称“阿基米德多面体”,是由边数不全相同的正多边形围成的多面体,体现了数学的对称美.二十四等边体就是一种半正多面体,是由正方体切截而成的,它由八个正三角形和六个正方形构成(如图所示),则()A.BF ⊥平面EABB.该二十四等边体的体积为203C.该二十四等边体外接球的表面积为6πD.PN 与平面EBFN 所成角的正弦值为2【答案】BD 【解析】【分析】A 用反证法判断;B 先补齐八个角成正方体,再计算体积判断;C 先找到球心与半径,再计算表面积判断;D 先找到直线与平面所成角,再求正弦值判断.【详解】对于A ,假设A 对,即BF ⊥平面EAB ,于是BF AB ⊥,90ABF ∠=︒,但六边形ABFPQH 为正六边形,120ABF ∠=︒,矛盾,所以A 错误;对于B ,补齐八个角构成棱长为2的正方体,则该二十四等边体的体积为3112028111323-⋅⋅⋅⋅⋅=,所以B 对;对于C ,取正方形A C P M 对角线交点O ,即为该二十四等边体外接球的球心,其半径为R =24π8πR =,所以C 错误;对于D ,因为PN 在平面EBFN 内射影为NS ,所以PN 与平面EBFN 所成角即为PNS ∠,其正弦值为2PS PN ==,所以D 对.故选:BD .三、填空题(本大题共3小题,每小题5分,共计15分)12.如下图,三角形A'B'C'是三角形ABC 的直观图,则三角形ABC 的面积是_______.【答案】2【解析】【分析】画出原图形可得答案.【详解】由直观图画出原图,如图,可得ABC V 是等腰三角形,且2,2BC OA ==,所以三角形ABC 的面积12222S =⨯⨯=.故答案为:2.13.圆柱的底面半径为1,侧面积为10π,则该圆柱外接球的表面积为______.【答案】29π【解析】【分析】先利用侧面积求出圆柱的高,再求出球的半径可得表面积.【详解】设圆柱的高为h ,其外接球的半径为R ,由圆柱的底面半径为1,侧面积为10π,得2π10πh =,解得5h =,由圆柱和球的对称性可知,球心位于圆柱上下底面中心连线的中点处,因此2R ==,所以球的表面积为24π29πS R ==.故答案为:29π14.球面被平面所截得的一部分叫做球冠,截得的圆叫做球冠的底,垂直于截面的直径被截得的一段叫做球冠的高.球被平面截下的一部分叫做球缺,截面叫做球缺的底面,垂直于截面的直径被截下的线段长叫做球缺的高,球缺是旋转体,可以看做是球冠和其底所在的圆面所围成的几何体.如图1,一个球面的半径为R ,球冠的高是h ,球冠的表面积公式是2πS Rh =.如图2,已知,C D 是以AB 为直径的圆上的两点,ππ,63AOC BOD ∠∠==,扇形COD 的面积为π,则扇形COD 绕直线AB 旋转一周形成的几何体的表面积为__________.【答案】)61π【解析】【分析】首先求出DOC ∠,再根据扇形面积公式求出圆的半径,过点C 作CE AB ⊥交AB 于点E ,过点D 作DF AB ⊥交AB 于点F ,即可求出,,,,,CE OE AE OF BF DF ,将扇形DOC 绕直线AB 旋转一周形成的几何体为一个半径R 的球中上下截去两个球缺所剩余部分再挖去两个圆锥,再根据所给公式分别求出表面积.【详解】因为ππ,63AOC BOD ∠∠==,所以π2DOC ∠=,设圆的半径为R ,又2COD 1ππ22S R =⨯⨯=扇形,解得2R =(负值舍去),过点C 作CE AB ⊥交AB 于点E ,过点D 作DF AB ⊥交AB 于点F ,则ππsin 1,cos 66CE OC OE OC ====,所以2AE R OE =-=-,同理可得1DF OF ==,将扇形COD 绕直线AB 旋转一周形成的几何体为一个半径2R =的球中,上下截去两个球缺所剩余部分再挖去两个圆锥,其中上面球缺的高12h =,上面圆锥的底面半径11r =,高为1h =',下面球缺的高21h =,下面圆锥的底面半径2r =21h =',则上面球冠的表面积(112π2π228πs Rh ==⨯⨯-=-,下面球冠的表面积222π2π214πs Rh ==⨯⨯=,球的表面积24π16πS R ==球,上面圆锥的侧面积111ππ122πS rl ==⨯⨯=',下面圆锥的侧面积222ππ2S r l ===',所以几何体的表面积())''121116π8π4π2π61πS S S S S S =--++=---++=球.故答案为:)61π+.【点睛】关键点点睛:本题关键是弄清楚经过旋转之后得到的几何体是如何组成,对于表面积要合理转化.四、解答题(本题共5小题,共7分,解答应写出文字说明、证明过程或演算步骤.)15.如图,在正三棱柱111ABC A B C -中,E ,F ,G ,H 分别是AB ,AC ,11A B ,11A C 的中点.(1)求证:B ,C ,H ,G 四点共面;(2)求证:平面//BCHG 平面1A EF ;【答案】(1)证明见解析(2)证明见解析【解析】【分析】(1)证明出//GH BC ,得到四点共面;(2)先得到1//A E BG ,//GH EF ,证明出线面平行,面面平行.【小问1详解】∵G ,H 分别是11A B ,11A C 的中点,∴GH 是111A B C △的中位线,∴11//GH B C ,又在三棱柱111ABC A B C -中,11//B C BC ,∴//GH BC ,∴B ,C ,H ,G 四点共面.【小问2详解】∵在三棱柱111ABC A B C -中,11//A B AB ,11A B AB =,∴1//A G EB ,1111122A G AB AB EB ===,∴四边形1A EBG 是平行四边形,∴1//A E BG ,∵1A E ⊂平面1A EF ,BG ⊂/平面1A EF ,∴//BG 平面1A EF .又E ,F 是AB ,AC 的中点,所以//EF BC ,又//GH BC .所以//GH EF ,∵EF ⊂平面1A EF ,GH ⊂/平面1A EF ,∴//GH 平面1A EF .又BG GH G = ,,BG GH ⊂平面BCHG ,所以平面//BCHG 平面1A EF .16.如图,AB 为⊙O 的直径,PA 垂直于⊙O 所在的平面,M 为圆周上任意一点,AN ⊥PM ,N 为垂足.(1)若2PA AM BM ===,Q 为PB 的中点,求三棱锥Q ABM -的体积;(2)求证:AN ⊥平面PBM ;(3)若AQ ⊥PB ,垂足为Q ,求证:NQ ⊥PB.【答案】(1)23(2)证明见解析(3)证明见解析【解析】【分析】(1)先得到1433P AMB AMB V S PA -=⋅= ,根据Q 为PB 的中点,故1223Q ABM P AMB V V --==;(2)由线线垂直,得到线面垂直,即BM ⊥平面PAM .,故BM ⊥AN ,又AN ⊥PM ,从而得到线面垂直;(3)由(1)知AN ⊥平面PBM ,故AN ⊥PB ,又AQ ⊥PB ,故PB ⊥平面ANQ ,得到答案.【小问1详解】因为AB 为⊙O 的直径,所以AM ⊥BM ,又2AM BM ==,故122AMB S AM BM =⋅= ,又PA 垂直于⊙O 所在的平面,2PA =,故11422333P AMB AMB V S PA -=⋅=⨯⨯= ,因为Q 为PB 的中点,所以11422233Q ABM P AMB V V --==⨯=.【小问2详解】∵AB 为⊙O 的直径,∴AM ⊥BM .又PA ⊥平面ABM ,BM ⊂平面ABM ,∴PA ⊥BM .又∵PA AM A = ,PA ,AM ⊂平面PAM ,∴BM ⊥平面PAM .又AN ⊂平面PAM ,∴BM ⊥AN .又AN ⊥PM ,且BM PM M = ,BM ,PM ⊂平面PBM ,∴AN ⊥平面PBM .【小问3详解】由(1)知AN ⊥平面PBM ,PB ⊂平面PBM ,∴AN ⊥PB .又∵AQ ⊥PB ,AN ∩AQ =A ,AN ,AQ ⊂平面ANQ ,∴PB ⊥平面ANQ .又NQ ⊂平面ANQ ,∴PB ⊥NQ .17.我国古代数学名著《九章算术》中,称四面都为直角三角形的三棱锥为“鳖臑”.如图,在三棱锥A BCD -中,AB ⊥平面,BCD BC CD ⊥.(1)证明:三棱锥A BCD -为鳖臑;(2)若E 为AD 上一点,点,P Q 分别为,BC BE 的中点.平面DPQ 与平面ACD 的交线为l .①证明:直线//PQ 平面ACD ;②判断PQ 与l 的位置关系,并证明你的结论.【答案】(1)证明见解析;(2)①证明见解析;②平行,证明见解析.【解析】【分析】(1)利用线面垂直的性质及判定定理即可求解;(2)①利用三角形的中位线定理及线面平行的判定定理即可求解;②利用①的结论及线面平行的性质定理即可求解.【小问1详解】∵BC CD ⊥,∴BCD △为直角三角形,∵AB ⊥平面BCD ,且BD ⊂平面BCD ,⊂BC 平面BCD ,CD ⊂平面BCD ,∴AB BC ⊥,AB BD ⊥,AB CD ⊥,∴ABC V 和ABD △为直角三角形,∵BC AB B ⋂=,BC ⊂平面ABC ,AB ⊂平面ABC ,∴CD ⊥平面ABC ,又∵AC ⊂平面ABC ,∴CD AD ⊥,∴ACD 为直角三角形,∴三棱锥A BCD -为鳖曘.【小问2详解】①连接CE ,∵点,P Q 分别为,BC BE 的中点,∴//PQ CE ,且PQ ⊄平面ACD ,CE ⊂平面ACD ,所以直线//PQ 平面ACD ,②平行,证明://PQ 平面ACD ,PQ ⊂平面DPQ ,平面DPQ ⋂平面ACD =l ,所以//PQ l .18.一块四棱锥木块如图所示,SD ⊥平面ABCD ,四边形ABCD 为平行四边形,且60BAD ∠=︒,224AB BC SD ===.(1)要经过点B 、D 将木料锯开,使得截面平行于侧棱SA ,在木料表面该怎样画线?并说明理由;(2)计算(1)中所得截面的面积;(3)求直线SC 与(1)中截面所在平面所成角的正弦值.【答案】(1),ED EB 即为要画的线,理由见解析;(2(3)5【解析】【分析】(1)要使截面与SA 平行,考虑构造线线平行,取SC 的中点E ,取ABCD 的对称中心O ,连接OE ,证明//SA OE 即得截面BDE ;(2)分别计算BDE 的三边,再利用三角形面积公式计算即得;(3)利用等体积求出点C 到平面BDE 的距离,再由线面所成角的定义即可求得.【小问1详解】如图,取SC 的中点E ,连接,,ED EB ,则,ED EB 即为要画的线.理由如下:连接BD 与AC 交于点O ,连接OE .因四边形ABCD 为平行四边形,则点O 为AC 的中点,故//SA OE ,又因SA ⊄平面BDE ,OE ⊂平面BDE ,故有SA ∥平面BDE ;【小问2详解】如图中,过点E 作EF DC ⊥于点F ,连接BF ,因SD ⊥平面ABCD ,CD ⊂平面ABCD ,则SD CD ⊥,故//EF SD ,⊥EF 平面ABCD ,112EF SD ==,12DE SC ===因12,60,22CF DC DCB BC ==∠== ,则2BF =,因BF ⊂平面ABCD ,则EF FB ⊥,故BE ==,又由余弦定理,22224224cos6012BD =+-⨯⨯=,故得BD =.又DE DB =,O 为BD 中点,则OE BD ⊥,于是截面的面积为12BDE S =⨯ ;【小问3详解】过点C 作CH ⊥平面BDE ,交平面BDE 于点H ,连接EH,则CEH ∠即直线SC 与截面BDE 所成的角.由E BCD C BED V V --=可得,1133BCD BED S EF S CH ⨯=⨯ ,即得:124sin 6012BCD BED S EF CH S ⨯⨯⨯⨯==,则sin 5CH CEH EC ∠===,即直线SC 与平面BDE所成角的正弦值为5.【点睛】思路点睛:本题主要考查运用线面平行的判定方法解决实际问题和线面所成角的求法,属于较难题.解题的思路在于充分利用平行四边形对角线性质、等腰三角形三线合一,三角形中位线性质等方法寻找线线平行;对于线面所成角问题,除了定义法作图求解外,对于不易找到点在平面的射影时,可考虑运用等体积转化求解.19.空间的弯曲性是几何研究的重要内容,用曲率刻画空间的弯曲性,规定:多面体顶点的曲率等于2π与多面体在该点的面角之和的差,其中多面体的面的内角叫做多面体的面角,角度用弧度制.例如:正四面体每个顶点均有3个面角,每个面角均为π3,故其各个顶点的曲率均为π2π3π3-⨯=.如图,在直三棱柱111ABC A B C -中,点A 的曲率为2π3,M 为1CC 的中点,且AB AC =.(1)判断ABC V 的形状,并说明理由;(2)若124A A AB ==,求点B 到平面1AB M 的距离;(3)表面经过连续变形可以变为球面的多面体称为简单多面体.关于简单多面体有著名欧拉定理:设简单多面体的顶点数为D ,棱数为L ,面数为M ,则有:2D L M -+=.利用此定理试证明:简单多面体的总曲率(多面体有顶点的曲率之和)是常数.【答案】(1)ABC V 为等边三角形,理由见解析(2)5(3)证明见解析【解析】【分析】(1)根据线面垂直的性质可得1AA AC ⊥,1AA AB ⊥,即可根据曲率的定义求解,(2)利用等体积法,结合锥体体积公式即可求解,(3)根据则多面体的棱数,顶点数,以及内角之和,即可根据曲率的定义求解.【小问1详解】因为在直三棱柱111ABC A B C -中,1AA ⊥平面ABC ,,AC AB ⊂平面ABC ,所以1AA AC ⊥,1AA AB ⊥,所以点A 的曲率为π2ππ2232BAC -⨯-∠=,得π3BAC ∠=,因为AB AC =,所以ABC V 为等边三角形.【小问2详解】取BC 中点D ,连接AD 、AM ,因为D 为BC 的中点,所以AD BC ⊥,因为1BB ⊥平面ABC ,AD ⊂平面ABC ,所以1BB AD ⊥,因为1BB BC B = ,1,AA AB ⊂平面11ABB A ,所以AD ⊥平面11BB C C ;所以AD 是三棱锥1A BB M -的高.设点B 到平面1AB M 的距离为h ,则有11B AB M A BB M V V --=,即11AB M BB M S h S AD =⋅.在11Rt AA B △中有1AB ==,同理计算得1AM B M BM ===,AD =.所以112AB M S =⨯=,114242BB M S =⨯⨯=,所以455h ==.【小问3详解】证明:设多面体有M 个面,给组成多面体的多边形编号,分别为1,2,,M ⋅⋅⋅号,设第i 号()1i M ≤≤多边形有i L 条边,则多面体共有122M L L L L ++⋅⋅⋅+=条棱,由题意,多面体共有12222M L L L D M L M ++⋅⋅⋅+=-+=-+个顶点,i 号多边形的内角之和为π2πi L -,所以所有多边形的内角之和为()12π2πM L L L M ++⋅⋅⋅+-,所以多面体的总曲率为()122ππ2πM D L L L M ⎡⎤-++⋅⋅⋅+-⎣⎦()12122π2π2π4π2M M L L L M L L L M ++⋅⋅⋅+⎛⎫⎡⎤=-+-++⋅⋅⋅+-= ⎪⎣⎦⎝⎭.所以简单多面体的总曲率为4π.。

高一第三次月考英语试卷 Word版含答案

高一第三次月考英语试卷 Word版含答案

普兰店区第38中学2020-2021学年第一学期第三次考试高一英语试卷总分:150分时间:120分钟命题人:高二命题组2020.11第一部分听力(共两节,满分30 分)第一节(共5 小题,每小题1.5 分,满分7.5 分)听下面5 段对话,每段对话后有一个小题,从题中所给的A、B、C 三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有10 秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1.What did the woman get from her mother for her birthday?A. A birthday cardB. A cakeC. A new bike2. What’s the relationship between the speakers?A. AcquaintancesB.Friends .C.Strangers3.What will the master do for Wukong ?A. He will refuse to teach him.B. He will agree to teach him..C. He will cause more troubles for him4.Where does the conversation probably take place ?A. In a restaurantB. In a shopC. In a hotel5. What does the woman ask the man to do ?A. Buy some breadB.Pick up Emma .C. Get home early第二节(共15 小题;每小题1.5 分,满分22.5 分)听下面5 段对话或独白。

每段对话或独白后有几个小题,从题中所给的A、B、C 三个选项中选出最佳选项,并标在试卷的相应位置。

听每段对话或独白前,你将有时间阅读各个小题,每小题5 秒钟;听完后,各小题给出5 秒钟的作答时间。

八年级上学期第三次月考数学试卷附答案

八年级上学期第三次月考数学试卷附答案

八年级上学期第三次月考数学试卷一、选择题(每小题2分,共16分)1.(2分)下列关于两个三角形全等的说法:①三个角对应相等的两个三角形全等;②三条边对应相等的两个三角形全等;③有两角和其中一个角的对边对应相等的两个三角形全等;④有两边和一个角对应相等的两个三角形全等.正确的说法个数是()A.1个B.2个C.3个D.4个2.(2分)如图,在△ABC中,AB=AC,AD平分∠BAC,DE⊥AB,DF⊥AC,E、F为垂足,则下列四个结论:(1)∠DEF=∠DFE;(2)AE=AF;(3)AD平分∠EDF;(4)EF垂直平分AD.其中正确的有()A.1个B.2个C.3个D.4个3.(2分)已知实数x,y满足,则x﹣y等于()A.3B.﹣3 C.1D.﹣14.(2分)已知等腰三角形两边a,b,满足|2a﹣3b+5|+(2a+3b﹣13)2=0,则此等腰三角形的周长为()A.7或8 B.6或10 C.6或7 D.7或105.(2分)在下列实数中,无理数是()A.2B.0C.D.6.(2分)在实数,0,,π,中,无理数有()A.1个B.2个C.3个D.4个7.(2分)下列各式中,正确的是()A. B.C. D.8.(2分)不等式组的解集在数轴上表示为()A.B.C.D.二、填空题(每小题2分,共16分)9.(2分)化简(+)÷(m+2)的结果是.10.(2分)若分式方程的解为正数,则a的取值范围是.11.(2分)已知:一个正数的两个平方根分别是2a﹣2和a﹣4,则a的值是.12.(2分)不等式的最小整数解是.13.(2分)如图,已知△ABC和△BDE均为等边三角形,连接AD、CE,若∠BAD=39°,那么∠BCE=度.14.(2分)如图所示,在边长为2的正三角形ABC中,E、F、G分别为AB、AC、BC的中点,点P为线段EF上一个动点,连接BP、GP,则△BPG的周长的最小值是.15.(2分)已知a、b为有理数,m、n分别表示的整数部分和小数部分,且amn+bn2=1,则2a+b=.16.(2分)已知一个等腰三角形两内角的度数之比为1:4,则这个等腰三角形顶角的度数为.三、解答题(共68分)17.(4分)解不等式组并求它的所有的非负整数解.18.(6分)解方程:(1)﹣1=.(2)=3.19.(3分)计算:.20.(10分)(1)解不等式:x﹣>﹣(2)如果不等式组有解,求m的取值范围.21.(5分)先将代数式化简,再从﹣1,1两数中选取一个适当的数作为x的值代入求值.22.(8分)如图:已知BD=CD,BF⊥AC,CE⊥AB,求证:点D在∠BAC的平分线上.23.(8分)如图,△ABC是等腰三角形,D,E分别是腰AB及AC延长线上的一点,且BD=CE,连接DE交底BC于G.求证GD=GE.24.(8分)为了创建全国卫生城市,某社区要清理一个卫生死角内的垃圾,租用甲、乙两车运送,两车各运12趟可完成,需支付运费4800元.已知甲、乙两车单独运完此堆垃圾,乙车所运趟数是甲车的2倍,且乙车每趟运费比甲车少200元.(1)求甲、乙两车单独运完此堆垃圾各需运多少趟?(2)若单独租用一台车,租用哪台车合算?25.(8分)如图,已知△ABC中,AB=AC,D是△ABC外一点且∠ABD=60°,∠ADB=90°﹣∠BDC.求证:AC=BD+CD.26.(8分)如图,在四边形ABCD中,AD∥BC,E为CD的中点,连接AE、BE,BE⊥AE,延长AE交BC的延长线于点F.求证:(1)FC=AD;(2)AB=BC+AD.参考答案与试题解析一、选择题(每小题2分,共16分)1.(2分)下列关于两个三角形全等的说法:①三个角对应相等的两个三角形全等;②三条边对应相等的两个三角形全等;③有两角和其中一个角的对边对应相等的两个三角形全等;④有两边和一个角对应相等的两个三角形全等.正确的说法个数是()A.1个B.2个C.3个D.4个考点:全等三角形的判定.分析:根据全等三角形的判定方法,此题应采用排除法,对选项逐个进行分析从而确定正确答案.解答:解:A、不正确,因为判定三角形全等必须有边的参与;B、正确,符合判定方法SSS;C、正确,符合判定方法AAS;D、不正确,此角应该为两边的夹角才能符合SAS.所以正确的说法有两个.故选B.点评:主要考查全等三角形的判定方法,常用的方法有SSS,SAS,AAS,HL等,应该对每一种方法彻底理解真正掌握并能灵活运用.而满足SSA,AAA是不能判定两三角形是全等的.2.(2分)如图,在△ABC中,AB=AC,AD平分∠BAC,DE⊥AB,DF⊥AC,E、F为垂足,则下列四个结论:(1)∠DEF=∠DFE;(2)AE=AF;(3)AD平分∠EDF;(4)EF垂直平分AD.其中正确的有()A.1个B.2个C.3个D.4个考点:等腰三角形的判定与性质;线段垂直平分线的性质.专题:几何图形问题;综合题.分析:利用等腰三角形的概念、性质以及角平分线的性质做题.解答:解:∵AB=AC,AD平分∠BAC,DE⊥AB,DF⊥AC∴△ABC是等腰三角形,AD⊥BC,BD=CD,∠BED=∠DFC=90°∴DE=DF∴AD垂直平分EF∴(4)错误;又∵AD所在直线是△ABC的对称轴,∴(1)∠DEF=∠DFE;(2)AE=AF;(3)AD平分∠EDF.故选C.点评:有两边相等的三角形是等腰三角形;等腰三角形的两个底角相等;(简写成“等边对等角”)等腰三角形的顶角的平分线,底边上的中线,底边上的高的重合(简写成“三线合一”).3.(2分)已知实数x,y满足,则x﹣y等于()A.3B.﹣3 C.1D.﹣1考点:非负数的性质:算术平方根;非负数的性质:偶次方.专题:常规题型.分析:根据非负数的性质列式求出x、y的值,然后代入代数式进行计算即可得解.解答:解:根据题意得,x﹣2=0,y+1=0,解得x=2,y=﹣1,所以,x﹣y=2﹣(﹣1)=2+1=3.故选A.点评:本题考查了算术平方根非负数,平方数非负数的性质,根据几个非负数的和等于0,则每一个算式都等于0列式是解题的关键.4.(2分)已知等腰三角形两边a,b,满足|2a﹣3b+5|+(2a+3b﹣13)2=0,则此等腰三角形的周长为()A.7或8 B.6或10 C.6或7 D.7或10考点:等腰三角形的性质;非负数的性质:绝对值;非负数的性质:偶次方;解二元一次方程组.专题:计算题.分析:先根据非负数的性质求出a,b的值,再分两种情况确定第三边的长,从而得出三角形的周长.解答:解:∵|2a﹣3b+5|+(2a+3b﹣13)2=0,∴,解得,当a为底时,三角形的三边长为2,3,3,则周长为8;当b为底时,三角形的三边长为2,2,3,则周长为7;故选A.点评:本题考查了非负数的性质、等腰三角形的性质以及解二元一次方程组,是基础知识要熟练掌握.5.(2分)在下列实数中,无理数是()A. 2 B.0C.D.考点:无理数.专题:存在型.分析:根据无理数的定义进行解答即可.解答:解:∵无理数是无限不循环小数,∴是无理数,2,0,是有理数.故选C.点评:本题考查的是无理数的定义,即初中范围内学习的无理数有:π,2π等;开方开不尽的数;以及像0.1010010001…,等有这样规律的数.6.(2分)在实数,0,,π,中,无理数有()A.1个B.2个C.3个D. 4个考点:无理数.分析:由于无理数就是无限不循环小数.初中范围内学习的无理数有:π,2π等;开方开不尽的数;以及像0.1010010001…,等有这样规律的数.由此即可判定选择项.解答:解:﹣是分数,0是整数,=3是整数,这三个数都是有理数,和π是无理数,故选B.点评:此题主要考查了无理数的定义.注意带根号的数与无理数的区别:带根号的数不一定是无理数,带根号且开方开不尽的数一定是无理数.本题中是有理数中的整数.7.(2分)下列各式中,正确的是()A. B.C. D.考点:算术平方根.专题:计算题.分析:算术平方根的定义:一个非负数的正的平方根,即为这个数的算术平方根,由此即可求出结果.解答:解:A、=|﹣3|=3;故A错误;B、=﹣|3|=﹣3;故B正确;C、=|±3|=3;故C错误;D、=|3|=3;故D错误.故选:B.点评:此题主要考查了算术平方根的定义,算术平方根的概念易与平方根的概念混淆而导致错误.8.(2分)不等式组的解集在数轴上表示为()A.B.C.D.考点:在数轴上表示不等式的解集;解一元一次不等式组.分析:求出每个不等式的解集,找出不等式组的解集,再在数轴上把不等式组的解集表示出来,即可得出选项.解答:解:,∵解不等式①得:x>1,解不等式②得:x≤2,∴不等式组的解集为:1<x≤2,在数轴上表示不等式组的解集为:,故选A.点评:本题考查了在数轴上表示不等式组的解集,解一元一次不等式(组)的应用,关键是能正确在数轴上表示不等式组的解集.二、填空题(每小题2分,共16分)9.(2分)化简(+)÷(m+2)的结果是1.考点:分式的混合运算.专题:计算题.分析:原式括号中两边变形后,利用同分母分式的减法法则计算,同时利用除法法则变形,约分即可得到结果.解答:解:原式=•=1.故答案为:1点评:此题考查了分式的混合运算,熟练掌握运算法则是解本题的关键.10.(2分)若分式方程的解为正数,则a的取值范围是a<8,且a≠4.考点:分式方程的解.专题:计算题.分析:分式方程去分母转化为整式方程,求出整式方程的解得到x的值,根据分式方程解为正数求出a的范围即可.解答:解:分式方程去分母得:x=2x﹣8+a,解得:x=8﹣a,根据题意得:8﹣a>0,8﹣a≠4,解得:a<8,且a≠4.故答案为:a<8,且a≠4.点评:此题考查了分式方程的解,需注意在任何时候都要考虑分母不为0.11.(2分)已知:一个正数的两个平方根分别是2a﹣2和a﹣4,则a的值是2.考点:平方根.专题:计算题.分析:根据正数有两个平方根,它们互为相反数.解答:解:∵一个正数的两个平方根分别是2a﹣2和a﹣4,∴2a﹣2+a﹣4=0,整理得出:3a=6,解得a=2.故答案为:2.点评:本题考查了平方根的概念.注意一个正数有两个平方根,它们互为相反数;0的平方根是0;负数没有平方根.12.(2分)不等式的最小整数解是x=3.考点:一元一次不等式组的整数解.分析:先求出一元一次不等式组的解集,再根据x是整数得出最小整数解.解答:解:,解不等式①,得x≥1,解不等式②,得x>2,所以不等式组的解集为x>2,所以最小整数解为3.故答案为:x=3.点评:此题考查的是一元一次不等式组的整数解,正确解出不等式组的解集是解决本题的关键.求不等式组的解集,应遵循以下原则:同大取较大,同小取较小,小大大小中间找,大大小小解不了.13.(2分)如图,已知△ABC和△BDE均为等边三角形,连接AD、CE,若∠BAD=39°,那么∠BCE=39度.考点:全等三角形的判定与性质;等边三角形的性质.专题:几何图形问题.分析:因为△ABC和△BDE均为等边三角形,由等边三角形的性质得到AB=BC,∠ABC=∠EBD,BE=BD.再利用角与角之间的关系求得∠ABD=∠EBC,则△ABD≌△EBC,故∠BCE可求.解答:解:∵△ABC和△BDE均为等边三角形,∴AB=BC,∠ABC=∠EBD=60°,BE=BD,∵∠ABD=∠ABC+∠DBC,∠EBC=∠EBD+∠DBC,∴∠ABD=∠EBC,∴△ABD≌△EBC,∴∠BAD=∠BCE=39°.故答案为39.点评:本题考查三角形全等的判定方法,判定两个三角形全等的一般方法有:SSS、SAS、ASA、AAS、HL.注意:AAA、SSA不能判定两个三角形全等,判定两个三角形全等时,必须有边的参与,若有两边一角对应相等时,角必须是两边的夹角.14.(2分)如图所示,在边长为2的正三角形ABC中,E、F、G分别为AB、AC、BC的中点,点P为线段EF上一个动点,连接BP、GP,则△BPG的周长的最小值是3.考点:轴对称-最短路线问题;等边三角形的性质;平行线分线段成比例.专题:计算题.分析:连接AG交EF于M,根据等边三角形的性质证明A、G关于EF对称,得到P,△PBG周长最小,求出AB+BG即可得到答案.解答:解:要使△PBG的周长最小,而BG=1一定,只要使BP+PG最短即可,连接AG交EF于M,∵等边△ABC,E、F、G分别为AB、AC、BC的中点,∴AG⊥BC,EF∥BC,∴AG⊥EF,AM=MG,∴A、G关于EF对称,即当P和E重合时,此时BP+PG最小,即△PBG的周长最小,AP=PG,BP=BE,最小值是:PB+PG+BG=AE+BE+BG=AB+BG=2+1=3.故答案为:3.点评:本题主要考查对等边三角形的性质,轴对称﹣最短路线问题,平行线分线段成比例定理等知识点的理解和掌握,能求出BP+PG的最小值是解此题的关键.15.(2分)已知a、b为有理数,m、n分别表示的整数部分和小数部分,且amn+bn2=1,则2a+b=2.5.考点:二次根式的混合运算;估算无理数的大小.专题:计算题;压轴题.分析:只需首先对估算出大小,从而求出其整数部分a,其小数部分用﹣a 表示.再分别代入amn+bn2=1进行计算.解答:解:因为2<<3,所以2<5﹣<3,故m=2,n=5﹣﹣2=3﹣.把m=2,n=3﹣代入amn+bn2=1得,2(3﹣)a+(3﹣)2b=1化简得(6a+16b)﹣(2a+6b)=1,等式两边相对照,因为结果不含,所以6a+16b=1且2a+6b=0,解得a=1.5,b=﹣0.5.所以2a+b=3﹣0.5=2.5.故答案为:2.5.点评:本题主要考查了无理数大小的估算和二次根式的混合运算.能够正确估算出一个较复杂的无理数的大小是解决此类问题的关键.16.(2分)已知一个等腰三角形两内角的度数之比为1:4,则这个等腰三角形顶角的度数为120°或20°.考点:等腰三角形的性质.分析:设两个角分别是x,4x,根据三角形的内角和定理分情况进行分析,从而可求得顶角的度数.解答:解:设两个角分别是x,4x①当x是底角时,根据三角形的内角和定理,得x+x+4x=180°,解得,x=30°,4x=120°,即底角为30°,顶角为120°;②当x是顶角时,则x+4x+4x=180°,解得,x=20°,从而得到顶角为20°,底角为80°;所以该三角形的顶角为120°或20°.故答案为:120°或20°.点评:本题考查了等腰三角形的性质;若题目中没有明确顶角或底角的度数,做题时要注意分情况进行讨论,这是十分重要的,也是解答问题的关键.已知中若有比出现,往往根据比值设出各部分,利用部分和列式求解.三、解答题(共68分)17.(4分)解不等式组并求它的所有的非负整数解.考点:解一元一次不等式组;一元一次不等式组的整数解.专题:计算题.分析:先求出两个不等式的解集,再求其公共解,然后写出范围内的非负整数解即可.解答:解:,由①得x>﹣2,…(1分)由②得x≤,…(3分)所以,原不等式组的解集是﹣2<x≤,…(4分)所以,它的非负整数解为0,1,2.…(5分)点评:本题主要考查了一元一次不等式组解集的求法,其简便求法就是用口诀求解.求不等式组解集的口诀:同大取大,同小取小,大小小大中间找,大大小小找不到(无解).18.(6分)解方程:(1)﹣1=.(2)=3.考点:解分式方程.专题:计算题.分析:两分式方程去分母转化为整式方程,求出整式方程的解得到x的值,经检验即可得到分式方程的解.解答:解:(1)去分母得:x2+2x﹣x2﹣x+2=3,解得:x=1,经检验x=1是分式方程的解;(2)去分母得:3+3x=3x+3,即0=0,经检验分式方程的解为x≠﹣1.点评:此题考查了解分式方程,解分式方程的基本思想是“转化思想”,把分式方程转化为整式方程求解.解分式方程一定注意要验根.19.(3分)计算:.考点:分式的加减法.专题:计算题.分析:原式通分并利用同分母分式的加法法则计算即可得到结果.解答:解:原式=.点评:此题考查了分式的加减法,熟练掌握运算法则是解本题的关键.20.(10分)(1)解不等式:x﹣>﹣(2)如果不等式组有解,求m的取值范围.考点:解一元一次不等式;不等式的解集.分析:(1)先去分母,然后移项合并同类项,系数化为1求解;(2)根据不等式组有解,可得m<5.解答:解:(1)去分母得:12x﹣3x﹣6>8x﹣12﹣4+6x,移项合并同类项得:5x<10,系数化为1得:x<2;(2)∵不等式组有解,∴m<5.点评:本题考查了解简单不等式的能力,解不等式要依据不等式的基本性质:(1)不等式的两边同时加上或减去同一个数或整式不等号的方向不变;(2)不等式的两边同时乘以或除以同一个正数不等号的方向不变;(3)不等式的两边同时乘以或除以同一个负数不等号的方向改变.21.(5分)先将代数式化简,再从﹣1,1两数中选取一个适当的数作为x的值代入求值.考点:分式的化简求值.专题:开放型.分析:根据本题须先对要求的式子进行化简,再选取一个数代入即可求出结果.解答:解:原式=x(x+1)×=x,当x=﹣1时,分母为0,分式无意义,故不满足,当x=1时,成立,代数式的值为1.故答案为:1.点评:本题主要考查了分式的化简求值,解题时要注意先对括号里边进行化简,再约分,注意分母不能为0,难度适中.22.(8分)如图:已知BD=CD,BF⊥AC,CE⊥AB,求证:点D在∠BAC的平分线上.考点:角平分线的性质;全等三角形的判定与性质.专题:证明题.分析:此题容易根据条件证明△BED≌△CFD,然后利用全等三角形的性质和角平分线的性质就可以证明结论.解答:证明:∵BF⊥AC,CE⊥AB,∴∠BED=∠CFD=90°,在△BED和△CFD中,,∴△BED≌△CFD(AAS),∴DE=DF,又∵DE⊥AB,DF⊥AC,∴点D在∠BAC的平分线上.点评:常用主要考查了全等三角形的判定与性质,角平分线的性质.由全等等到DE=DF 是解答本题的关键.23.(8分)如图,△ABC是等腰三角形,D,E分别是腰AB及AC延长线上的一点,且BD=CE,连接DE交底BC于G.求证GD=GE.考点:等腰三角形的性质;全等三角形的判定与性质.专题:证明题.分析:过E作EF∥AB交BC延长线于F,根据等腰三角形的性质及平行线的性质可推出∠F=∠FCE,从而可得到BD=CE=EF,再根据AAS判定△DGB≌△EGF,根据全等三角形的性质即可证得结论.解答:证明:过E作EF∥AB交BC延长线于F.∵AB=AC,∴∠B=∠ACB,∵EF∥AB,∴∠F=∠B,∵∠ACB=∠FCE,∴∠F=∠FCE,∴CE=EF,∵BD=CE,∴BD=EF,在△DBG与△GEF中,,∴△DGB≌△EGF(AAS),∴GD=GE.点评:此题主要考查等腰三角形的性质及全等三角形的判定与性质的综合运用.24.(8分)为了创建全国卫生城市,某社区要清理一个卫生死角内的垃圾,租用甲、乙两车运送,两车各运12趟可完成,需支付运费4800元.已知甲、乙两车单独运完此堆垃圾,乙车所运趟数是甲车的2倍,且乙车每趟运费比甲车少200元.(1)求甲、乙两车单独运完此堆垃圾各需运多少趟?(2)若单独租用一台车,租用哪台车合算?考点:分式方程的应用;一元一次方程的应用.专题:压轴题.分析:(1)假设甲车单独运完此堆垃圾需运x趟,则乙车单独运完此堆垃圾需运2x趟,根据工作总量=工作时间×工作效率建立方程求出其解即可;(2)分别表示出甲、乙两车单独运每一趟所需费用,再根据关键语句“两车各运12趟可完成,需支付运费4800元”可得方程,再解出方程,再分别计算出利用甲或乙所需费用进行比较即可.解答:解:(1)设甲车单独运完此堆垃圾需运x趟,则乙车单独运完此堆垃圾需运2x趟,根据题意得出:12(+)=1,解得:x=18,经检验得出:x=18是原方程的解,则乙车单独运完此堆垃圾需运:2x=36,答:甲车单独运完需18趟,乙车单独运完需36趟;(2)设甲车每一趟的运费是a元,由题意得:12a+12(a﹣200)=4800,解得:a=300,则乙车每一趟的费用是:300﹣200=100(元),单独租用甲车总费用是:18×300=5400(元),单独租用乙车总费用是:36×100=3600(元),3600<5400,故单独租用一台车,租用乙车合算.答:单独租用一台车,租用乙车合算.点评:此题主要考查了分式方程的应用以及一元一次方程的应用,关键是正确理解题意,找出题目中的等量关系,列出方程.25.(8分)如图,已知△ABC中,AB=AC,D是△ABC外一点且∠ABD=60°,∠ADB=90°﹣∠BDC.求证:AC=BD+CD.考点:轴对称的性质;等边三角形的判定与性质.专题:证明题.分析:以AD为轴作△ABD的对称△AB′D,后证明C、D、B′在一条直线上,及△ACB′是等边三角形,继而得出答案.解答:证明:以AD为轴作△ABD的对称△AB′D(如图),则有B′D=BD,AB′=AB=AC,∠B′=∠ABD=60°,∠ADB′=∠ADB=90°﹣∠BDC,所以∠ADB′+∠ADB+∠BDC=180°﹣∠BDC+∠BDC=180°,所以C、D、B′在一条直线上,所以△ACB′是等边三角形,所以CA=CB′=CD+DB′=CD+BD.点评:本题考查了轴对称的性质及全等三角形的判定与性质,有一定难度,准确作出合适的辅助线是关键.26.(8分)如图,在四边形ABCD中,AD∥BC,E为CD的中点,连接AE、BE,BE⊥AE,延长AE交BC的延长线于点F.求证:(1)FC=AD;(2)AB=BC+AD.考点:线段垂直平分线的性质;全等三角形的判定与性质.专题:证明题.分析:(1)根据AD∥BC可知∠ADC=∠ECF,再根据E是CD的中点可求出△ADE≌△FCE,根据全等三角形的性质即可解答.(2)根据线段垂直平分线的性质判断出AB=BF即可.解答:证明:(1)∵AD∥BC(已知),∴∠ADC=∠ECF(两直线平行,内错角相等),∵E是CD的中点(已知),∴DE=EC(中点的定义).∵在△ADE与△FCE中,,∴△ADE≌△FCE(ASA),∴FC=AD(全等三角形的性质).(2)∵△ADE≌△FCE,∴AE=EF,AD=CF(全等三角形的对应边相等),∴BE是线段AF的垂直平分线,∴AB=BF=BC+CF,∵AD=CF(已证),∴AB=BC+AD(等量代换).点评:此题主要考查线段的垂直平分线的性质等几何知识.线段的垂直平分线上的点到线段的两个端点的距离相等.。

广东省三校2025届高三上学期8月开学摸底考化学试题+答案

广东省三校2025届高三上学期8月开学摸底考化学试题+答案

广东省三校2025届8月新高三年级摸底考试化学试题卷学校:建文外国语学校、广东碧桂园学校、广州亚加达外国语高级中学注意事项:1.答卷前,考生务必将自己的学校、班级、姓名、考场号、座位号和准考证号填写在答题卡上,将条形码横贴在答题卡“条形码粘贴处”。

2.作答选择题时,选出每小题答案后,用2B铅笔在答题卡上将对应题目选项的答案信息点涂黑;如需改动,用橡皮擦干净后,再选涂其他答案。

答案不能答在试卷上。

3.非选择题必须用黑色字迹的钢笔或签字笔作答,答案必须写在答题卡各题目指定区域内相应位置上;如需改动,先划掉原来的答案,然后再写上新答案;不准使用铅笔和涂改液。

不按以上要求作答无效。

4.考生必须保证答题卡的整洁。

考试结束后,将试卷和答题卡一并交回。

一、单选题:本大题共16小题,共48分。

1.下列关于常见有机物的说法正确的是( )A. 乙烯和苯都能和溴水发生化学反应而使溴水褪色B. 乙烷可使酸性高锰酸钾溶液褪色C. 糖类和蛋白质都是高分子化合物D. 乙酸和油脂都能与氢氧化钠溶液反应2.能源可划分为一级能源和二级能源,直接来自自然界的能源称为一级能源;需依靠其他能源的能量间接制取的能源称为二级能源.下列叙述正确的是( )A. 水煤气(CCCC,HH2)是二级能源B. 水力是二级能源C. 天然气是二级能源D. 电能是一级能源3.下列各项叙述中,正确的是( )A. NN、PP、AAAA的电负性随原子序数的增大而增大B. 价电子排布为4AA24pp3的元素位于第4周期第ⅤAA族,是pp区元素C. 2pp和3pp轨道形状均为哑铃形,能量也相等D. CCCC原子的价电子表示式3dd54SS1不符合洪特规则4.下列金属中,通常采用电解法冶炼的是( )A. NNNNB. CCCCC. FFFFD. AAAA5.下列反应属于氧化还原反应的是( )A. AAAA(CCHH)3+NNNNCCHH=NNNNAAAACC2+2HH2CCB. 2NNNN+2HH2CC=2NNNNCCHH+HH2↑C. NNNN2CC+CCCC2=NNNN2CCCC3D. 2NNNNHHCCCC3=NNNN2CCCC3+CCCC2↑+HH2CC6.下列关于有机化合物的说法正确的是( )A. 乙烯能使溴水和酸性高锰酸钾溶液褪色,且褪色原理相同B. 葡萄糖和果糖的分子式都是CC6HH12CC6,二者互为同分异构体C. 不含其他杂质的油脂属于纯净物D. 石油裂解和油脂皂化都是化学变化,而石油的分馏和煤的干馏都是物理变化7.工业上合成的反应为+→一定条件,下列有关说法不正确的是( )A. XX、YY、ZZ三种有机物均易发生加成反应B. XX、YY、ZZ三种有机物分子中所有原子均有可能共平面C. XX、YY、ZZ三种有机物使酸性高锰酸钾溶液褪色的原理相同D. ZZ的二氯代物有7种(不考虑立体异构)8.下列离子方程式正确的是( )A. AAAA CC AA3溶液中加入足量的氨水:AAAA3++3CCHH−=AAAA(CCHH)3↓B. 将一小块金属钠投入到硫酸铜溶液中:2NNNN+CCCC2+=CCCC+2NNNN+C. 氢氧化钡溶液与硫酸反应:2HH+SSCC42−+BBNN2++2CCHH−=BBNNSSCC4↓+2HH2CCD. 将CCAA2通入FFFFCCAA2溶液中:FFFF2++CCAA2=FFFF3++2CCAA−9.短周期主族元素WW、XX、YY、ZZ的原子序数依次增加。

八年级第一学期第三次月月考

八年级第一学期第三次月月考

八年级第一学期第三次月月考初二语文试题一、积累运用(20分)(一)、积累1、根据拼音写汉字或注音。

(4分)①风mǐ( ) ②池zhǎo( ) ③quán( )曲④恐hè( )2、默写。

(5分)(6题选5题,多做不给分。

)(1)商女不知亡国恨,。

(2),乌蒙磅礴走泥丸。

(3)、,儿女共沾巾。

(4), 绿杨阴里白沙堤。

(5),化作春泥更护花。

(6),铁马冰河入梦来。

(二)、运用(11分)3、下列句子中加点的成语,使用不当的一句是()(2分)之际,后援部队终于赶A、我军坚守阵地一天一夜,牺牲惨重,在这千钧一发....到了。

的,但又是不可缺少的,因为B、这个扫大街的老妈妈的劳动也许是微不足道....我们的城市需要“美容师”。

的毅力和辛勤的汗水,铺设了这道宽C、北国人民连续30年用他们坚忍不拔....达100公里的防护林。

,不考虑后果,常常妨害大家的利益。

D、他办事不与人商量,喜欢自出心裁....4、下列各句没有语病、句意明确的一项是()(2分)A.想起它,就像想起旅伴,想起战友,心里充满着深切的怀念。

B.中学生理解和阅读大量的文学名著,有利于开阔视野,陶冶情操。

C.二十一世纪需要德才兼备的一大批知识分子去建设祖国。

D.介绍菲律宾的一种权威著作。

5、下列有关对课文的理解说法不正确的是:()(2分)A、《枣核》这篇散文以“枣核”为贯穿全文的线索,其实,枣核既是叙事线索,更是一条凝聚着乡情的感情线索。

B、《老山界》作者陆定一,全文运用时间的变化和地点的转移记叙了翻越老山界的经过,表现了红军不畏牺牲、英勇顽强的革命精神。

C、《我的母亲》作者邹韬奋,文中运用多处细节刻画了一个伟大的母亲,表达作者对母亲的思念。

D、《阿里山纪行》作者运用定点换景的手法,向我们描绘了阿里山美丽如画的风景。

6、2008年奥运会在北京举行,假如你作为北京的青年志愿者迎接四面八方的客人,并向他们致以美好的祝愿,在下列两种情况下你该怎么说?(要求简明、连贯、得体(2分)(1)接待外国观众:(2)接待中国运动员:7、根据两个表格的内容,各用一句话概括其表明的意思,分别填写在横线上,并用一句话归纳出一个结论。

苏科版物理八年级上册三校第一学期第三次月度联考

苏科版物理八年级上册三校第一学期第三次月度联考

三校2015~2016学年度第一学期第三次月度联考八 年 级 物 理 试 题(考试时间:90分钟,满分:100分) 成绩第一部分 选择题(共26分)题号 12345678910111213答案 1.以下是小明估计的常见温度值,其中合理的是( )A .中考考场的室温约为50℃B .冰箱保鲜室中矿泉水的温度约为﹣5℃C .洗澡时淋浴水温约为70℃D .健康成年人的腋下体温约为37℃ 2.下列说法正确的是A .在道路旁安装隔音板是从声源处控制噪声的B .雨后天空中出现的彩虹,是由光的折射产生的C .用久了的电灯泡玻璃壁会变黑,是因为灯丝发生了汽化和凝固现象D .雾、露是水的气态,霜是水的固态。

3.五一佳节,在常州紫荆公园月季花展上,小明将红色滤色镜(即红色玻璃)挡在照相机镜头前给一株绿叶黄花的月季拍照,照片上该花卉的颜色是( ) A .绿叶黄花 B .黑叶红花 C .黑叶黑花 D .红叶红花4.南宋著名诗人辛弃疾的诗句“溪边照影行,天在清溪底。

天上有行云,人在云里行。

”中所描写的“天在清溪底”和“人在云里行”两种情景,是光的 ( ) A .反射现象 B .折射现象 C .反射现象和折射现象D .折射现象和反射现象 5.雨后初晴的夜晚,地上有积水,当我们背着月光走时,地上发亮处是干净的路面,这是因为( )A.地上暗处是光发生镜面发射B.地上发亮处是光发生镜面发射C. 地上发亮处是光发生漫反射D. 地上发亮处和暗处都发生镜面反射 6.如图所示为小明用透镜观察字的情况,下列说法正确的是( ) A .字与透镜的距离大于透镜的焦距B .此透镜只能成虚像C .此透镜可以用作近视眼镜D .此透镜可以用来做照相机的镜头7.将一个凸透镜正对太阳,在距凸透镜20cm 处得到一个最小、最亮的光斑.将一个物体放在此透镜前40cm 处,则可在凸透镜的另一侧得到一个( ) A .倒立、放大的实像 B .倒立、缩小的实像 C .倒立、等大的实像 D .正立、放大的虚像8.如右图所示,烛焰在光屏上成清晰的像.下列哪个光学器材的成像原理与其相同( ) A .投影仪 B .照相机 C .放大镜 D .近视眼镜班级 姓名 学号_________ 试场号_________密 封 线 内 不 要 答 卷……………………………………………装………………订…………………线…………………………………………………………9.在探究凸透镜成像的规律实验中,当烛焰位置距离凸透镜一倍焦距以内时,人眼观察到烛焰成像的情形是图中的( )10.在探究凸透镜成像的实验中,先用焦距为10㎝的透镜甲进行实验,在透镜右侧的光屏上得到了清晰缩小的实像,接下来不改变发光体和凸透镜的位置,改用焦距为20㎝的透镜乙继续进行实验。

2022初二英语上学期第三次联考试卷带答案

2022初二英语上学期第三次联考试卷带答案

2022初二英语上学期第三次联考试卷带答案(学习版)编制人:__________________审核人:__________________审批人:__________________编制学校:__________________编制时间:____年____月____日序言下载提示:该文档是本店铺精心编制而成的,希望大家下载后,能够帮助大家解决实际问题。

文档下载后可定制修改,请根据实际需要进行调整和使用,谢谢!并且,本店铺为大家提供各种类型的学习资料,如英语资料、语文资料、数学资料、物理资料、化学资料、生物资料、地理资料、历史资料、政治资料、其他资料等等,想了解不同资料格式和写法,敬请关注!Download tips: This document is carefully compiled by this editor.I hope that after you download it, it can help you solve practical problems. The document can be customized and modified after downloading, please adjust and use it according to actual needs, thank you!In addition, this shop provides various types of learning materials for everyone, such as English materials, language materials, mathematics materials, physical materials, chemical materials, biological materials, geographic materials, historical materials, political materials, other materials, etc. Please pay attention to the data format and writing method!2022初二英语上学期第三次联考试卷带答案2022-2022学年八年级英语上学期第三次联考试卷(人教新目标带答案)亲爱的考生:欢迎参加考试!请你认真审题,积极思考,仔细答题,发挥最佳水平。

  1. 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
  2. 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
  3. 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。

三校 八英 第一学期第三次月度联考答题卡 共2页 第1页 八 年 级 英 语 答 题 卡 得分_______ 第I 卷 选择题(共95分)
第Ⅱ卷 非选择题 (共55分) 五 词汇运用 用所给词的适当形式填空(共15小题;每小题1分,满分15分) 1.The baby panda ______________ about 100 grams at birth but it was very healthy. (weigh) 2.When autumn comes, farmers are busy ______________ crops.(harvest) 3.Don ’t keep me ______________ so long because I am very busy these days . (wait) 4.All of us should do something ______________ the lovely giant pandas. (protect) 5.I heard your daughter ______________ in the next room at ten last night.(sing) 6.The ______________ names are all on the paper. (tour) 7.Bees and ______________ play among flowers(butterfly) 8.Many wild animals ’ ______________ areas are becoming farmlands. (live) 9.Soon the ______________ season will begin.(snow) 10.Do wolves spend a lot of time ______________ for food every day ?( look ) 11.The zookeeper told us ______________ any noise in the zoo. (not make) 12.The little panda had to look after ____________ when her mother had another baby. (her) 13.Could you please______________ wild animals.(not eat) 14.The horse is standing with its eyes ______________.(close) 15.They count the birds in Zhalong ______________ a year.(one) 六 任务型阅读 请根据材料内容,完成各题。

(共5小题;每小题2分,满分10分) If you want to live for 100 years, you need healthy food and lots of exercise. In the morning, you have a big and nice breakfast. A big and nice breakfast can help you start a day. For lunch and dinner, you can have rice and vegetables. ⑵Healthy people seldom eat much snacks. Sweet snacks give you energy but they’re not healthy. You can have an orange or an apple after each meal. Except for keeping a healthy ⑴______, you need lots of ⑴______ too. You can do sports for 30 minutes every day. After a month, you will feel healthier. You can swim. You can run. You can play ball games. But don’t play computer gam es or chat with friends on the Internet for many hours every day. This is not exercise. Get up from your chair, everyone! Change your diet and lifestyle today! You can live for 100 years. You can always feel young with right food and exercise. 1. 将⑴处分别填上适当的单词。

_____________ 2. 将⑵画线处翻译成汉语: ______________________ 3. How long can we exercise every day? 4. 在文中找出下面句子的同义句。

You can do some exercise for half an hour every day. _____________________________________ 5. 找出文中最能表达中心意思的句子。

_______________________________________________. 七 缺词填空 根据短文内容及首字母提示补全单词(共10小题;每小题1分,共10分) The Internet says people killed a lot of wild animals for m____1____ money. Some people like e____2_____ the wild animals, but they don’t know wild animals can also b ___3___ diseases to people. We must do something to s____4____ the diseases from harming our h____5____. A____6____, many people don’t know the i_____7_____ of protecting the wild animals. The n____8____of wild animals is smaller and smaller. In order to k___9____ the balance of nature, we should try our best to p____10______ wild animals and our environment. 1.m_________ 2. e_________ 3. b_________ 4. s_________ 5.h_________ 6. A_________ 7. i_________ 8. n_________ 9. k_________ 10. p_________ 八 写作(20分) 写一篇请求加入观鸟协会的申请(80字左右) 1. 你叫李明,阳光中学八年级学生 2. 最喜爱的科目是生物,对各种鸟类植物感兴趣 3. 想成为观鸟协会的一员,了解更多鸟类知识 4. 做点事保护鸟类,星期日下午2-5点可以参加活动 Dear Chairperson , I would like to join the Birdwatching Society. First,______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ I will be very happy if I can become a member of the Birdwatching Society. My telephone number is 5558-6930. yours sincerely,
Liming
学校 班级 姓名 考场(考试)号 座位号 密封线内不要答题 ………………
………………装………………………………订………………………………………线
………………………………………………。

相关文档
最新文档