广东省佛山市南海区南海外国语学校2019-2020i年九年级上学期第三次月测数学试卷(无答案)
2019-2020学年佛山市南海区南海中学高三英语第三次联考试卷及参考答案
2019-2020学年佛山市南海区南海中学高三英语第三次联考试卷及参考答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AThank you. It’s my great honor to be given this award.You cannot imagine that I have always been a late starter. Years ago, when I was 16, I took an important exam — GCE(General Certificate of Education), which turned out to be a failure. My dad was reading my report card and saw that my position in class was 29th, but the number in class was 29. It meant that I had achieved the distinction of being bottom of my class.I wasn’t lazy, and I was really trying. You can picture how I felt. Dad put his hand on my shoulder and said, “You can only do the best you can, but whatever you decide to do, make sure you love it.” He was a really sweet guy and a great man. I knew his attempt to hide his disappointment with some of his encouraging words. I was depressed for a week, but his advice was a wake-up call.Fortunately I love working with my hands, and I was good at two things: woodwork and art, and I really loved to draw and paint. I was quite talented. Dad strongly encouraged me to go to art school, which in those days wasn’t the obvious place that a father would suggest.So I got into Hartlepool College of Art. The college was a revelation (出乎意料), the passionate teachers there, who were extremely interested in the students, not just tolerating them but actually engaging with them. It was a world apart from my schooling until then. It’s extraordinary what an enthusiastic teacher can do, drawing the student out, lighting independence, and encouraging a design of your own future, rather than waiting for something to happen. I’m honored to have become one of these passionate teachers years later.My teachers inspired me, and thanks to my dad, here I am tonight. I think I should mention all the talents I have worked with over time, and to my kids and my wife Giannina, thank you.Thank you for this great award. I shall find a very special place for it.1. How did the author feel after taking GCE?A. Happy.B. Upset.C. Tired.D. Relieved.2. What didHartlepoolCollege of Art impress the author most?A. The teachers were strict with students.B. The students set good examples for each other.C. The teachers inspired students’ passion for learning.D. The students got prepared for their lessons independently.3. The author gave this speech to ________.A. share his career choiceB. explain his teaching methodsC. describe his life experienceD. show his appreciationBFairy tales perform many functions. They entertain, encourage imagination and teach problem—solving skills. They can also provide moral lessons, highlighting the dangers of failing to follow the social codes that let human beings coexist in harmony. Such moral lessons may not mean much to a robot, but a team of researchers at Georgia Institute of Technology believes it has found a way to use the fairy tales as moral lessons that AI (artificial intelligence) can take to its cold, mechanical heart.The collected stories of different cultures teach children how to behave in socially acceptable ways with examples of proper and improper behavior in fables, novels and other literature. We believe story comprehension in robots can prevent the intelligent robots from killing humanity which was predicted and feared by some of the biggest names in technology including Stephen Hawking and Bill Gates. This system is called “Quixote” (堂吉诃德). It collects story plotsfrom the Internet and then uses those stories to teach robots how to behave.The experiment done by the designers involves going to a drugstore to purchase some medicine for a human who needs to get it as soon as possible. The robot has three options. It can wait in line; it can interact with the store keeper politely and purchase the medicine with priority; or it can steal the medicine and escape. Without any further directives(指令), the robot will come to the conclusion that the most efficient means of obtaining the medicine is to steal it. But Quixote offers a reward for waiting in line and politely purchasing the medicine and a punishment for stealing it. In this way, the robotwill learn the moral way to behave on that occasion.Quixote would work best on a robot that has a very limited function. It’s a baby step in the direction of teaching more moral lessons into robots. We believe that AI has to be trained to adopt the values of a particular society, and in doing so, it will strive to avoid unacceptable behavior. Giving robots the ability to read and understand our stories may be the most efficient means.4. What function do fairy tales perform in the robots?A. They entertain robots.B. They highlight dangers.C. They make robots more intelligent.D. They enable robots to behave morally.5. What is “Quixote” in the text?A. A punishment systemB. A character in literatureC. A big name in technologyD. A software educating robots.6. What does the designer expect robot to do in the experiment?A. To take advantage of its privilege.B. To finish the task most efficiently.C. To perform in a good mannered way.D. To be rewarded by the storekeeper17. Which of the follow can bestexpress the author’s opinion?A. Robots will definitely have more functions.B. Robots with human’s emotions are perfect.C. Training robots to be socially acceptable is necessary.D. The development of robots is still in a baby step.CWith graduation days being celebrated all over the country, a student who has to use a wheelchair honored his mother on his graduation day in a special way. Easley High School graduate, Alex Mays surprised people present when he got up and walked across the stage at Clemson's Littlejohn Coliseum.“I was really happy—it made me feel good,” Alex said.Alex was not given a chance to live right from his birth. He was born at 25 weeks and weighed just 1 pound, 10 ounces at birth. When he was very young, he had a disease and lost the ability to walk. After his mother's death in 2013, Alex had several other difficult life changes until he came to live with his grandparents, Dousay and her husband, Dewayne. Dousay said that when Alex came to live with them, they decided to bring him up in the best possible way they could.Last fall, Alex said that he would walk across the stage to get his diploma to honor his late mother. Hepracticed hard and worked with a physical therapist for 9 months to complete his plan.The only help Alex got was from his mom's best friend, Tonya Johnson, who pushed his wheelchair to the stage wearing one of his mother's favorite shirts. “I had support from my family. I couldn't have done it without them,” Alex said.“Alex made everyone in the building feel encouraged that day” Pickens County School District public information specialist John Eby said. “The school teachers knew he was going to get up to get his diploma, but the distance he walked was a surprise, even to them,” Eby said.“Some of life's most important tests aren’t given in a classroom; Alex tested himself and passed with flying color1 s,” Eby added.8. In what way did Alex honor his late mother on his graduation day?A. By dressing like her.B. By saying sorry to her.C. By inviting her best friend.D. By walking to get his diploma.9. What can we learn from Paragraph 3?A. Alex was born healthy.B. Alex went through a lot.C. Alex had a purpose in life as a child.D. Alex has lived with his grandparents all the time.10. What did Alex also express on his graduation day?A. His big regret in life.B. His feelings for hisschool.C. His thanks for his family.D. His will to complete his study.11. Which of the following words can best describe Alex?A. Strong-minded.B. Warm-hearted.C. Cool-headed.D. Easy-going.DJapan is known to have higher than average rates of stomach cancer. Recently, the town of Kaneyama in Yamagata Prefecture decided to get its 6, 000 residents (居民) tested.However, the frozen urine samples (尿样) are not tested in conventional ways. Instead, Professor Masao Miyashita and his team are using them in a trial to determine if specially trained cancer-sniffing dogs can accurately detect the disease. Though the study is still in its early stages, Miyashita is thrilled with the results. He said, “In our research so far, cancer detection dogs have been able to find signs of cancer with an accuracy of nearly 100 percent.”Researchers have known about the animals’ superior sensory skills for decades. However, their ability to detect cancer in humans came to light in 1989, after a dog sniffed out early-stage malignant melanoma (恶性黑色素瘤) on a patient’s leg in London. Since then, scientists from many countries have conducted studies to test dogs’ great skill at identifying cancer chemicals.While most dogs can be trained for the task researchers say the best candidates are dogs that are precise, quiet, and perhaps even a little shy. The training process is similar to how dogs are taught to learn any trick — by rewarding them with treats! However, it takes much longer because the dogs have to learn to separate the “cancer scent (气味)”from the thousands of organic compounds (有机化合物) in the human body. Researchers begin by exposing the dogs to urine samples from people with cancer, people with other diseases, and patients with no health issues, Once the dogs are able to accurately identify cancer, they are further trained to detect particular kinds of cancer.Successful as they may be, experts think dogs are unlikely to replace conventional tests. For one, it takes about seven years and costs as much as $45,000 to train a single dog. Klaus Hackner, a researcher and physician who studies dogs detecting cancer in breath samples at Krems University Hospital in Austria, is also not convinced dogs can be relied upon alone. Patients, therefore, have to receive further tests to confirm if they have the disease.12. What do we know about the cancer-sniffing dogs mentioned in Paragraph 2?A. They have done a great job.B. They are trained in a special way.C. They can easily learn to distinguish cancer.D. They can be seen in many Japanese hospitals.13. What does the author intend to do in Paragraph 3?A. Offer readers some advice.B. Add some background information.C. Summarize the previous paragraphs.D. Introduce a new topic for discussion.14. What kind of dog is suitable for the cancer-sniffing job?A. Smart and brave.B. Active and faithful.C. Strong and patient.D. Careful and peaceful.15. What is Klaus Hackner’s opinion on cancer-sniffing dogs?A. They should work as a team.B. They need to receive more training.C. They can replace doctors in detecting cancer.D. They should be used together with traditional tests.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
2019-2020学年广东佛山九年级上英语月考试卷
2019-2020学年广东佛山九年级上英语月考试卷一、单项选择1. As we know, England is ______ European country and Singapore is ______ Asian country.A.an; aB.an; anC.a; anD.a; a2.—Mrs Zhang was sent to teach English in a poor mountain village last year.—She said she would never forget some pleasant ______ while working there.A.experiencesB.experimentsC.expectationsD.expressions3. Let's do it ______. There is only five minutes left.A.quicklyB.hardlyC.politelyD.slowly4. —______ do you play football?—Once a week.A.How longB.How muchC.How farD.How often5. —May I use your cup, Tom?—Sorry, it ______ by my sister just now.A.brokenB.was brokenC.breaksD.is broken6. The teachers ______ came for a visit are foreigners.A.whichB./C.whoD.whom7. Jenny is on holiday now. I wonder ______.A.when will she come backB.when she will come backC.when did she come backD.when she came back8. I don't know if she ______ to my birthday party tomorrow. If she ______, I'll be very happy.es; will comees; comesC.will come; will comeD.will come; comes 9. —______ wonderful dictionary it is! Thank you for buying me such a useful present.—I'm glad you like it.A.HowB.What aC.How aD.What10. —Jerry, do you mind you pointing out your mistakes?—______. Your advice is of great value to me.A.Not at allB.Of courseC.It's my pleasureD.You'd better not二、完形填空In a small town in France, there was a farmer who lived alone. Every day he（1）______ a pound of butter to his neighbour, who was a baker（面包师）. One day the baker decided to（2）______ the butter to see if he was getting a pound. After he weighed it, he found that he wasn't. The baker was very（3）______ and then he took the farmer to the judge（法官）.The judge asked the farmer if he had any way to weigh the butter. The farmer replied. "I am so（4）______ that I do not have enough money to buy anything to weigh it, but I do have a kind of scale（天平）." The judge asked, "Then （5）______ do you weigh the butter?" The farmer replied, "Before the baker started buying butter from mine, I had bought bread from him. So now every time when I bring home the bread from the baker, I（6）______ the scale and give him the butter of the same weight."We（7）______ what we give to others in life. Whenever you take action, ask yourself this（8）______. Am I honest?"Honesty or dishonesty can become a（9）______. Some dishonest people can lie（说谎）without a red face. Others lie so much that they do not even what the truth is any more. But who is it bad for? As a matter of fact, those who lie will hurt（10）______ by their own dishonest behaviour.（1）A.threwB.gaveC.soldD.fed（2）A.cutB.weigheD.watch（3）A.angryB.happyC.pleasedD.excited（4）A.honestB.richC.quietD.poor（5）A.whenB.whatC.howD.why（6）A.put it offB.put it upC.put it downD.put it on（7）A.give backB.get backC.turn backD.look back（8）A.questionB.answerC.ideaD.help（9）A.mistakeB.differenceC.difficultyD.habit（10）A.himselfB.themselvesC.yourselvesD.myself三、阅读理解Let's explore nature!Are you interested in animals and nature? Do you want to enjoy your summer holiday and learn something interesting and useful?Here comes a good way! Youth Explore will provide you with a THREE-DAY SUMMER PROGRAM! You can take part in many activities and meet with different kinds of wild animals such as giraffes, pandas and tigers!Age: Students aged 8-16Group: There are two programs for you to choose from. Each program needs at most 15 studentsProgram costs: 230formembers; 250 for non-membersHow to do: Just come to our center to get an application form. Send your form on or before 12th July, 2019（Friday）. Note:NO FREE LUNCHES ARE PROVIDED because of personal food difference. You may either bring a lunch-box or pay for lunch at our restaurant.Both programs will start 1 DAY later if there is a sign of rainstorms.COME AND JOIN US NOW! DON'T MISS IT!（1）Who can take part in the program?A.Kevin, 21 years old.B.John, 14 years old.C.Sara, 17 years old.D.Kelly, 7 years old.（2）Program A will start on ______ if there is a rainstorm on July 15th.A.TuesdayB.SundayC.MondayD.Wednesday（3）Two members and one non-member should pay ______ for the program.A.$ 730B.$ 750C.$ 690D.$ 710（4）The program does NOT provide free lunches because ______.A.people have different favorite foodB.people will bring lunch-boxesC.the lunches cost too muchD.there are many restaurants（5）From the above information, we can know ______.A.we can only see three kinds of animals in the programB.Youth Explore is the name of the programC.you need to hand in the application form to the programD.more than 15 students are in each programTim Berners-Lee is not the most famous inventor in the world. However, his invention has changed our lives.He was born in London, England in 1955. When he was a small boy, Tim was interested in playing electrical things. He studied science at Oxford University. He made his first computer from an old television at the age of 21.Tim started working on early computers. At that time, they were much bigger than now. He worked in England thenSwitzerland. Tim was really interested in two things, computers and how the brain works. How could the brain connect so many facts so quickly? He had to work with people all over the world. They shared information about computers. It was hard to manage all the information. He answered the same questions again and again. It took a lot of time. It was even difficult for computers in the same office in Switzerland to share information. Tim also forgot things easily. Could a computer work like a brain? Could it "talk" to other computers?There was an Internet already but it was difficult to use. In 1989, Tim Berners-Lee invented the World Wide Web （WWW）all by himself. This had a special language that helped computers talk to each other on the Internet. When people wanted to share information with others, they used the World Wide Web. The Internet grew quickly after that.Tim Berners-Lee doesn't think he did anything special. He says that all of the idea about the Internet were already there. All he did was to put them together. He says that many other people worked together to make the Internet what it is today.Most inventors want to become rich. But Tim gave away the World Wide Web for nothing. He now works in America. He helps people share technology and wants the Internet to be free for everyone to use. Maybe he is the most important but least famous inventor in the world today!（1）What was Tim Berners-Lee interested in?A.Studying how to connect computers.B.Looking for jobs in different cities.C.Exploring how to improve memory.D.Talking to people around the world.（2）The underline word "it" in paragraph 3 refers to "______".A.travelling to the office in SwitzerlandB.working on early computersC.repeating the answers to the same questionsD.connecting different facts together（3）Why is Tim Berners-Lee one of the most important men in the world?A.He helped people understand better how the brain works.B.He made information sharing on the Internet possible.C.He invented the Internet and made it free for everyone to use.D.He made the first computer when he was 21 years old.（4）In what order did the following events take place?a. Tim worked in England.b. Tim worked in Switzerland.c. Tim made his first computer.d. Tim invented the World Wide Web.e. Tim studied science at Oxford University.A.c-d-a-e-bB.c-e-d-a-bC.e-c-a-b-dD.e-b-a-c-d（5）What does the writer think of Tim Berners-Lee?A.He did nothing special but make people a good life.B.He is not famous because he is not rich.C.He has made great achievements in memory research.D.He has changed our lives and he is great.四、六选五阅读从下面选项中选出能填入文中空缺处的最佳选项，使短文意思通顺，内容完整。
广东省南海学校2019-2020学年九年级上月考试卷(数学解析版)
2019-2020学年九年级（上）月考数学试卷一．选择题（共11小题）1．下列选项中是一元二次方程的是（）A．x﹣2y＝3 B．2（x+1）＝3 C．2x2+x﹣4 D．x2+3x﹣4＝0选D．2．一元二次方程x2＝2x的根是（）A．x1＝0，x2＝2 B．x＝0 C．x＝2 D．x1＝0，x2＝﹣2 选A．3．如图，在▱ABCD中，下列说法能判定ABCD是菱形的是（）A．AC＝BD B．BA⊥BD C．AB＝CD D．AD＝BC【解析】∵对角线相等的平行四边形是菱形，或一组邻边相等的平行四边形是平行四边形，∴当AC＝BD或AB＝BC或AB＝AD或AD＝CD或BC＝CD时，平行四边形ABCD是菱形，选A．4．直角三角形两条直角边长分别是6和8，则斜边上的中线长为（）A．3 B．4 C．5 D．6【解析】∵直角三角形两条直角边长分别是6和8，∴斜边＝＝10，∴斜边上的中线长＝×10＝5．选C．5．正方形具有而矩形不一定具有的性质是（）A．对角线垂直B．对边相等C．对角相等D．对边平行【解析】正方形和矩形都是特殊的平行四边形，所以具有平行四边形所有的性质，即对边相等，对角相等，对边平行，正方形的对角线互相垂直，矩形的对角线只是相等不垂直．选A．6．一元二次方程x2+2x+4＝0的根的情况是（）A．有两个相等的实数根B．有两个不相等的实数根C．只有一个实数根D．没有实数根【解析】△＝b2﹣4ac＝22﹣4×1×4＝﹣12，∵﹣12＜0，∴原方程没有实数根．选D．7．根据下列表格对应值：判断关于x的方程ax2+bx+c＝0（a≠0）的一个解x的范围是（）A．x＜3.24 B．3.24＜x＜3.25C．3.25＜x＜3.26 D．3.25＜x＜3.28【解析】由图表可知，ax2+bx+c＝0时，3.24＜x＜3.25．选B．8．如果关于x的一元二次方程ax2+bx+1＝0的一个解是x＝﹣1，则2018﹣a+b＝（）A．2016 B．2017 C．2018 D．2019【解析】把x＝﹣1代入方程ax2+bx+1＝0得a﹣b+1＝0，所以a﹣b＝﹣1，所以2018﹣a+b＝2018﹣（a﹣b）＝2018+1＝2019．选D．9．如图，顺次连接四边形ABCD各边中点得四边形EFGH，要使四边形EFGH为矩形，应添加的条件是（）A．AB∥DC B．AC＝BD C．AC⊥BD D．AB＝DC【解析】依题意得，四边形EFGH是由四边形ABCD各边中点连接而成，连接AC、BD，故EF∥AC∥HG，EH∥BD∥FG，所以四边形EFGH是平行四边形，要使四边形EFGH为矩形，根据矩形的判定（有一个角为直角的平行四边形是矩形）故当AC⊥BD时，∠EFG＝∠EHG＝90度．四边形EFGH为矩形．选C．10．如图，矩形ABCD的对角线AC，BD相交于点O，CE∥BD，DE∥AC，AD＝2，DE＝2，则下列结论错误的是（）A．AB＝2 B．∠E＝60°C．四边形OCED是菱形D．四边形OCED的面积是4【解析】∵四边形ABCD是矩形，∴∠BAD＝90°，AO＝CO＝BO＝DO，∵CE∥BD，DE∥AC，∴四边形OCED是平行四边形，且OC＝OD，∴四边形OCED是菱形，故D选项不符合题意，∴DE＝OC＝OD＝2，∠E＝∠COD，∴BD＝4，∴AB＝＝＝2，故A选项不符合题意，∴AB＝AO＝BO＝2，∴△AOB是等边三角形，∴∠AOB＝60°＝∠COD＝∠E，故B选项不符合题意，∵矩形ABCD的面积＝AB×AD＝4，∴四边形OCED的面积为2，故D选项符合题意选D．11．如图，正方形ABCD中，点E、F分别在BC、CD上，△AEF是等边三角形，连接AC交EF于G，下列结论：①BE＝DF；②∠DAF＝15°；③AC垂直平分EF；④BE+DF＝EF；⑤S△CEF＝2S△ABE，其中正确结论有（）A．2个B．3个C．4个D．5个【解析】∵四边形ABCD是正方形，∴AB＝BC＝CD＝AD，∠B＝∠BCD＝∠D＝∠BAD＝90°．∵△AEF等边三角形，∴AE＝EF＝AF，∠EAF＝60°．∴∠BAE+∠DAF＝30°．在Rt△ABE和Rt△ADF中，，Rt△ABE≌Rt△ADF（HL），∴BE＝DF（故①正确）．∠BAE＝∠DAF，∴∠DAF+∠DAF＝30°，即∠DAF＝15°（故②正确），∵BC＝CD，∴BC﹣BE＝CD﹣DF，即CE＝CF，∵AE＝AF，∴AC垂直平分EF．（故③正确）．设EC＝x，由勾股定理，得EF＝x，CG＝x，AG＝AE sin60°＝EF sin60°＝2×CG sin60°＝x，∴AC＝，∴AB＝，∴BE＝﹣x＝，∴BE+DF＝x﹣x≠x，（故④错误），∵S△CEF＝x2，S△ABE＝x2，∴2S△ABE＝x2＝S△CEF，（故⑤正确）．综上所述，正确的有4个，选C．二．填空题（共6小题）12．将方程3x（x﹣1）＝5（x+1）化为一元二次方程的一般式为3x2﹣8x﹣5＝0 ．【解析】3x（x﹣1）＝5（x+1），3x2﹣3x＝5x+5，3x2﹣8x﹣5＝0．答案：3x2﹣8x﹣5＝0．13．已知菱形两条对角线长分别为4cm和6cm，则菱形的面积等于12cm2．【解析】如图，菱形ABCD的对角线AC与BD相交于点O，且AC＝6cm，BD＝4cm，∵四边形ABCD为菱形，∴AC⊥BD，∴S菱形＝AC•OD+AC•BD＝AC•BD＝×6×4＝12（cm2）．答案12cm2．14．关于x的一元二次方程x2+kx+k﹣2＝0，方程的一个根为x＝﹣2，则方程的另一个根为0 ．【解析】把x＝﹣2代入一元二次方程x2+kx+k﹣2＝0得：4﹣2k+k﹣2＝0，解得：k＝2，即原方程为：x2+2x＝0，解得：x1＝﹣2，x2＝0，即方程的另一个根为0，答案：0．15．某制药厂两年前生产1吨某种药品的成本是100万元，随着生产技术的进步，现在生产1吨这种药品的成本为81万元．则这种药品的成本的年平均下降率为10 %．【解析】设这种药品的成本的年平均下降率为x，则今年的这种药品的成本为100（1﹣x）2万元，根据题意得，100（1﹣x）2＝81，解得x1＝1.9（舍去），x2＝0.1＝10%．故这种药品的成本的年平均下降率为0.1，即10%．16．如图，已知P是正方形ABCD对角线BD上一点，且BP＝BC，则∠ACP度数是22.5 度．【解析】∵ABCD是正方形，∴∠DBC＝∠BCA＝45°，∵BP＝BC，∴∠BCP＝∠BPC＝（180°﹣45°）＝67.5°，∴∠ACP度数是67.5°﹣45°＝22.5°．17．如图，已知矩形ABCD的长和宽分别为16cm和12cm，连接其对边中点，得到四个矩形，顺次连接矩形AEFG各边中点，得到菱形l1；连接矩形FMCH对边中点，又得到四个矩形，顺次连接矩形FNPQ各边中点，得到菱形l2；…如此操作下去，则l4的面积是cm2．【解析】∵矩形ABCD的长和宽分别为16cm和12cm，∴EF＝8cm，AE＝6cm，∴菱形l1的面积＝×8×6＝24cm2，同理，菱形l2的面积＝×4×3＝6cm2，则菱形l3的面积＝×2×＝cm2，∴菱形l4的面积＝×1×＝cm2，答案：．三．解答题（共8小题）18．解不等式组【分析】分别求出不等式组中两不等式的解集，找出解集的公共部分即可．【解析】由①得：x＜4，由②得：x≥0，∴原不等式组的解集为：0≤x＜4．19．解方程：x2﹣8x+7＝0【解析】分解因式可得（x﹣1）（x﹣7）＝0，∴x﹣1＝0或x﹣7＝0，∴x＝1或x＝7．20．先化简，再求值：（﹣）÷，其中x＝．【解析】原式＝＝当x＝时，原式＝＝21．如图，在△ABC中，AB＝BC，点D为AC的中点，四边形ABED是平行四边形，DE交BC于点F，连接CE．求证：四边形BECD是矩形．【解答】证明：∵AB＝BC，点D为AC的中点∴BD⊥AC，AD＝CD．∵四边形ABED是平行四边形，∴BE∥AD，BE＝AD，∴BE＝CD，∴四边形BECD是平行四边形．∵BD⊥AC，∴∠BDC＝90°，∴四边形BECD是矩形．22．某商场销售一批名牌衬衫，平均每天可销售20件，每件盈利40元，为了扩大销售，增加盈利，尽量减少库存，商场决定采取适当的降价措施，经调査发现，如果每件衬衫每降价5元，商场平均每天可多售出10件，求：若商场平均每天要盈利1200元，每件衬衫应降价多少元？【解析】设每件衬衫应降价x元．根据题意，得（40﹣x）（20+2x）＝1200整理，得x2﹣30x+200＝0解得x1＝10，x2＝20．∵“扩大销售量，减少库存”，∴x1＝10应略去，∴x＝20．答：每件衬衫应降价20元．23．如图，△ABC中，∠C＝90°，AC＝8cm，BC＝4cm，一动点P从C出发沿着CB边以1cm/s的速度运动，另一动点Q从A出发沿着AC边以2cm/s的速度运动，P，Q两点同时出发，运动时间为t（s）．（1）当t为几秒时，△PCQ的面积是△ABC面积的？（2）△PCQ的面积能否为△ABC面积的一半？若能，求出t的值；若不能，说明理由．【解析】（1）∵S△PCQ＝t（8﹣2t），S△ABC＝×4×8＝16，∴t（8﹣2t）＝16×，整理得t2﹣4t+4＝0，解得t＝2．答：当t＝2s时△PCQ的面积为△ABC面积的；（2）当S△PCQ＝S△ABC时，t（8﹣2t）＝16×，整理得t2﹣4t+8＝0，△＝（﹣4）2﹣4×1×8＝﹣16＜0，∴此方程没有实数根，∴△PCQ的面积不可能是△ABC面积的一半．24．如图，在Rt△ABC中，∠ACB＝90°，过点C的直线MN∥AB，D为AB边上一点，过点D作DE⊥BC，垂足为F，交直线MN于E，连接CD、BE．（1）求证：CE＝AD；（2）当D为AB中点时，四边形BECD是什么特殊四边形？说明你的理由；（3）在（2）的条件下，当∠A的大小满足什么条件时，四边形BECD是正方形？请说明你的理由．【解答】（1）证明：∵DE⊥BC，∴∠DFB＝90°，∵∠ACB＝90°，∴∠ACB＝∠DFB，∴AC∥DE，∵MN∥AB，即CE∥AD，∴四边形ADEC是平行四边形，∴CE＝AD；（2）：四边形BECD是菱形，理由是：∵D为AB中点，∴AD＝BD，∵CE＝AD，∴BD＝CE，∵BD∥CE，∴四边形BECD是平行四边形，∵∠ACB＝90°，D为AB中点，∴CD＝BD，∴四边形BECD是菱形；（3）当∠A＝45°时，∵∠ACB＝90°，∴∠ABC＝45°，由（2）可知，四边形BECD是菱形，∴∠ABC＝∠CBE＝45°，∴∠DBE＝90°，∴四边形BECD是正方形．25．如图，在△ABC中，AB＝AC，AD⊥BC于点D，BC＝10cm，AD＝8cm，E，F分别是AB，AC的中点．（1）求证：四边形AEDF是菱形；（2）求菱形AEDF的面积；（3）若H从F点出发，沿线段FE以每秒2cm的速度向E点运动，点P从B点出发，在线段BC上以每秒3cm的速度向C点运动，设运动时间为t．①当t为何值时，四边形BPHE是平行四边形？②是否存在t的值，使四边形PCFH是菱形？若存在，求出t的值，若不存在，说明理由．【解答】（1）证明：∵AB＝AC，AD⊥BC，∴D为BC的中点．∵E、F分别为AB、AC的中点，∴DE和DF是△ABC的中位线，∴DE∥AC，DF∥AB，∴四边形AEDF是平行四边形．∵E，F分别为AB，AC的中点，AB＝AC，∴AE＝AF，∴四边形AEDF是菱形，（2）：∵EF为△ABC的中位线，∴EF＝BC＝5．∵AD＝8，AD⊥EF，∴S菱形AEDF＝AD•EF＝×8×5＝20．（3）：①∵EF∥BC，∴EH∥BP．若四边形BPHE为平行四边形，则EH＝BP，∴5﹣2t＝3t，解得：t＝1，∴当t＝1秒时，四边形BPHE为平行四边形．②不存在t的值，使四边形PCFH是菱形；理由如下：∵EF∥BC，∴FH∥PC．若四边形PCFH为菱形，则须FH＝PC＝CF，当FH＝PC时，2t＝10﹣3t，解得：t＝2，∴FH＝PC＝4，∵AB＝AC，AD⊥BC，∴BD＝CD＝BD＝5，∴AC＝＝＝，∵F是AC的中点，∴CF＝AC＝，∴FH＝PC≠CF，∴四边形PCFH是平行四边形，不是菱形；∴不存在t的值，使四边形PCFH是菱形．。
2023年广东省佛山市南海外国语学校中考三模数学试题(含答案解析)
2023年广东省佛山市南海外国语学校中考三模数学试题学校:___________姓名：___________班级：___________考号：___________45．．．．．下列从左到右的变形中，属于因式分解的是（）2y B 432186x y x -=-()()22x y x y =+-D ．(221x y x -+=，265∠=︒，125∠=，则ABC ∠度数是（A ．20°B ．30°7．如图，AB x ∥轴交反比例函数3OAB S = ，则k 的值是（A ．3-B ．38．如图，()2,0A ，()0,4C ，将线段A ．()6,2B ．(2,69．关于x 一元二次方程2x -A ．2022x =B ．x 10．如图，在Rt BDF △中，∠沿着CB 方向平移到点B ，使得点O ，则下列结论：①四边形面积为四边形ABCD 面积的一半．其中正确结论的个数为（A ．4B ．3C ．2D ．1二、填空题13．比较大小：23______14．不等式组240380x x -≤⎧⎨-->⎩的解集为15．如图，在矩形ABCD 中，则图中阴影部分的面积为______三、解答题16．先化简211 12a a a a +⎛+ --+⎝17．如图，在ABC 中，B ∠(1)请用尺规作图法，作B ∠的角平分线痕迹）(2)如果8AB =，求BD 的长．(1)求证：四边形AEBD 是菱形；(2)若210DC BD ==，，求四边形21．如图，平行于y 轴的直尺（一部分点B 和D ，点A 和B 的刻度分别为面直角坐标系内一个单位长度为(1)求反比例函数解析式；(2)若经过A ，C 两点的直线关系式为集；(3)求梯形ABCD 的面积．22．如图1，在矩形ABCD 中，点AC 分别交于点E ，F ，且ACB ∠(1)求证：CAB CED ∽△△；(2)判断直线CE 与O 的位置关系，并证明你的结论；(3)如图2，若点E 落在线段AC 的垂直平分线上，2CD =，求23．如图，抛物线2y x bx c =-++与x 轴交于A ，B 两点，与的坐标为()3,0，C 点的坐标为()0,3．(1)求抛物线的解析式；(2)图1中，点P 为抛物线上的动点，且位于第二象限，过P ，B 两点作直线l 交y 轴于点D ，交直线AC 于点E ．是否存在这样的直线l ：以C ，D ，E 为顶点的三角形与BOD 相似？若存在，请求出这样的直线l 的解析式；若不存在，请说明理由．(3)图2中，点C 和点C '关于抛物线的对称轴对称，点M 在抛物线上，且MBA CBC '∠=∠，求M 点的横坐标．参考答案：∵反比例函数图象在第二象限，∴0k <，∴6k =-，故选：D ．【点睛】本题考查了反比例函数k 的几何意义，熟练掌握反比例函数k 的几何意义是解题的关键．8．A【分析】过点B 作BD x ⊥轴于点D ，则90AOC BDA ∠=∠=︒，结合()2,0A ，()0,4C ，可得4OC =，2OA =，根据旋转的性质可得AC BA =，90BAC ∠=︒，易得OCA DAB ∠=∠，然后证明OAC DBA ≌，由全等三角形的性质可得4OC DA ==，2OA DB ==，结合点B 在第一象限，即可获得B 点坐标．【详解】解：过点B 作BD x ⊥轴于点D ，如下图，则90AOC BDA ∠=∠=︒，∵()2,0A ，()0,4C ，∴4OC =，2OA =，根据题意，将线段AC 绕点A 顺时针旋转90︒到AB ，∴AC BA =，90BAC ∠=︒，∴90OAC OCA OAC DAB ∠+∠=∠+∠=︒，∴OCA DAB ∠=∠，在OAC 和DBA 中，AOC BDA OCA DAB AC BA ∠=∠⎧⎪∠=∠⎨⎪=⎩，∴(AAS)OAC DBA ≌，∴4OC DA ==，2OA DB ==，（2）解：∵60ABC ∠=︒，45C ∠=︒，D 点坐标为()4,0，CD y ∥∴当4x =时，6342y ==，C ∴点坐标为34,2⎛⎫ ⎪⎝⎭，3AB ∴=，422BD =-=，CD ()12ABDC S CD AB BD ∴=+⋅梯形133222⎛⎫=+⨯ ⎪⎝⎭92=．【点睛】本题考查了待定系数法求反比例函数解析式，比例函数与不等式的关系，面积问题等，掌握求法是解题的关键．22．(1)见解析(2)直线CE 与O 相切，理由见解析(3)43OA =【分析】（1）证明90B D ∠=∠=（2）连接OE ，证明DAC ∠=90DCE DEC ∠+∠=︒，可得AEO ∠【详解】（1）证明：四边形ABCD 为矩形．∴90B D ∠=∠=︒，∵ACB DCE ∠=∠，∴CAB CED ∽△△．（2）判断：直线CE 与O 相切．证明：连接OE ，∵OA OE =，∴DAC AEO ∠=∠，∵四边形ABCD 是矩形，∴BC AD ∥，∴ACB DAC ∠=∠，∵ACB DCE ∠=∠，∴DAC DCE ∠=∠，∴AEO ACB DCE ∠=∠=∠，∵四边形ABCD 是矩形，∴90D Ð=°，∴90DCE DEC ∠+∠=︒∴90AEO DEC ∠+∠=︒，∴1809090OEC ∠=︒-︒=︒，即OE EC ⊥，∵OE 为半径，∴直线CE 与O 相切；（3）∵点E 落在线段AC 的垂直平分线上，∴AE CE =，∴DAC ECA ∠=∠，2⎝⎭由点、B N的坐标得，直线BN解方程213 2322x x x-++=-+，解得：12x=-或3（舍去），∴M的横坐标为1 2-；当30,2N⎛⎫-⎪⎝⎭，如图：同理可得，直线BN解析式为：。
广东省佛山市2019-2020学年中考三诊语文试题含解析
广东省佛山市2019-2020学年中考三诊语文试题一、选择题1．下列各句中，加点成语使用不正确的一项是（）A．一些西方媒体别有用心地宣称“中印会陷入冷战”，这是西方有意渲染和扩大中印在地缘政治方面的矛盾，有挑拨离间．．．．的嫌疑。
B．讲好中国故事，不能一味地选择富丽堂皇．．．．的镜头或夸张的语言来表现中国的繁荣富强，而是要用历史的眼光讲述中国人的时代变迁及其主题。
C．博鳌亚洲论坛2019年年会即将举行，而博鳌各主要路段被装扮一新，路面干干净净，路旁三角梅等花朵鲜艳绽放，彩旗花枝招展．．．．，迎接博鳌亚洲论坛年会的到来。
D．评论写作不能关起门来坐而论道、对空言说，而要有天下关怀，胸中有乾坤，笔下有千钧，对社会保持足够的敏感和洞见，才能倚马可待．．．．，百步穿杨。
2．下列各句中有语病的一项是（）A．网络文学领域实绩与泡沫共存，竞争与机遇同在，同时有很大的提升空间，网络文学的健康发展尤其需要良好的“生态系统”。
B．当老师交给我们任务的时候，我们大家有既光荣又愉快的感觉是颇难形容的。
C．在互联网高速发展、生活节奏越来越快的社会转型期，传统文化的传承和弘扬也面临着挑战。
D．世界上的事情都是干出来的，人世间的一切幸福都来自辛勤的劳动，只有通过不懈奋斗才能实现人世间的美好梦想。
3．下面加点字注音完全正确的一项是（）A．丘壑．hè谰．nán越俎代庖．bāo广袤无垠．yínB．疟．子yào菜畦．qí断壁残垣．yuán歇．斯底里xiēC．阴霾．mái纳罕．hǎng怒不可遏．jiè味同嚼．蜡jiáoD．喑．哑àn篡．夺cuàn鳞次栉．比jié如坐针毡．zhān4．下列句子中，没有语病的一项是（）A．在信息时代，一个人是否具有快速阅读、捕捉有效信息，决定着一个人成就的大小。
B．《三体》的英文译者刘宇昆被誉为“将中国科幻推向世界的引路人”，这确立了中国科幻小说在国际上的地位。
2019-2020年九年级(上)第三次月考数学试卷(解析版)(I)
2019-2020年九年级（上）第三次月考数学试卷（解析版）(I)一、选择题：（1-10每小题2分，11-16每小题2分，共38分）1．在下列关于x的函数中，一定是二次函数的是（）A．y=x2B．y=ax2+bx+c C．y=8x D．y=x2（1+x）2．一个点到圆的最小距离为6cm，最大距离为9cm，则该圆的半径是（）A．1.5cm B．7.5cm C．1.5cm或7.5cm D．3cm或15cm3．如图，AB、AC是⊙O的两条弦，∠A=30°，过点C的切线与OB的延长线交于点D，则∠D的度数为（）A．30°B．35°C．40°D．45°4．已知抛物线的解析式为y=（x﹣2）2+1，则这条抛物线的顶点坐标是（）A．（﹣2，1） B．（2，1）C．（2，﹣1） D．（1，2）5．在数轴上，点A所表示的实数为3，点B所表示的实数为a，⊙A的半径为2．下列说法中不正确的是（）A．当a＜5时，点B在⊙A内B．当1＜a＜5时，点B在⊙A内C．当a＜1时，点B在⊙A外D．当a＞5时，点B在⊙A外6．把二次函数y=﹣x2﹣x+3用配方法化成y=a（x﹣h）2+k的形式（）A．y=﹣（x﹣2）2+2 B．y=（x﹣2）2+4 C．y=﹣（x+2）2+4 D．y=2+37．圆I是三角形ABC的内切圆，D，E，F为3个切点，若∠DEF=52°，则∠A 的度数为（）A．68°B．52°C．76°D．38°8．抛物线y=﹣2（x﹣1）2上有三点A（﹣1，y1），B（，y2），C（2，y3），则y1，y2，y3从小到大是（）A．y1＜y2＜y3B．y2＜y3＜y1C．y2＜y1＜y3D．y1＜y3＜y29．如图，PA、PB、DE分别切⊙O于A、B、C，DE分别交PA，PB于D、E，已知P 到⊙O的切线长为8CM，则△PDE的周长为（）A．16cm B．14cm C．12cm D．8cm10．如图，Rt△ABC中，∠ACB=90°，AC=4，BC=6，以斜边AB上的一点O为圆心所作的半圆分别与AC、BC相切于点D、E，则AD的长为（）A．2.5 B．1.6 C．1.5 D．111．函数y=ax+1与y=ax2+bx+1（a≠0）的图象可能是（）A．B． C． D．12．有一个内角为120°的菱形的内切圆半径为，则该菱形的边长是（）A．B．C．4 D．613．已知二次函数y=ax2+bx+c（a≠0）的图象如图所示，给出以下结论：①a+b+c ＜0；②a﹣b+c＜0；③b+2a＜0；④abc＞0．其中所有正确结论的序号是（）A．③④B．②③C．①④D．①②③14．如图，已知平行四边形ABCD中，AB=5，BC=8，cosB=，点E是BC边上的动点，当以CE为半径的⊙C与边AD不相交时，半径CE的取值范围是（）A．0＜CE≤8 B．0＜CE≤5C．0＜CE＜3或5＜CE≤8 D．3＜CE≤515．抛物线y=ax2+bx+c（a≠0）部分点的横坐标x，纵坐标y的对应值如下表x…﹣2﹣1012…y…04664…从上表可知，下列说法错误的是（）A．抛物线与x轴的一个交点坐标为（3，0）B．函数y=ax2+bx+c的最大值为6C．抛物线的对称轴是直线x=D．在对称轴左侧，y随x增大而增大16．如图，等腰直角三角形ABC（∠C=90°）的直角边长与正方形MNPQ的边长均为4cm，CA与MN在同一直线上，开始时A点与M点重合，让△ABC向右平移，直到C点与N点重合时为止，设△ABC与正方形MNPQ的重叠部分（图中阴影部分）的面积为ycm2，MA的长度为xcm，则y与x之间的函数关系大致为（）A．B．C．D．二、填空题（每小题4分，共16分）17．已知二次函数的图象经过原点及点（，），且图象与x轴的另一交点到原点的距离为1，则该二次函数解析式为．18．如图，已知两同心圆，大圆的弦AB切小圆于M，若环形的面积为9π，则AB的长是．19．将抛物线y=x2﹣4x+9向平移个单位，向平移个单位，得到抛物线y=x2﹣6x+5．20．如图，∠AOB=60°，点M是射线OB上的点，OM=4，以点M为圆心，2cm为半径作圆．若OA绕点O按逆时针方向旋转，当OA和⊙M相切时，OA旋转的角度是．三、解答题21．某公司草坪的护栏是由50段形状相同的抛物线组成的，为牢固起见，每段护栏需按间距0.4m 加设不锈钢管（如图）做成立柱．为了计算所需不锈钢管立柱的总长度，设计人员测得如图所示的数据．（1）求该抛物线的表达式（2）计算所需不锈钢管的总长度．22．某贸易公司购进“长青”胶州大白菜，进价为每棵20元，物价部门规定其销售单价每棵不得超过80元，也不得低于30元．经调查发现：日均销售量y（棵）与销售单价x（元/棵）满足一次函数关系，并且每棵售价60元时，日均销售90棵；每棵售价30元时，日均销售120棵．（1）求日均销售量y与销售单价x的函数关系式；（2）在销售过程中，每天还要支出其他费用200元，求销售利润w（元）与销售单价x之间的函数关系式；并求当销售单价为何值时，可获得最大的销售利润？最大销售利润是多少？23．已知：如图，△ABC内接于⊙O，点D在OC的延长线上，sinB=，∠CAD=30°．（1）求证：AD是⊙O的切线；（2）若OD⊥AB，BC=5，求AD的长．24．如图，△ABC是一块铁皮余料．已知底边BC=160cm，高AD=120cm．在铁皮余料上截取一个矩形EFGH，使点H在AB上，点G在AC上，点E、F在BC上，AD交HG于点M．（1）设HG=y cm，HE=x cm，试确定用x表示y的函数表达式．（2）当x为何值时，矩形EFGH的面积S 最大？25．在平面直角坐标系xOy中，⊙O的半径为1，P是坐标系内任意一点，点P到⊙O的距离S P的定义如下：若点P与圆心O重合，则S P为⊙O的半径长；若点P与圆心O不重合，作射线OP交⊙O于点A，则S P为线段AP的长度．图1为点P在⊙O外的情形示意图．（1）若点B（1，0），C（1，1），，则S B= ；S C= ；S D= ；（2）若直线y=x+b上存在点M，使得S M=2，求b的取值范围；（3）已知点P，Q在x轴上，R为线段PQ上任意一点．若线段PQ上存在一点T，满足T在⊙O内且S T≥S R，直接写出满足条件的线段PQ长度的最大值．xx学年河北省石家庄二十二中九年级（上）第三次月考数学试卷参考答案与试题解析一、选择题：（1-10每小题2分，11-16每小题2分，共38分）1．在下列关于x的函数中，一定是二次函数的是（）A．y=x2B．y=ax2+bx+c C．y=8x D．y=x2（1+x）【考点】二次函数的定义．【分析】根据二次函数的定义：y=ax2+bx+c（a≠0．a是常数），可得答案．【解答】解：A、y=x2是二次函数，故A符合题意；B、a=0时是一次函数，故B不符合题意，C、y=8x是一次函数，故C不符合题意；D、y=x2（1+x）不是二次函数，故D不符合题意；故选：A．2．一个点到圆的最小距离为6cm，最大距离为9cm，则该圆的半径是（）A．1.5cm B．7.5cm C．1.5cm或7.5cm D．3cm或15cm【考点】点与圆的位置关系．【分析】点P应分为位于圆的内部于外部两种情况讨论．当点P在圆内时，直径=最小距离+最大距离；当点P在圆外时，直径=最大距离﹣最小距离．【解答】解：分为两种情况：①当点P在圆内时，最近点的距离为6cm，最远点的距离为9cm，则直径是15cm，因而半径是7.5cm；②当点P在圆外时，最近点的距离为6cm，最远点的距离为9cm，则直径是3cm，因而半径是 1.5cm．故选C．3．如图，AB、AC是⊙O的两条弦，∠A=30°，过点C的切线与OB的延长线交于点D，则∠D的度数为（）A．30°B．35°C．40°D．45°【考点】切线的性质．【分析】由于CD是切线，可知∠OCD=90°，而∠A=35°，利用圆周角定理可求∠COD，进而可求∠D．【解答】解：连接OC，∵CD是切线，∴∠OCD=90°，∵∠A=30°，∴∠COD=2∠A=60°，∴∠D=90°﹣60°=30°．故选：A．4．已知抛物线的解析式为y=（x﹣2）2+1，则这条抛物线的顶点坐标是（）A．（﹣2，1） B．（2，1）C．（2，﹣1） D．（1，2）【考点】二次函数的性质．【分析】直接根据顶点式的特点写出顶点坐标．【解答】解：因为y=（x﹣2）2+1为抛物线的顶点式，根据顶点式的坐标特点可知，顶点坐标为（2，1）．故选B．5．在数轴上，点A所表示的实数为3，点B所表示的实数为a，⊙A的半径为2．下列说法中不正确的是（）A．当a＜5时，点B在⊙A内B．当1＜a＜5时，点B在⊙A内C．当a＜1时，点B在⊙A外D．当a＞5时，点B在⊙A外【考点】点与圆的位置关系．【分析】先找出与点A的距离为2的点1和5，再根据“点与圆的位置关系的判定方法”即可解．【解答】解：由于圆心A在数轴上的坐标为3，圆的半径为2，∴当d=r时，⊙A与数轴交于两点：1、5，故当a=1、5时点B在⊙A上；当d＜r即当1＜a＜5时，点B在⊙A内；当d＞r即当a＜1或a＞5时，点B在⊙A外．由以上结论可知选项B、C、D正确，选项A错误．故选：A．6．把二次函数y=﹣x2﹣x+3用配方法化成y=a（x﹣h）2+k的形式（）A．y=﹣（x﹣2）2+2 B．y=（x﹣2）2+4 C．y=﹣（x+2）2+4 D．y=2+3【考点】二次函数的三种形式．【分析】利用配方法先提出二次项系数，在加上一次项系数的一半的平方来凑完全平方式，把一般式转化为顶点式．【解答】解：y=﹣x2﹣x+3=﹣（x2+4x+4）+1+3=﹣（x+2）2+4故选C．7．圆I是三角形ABC的内切圆，D，E，F为3个切点，若∠DEF=52°，则∠A 的度数为（）A．68°B．52°C．76°D．38°【考点】三角形的内切圆与内心．【分析】先利用切线的性质得∠IDA=∠IFA=90°，则根据四边形的内角和得到∠A+∠DIF=180°，再根据圆周角定理得到∠DIF=2∠DEF=104°，然后利用互补计算∠A的度数即可．【解答】解：∵圆I是三角形ABC的内切圆，∴ID⊥AB，IF⊥AC，∴∠IDA=∠IFA=90°，∴∠A+∠DIF=180°，∵∠DIF=2∠DEF=2×52°=104°，∴∠A=180°﹣104°=76°．故选C．8．抛物线y=﹣2（x﹣1）2上有三点A（﹣1，y1），B（，y2），C（2，y3），则y1，y2，y3从小到大是（）A．y1＜y2＜y3B．y2＜y3＜y1C．y2＜y1＜y3D．y1＜y3＜y2【考点】二次函数图象上点的坐标特征．【分析】根据二次函数的性质求出抛物线的对称轴，根据二次函数的增减性解答．【解答】解：∵抛物线y=﹣2（x﹣1）2的对称轴是x=1，∴x=﹣1时的函数值与x=3时的函数值相等，当x＞1时，y随x的增大而减小，∵＜2＜3，∴y1＜y3＜y2，故选：D．9．如图，PA、PB、DE分别切⊙O于A、B、C，DE分别交PA，PB于D、E，已知P 到⊙O的切线长为8CM，则△PDE的周长为（）A．16cm B．14cm C．12cm D．8cm【考点】切线的性质．【分析】由切线长定理可知AD=CD、BE=CE，PA=PB，则可求得△PDE的周长=PA+PB，可求得答案．【解答】解：∵PA、PB、DE分别切⊙O于A、B、C，∴PA=PB=8cm，AD=CD，BE=CE，（cm），∴PD+DE+PE=PD+DC+CE+PE=PD+AD+BE+PE=PA+PB=8+8=16故选A．10．如图，Rt△ABC中，∠ACB=90°，AC=4，BC=6，以斜边AB上的一点O为圆心所作的半圆分别与AC、BC相切于点D、E，则AD的长为（）A．2.5 B．1.6 C．1.5 D．1【考点】切线的性质．【分析】连结OD、OE，如图，先根据切线的性质得OD⊥AC，OE⊥BC，再判断四﹣AD=4﹣AD，接着证明Rt△AOD∽Rt△ABC，然边形ODCE为正方形得到OD=CD=AC后利用相似比计算AD的长．【解答】解：连结OD、OE，如图，∵以点O为圆心所作的半圆分别与AC、BC相切于点D、E，∴OD⊥AC，OE⊥BC，而∠ACB=90°，∴四边形ODCE为矩形，∵OD=OE，∴四边形ODCE为正方形，∴OD=CD=AC﹣AD=4﹣AD，∵∠OAD=∠BAC，∴Rt△AOD∽Rt△ABC，∴=，即=，∴AD=1.6．故选B．11．函数y=ax+1与y=ax2+bx+1（a≠0）的图象可能是（）A．B． C． D．【考点】二次函数的图象；一次函数的图象．【分析】根据a的符号，分类讨论，结合两函数图象相交于（0，1），逐一排除；【解答】解：当a＞0时，函数y=ax2+bx+1（a≠0）的图象开口向上，函数y=ax+1的图象应在一、二、三象限，故可排除D；当a＜0时，函数y=ax2+bx+1（a≠0）的图象开口向下，函数y=ax+1的图象应在一二四象限，故可排除B；当a=0时，两个函数的值都为1，故两函数图象应相交于（0，1），可排除A．正确的只有C．故选C．12．有一个内角为120°的菱形的内切圆半径为，则该菱形的边长是（）A．B．C．4 D．6【考点】菱形的性质；勾股定理；切线长定理．【分析】根据菱形的内切圆半径为即可求菱形的高，菱形的一个内角为120°则其邻角为60°，在直角三角形ABE中即可求的AB即菱形的边的长．【解答】解：过A作AE⊥BC，∵内切圆半径为，∴AE的长度为2，∵∠BAD=120°，则∠ABC=60°，在Rt△ABC中，AE=2，∠ABC=60°，∴AB=4，故选 C．13．已知二次函数y=ax2+bx+c（a≠0）的图象如图所示，给出以下结论：①a+b+c ＜0；②a﹣b+c＜0；③b+2a＜0；④abc＞0．其中所有正确结论的序号是（）A．③④B．②③C．①④D．①②③【考点】二次函数图象与系数的关系．【分析】由抛物线的开口方向判断a的符号，由抛物线与y轴的交点判断c的符号，然后根据对称轴及抛物线与x轴交点情况进行推理，进而对所得结论进行判断．【解答】解：①当x=1时，y=a+b+c=0，故①错误；②当x=﹣1时，图象与x轴交点负半轴明显大于﹣1，∴y=a﹣b+c＜0，故②正确；③由抛物线的开口向下知a＜0，∵对称轴为0＜x=﹣＜1，∴2a+b＜0，故③正确；④对称轴为x=﹣＞0，a＜0∴a、b异号，即b＞0，由图知抛物线与y轴交于正半轴，∴c＞0∴abc＜0，故④错误；∴正确结论的序号为②③．故选：B．14．如图，已知平行四边形ABCD中，AB=5，BC=8，cosB=，点E是BC边上的动点，当以CE为半径的⊙C与边AD不相交时，半径CE的取值范围是（）A．0＜CE≤8 B．0＜CE≤5C．0＜CE＜3或5＜CE≤8 D．3＜CE≤5【考点】直线与圆的位置关系；平行四边形的性质；解直角三角形．【分析】过A作AM⊥BC于N，CN⊥AD于N，根据平行四边形的性质求出AD∥BC，AB=CD=5，求出AM、CN、AC、CD的长，即可得出符合条件的情况．【解答】解：如图，过A作AM⊥BC于N，CN⊥AD于N，∵四边形ABCD是平行四边形，∴AD∥BC，AB=CD=5，∴AM=CN，∵AB=5，cosB=，∴BM=4，∵BC=8，∴CM=4=BC，∵AM⊥BC，∴AC=AB=5，由勾股定理得：AM=CN==3，∴当以CE为半径的圆C与边AD有两个交点时，半径CE的取值范围是3＜CE≤5，故选D．15．抛物线y=ax2+bx+c（a≠0）部分点的横坐标x，纵坐标y的对应值如下表x…﹣2﹣1012…y…04664…从上表可知，下列说法错误的是（）A．抛物线与x轴的一个交点坐标为（3，0）B．函数y=ax2+bx+c的最大值为6C．抛物线的对称轴是直线x=D．在对称轴左侧，y随x增大而增大【考点】抛物线与x轴的交点；二次函数的最值．【分析】根据表格的数据首先确定抛物线的对称轴，然后利用抛物线的对称性可以确定抛物线与x轴的另一个交点坐标，也可以确定抛物线的最大值的取值范围，也可以确定开口方向．【解答】解：根据表格数据知道：抛物线的开口方向向下，∵x=0，x=1的函数值相等，∴对称轴为x=，所以选项C正确，不符合题意；∴抛物线与x轴的另一个交点坐标为：（3，0），所以选项A正确，不符合题意；在对称轴左侧，y随x增大而增大，最大值大于6．所以选项D正确，不符合题意；选项B错误，符合题意；故选B．16．如图，等腰直角三角形ABC（∠C=90°）的直角边长与正方形MNPQ的边长均为4cm，CA与MN在同一直线上，开始时A点与M点重合，让△ABC向右平移，直到C点与N点重合时为止，设△ABC与正方形MNPQ的重叠部分（图中阴影部分）的面积为ycm2，MA的长度为xcm，则y与x之间的函数关系大致为（）A．B．C．D．【考点】动点问题的函数图象；二次函数的图象．【分析】首先确定每段与x的函数关系类型，根据函数的性质确定选项．【解答】解：当x≤4cm时，重合部分是边长是x的等腰直角三角形，面积y=x2，是一个开口向上的二次函数；当x＞4时，重合部分是直角梯形，面积y=8﹣（x﹣4）2，即y=﹣x2+4x，是一个开口向下的二次函数．故选B．二、填空题（每小题4分，共16分）17．已知二次函数的图象经过原点及点（，），且图象与x轴的另一交点到原点的距离为1，则该二次函数解析式为y=﹣x2+x或y=x2+x．．【考点】待定系数法求二次函数解析式．【分析】设二次函数的解析式为y=ax2+bx+c（a≠0），由图象与x轴的另一交点到原点的距离为1可得到抛物线与x轴的另一交点坐标为（1，0）或（﹣1，0），然后分别把（0，0）、（1，0）、（﹣，﹣）或（0，0）、（﹣1，0）、（﹣，﹣）代入解析式中得到两个方程组，解方程组即可确定解析式．【解答】解：设二次函数的解析式为y=ax2+bx+c（a≠0），当图象与x轴的另一交点坐标为（1，0）时，把（0，0）、（1，0）、（﹣，﹣）代入得，解方程组得，则二次函数的解析式为y=﹣x2+x；当图象与x轴的另一交点坐标为（﹣1，0）时，把得，解方程组得，则二次函数的解析式为y=x2+x．所以该二次函数解析式为y=﹣x2+x或y=x2+x．18．如图，已知两同心圆，大圆的弦AB切小圆于M，若环形的面积为9π，则AB的长是 6 ．【考点】切线的性质；垂径定理．【分析】环形的面积为9π，就是大圆面积﹣小圆的面积，根据圆的面积公式，可得π×OA2﹣π×OM2=9π，解得OA2﹣OM2=9，再根据勾股定理可知就是AM的平方，所以AM=3，AB=6．【解答】解：连接OA、OM，如图所示：∵大圆的弦AB切小圆于M，∴AB⊥OM，∴AM=BM，∵环形的面积为9π，根据圆的面积公式可得：π×OA2﹣π×OM2=9π，解得：OA2﹣OM2=9，根据勾股定理可知：AM2=OA2﹣OM2，∴AM=3，∴AB=2AM=6．19．将抛物线y=x2﹣4x+9向右平移 1 个单位，向下平移9 个单位，得到抛物线y=x2﹣6x+5．【考点】二次函数图象与几何变换．【分析】根据配方法，可得顶点式解析式，根据平移规律，可得到答案．【解答】解：y=x2﹣4x+9配方，得y=（x﹣2）2+5；y=x2﹣6x+5配方，得y=（x﹣3）2﹣4．抛物线y=x2﹣4x+9向右平移 1个单位，向下平移 9个单位，得到抛物线y=x2﹣6x+5，故答案为：右，1，下，9．20．如图，∠AOB=60°，点M是射线OB上的点，OM=4，以点M为圆心，2cm为半径作圆．若OA绕点O按逆时针方向旋转，当OA和⊙M相切时，OA旋转的角度是30°或90°．【考点】切线的性质．【分析】OA与⊙O相切时，有两种情况：①切线在OB右侧；②切线在OB左侧；解法相同，都是连接圆心与切点，通过构建的直角三角形求解．【解答】解：如图；①当OA旋转到OE位置时，与圆M相切于点E，连接ME；则ME=2，∠MEO=90°；Rt△OEM中，sin∠MOE==，∴∠MOE=30°，∴∠AOE=∠AOB﹣∠MOE=30°；②当OA旋转到OF位置时，与圆M相切于点F，连接MF；则MF=2，∠MFO=90°；Rt△OFM中，sin∠MOF==，∴∠MOF=30°，∴∠AOF=∠AOB+∠FOB=90°；故OA旋转的角度为30°或90°．三、解答题21．某公司草坪的护栏是由50段形状相同的抛物线组成的，为牢固起见，每段护栏需按间距0.4m 加设不锈钢管（如图）做成立柱．为了计算所需不锈钢管立柱的总长度，设计人员测得如图所示的数据．（1）求该抛物线的表达式（2）计算所需不锈钢管的总长度．【考点】二次函数的应用．【分析】（1）根据所建坐标系特点可设解析式为y=ax2+c的形式，结合图象易求B点和C点坐标，代入解析式解方程组求出a，c的值得解析式；（2）根据对称性求B3、B4的纵坐标后再求出总长度．【解答】解：（1）由题意得B（0，0.5）、C（1，0）设抛物线的解析式为：y=ax2+c代入得a=﹣0.5，c=0.5，故解析式为y=﹣0.5x2+0.5；（2）如图1所示：∵当x=0.2时，y=0.48，当x=0.6时，y=0.32，∴B1C1+B2C2+B3C3+B4C4=2×（0.48+0.32）=1.6米∴所需不锈钢管的总长度为： 1.6×50=80米．22．某贸易公司购进“长青”胶州大白菜，进价为每棵20元，物价部门规定其销售单价每棵不得超过80元，也不得低于30元．经调查发现：日均销售量y（棵）与销售单价x（元/棵）满足一次函数关系，并且每棵售价60元时，日均销售90棵；每棵售价30元时，日均销售120棵．（1）求日均销售量y与销售单价x的函数关系式；（2）在销售过程中，每天还要支出其他费用200元，求销售利润w（元）与销售单价x之间的函数关系式；并求当销售单价为何值时，可获得最大的销售利润？最大销售利润是多少？【考点】二次函数的应用．【分析】（1）设一次函数解析式为y=kx+b，把（60，90），（30，120）分别代入上式得到一次函数解析式；（2）根据题意得到W=（x﹣20）（﹣x+150）﹣200，配方后求最大值．【解答】解：（1）设一次函数解析式为设一次函数解析式为y=kx+b，把（60，90），（30，120）分别代入上式得，，解得．故y=﹣x+150，（30≤x≤80）．（2）根据题意得W=（x﹣20）（﹣x+150）﹣200=﹣x2+170x﹣3200=﹣（x2﹣170x+852﹣852）﹣3200=﹣（x﹣85）2+852﹣3200=﹣（x﹣85）2+852﹣3200=﹣（x﹣85）2+4025．当x=80时取得最大值，为W最大值=﹣（80﹣85）2+4025=4000元．23．已知：如图，△ABC内接于⊙O，点D在OC的延长线上，sinB=，∠CAD=30°．（1）求证：AD是⊙O的切线；（2）若OD⊥AB，BC=5，求AD的长．【考点】切线的判定．【分析】（1）连接OA，由于sinB=，那么可求∠B=30°，利用圆周角定理可求∠AOC=60°，而OA=OB，那么△AOC是等边三角形，从而有∠OAC=60°，易求∠OAD=90°，即AD是⊙O的切线；（2）由于OC⊥AB，OC是半径，利用垂径定理可知OC是AB的垂直平分线，那么CA=CB，而∠B=30°，则∠BAC=30°，于是有∠DAE=60°，∠D=30°，在Rt△ACE 中，利用三角函数值可求AE，在Rt△ADE中利用30°的锐角所对的直角边等于斜边的一半，可求AD．【解答】证明：连接OA，（1）∵sinB=，∴∠B=30°，∠AOC=60°，又∵OA=OC，∴△AOC是等边三角形，∴∠OAC=60°，∴∠OAD=60°+30°=90°，∴AD是⊙O的切线；（2）∵OC⊥AB，OC是半径，∴BE=AE，∴OD是AB的垂直平分线，∴∠DAE=60°，∠D=30°，在Rt△ACE中，AE=cos30°×AC=，∴在Rt△ADE中，AD=2AE=5．24．如图，△ABC是一块铁皮余料．已知底边BC=160cm，高AD=120cm．在铁皮余料上截取一个矩形EFGH，使点H在AB上，点G在AC上，点E、F在BC上，AD交HG于点M．（1）设HG=y cm，HE=x cm，试确定用x表示y的函数表达式．（2）当x为何值时，矩形EFGH的面积S 最大？【考点】相似三角形的应用；二次函数的最值．【分析】（1）先表示出AM，再根据相似三角形对应高的比等于相似比列式整理即可；（2）根据矩形的面积公式列式整理，再根据二次函数的最值问题求解即可．【解答】解：（1）∵矩形EFGH，AD是高，∴MD=HE=x，HG∥BC，∴AM=AD﹣MD=120﹣x，△AHG∽△ABC，∴=，即=，∴y=﹣x+160；（2）矩形EFGH的面积S=xy=x（﹣x+160），=﹣x2+160x，=﹣（x2﹣120x+3600）+4800，=﹣（x﹣60）2+4800，所以，当x=60时，S取最大值4800．25．在平面直角坐标系xOy中，⊙O的半径为1，P是坐标系内任意一点，点P 到⊙O的距离S P的定义如下：若点P与圆心O重合，则S P为⊙O的半径长；若点P与圆心O不重合，作射线OP交⊙O于点A，则S P为线段AP的长度．图1为点P在⊙O外的情形示意图．（1）若点B（1，0），C（1，1），，则S B= 0 ；S C= ﹣1 ；S D= ；（2）若直线y=x+b上存在点M，使得S M=2，求b的取值范围；（3）已知点P，Q在x轴上，R为线段PQ上任意一点．若线段PQ上存在一点T，满足T在⊙O内且S T≥S R，直接写出满足条件的线段PQ长度的最大值．【考点】圆的综合题．【分析】（1）根据点的坐标和新定义解答即可；（2）根据直线y=x+b的特点，结合S M=2，根据等腰直角三角形的性质解答；（3）根据T在⊙O内，确定S T的范围，根据给出的条件、结合图形求出满足条件的线段PQ长度的最大值．【解答】解：（1）∵点B（1，0），∴S B=0，∵C（1，1），∴S C=﹣1，∵，∴S D=，故答案为：0；﹣1；；（2）设直线y=x+b与分别与x轴、y轴交于F、E，作OG⊥EF于G，∵∠FEO=45°，∴OG=GE，当OG=3时，GE=3，由勾股定理得，OE=3，此时直线的解析式为：y=x+3，∴直线y=x+b上存在点M，使得S M=2，b的取值范围是﹣3≤b≤3；（3）∵T在⊙O内，∴S T≤1，∵S T≥S R，∴S R≤1，∴线段PQ长度的最大值为1+2+1=4．xx年3月10日。
广东省佛山市南海外国语学校2019-2020学年第二学期九年级线上测试3数学试卷
⼴东省佛⼭市南海外国语学校2019-2020学年第⼆学期九年级线上测试3数学试卷南海外国语学校九年级线上模拟测试3数学试题全卷120分限时90分钟⼀、选择题（每⼩题3分，共30分）1.下列四个数中，最⼩的数是（）A. 0B. 1C. 12-D. -12.将如图所⽰的直⾓三⾓形ABC 绕直⾓边AB 旋转⼀周得到⼀个⼏何体，从正⾯看这个⼏何体得到的平⾯图形应为（）A. B. C. D.3.2020年4⽉1⽇上午，国电浙能宁东发电公司⽅家庄电⼚公布，第⼀季度该⼚新投产的百万千⽡级机组完成发电量19.36亿千⽡时，⽤科学记数法表⽰19.36亿正确的是（）A. 919.3610?B. 81.93610?C. 91.93610?D. 100.193610? 4.下列运算正确的是（）A. 235x y xy +=B. ()3236xy x y =C. ()2239x x +=+D. 1052x x x ÷=5.下列航空公司的标志中，是中⼼对称图形的是（）A. B. C. D.6.已知在12件相同的产品中，8件⼀等品，3件⼆等品，1件次品，任取1件产品是⼀等品的概率是（） A. 14 B. 13 C. 23 D. 127.如果关于x 的⼀元⼆次⽅程2410x x a ++-=有两个实数根，那么a 的取值范围是（）A. 5a >B. 5a <C. 5a ≥D. 5a ≤8.点O ，A ，B ，C 在数轴上的位置如图所⽰，O 为原点，2AC =，OA OB =.若点C 所表⽰的数为x ，则点B 所表⽰的数为（）A. ()2x -+B. ()2x --C. 2x +D. 2x -9.如图，在ABC △中，D 、E 分别是AB 、BC 上的点，且//DE AC ，若:1:4BDE CDE S S =△△，则:CDE ADC S S =△△（）A. 1:5B. 4:25C. 1:6D. 4:2310.如图，已知四边形ABCD 为矩形，点E 在BC 上（不与B ，C 重合），连接AE ，DE ，以AE 为⼀边作正⽅形AEFG ，使得点F 在边CD 上，给出以下结论：①ABE ECF ?△△；②AED CDE ABE S S S =+△△△；③2EC BC DF =+；④ADE AQG ∠=∠；⑤2EF AQ EC =?；其中正确的结论的个数是（）A.1个B.2个C.3个D.4个⼆、填空题（每⼩题4分，共28分）11.分解因式：39x x -=______.12.⼀个n 边形的内⾓和是它外⾓和的4倍，则n =______.13.不等式组：()13320x x x -15.如图，某商场停车场门⼝的柱⼦上⽅挂着⼀块收费标准牌CD ，收费标准牌的⼀侧⽤绳⼦AD 和BC 牵引着两排⼩彩旗，经过测量得到如下数据：4AM =⽶，8AB =⽶，45MAD ∠=?，30MBC ∠=?，则CD 的长度为______⽶.（结果保留根号）16.如图，点()1,2A 为双曲线k y x=上的⼀点，连接AO 并延长与双曲线在第三象限交于点B ，M 为y 轴正半轴上⼀点，连接MA 并延长与双曲线交于点N ，连接BM 、BN ，已知MBN △的⾯积为6，则点N 的坐标为______.17.将⼤⼩相同的正三⾓形按如图所⽰的规律拼图案，其中第①个图案中有6个⼩三⾓形和1个正六边形；第②个图案中有10个⼩三⾓形和2个正六边形；第③个图案中有14个⼩三⾓形和3个正六边形；…；按此规律排列下去，已知⼀个正六边形的⾯积为a ，⼀个⼩三⾓形的⾯积为b ，则第③个图案中所有的⼩三⾓形和正六边形的⾯积之和为______.（结果⽤含a 、b 的代数式表⽰）。
广东省佛山市南海区南海实验中学2023-2024学年九年级上学期月考数学试题(含答案)
2023~2024学年上学期初三第三次学情反馈数学一、选择题（每题3分，共30分）1．衢州莹白瓷以瓷质细腻、釉面柔和、透亮皎洁，似象牙又似羊脂白玉而名闻遐迩，被誉为瓷中珍品．如图是衢州莹白瓷的直口杯，它的左视图是（）A ．B ．C ．D ．2．若线段a ，b ，c满足，且，则b 的值为（ ）A ．4B ．6C ．9D ．363．在中，，且，则（ ）ABC ．D4．已知五边形，相似比为4：9．若五边形的周长为12，则五边形ABCDE 的周长为（ ）A ．B ．C ．12D ．275．某生态公园的人工湖周边修葺了3条湖畔小径，如图小径BC ，AC 恰好互相垂直，小径AB 的中点M 刚好在湖与小径相交处．若测得BC 的长为，AC的长为，则C ，M 两点间的距离为（）A ．B ．C ．D ．6．已知的边AB ，AD 长是关于x 的一元二次方程的两个实数根，若，则a bb c=4,9a c ==Rt ABC △90C ∠=︒3c b =cos A =1311111ABCDE A B C D E ∽五边形11111A B C D E 1632740.8km 0.6km 0.5km 0.6km 0.8km 1kmABCD 240x mx -+=AB =另一边AD 的长（ ）A ．2B ．C ．4D ．7．第19届亚运会会徽名为“潮涌”，吉祥物是一组名为“江南忆”的机器人，出自唐朝诗人白居易名句“江南忆，最忆是杭州”．小东收集了如图所示的四张小卡片（除正面内容不同外，其余均相同），现将四张卡片背面朝上，洗匀放好．小东从中随机抽取两张卡片，则他抽到的两张卡片恰好是“吉祥物莲莲”和“吉祥物底底”的概率是（）A．B ．C ．D ．8．关于反比例函数，下列说法不正确的是（ ）A ．点在它的图象上B ．此函数图象关于直线对称C ．当时，D ．每个分支上，y 随x 的增大而减少9．在认识特殊平行四边形时，小红用四根长度均为的木条首尾相接，钉成正方形ABCD ，转动这个四边形，使它的形状改变，当转动到四边形时，测得，则，C 之间的距离比变形前A ，C 之间的距离短（）A． B ．C ．D ．10．如图，矩形ABCD 对角线AC ，BD 相交于点O ，过点A 作于点M ，交BC 于点E ，过点C 作于点N ，交AD 于点F ，连接EN ，FM ，若，则下列结论：①；②；③；④四边形AECF 是菱形．正确的有（ ）116112161412y x=()2,6--y x =-4x <-3y >-13cm 11A BCD 124cm BD =1A 3cm ()5cm -5cm ⎫-⎪⎪⎭()10cm-AE BD ⊥CF BD ⊥tan AOB ∠=EN FM =2AM MD ND =⋅60AEN ∠=︒A ．①②③B ．①②④C ．②③④D ．①③④二、填空题（每题3分，共18分）11．小亮在解一元二次方程时，不小心把常数项丢掉了，已知这个一元二次方程有两个相等的实数根，则丢掉的常数项为_______．12．在测量旗杆高度的活动课上，某小组学生于同一时刻在阳光下对一根直立于平地的竹竿及其影长和旗杆的影长进行了测量，得到如图所示的数据，请根据这些数据计算出旗杆的高度为_______m ．13．在一个不透明的盒子有7枚黑棋子和若干枚白棋子，这些棋子除颜色外无其他差别．从盒子随机取出一枚棋子，记下颜色后再放回盒中．不断重复上述过程，一共取了300次，其中有100次取到黑棋子，由此估计盒中有______枚白棋子．14．如图，在长为52米，宽为20米的长方形地面上修筑宽度相同的小路（图中阴影部分），余下部分种植草坪．要使小路的面积为100平方米，设小路的宽为x 米，则根据题意可列的方程为_______．15．如图，在平行四边形ABCD 中，点E 为边AD 的黄金分割点，且，则_______．16．如图，在平面直角坐标系中，矩形OABC 边BC 取点E ，使，连接AE ，OB 交于点D ，已知的面积3．若反比例函数的图象恰好经过点D ，则_______．260x x -+=AE ED >CFAF=2BE CE =AOD △ky x=k =三、解答题（一）（第17题4分，第18、19、20题各6分，共22分）1718．为便于劳动课程的开展，学校打算在校园东北角建一个矩形生态园ABCD ．如图，生态园一面靠墙（墙足够长），另外三面用长的篱笆围成．若要使得生态园的面积为，则AB 的长为多少?19．如图，在平面直角坐标系中，的顶点的坐标分别为．（1）以点O 为位似中心，位似比为2：1，将放大得，请在网格中画出（不要超出方格区域）；（2）与的面积比为_______；20．如图，在等腰中，，D 是BC 边上的中点，E 点是AD 上一点，连接BE ，过C 作，交AD 延长线于点F ，连接BF ，CE ．试判断四边形BFCE 的形状，并证明你的结论．四、解答题（二）（第21、22题各8分，第23题10．分，共26分）21．心理学家研究发现，一般情况下，一节课40分钟中，学生的注意力随教师讲课的变化而变化：开始上课2452cos 603tan 30︒+︒-︒18m 236m ABC △()()()2,2,5,4,1,5A B C ------ABC △111A B C △111A B C △111A B C △ABC △ABC △AB AC =CF BE ∥时，学生的注意力逐步增强，中间有一段时间学生的注意力保持较为理想的稳定状态，随后学生的注意力开始分散．经过实验分析可知，学生的注意力指标数y 随时间x （分钟）的变化规律如图所示（其中AB ，BC 分别为线段，轴，CD 为双曲线的一部分），其中AB 段的关系式为．（1）点B 坐标为_______；（2）根据图中数据，求出CD 段双曲线的表达式：（3）一道数学竞赛题，需要讲20分钟，为了效果较好，要求学生的注意力指标数最低达到32，那么经过适当安排，老师能否在学生注意力达到所需的状态下讲解完这道题目?22．综合实践为了测量一条两岸平行的河流宽度，三个数学兴趣小组设计了不同的方案，他们在河的南岸点A 处测得北岸的树H 恰好在A 的正北方向．测量方案与数据如下表：课题测量河流宽度测量工具测量角度的仪器、皮尺等测量方案示意图说明点B ，C 在点A 的正东方向点B ，D 在点A 的正东方向点B 在点A 的正东方向点C 在点A 的正西方向测量数据，，．，，．，，．（1）哪个小组的数据无法计算出河宽?（2）请选择其中一个方案及其数据求出河宽（结果精确到）．BC x ∥220y x =+60m BC =70ABH ∠=︒35ACH ∠=︒20m BD =70ABH ∠=︒35ACH ∠=︒101m BC =70ABH ∠=︒35ACH ∠=︒1m（参考数据：）23．如图，已知中，E 为CD 上一点，且，连接AE 并延长，交BD 于点M ，交BC 的延长线于点N ．（1）若，求BN 的长；（2）求证：．五、解答题（三）（每题12分，共24分）24．综合运用如图，直线与x 轴交于C 点，与y 轴交于B 点，在直线上取点，过点A 作反比例函数的图象．（1）求a 的值及反比例函数的表达式；（2）点P 为反比例函数图象上的一点，看，求点P 的坐标．（3）在x 轴是否存在点Q ，使得，若存在请求出点Q 的坐标，若不存在请说明理由．25．综合探究如图①，在矩形ABCD 中，，点E 在边BC 上，且，动点P 从点E 出发，沿折线以每秒2个单位长度的速度运动．作，EQ 交边AD 或边DC 于点Q ，连接PQ ，当点Q 与点C 重合时，点P 停止运动，设点P 的运动时间为t秒．sin 700.94,sin 350.57,tan 70 2.75,tan 350.70︒≈︒≈︒≈︒≈ABCD 3DE CE =6AD =2AM ME MN =⋅22y x =+()2,A a ()0ky x x=>()0ky x x=>2POB AOB S S =△△BOA OAQ ∠=∠6,10AB AD ==4BE =EB BA AD --90PEQ ∠=︒()0t >（1）当点P 和点B 重合时，线段PQ 的长为_________；（2）当点Q 和点D 重合时，求的值；（3）当点P 在边AD 上运动时，的形状始终是等腰直角三角形，如图②，请说明理由；（4）将沿直线PQ 翻折到，点E 对称点为点F ，当点F 刚好在矩形ABCD 的边上（包括顶点），请直接写出t的值．tan PQE PQE △PQE △PQF △2023~2024学年上学期初三第三次学情反馈数学参考答案及评分标准一、选择题（每题3分，共30分）题号12345678910答案DBCAABCCDB二、填空题（每题3分，共18分）11．912．1213．1414．1516．三、解答题（一）（第17题4分，第18、19、20题各6分，共22分）17解：原式 3分4分18．解：∵四边形ABCD 是矩形，∴设，则 1分由题意得：，3分整理得：解得： 4分∴或6答：AB 的长为或． 6分（或直接设求解）19．（1）解：如图所示，即为所求．4分23220100x x x +-=1852452cos 603tan 30-︒︒+︒21232=⨯-⨯1111=+-=AD BC xm ==()182AB CD x m ==-()18236x x -=29180x x -+=123,6x x ==18212x -=12m 6m AB xm =111A B C △注： （没有不扣分）（2）4：16分20．解：四边形BFCE 是菱形，证明如下：1分∵，D 是BC 边上的中点∴∵，∴3分在和中，，．4分∴，又∴四边形BFCE 是平行四边形5分∵，∴四边形BFCE 是菱形6分四、解答题（二）（第21、22题各8分，第23题10分，共26分）21．（1）点B 坐标为； 1分（2）解：由图：点C 的坐标为， 2分设C 、D 所在双曲线的解析式为，把代入得，，∴．5分（3）令，∴．6分令，∴， 7分∵，∴经过适当安排，老师能在学生注意力达到所需的状态下讲解完这道题目． 8分22．（1）第二个小组的数据无法计算出河宽． 2分（2）第一个小组的解法：∵，∴，即∴，()()()1114,410,,,82,10A B C AB AC =,90BD CD ADB =∠=︒CF BE ∥,EBD FCD BED CFD ∠=∠∠=∠BDE △CDF △BED CFD EBD FCD BD CD ∠=∠⎧⎪∠=∠⎨⎪=⎩()BDE CDF AAS △≌△CF BE =CF BE ∥90ADB ∠=︒()10,40()24,402ky x=()24,40C 960k =()96024y x x=>22032y x =+=6x =96032y x==30x =3062420-=>70,35ABH ACH ∠=︒∠=︒35BHC ABH ACH ∠=∠-∠=︒BHC ACH ∠=∠60m BH BC ==中，，∴ 8分（只要选择一个方案计算出河宽即可）第三个小组的解法：设，则∵，∴解得：答：河宽为．23．（1）解：∵四边形ABCD 是平行四边形，∴∴，又，∴，∴，∴，∴ 4分（2）证明：由得：，又∴，∴6分．同理可证：，∴8分∴，则 10分五、解答题（三）（每题12分，共24分）24．解：（1）把代入得，，1分把代入，得，∴反比例函数的函数表达式为，3分Rt ABH △sin AHABH BH∠=sin 70600.9456m AH BH =⋅︒≈⨯≈AH xm =,tan 35tan 70AH AHCA AB ==︒︒CA AB CB +=101tan 35tan 70x x+=︒︒56mx ≈56m ,6AD BC BC AD ==∥DAE N ∠=∠NEC AED ∠=∠ADE NCE △∽△3AD DENC CE==2NC =8BN BC CN =+=AD BC ∥DAM N ∠=∠AMD NMB∠=∠AMD NMB △∽△AM DMMN BM=AMB EMD △∽△AM BMEM DM=AM EMMN AM=2AM ME MN =⋅()2,A a 22y x =+2226a =⨯+=()2,6A ky x=12k =12y x=（2）解：把代入，即4分∴，∴又 6分∴，代入，得∴点P 坐标为 7分（3）在x 轴存在点Q ，使得．当点Q 在x 轴正半轴上时，如图，过点A 作轴交x 轴于，则，∴点当点Q 在x 轴负半轴上时，如图，设与y 轴交于点∵，∴，则，解得：，∴设直线表达式为，把分别代入，∴，解得，∴直线的表达式为，当时，，即点的坐标为，0x =222y x =+=()0,2B 12,2212AOB OB S ==⨯⨯=△24POB AOB S S ==△△1242POB P S x =⨯⨯=△4P x =12y x =3y =()4,3BOA OAQ ∠=∠1AQ y ∥1Q 1BOA OAQ ∠=∠()2,0Q 2AQ ()0,D b 2BOA OAQ ∠=∠OD AD =2222(6)b b +-=103b =100,3D ⎛⎫ ⎪⎝⎭2AQ y mx n =+()102,6,0,3A D ⎛⎫ ⎪⎝⎭26103m n n +=⎧⎪⎨=⎪⎩43103m n ⎧=⎪⎪⎨⎪=⎪⎩2AQ 41033y x =+0y =52x =-2Q 5,02⎛⎫- ⎪⎝⎭综上所述，点Q 的坐标为或25．（1） 2分解析：如图所示，当点P 和点B 重合时，∴，在中，，即：﹔（2）当点Q 和点D 重合时，如图所示：∵，∴，∴，∴，∴，∵四边形ABCD 是矩形，∴，则，∴，∴，∴，∴， 6分（3）过P 作于点F ，则有，又∵矩形ABCD ，∴()2,05,02⎛⎫-⎪⎝⎭6,4QE AB BE ===Rt QBE△BQ ===PQ =90,90PEQ PBE ECD ∠=︒∠=∠=︒1290,2390∠+∠=︒∠+∠=︒13∠=∠PBE ECD △∽△PB BE EC CD =6,10AB CD AD BC ====6EC BC BE AD BE =-=-=466PB =4PB =PE EQ ====2tan 3PE PQE QE ∠===PF BC ⊥1390,6PFE PF AB ∠+∠=∠=︒==90,10B C AD BC ∠=∠=︒==又∵，∴，∴，∵，∴，∴∴，∴，又∵，∴是等腰直角三角形；（4）或①如图所示，当点P 在BE 上时，点F 落在AB 上∵，在中，，则，∵，∴，在中，，∴，解得：，②当P 点在AB 上时，当F ，A重合时符合题意，此时如图，90PEQ ∠=︒1290∠+∠=︒23∠=∠4,10BE BC ==6CE =PF CE=PFE ECQ △≌△PE EQ =90PEQ ∠=︒PQE △t =176t =7t =6,4QE QF AQ BE ====Rt AQF△AF ===6BF =-2PE t =42,2BP t PF PE t =-==Rt PBF △222PF PB FB =+()(()2222642t t =-+-t =则，在中，∴，解得，③当点P 在AD 上，当F ，∴D 重合时，此时点Q 与点C 重合，则PFQE 是正方形，此时；综上所述，或()224,624102PB t BE t PE AP AB PB t t =-=-==-=--=-Rt PBE △222PE PB BE =+222(102)(24)4t t -=-+176t =2327t =++=t =176t =7t =。
2019-2020学年佛山市南海区南海中学高三英语第三次联考试题及答案
2019-2020学年佛山市南海区南海中学高三英语第三次联考试题及答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项ANothing beats live music, but the venue makes a difference. When you're able to score tickets to an incredible concert in an incredible place, you won't forget the experience. Here are some of the coolest music venues from around the world. If you haven't been to any of these, you've got some traveling to do.Red Rocks, Morrison, the United StatesRed Rocks might be the most beautiful and famous venue in the United States. At 6,450 feet above sea level, Red Rocks is a geologically formed natural stage. Its massive sandstone provides a perfect stage for jam bands. If you're into the blues and jazz, you'll have no trouble finding something in line with your interests.Meet Factory, Prague, Czech RepublicSmallest venues on this list, Meet Factory is an art gallery, theater, and music venue. The venue only accommodates 1,000 people, so you won't see any huge names come through. Still, it's a great place to see up-and-coming local acts, and if you've got an eye for contemporary art, you'll love your time here.Arena of NÎmes, NÎmes, FranceOriginally built around A. D. 70, the Arena of Nimes presents concertgoers with an interesting question: Should they enjoy the music, or marvel at the architecture? The Arena is, after all, one of the world's best-preserved Roman theaters. Many major touring acts plan stops at the Arena of Nimes, especially during the venue's annual festival.Sydney Opera House, Sydney, AustraliaThe Sydney Opera House is one of the world's most famous performing venues. I's also one of the most distinctive buildings in Sydney, thanks to the breathtaking design by Danish architect Utzon. It hosts about 40 events per week, so whether you're into jazz, rock, classical music, or opera, you'll find something to watch.1.Where can you enjoy music in natural beauty?A.At Red Rocks.B.At Meet Factory.C.At Arena of Nimes.D.At Sydney Opera House.2.What is special about Meet Factory?A.It enjoys breathtaking scenery.B.It hosts both musical and artistic events.C.It is the largest venue of all.D.It is famous for contemporary music.3.What do the listed music venues have in common?A.They have a long history.B.They are built near the sea.C.They accommodate thousands of people.D.They are beautiful tourist attractions.BIn recent years,people have been focusing on the quality of food that children are fed in schools. Former First Lady Michelle Obama worked hard to make school lunches healthier, resulting in new menus that featured less fat and salt, more fruits and vegetables.But high-quality nutrients count for little when there is no time to eat them. Amy Ettinger reports, "There is no national standard on how much time kids get to eat that meal. " And with schools being occupied with test scores, teachers are using every available minute for lesson time, which often leaves kids without enough eating time.This is a problem because the length of the school lunch period is a key factor (因素) in how much nutrition children actually gel. Research has found that having less than 20 minutes for lunch results in children consuming much less of their lunch than those with more than 20 minutes.This is really terrible. For many low-income kids, that cafeteria lunch can represent half their daily energy intake. There's also another terrible message that it's acceptable to wolf down food as fast as possible before rushing off to your next class. Cafeteria time should be a chance to interact with friends, to learn important social skills, to observe and share varieties of food. It should be a break in day, a chance to relax before heading into the afternoon.As Ettinger explains,some parents are hoping the National Parent Teacher Association will address this issue. This, in turn, would help parents push their kids' schools for better lunch time standards. Meanwhile, if you have a kid in this situation, you can help by packing a healthy lunch to spare them the cafeteria lineup. Make the foods easy to eat, provide non-messy snacks that can be eaten in class, put great effort into serving a hearty breakfast, and sit down as a family for dinner whenever possible.4. What did Michelle Obama make efforts to improve?A. The quality of school lunches.B. The performance of school kids.C. The school lunch time kids have.D. The eating habits of school kids.5. What happens to children in American schools?A. They are occupied with many tests.B. They fail to get along with each other.C. They consume more meat than before.D. They have less lunch time than before.6. How are low-income kids influenced by the problem at school?A. They can't go to classes on time.B. They can't have enough energy.C. They can't share different kinds of food.D. They can't hold a positive attitude toward life.7.What can parents do to solve the problem?A. Prepare a better lunch for their kids.B. Stop their kids going to the cafeteria.C. Force schools to make adjustments to lunch.D. Guide their kids on how to pack their own lunch.CThe Great Barrier Reef's outlook remains “very poor” despite coral (珊瑚) recovery over the past year, Australian government scientistssaid Monday, just days before a UNESCO ruling on the site's world heritage (遗产) status.The United Nations cultural agency recommended last month that the world's largest reef (珊瑚礁) system be placed on its endangered list because of damage to the corals largely caused by climate change.The Australian Institute of Marine Science (AIMS) said the corals were now in a “recovery window” after a decade of harmful heat stress and cyclones (旋风). But such opportunities were becoming rarer due to the influence ofclimate change, the government agency, which has monitored the reef for 35 years, said in its annual report released today. “The increasing emergence of climate-related extreme weather events and starfish outbreaks is causing more severe and frequent pressures, giving the reef fewer opportunities like this to recover,”CEO Paul Hardisty said. The scientists surveyed 127 reef sites in 2021 and found hard coral cover had increased at 69 of the 81 locations surveyed in the past two years.Separate scientific research released last October found the 2, 300-kilometre (1, 400 miles) system had lost half its corals since 1995, with a series of ocean heatwaves causing mass coral death.Britta Schaffelke, research program director at AIMS, said the latest findings provided a slight hope that thereef still has the power of recovering. But she added that its future is still very poor because of the dangers of climate change and other factors that are affecting the reef.UNESCO has urged Australia to take urgent climate action but the government has long resisted calls to commit to net zero emissions (排放) by 2050. The government has said it hopes to meet the target “as soon as possible” without harming its economy, insisting dealing with climate change requires a global effort. The reef was worth about US $4. 8 billion a year in tourism for the Australian economy and there are fears that an “in danger” listing could weaken its tourist appeal.8. What is the major cause of the damage to the corals?A. The climate change.B. Lack of money.C. Over development.D. Too many tourists.9. What is mainly talked about in Paragraph 3?A. The result of the survey.B. The efforts AIMS has made.C. The slight chance of the recovery.D. The terrible situation of the climate.10. What is Britta Schaffelke's attitude towards the future of the reef?A. Unclear.B. Positive.C Intolerant. D. Anxious.11. What can we infer from the last paragraph?A. Australia wants to put the reef on the endangered list.B. The Australian government has ignored UNESCO's demand.C. Australia hopes to keep a balance between emission target and its economy.D. The Australian government refuses to take its share of responsibility of climate change.DOwning a dog is associated with a significantly lower risk of heart disease and death, according to a comprehensive new study published by a team of Swedish researchers on Friday in the journal Scientific Reports.The scientists followed 3.4 million people over the course of 12 years and found that adults who lived alone and owned a dog were 33 percent less likely to die during the study than adults who lived alone without dogs. In addition, the single adults with dogs were 36 percent less likely to die from heart disease.“Dog ownership was especiallyprominentas a protective factor in persons living alone, which is a groupreported previously to be at higher risk of heart disease and death than those living in a multi-person household,” Mwenya Mubanga, a Ph.D. student at Uppsala University in Uppsala, Sweden, and the lead junior author of the study, said in a statement announcing its findings. The link between dog ownership and lower mortality(死亡率)was less pronounced in adults who lived either with family members or partners, but still present, according to the study. “Perhaps a dog may stand in as an important family member in the single households,” Mubanga added. “Another interesting findingwas that owners of dogs which were intended originally for hunting were most protected.”The study, which is the largest to date on the health relations of owning a dog, suggested that some of the reasons dog owners may have a lower risk of mortality and heart disease were because dog owners walk more. “These kind of epidemiological (流行病学的)studies look for associations in large populations but do not provide answers on whether and how dogs could protect their owners from heart disease,” Tove Fall, a senior author of the study and a professor at Uppsala University, said in a statement“We know that dog owners in general have a higher level of physical activity, which could be one explanation to the observed results,” Fall added. “Other explanations include an increased well-being and social contacts or effects of the dog on the bacterial microbiome(微生物菌群) in the owner.” Fall added that because all participants of dog owners in Sweden or other “European populations with similar culture regarding dog ownership.”12. Why did the researchers do the study related to 3.4 million people’s health and the dogs?A. To help Europeans,B. To find their association.C. To protect unhealthy adults.D. To reduce risk of heart disease.13. What does the underlined word “prominent” probably mean in Para.3?A. Universal.B. Confusing.C. Appealing.D. Important14. What’s the main idea of the text?A. Adults living with dogs are less likely to die.B. Swedish people are very fond of animal pets.C. Keeping a dog is a popular and healthy hobby.D. Owning dogs reduces the risk of heart disease.15. What’s the writer’s attitude towards owning a dog?A. Positive.B. Negative.C. Objective.D. Contradictory.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
广东省佛山市南海实验中学2022-2023学年九年级上学期 数学第三次月考测试题
广东省佛山市南海实验中学2022-2023学年第一学期九年级数学第三次月考测试题（附答案）一、选择题（每小题3分，共30分）1．用配方法解方程x2﹣6x+4＝0，下列配方正确的是（）A．（x﹣3）2＝13B．（x+3）2＝13C．（x﹣3）2＝5D．（x+3）2＝52．关于x的方程x2+mx+6＝0的一个根为﹣2，则另一个根是（）A．﹣3B．﹣6C．3D．63．如果分式的值为0，那么x的值是（）A．±1B．1C．﹣2D．﹣14．如图所示，在矩形ABCD中，已知AE⊥BD于E，∠DBC＝30°，BE＝1cm，则AE的长为（）A．3cm B．2cm C．2cm D．cm5．菱形ABCD的对角线AC，BD相交于点O，E，F分别是AD，CD边上的中点，连接EF．若EF＝，BD＝2，则菱形ABCD的面积为（）A．2B．C．6D．86．如图，正方形ABCD的面积为256，点F在AD上，点E在AB的延长线上，Rt△CEF 的面积为200，则BE的长为（）A．10B．11C．12D．157．在“﹣3，﹣2，﹣1，0，1，2，3”七个数中，任取一个数等于a，恰好使方程（a2﹣1）x2+（a+2）x+a﹣3＝0是一元二次方程的概率是（）A．B．C．D．18．如图，在▱ABCD中，E是AD上一点，且EM∥AD，EN∥CD，则下列式子中错误的是（）A．B．C．D．9．如图所示几何体的左视图是（）A．B．C．D．10．如图，点B是反比例函数y＝（x＞0）图象上一点，过点B分别向坐标轴作垂线，垂足为A，C．反比例函数y＝（x＞0）的图象经过OB的中点M，与AB，BC分别相交于点D，E．连接DE并延长交x轴于点F，点G与点O关于点C对称，连接BF，BG．则△BDF的面积为（）A．B．C．2D．3二、填空题（每小题4分，共28分）11．若直角三角形斜边上的高和中线长分别是3cm和4cm，则它的面积是．12．设α、β是方程x2+2x﹣2021＝0的两根，则α2+3α+β的值为．13．一个口袋中装有8个黑球和若干个白球，现从口袋中随机摸出一球，记下其颜色，再把它放回口袋中．不断重复上述过程，若共摸了200次，其中有50次摸到黑球，因此可估计口袋中大约有白球个．14．如图，在△ABC中，DE∥BC，AD＝9，DB＝3，CE＝2，则AC的长为．15．如图，一棵树（AB）的高度为7.5米，下午某一个时刻它在水平地面上形成的树影长（BE）为10米，现在小明想要站这棵树下乘凉，他的身高为 1.5米，那么他最多离开树干米才可以不被阳光晒到？16．如图，点A是反比例函数y＝在第四象限上的点，AB⊥x轴，若S△AOB＝1，则k的值为．17．如图，AD是△ABC的中线，E是AD上的一点，且DE＝2AE，CE的延长线交AB于点F．若AF＝1.6cm，则AB＝cm．三、解答题（共62分）18．解方程：3x（x﹣3）＝2x﹣6．19．在不透明的袋子里装有2个红球、1个蓝球（除颜色外其余都相同），（1）第一次任意摸出一个球（不放回），第二次再摸出一个球，请用画树状图或列表的方法，求两次摸到一红一蓝的概率．（2）若向袋中再放入若干个同样的蓝球，搅拌均匀后，使从袋中摸出一个蓝球的概率为，求后来放入袋中的蓝球个数．20．如图1，是一个长方体截成的几何体，请在网格中依次画出这个几何体的三视图．21．如图，矩形ABCD中，AB＝8厘米，BC＝12厘米，P、Q分别是AB、BC上运动的两点．若点P从点A出发，以1厘米/秒的速度沿AB方向运动，同时，点Q从点B出发以2厘米/秒的速度沿BC方向运动，设点P，Q运动的时间为x秒．（1）当x为何值时，△PBQ的面积等于12厘米2；（2）当x为何值时，以P，B，Q为顶点的三角形与△BDC相似？22．“天下面食，尽在三晋”，山西面食历史悠久．太原一家特色小面店希望在旅游旺季期间获得较好的收益，经测算知，该小面的成本价为每碗6元，借鉴以往经验，若每碗售价为25元，平均每天可销售300碗，售价每降低1元，平均每天可多销售30碗．设每碗售价降低x元．（1）平均每天可销售碗（用含x的代数式表示）；（2）为了维护城市形象，规定每碗售价不得超过20元，那么当每碗售价定为多少元时，店家才能每天盈利6300元？23．如图，在△ABD中，AB＝AD，AO平分∠BAD，过点D作AB的平行线交AO的延长线于点C，连接BC．（1）求证：四边形ABCD是菱形；（2）如果OA，OB（OA＞OB）的长（单位：米）是一元二次方程x2﹣7x+12＝0的两根，求AB的长以及菱形ABCD的面积；（3）在（2）的条件下，若动点M从A出发，沿AC以2米/秒的速度匀速直线运动到点C，动点N从B出发，沿BD以1米/秒的速度匀速直线运动到点D，当M运动到C点时，运动停止．若M、N同时出发，问出发几秒钟后，△MON的面积为2平方米．24．如图，在平面直角坐标系中，点B在x轴上，∠ABO＝90°，AB＝BO，直线y＝kx﹣4与反比例函数y＝（x＞0）的图象交于点A，与y轴分别交于点C．（1）求k的值；（2）点D与点O关于AB对称，连接AD，CD．证明：△ACD是直角三角形；（3）在（2）的条件下，点E在反比例函数的图象上，若S△ECD＝S△OCD，直接写出点E 的坐标．25．△ABC中，∠BAC＝90°，AB＝AC，点D为直线BC上一动点（点D不与B，C重合），以AD为边在AD右侧作正方形ADEF，连接CF．（1）探究猜想：如图1，当点D在线段BC上时，①BC与CF的位置关系为：；②BC、CD、CF之间的数量关系为：；（2）深入思考：如图2，当点D在线段CB的延长线上时，结论①、②是否仍然成立？若成立，请给予证明；若不成立，请你写出正确结论再给予证明．（3）拓展延伸：如图3，当点D在线段BC的延长线上时，正方形ADEF对角线交于点O．若已知AB＝4，CD＝BC，请求出OC的长．参考答案一、选择题（每小题3分，共30分）1．解：x2﹣6x＝﹣4，x2﹣6x+32＝5，（x﹣3）2＝5．故选：C．2．解：设方程的另一根为x1，又∵x2＝﹣2，∴根据根与系数的关系可得：，解得：x1＝﹣3，m＝﹣5．故选：A．3．解：由题意得：|x|﹣1＝0，且x2+x﹣2≠0，解得：x＝﹣1，故选：D．4．解：∵四边形ABCD是矩形，∴∠ABC＝∠BAD＝90°，∵∠DBC＝30°，∴∠ABE＝60°，∵AE⊥BD，∴∠BAE＝30°，∴AE＝BE＝cm，故选：D．5．解：∵E，F分别是AD，CD边上的中点，EF＝，∴AC＝2EF＝2，又∵BD＝2，∴菱形ABCD的面积S＝×AC×BD＝×2×2＝2，故选：A．6．解：∵∠ECF＝90°，∠DCB＝90°，∴∠BCE＝∠DCF，∵，∴△CDF≌△CBE，故CF＝CE．因为Rt△CEF的面积是200，即•CE•CF＝200，故CE＝20．正方形ABCD的面积＝BC2＝256，得BC＝16．根据勾股定理得：BE＝＝12．故选：C．7．解：当a2﹣1≠0，即a≠±1时，方程（a2﹣1）x2+（a+2）x+a﹣3＝0是一元二次方程，∴在“﹣3，﹣2，﹣1，0，1，2，3”七个数中有5个数使方程（a2﹣1）x2+（a+2）x+a ﹣3＝0是一元二次方程，∴恰好使方程（a2﹣1）x2+（a+2）x+a﹣3＝0是一元二次方程的概率是，故选：C．8．解：A、∵EM∥AD，∴，故正确；B、∵EM∥AD，EN∥CD，∴＝，＝，∴＝，故正确；C、∵EM∥AD，EN∥CD，∴，，∵四边形ABCD是平行四边形，∴AD＝BC，AB＝CD，∴，故正确；D、∵EN∥CD，∴，故错误．故选：D．9．解：根据左视图的定义可知，这个几何体的左视图是选项D，故选：D．10．解：（1）设点B（s，t），st＝8，则点M（s，t），则k＝s t＝st＝＝2，∴△BDF的面积＝△OBD的面积＝S△BOA﹣S△OAD＝×8﹣×2＝3；故选：D．二、填空题（每小题4分，共28分）11．解：∵在Rt△ACB中，∠ACB＝90°，CE是△ACB中线，CE＝4cm，∴AB＝2CE＝8cm，∴△ACB的面积是×AB×CD＝×8cm×3cm＝12cm2，故答案为：12cm2．12．解：根据题意知，α2+2α﹣2021＝0，即α2+2α＝2021．又∵α+β＝﹣2．所以α2+3α+β＝α2+2α+（α+β）＝2021﹣2＝2019．故答案是：2019．13．解：200次中摸到黑球的频率为＝0.25，而这个口袋中有黑球8个，则总球数为8÷0.25＝32个，所以白球的个数为32﹣8＝24．14．解：∵DE∥BC，AD＝9，DB＝3，CE＝2，∴＝，即＝，解得，AE＝6，∴AC＝AE+EC＝8，故答案为：8．15．解：设小明这个时刻在水平地面上形成的影长为x米，根据题意得＝，解得x＝2，小明这个时刻在水平地面上形成的影长为2米，因为10﹣2＝8（米），所以他最多离开树干8米才可以不被阳光晒到．故答案为8．16．解：设A（x，y），则OB＝x，AB＝﹣y，∵S△AOB＝1，∴OB×AB＝1，∴﹣xy＝2，∴xy＝﹣2，∵点A在y＝上，∴k＝xy＝﹣2，故答案为：﹣2．17．解：作DH∥CF交AB于H，则＝＝1，＝＝，∴FH＝HB，FH＝2AF，∵AF＝1.6cm，∴FH＝3.2cm，∴HB＝FH＝3.2cm，∴AB＝AF+FH+HB＝1.6+3.2+3.2＝8cm．故答案为：8．三、解答题（共62分）18．解：∵3x（x﹣3）＝2x﹣6，∴3x（x﹣3）﹣2（x﹣3）＝0，则（x﹣3）（3x﹣2）＝0，则x﹣3＝0或3x﹣2＝0，解得x1＝3，x2＝．19．解：（1）列表如下：红红蓝红（红，红）（蓝，红）红（红，红）（蓝，红）蓝（红，蓝）（红，蓝）由表知，共有6种等可能结果，其中两次摸到一红一蓝的有4种结果，所以两次摸到一红一蓝的概率为＝；（2）设后来放入的篮球有x个，根据题意，得：＝，解得x＝3，经检验：x＝3是分式方程的解，所以后来放入袋中的蓝球有3个．20．解：三视图，如图所示．21．解：（1）由题意得：×BQ×BP＝12，即•2x•（8﹣x）＝12，整理得：x2﹣8x+12＝0，解得：x＝2或6，即当x为2或6时，△PBQ的面积等于12厘米2；（2）①当∠1＝∠2时，由∠PBQ＝∠BCD＝90°，所以△QBP∽△BCD，则＝，即＝，解得：x＝；②当∠1＝∠3时，由∠PBQ＝∠BCD＝90°，所以△PBQ∽△BCD，所以＝，即＝，解得：x＝2；即x＝或x＝2时，以P、B、Q为顶点的三角形与△BDC相似．22．解：（1）设每碗售价降低x元．平均每天可销售（300+30x）碗．故答案为：（300+30x）；（2）设每碗售价降低x元．店家才能实现每天利润6300元，依题意有：（25﹣x﹣6）（300+30x）＝6300，解得x1＝4，x2＝5，当x＝4时，售价为21元，当x＝5时，售价为20元，∵每碗售价不得超过20元，∴x＝5．答：当每碗售价定为20元时，店家才能实现每天利润6300元．23．（1）证明：∵AO平分∠BAD，AB∥CD，∴∠DAC＝∠BAC＝∠DCA，∴△ACD是等腰三角形，AD＝DC，又∵AB＝AD，∴AB＝CD，∴四边形ABCD为平行四边形，又∵AB＝AD，∴四边形ABCD是菱形；（2）解：解方程x2﹣7x+12＝0，得，OA＝4，OB＝3，利用勾股定理AB＝＝5，S菱形ABCD＝AC×BD＝×8×6＝24平方米；（3）解：在第（2）问的条件下，设M、N同时出发x秒钟后，△MON的面积2m2，当点M在OA上时，x＜2，S△MON＝（4﹣2x）（3﹣x）＝2，解得x1＝1，x2＝4（大于2，舍去）；当点M在OC上且点N在OB上时，2＜x＜3，S△MON＝（3﹣x）（2x﹣4）＝2，整理得，x2+5x+8＝0，方程无解．当点M在OC上且点N在OD上时，即3＜x≤4，S△MON＝（2x﹣4）（x﹣3）＝2，解得x1＝4，x2＝1（小于3，舍去）．综上所述：M，N出发1秒或4秒钟后，△MON的面积为2m2．24．（1）解：令AB＝BO＝m，∵∠ABO＝90°，∴AB⊥x轴，则设点A的坐标为（m，m），∵反比例函数y＝（x＞0）的图象交于点A，∴＝m，解得m＝±2，∵m＞0，∴m＝2，∵点A（2，2）在直线y＝kx﹣4上，∴2＝2k﹣4，∴k＝3；（2）证明：由（1）可知B（2，0），AB＝2，∵AB⊥BO，点D与点O关于AB对称，∴D（4，0），BD＝2，∴AD2＝AB2+BD2＝22+22＝8，过点A作AE⊥y轴，垂足为E，则点E（0，2），AE＝2，∵直线y＝3x﹣4与y轴交于点C，∴C（0，﹣4）则CE＝6，∴AC2＝AE2+CE2＝22+62＝40，∵∠OCD＝90°，OD＝4，OC＝4，∴CD2＝OD2+OC2＝42+42＝32，∵8+32＝40，∴AD2+CD2＝AC2，∴△ACD是直角三角形；（3）解：①当点E在CD上方时，如下图，过点O、A作直线m，由点O、A的坐标知，直线OA的表达式为y＝x，由点C、D的坐标知，直线CD的表达式为y＝x﹣4，则直线CD∥m，即OA∥CD，∵S△ECD＝S△OCD，即两个三角形同底，则点E与点A重合，故点E的坐标为（2，2）；②当点E（E′）在CD下方时，在y轴负半轴取CH＝OC＝4，则点H（0，﹣8），∵则S△ECD＝S△OCD，∴过点H作直线m′∥CD，则直线m′与反比例函数的交点即为点E，∴直线m′的表达式为y＝x﹣8，联立y＝x﹣8和y＝并解得（不合题意值已舍去），故点E的坐标为（4+2，2﹣4），综上，点E的坐标为（4+2，2﹣4）或（2，2）．25．解：（1）如图1，∵AB＝AC，∠BAC＝90°，∴∠ABD＝∠ACB＝45°，∵四边形ABCD是正方形，∴AD＝AF，∠DAF＝90°，∴∠BAD＝∠CAF＝90°﹣∠DAC，∴△BAD≌△CAF（SAS），∴∠ABD＝∠ACF＝45°，BD＝CF，①∵∠BCF＝∠ACB+∠ACF＝45°+45°＝90°，∴BC⊥CF，故答案为：BC⊥CF．②∵BC＝CD+BD＝CD+CF，∴BC＝CD+CF，故答案为：BC＝CD+CF．（2）结论①，即BC⊥CF成立，结论②，即BC＝CD+CF不成立，BC＝CD﹣CF，证明：如图2，∵AB＝AC，∠BAC＝90°，∴∠ABC＝∠ACB＝45°，∵四边形ABCD是正方形，∴AD＝AF，∠DAF＝90°，∴∠BAD＝∠CAF＝90°﹣∠BAF，∴△BAD≌△CAF（SAS），∴∠ABD＝∠ACF，BD＝CF，∵∠ACF＝∠ABD＝180°﹣∠ABC＝180°﹣45°＝135°，∴∠BCF＝∠ACF﹣∠ACB＝135°﹣45°＝90°，∴BC⊥CF，∴结论①成立；∵BC＝CD﹣BD＝CD﹣CF，∴结论②不成立，BC＝CD﹣CF成立．（3）如图3，∵AB＝AC，∠BAC＝90°，∴∠ABD＝∠ACB＝45°，∵四边形ABCD是正方形，∴AD＝AF，∠DAF＝90°，∴∠BAD＝∠CAF＝90°+∠DAC，∴△BAD≌△CAF（SAS），∴∠ABD＝∠ACF＝45°，BD＝CF，∵∠BCF＝∠ACB+∠ACF＝45°+45°＝90°，∴BC⊥CF，∴∠FCD＝90°，∵AB＝AC＝4，∠BAC＝90°，∴BC＝＝＝8，∵CD＝BC＝×8＝2，∴CF＝BD＝BC+CD＝8+2＝10，∴DF＝＝＝2，∵OD＝OF，∴OC＝DF＝×2＝，∴OC的长为．。
广东省佛山市南海外国语学校2023-2024学年九年级上学期第三次月考数学试题(解析版)
广东省佛山市南海区南海外国语学校2023-2024学年第一学期九年级第三阶段素养评价——数学科一、单选题1. 如图几何体的主视图是（ ）A. B. C. D.【答案】A【解析】【分析】根据题意可得：从正面看，主视图是两个长方形，即可求解．【详解】解：从正面看，主视图是两个长方形．故选：A【点睛】本题主要考查了几何体的三视图，熟练掌握几何体的三视图的特征是解题的关键． 2. 若关于x 的一元二次方程220x x m −+=一个根为－1，则m 的值为（ ）A. 2B. －1C. 1D. －3【答案】D【解析】【分析】将x =-1代入方程求出m 的值即可．【详解】解：将x =-1代入方程x 2-2x +m =0得1+2+m =0，解得：m =-3，故选：D ．【点睛】本题考查一元二次方程解的意义，使一元二次方程成立的未知数的值叫一元二次方程的解． 3. 如图，在Rt ABC 中，=4AC ，=3BC ，90C ∠=°，则sin A 的值为（ ）A. 34B. 53C. 43D. 35【答案】D【解析】【分析】根据勾股定理求出AB ，再根据正弦的定义：对边比斜边，进行计算即可．【详解】解：∵=4AC ，=3BC ，90C ∠=°，∴5AB ， ∴3sin5BC A AB ==； 故选D ．【点睛】本题考查正弦的定义．熟练掌握正弦的定义是解题的关键．4. 下列判断错误的是（ ）A. 两组对边分别相等的四边形是平行四边形B. 四个内角都相等的四边形是矩形C. 四条边都相等的四边形是菱形D. 两条对角线垂直且平分的四边形是正方形 【答案】D【解析】【分析】根据平行四边形、矩形、菱形和正方形的判定方法逐一进行判断即可．【详解】A 、两组对边分别相等的四边形是平行四边形，说法正确，不符合题意；B 、四个内角都相等的四边形是矩形，说法正确，不符合题意；C 、四条边都相等的四边形是菱形，说法正确，不符合题意；D 、两条对角线垂直且平分且相等的四边形是正方形，原说法错误，符合题意；故选：D ．【点睛】本题考查平行四边形、矩形、菱形和正方形的判定方法．熟练掌握平行四边形、矩形、菱形和正方形的判定方法是解题的关键．5. 关于反比例函数2y x=的图象性质，下列说法正确的是（ ） A. 图象经过点()12，B. 图象位于第二、四象限C. 当0x >时，y 随x 的增大而增大D. y 随x 的增大而减小【答案】A【解析】【分析】根据反比例函数图象上点的坐标特征对A 进行判断；根据反比例函数的性质对B 、C 、D 进行判断．【详解】解：A 、把1x =代入2y x =得2y =，则点()12，在2y x =的图象上，所以本选项符合题意； B 、20k =>，则双曲线2y x=的两支分别位于第一、第三象限，所以本选项不符合题意； C 、当0x >时，y 随x 的增大而减小，所以本选项不符合题意；D 、20k =>，在每一象限内y 随x 的增大而减小，所以本选项不符合题意．故选：A ．【点睛】此题考查反比例函数的性质，解题关键在于掌握反比例函数(0)k y k x=≠的图象是双曲线；当0k >，双曲线的两支分别位于第一、第三象限，在每一象限内y 随x 的增大而减小；当0k <，双曲线的两支分别位于第二、第四象限，在每一象限内y 随x 的增大而增大．6. 如图，小颖同学用自制的直角三角形纸板DEF 测量树的高度AB ，他调整自己的位置，设法使斜边DF 保持水平，并且边DE 与点B 在同一直线上，已知纸板的两条直角边DE =30cm ，EF =15cm ，测得边DF 离地面的高度AC =1.5m ，CD =7m ，则树高AB =（ ）m ．A. 3.5B. 4C. 4.5D. 5【答案】D【解析】 【分析】利用直角三角形DEF 和直角三角形DCB 相似求得BC 的长，再加上AC 的长即可求得树高AB ．【详解】∵∠DEF=∠BCD=90°，∠D=∠D ，∴△DEF ∽△DCB ，∴BC ：EF=DC ：DE ，∵DE=30cm=0.3m ，EF=15cm=1.5m ，AC=1.5m ，CD=7m ， ∴71.50.3BC ＝， ∴BC=3.5米，∴AB=AC+BC=1.5+3.5=5m ，故选D ．【点睛】本题考查了相似三角形的应用，解题的关键是从实际问题中整理出相似三角形的模型． 7. 已知关于x 的方程2210x x −−=，则下列关于该方程根的判断，正确的是（ ）A. 没有实数根B. 有两个相等的实数根C. 有两个不相等的实数根D. 不能确定 【答案】C【解析】【分析】先求出“Δ”的值，再根据根的判别式判断即可．【详解】解：x 2-2x -1=0，∵1a =，2b =−，1c =−，∴Δ=（-2）2-4×1×（-1）=8＞0，∵Δ＞0，∴方程有两个不相等的实数根，故选：C ．【点睛】本题考查了根的判别式，能熟记根的判别式的内容是解此题的关键，注意：一元二次方程ax 2+bx +c =0（a 、b 、c 为常数，a ≠0），当b 2-4ac ＞0时，方程有两个不相等的实数根；当b 2-4ac =0时，方程有两个相等的实数根；当b 2-4ac ＜0时，方程没有实数根．8. “敬老爱老”是中华民族的优秀传统美德．小刚、小强计划利用暑期从A ，B ，C 三处养老服务中心中，随机选择一处参加志愿服务活动，则两人恰好选到同一处的概率是（ ） A. 12 B. 13 C. 16 D. 29【答案】B【解析】【分析】画出树状图展示所有9种等可能的结果数，找出两人恰好选择同一场所的结果数，然后根据概率公式求解．【详解】解：画树状图如图：共有9种等可能的结果数，其中两人恰好选择同一场所的结果数为3，∴明明和亮亮两人恰好选择同一场馆的概率3193==, 故选：B ． 【点睛】本题考查了列表法与树状图法：利用列表法或树状图法展示所有等可能的结果n ，再从中选出符合事件A 或B 的结果数目m ，然后利用概率公式计算事件A 或事件B 的概率．9. 如图，关于x 的函数y kx k =−和()0ky k x=−≠，它们在同一坐标系内的图象大致是（ ）A. B. C.D.【答案】B【解析】【分析】首先根据反比例函数图象所经过的象限判断出k 的符号；然后由k 的符号判定一次函数图象所经过的象限，图象一致的选项即为正确选项．【详解】解：A 、反比例函数()0ky k x=−≠的图象经过第一、三象限，则0<k ． 所以一次函数y kx k =−的图象经过第二、四象限，且与y 轴交于正半轴，故本选项错误； B 、反比例函数()0ky k x=−≠的图象经过第二、四象限，则>0k ． 所以一次函数y kx k =−的图象经过第一、三象限，且与y 轴交于负半轴，故本选项正确； C 、反比例函数()0ky k x=−≠的图象经过第一、三象限，则0<k ． 所以一次函数y kx k =−的图象经过第二、四象限，且与y 轴交于正半轴，故本选项错误； D 、反比例函数()0ky k x=−≠的图象经过第二、四象限，则>0k ．所以一次函数y kx k =−的图象经过第一、三象限，且与y 轴交于负半轴，故本选项错误； 故选：B ． 【点睛】本题考查反比例函数与一次函数的图象特点：①反比例函数k y x=的图象是双曲线；②当>0k 时，它的两个分支分别位于第一、三象限；③当0<k 时，它的两个分支分别位于第二、四象限． 10. 如图，已知E ，F 分别为正方形ABCD 的边AB 、BC 的中点，AF 与DE 交于点M ，则下列结论：①AF ⊥DE ；②AE EG =；③AM =23MF ；④14AEM ADM S S ∆∆=．其中正确的结论有（ ）A. 4个B. 3个C. 2个D. 1个【答案】B【解析】 【分析】先由E ，F 分别为正方形ABCD 的边AB 、BC 的中点得到AE =BE =BF 、∠DAE =∠ABF =90°、AD =AB ，从而得证△DAE ≌△ABF ，进而利用全等三角形的性质得到∠BAM +∠AEM =90°判定①；假设AE =EG ，则AE =BE =EG ，则∠EBG =∠EGB ，∠EAG =∠EGA ，从而推出∠EAG =45°判定②；由BF =AE =BE 得到AF，然后证明△AEM ∽△AFB ，进而利用相似三角形的性质得到AM =23MF 判定③；先证明△AEM ∽△DAM ，然后利用AD =2AE 得到14AEM ADM S S ∆∆=判定④． 【详解】解：∵E ，F 分别为正方形ABCD 的边AB 、BC 的中点，∴AE =BE =BF ，∠DAE =∠ABF =90°，AD =AB ，∴△DAE ≌△ABF （SAS ），∴∠BAF =∠ADE ，∵∠ADE +∠AED =90°，∴∠BAM +∠AEM =90°，∴∠AME =90°，故①正确，符合题意；假设AE =EG ，则AE =BE =EG ，∴∠EBG =∠EGB ，∠EAG =∠EGA ，∵四边形ABCD 是正方形，∴∠ABD =45°，∴∠EBG =∠EGB =45°，∴∠BEG =∠EAG +∠EGA =90°，∴∠EAG =45°，又∵∠EAG ≠45°，∴AE ≠EG ，故②错误，不符合题意∵BF =AE =BE ，AB =2AE ，∴AF ，∵∠EAM +∠AEM =90°，∠BAF +∠AFB =90°，∴∠AEM =∠AFB ，∵∠AME =∠ABF =90°，∴△AEM ∽△AFB ， ∴AM AE EM AB AF BF==，即2AM AE =， ∴AMAE ， ∴MF =AF -AMAE-AEAE ， ∴AM =23MF ，故③正确，符合题意； ∵∠AEM +∠EAM =90°，∠EAM +∠DAM =90°，∴∠AEM =∠DAM ，∵∠EMA =∠AMD =90°，∴△AEM ∽△DAM ， ∴2211()()24AEM ADM S AE S AD ∆∆===，故④正确，符合题意； 故选：B ．【点睛】本题考查了正方形的性质、全等三角形的判定与性质、相似三角形的判定与性质、勾股定理，解题的关键是熟知相关知识．二、填空题11. 南外九年级某音乐生发现：音乐也可以有数学问题！如图，五线谱是由等距离的五条平行横线组成的，同一条直线上的三个点A ，B ，C 都在横线上．若线段6AB =，则线段AC 的长是________．【答案】9【解析】【分析】本题考查平行线分线段成比例定理，解题的关键是找准对应关系．过点A 作平行横线的垂线，交点B 所在的平行横线于D ，交点C 所在的平行横线于E ，根据平行线分线段成比例定理列出比例式，计算即可．【详解】解：过点A 作等距离平行横线的垂线，交点B 所在的平行横线于D ，交点C 所在的平行横线于E ， ∴AB AD AC AE =，即23AB AC =， ∴32AC AB =， ∵6AB =， ∴336922AC AB ==×=， ∴线段AC 的长是9．故答案：9．12. 小明身高1.76米，小亮身高1.6米，同一时刻他们站在太阳光下．小明的影子长为1.1米，则小亮的影长是________米．【答案】1【解析】【分析】本题考查相似三角形的应用，解题的关键理解：同一时间，同一地点测得物体与影子的比值相为等，也就是两人的身高比等于影长比．据此解答即可．【详解】解：设小亮的影长是x 米， 依题意，得：1.61.76 1.1x =， 解得：1x =，∴小亮的影长是1米．故答案为：1．13. Rt △ABC 中，∠C ＝90°，如果3a，那么sin A ＝________. 【答案】12.【解析】【分析】根据特殊角三角函数值计算．【详解】∵3a，∴a b＝； 令a，则b ＝3k ；c．∴sin A＝12， 故答案为12.【点睛】本题考查特殊角三角函数值的计算，特殊角三角函数值计算在中考中经常出现，题型以选择题、填空题为主．14. 关于x 的一元二次方程2310170x x −−=的两个根分别为1x 和2x ，则1211x x +=_________． 【答案】1017−【解析】 【分析】根据一元二次方程根与系数的关系可得12103x x +=，12173x x =−，再由12121211x x x x x x ++=进行求解即可．【详解】解：∵一元二次方程2310170x x −−=的两根是1x ，2x ， ∴12103x x +=，12173x x =−， 在的∴121212111017031173x x x x x x ++===−−． 故答案是：1017− ． 【点睛】本题主要考查了一元二次方程根与系数的关系，解题的关键在于能够熟练掌握一元二次方程根与系数的关系．15. 如图，1P 是反比例函数()00k y x k x=>≠，在第一象限图像上的一点，点1A 的坐标为()20，．若11POA 与212PA A △均为等边三角形，则点2A 的横坐标表示的数为________【答案】【解析】【分析】此题综合考查了反比例函数的性质，利用待定系数法求函数的解析式，正三角形的性质等多个知识点，此题难度稍大，综合性比较强，注意对各个知识点的灵活应用．作11PC OA ⊥于C ，作212P D A ⊥于D ，由等边三角形的性质结合点1A 的坐标为()20，，得出(11P，将(11P代入反比例函数解析式得出k =1A D a =，同理可得出()22P a +，代入反比例函数解析式求出1a =−122A A =−+2112OA OA A A =+进行计算即可得解． 【详解】解：如图，作11PC OA ⊥于C ，作212P D A A ⊥于D ， ，点1A 的坐标为()20，， 12OA ∴=，11POA 是等边三角形，，11PC OA ⊥，1OC ∴=，112PO OA ==，1PC ∴=，(1P ∴， 1P 是反比例函数()00k y x k x=>≠，在第一象限图像上的一点，1k =，k ∴y ∴ 设1A D a =，212P A A △为等边三角形，212P D A A ⊥，2112122P A A A A D a ∴===，2P D ∴=，112OD OA A D a =+=+， ()22P a ∴+，将()22P a +代入y ==，解得：1a =−1a =−−，0a > ，1a ∴=−+，1222A A a ∴==−+，(211222OA OA A A =+=+−+=∴2A 的横坐标表示的数为故答案为： 三、解答题16. 解方程：233x x x +=+．【答案】13x =−，21x =．【解析】【分析】本题考查了解一元二次方程，用因式分解法进行求解即可，解题的关键是熟练掌握解一元二次方程的各个方法．【详解】解：∵233x x x +=+，∴()2330x x x +−+=， ∴()()330x x x +−+=， ∴()()310x x +−=， ∴30x +=或10x −=，∴13x =−，21x =．17. 数学活动小组到某景点测量标志性建筑CD 的高度．如图，他们在地面上A 处仰望塔顶，测得仰角为30°，再往塔的方向前进50m 至B 处，测得仰角为60°，点A ，C ，B 在同一直线上，则求塔高CD ．（身高忽略不计，结果不取近似值）【答案】【解析】【分析】先根据三角形外角的性质得到30A ADB ∠=∠=°， 则BD AB =，再解Rt DBC △求出CD 即可得到答案．【详解】解：∵3060DAB DBC A ADB ∠=°∠=∠+∠=°，，∴30A ADB ∠=∠=°，∴BD AB =；∵50cm AB =，∴50cm BD =，又∵90DCB ∠=°，∴sin 50CD BD CBD =⋅==∠∴该塔高CD 为．【点睛】本题主要考查了解直角三角形的实际应用，三角形外角的性质，等腰三角形的判定，求出50cm BD =是解题的关键．18. 在ABC 中，90ACB ∠=°，D ，E 分别为AC ，AB 的中点，BF CE ∥交DE 的延长线于点F ．当30A ∠=°时，求证：四边形ECBF 是菱形．【答案】证明见解析【解析】【分析】根据三角形中位线定理可得EF BC ∥，继而证得四边形ECBF 是平行四边形，再根据直角三角形的性质可得CB CE =【详解】证明：∵D ，E 分别为AC ，AB 的中点，∴DE BC ∥，即EF BC ∥，又∵BF CE ∥，∴四边形ECBF 是平行四边形，∵90ACB ∠=°，30A ∠=°，E 为AB 的中点， ∴12CB AB =，12CE AB =， ∴CB CE =，∴四边形ECBF 是菱形．【点睛】本题考查平行四边形的判定以及菱形的判定，解题的关键是熟练掌握平行四边形以及菱形的判定．19. 如图，在网格图中，每个小正方形边长均为1，点O 和ABC 的顶点均在小正方形的格点上．（1）以O 为位似中心，在网格图中作A B C ′′′ ，使A B C ′′′ 和ABC 位似，且相似比为1:2；（2）连接（1）中的BB ′、CC ′，求四边形BB C C ′′的周长．（结果保留根号） 【答案】（1）作图见解析（2）5+【解析】【分析】本题考查位似变换，勾股定理的应用，（1）直接利用位似图形的性质得出对应点位置进而得出答案；（2）利用勾股定理得出各线段长，进而得出答案；正确得出对应点位置是解题的关键．【小问1详解】解：∵在网格图中，每个小正方形边长均为1，A B C ′′′ 和ABC 位似，且相似比为1:2，∴2OA =，6OC =，4OB =，OA C OAC ′′△∽△，OB C OBC ′′∽△△， ∴12OA OC OA OC ′′==，12OB OB ′=， 即1262OA OC ′′==，142OB ′=， ∴1OA ′=，3OC ′=，2OB ′=，连接A C ′′，B C ′′，则A B C ′′′ 即为所作；【小问2详解】∵在网格图中，每个小正方形边长均为1，∴2BB ′=，3CC ′=，B C ′′=BC ，∴四边形BB C C ′′的周长：235BB B C CC BC ′′′′+++=++=+．∴四边形BB C C ′′的周长为5．20. 2022年北京冬季奥运会于2月4日至2月20日在北京市和河北省张家口市联合举行，冬奥会吉祥物为“冰墩墩”．（1）据市场调研发现，某工厂今年二月份共生产500个“冰墩墩”，该工厂连续两个月增加生产量后四月份生产720个“冰墩墩”，求平均每月的增长率是多少？（2）已知某商店“冰墩墩”平均每天可销售20个，每个盈利20元，在每个降价幅度不超过8元的情况下，每下降2元，则每天可多售10件．如果每天要盈利700元，则每个“冰墩墩”应降价多少元？【答案】（1）20%（2）6元【解析】【分析】（1）设该工厂平均每月生产量增长率为x ，利用该工厂四月份生产“冰墩墩”的数量=该工厂二月份生产“冰墩墩”的数量×（1+该工厂平均每月生产量的增长率）的平方，即可得出关于x 的一元二次方程，解之取其正值即可得出结论；（2）设每个“冰墩墩”降价y 元，则每个盈利()20y −元，平均每天可售出(20)5y +个，利用该商店每天销售“冰墩墩”获得的利润=每个的销售利润×平均每天的销售量，即可得出关于y 的一元二次方程，解之取其符合题意的值即可得出结论．【小问1详解】解：设平均每月的增长率是x ，2500(1)720x ×+=（个）， 解得10.220%==x ，2 2.2x =−（舍） 答：平均每月的增长率是20%．【小问2详解】设每个“冰墩墩”降价y 元，则每个盈利()20y −元，平均每天可售出()20102052y y +×=+个，依题意得：()()20205700y y −+=， 整理得：216600y y −+=，解得：126,10y y ==（不符合题意，舍去） 答：每个“冰墩墩”应降价6元．【点睛】本题考查了一元二次方程的应用，找准等量关系，正确列出一元二次方程是解题的关键．（1）请用表格或树状图求出：一次随机取出2个小球，颜色不同的概率．（2）如果老师在布袋中加入若干个红色小球，然后小明通过做实验的方式猜测加入的小球数，小明每次摸出一个小球记录下颜色并放回，实验数据如下表：请你帮小明算出老师放入了多少个红色小球．【答案】（1）23 （2）5【解析】【分析】本题考查用列表法或树状图法求概率，列表法可以不重复不遗漏的列出所有可能的结果，适合于两步完成的事件；树状图法适合两步或两步以上完成的事件，（1）根据题意列出图表得出所有等情况数和颜色不同情况数，然后根据概率公式即可得出答案； 的（2）根据图表求出红球概率，设加入了x 个红球，根据概率公式列出算式，然后求解即可；解题的关键要注意是放回实验还是不放回实验，概率的计算公式：概率＝所求情况数与总情况数之比．也考查了利用频率估计概率．【小问1详解】解：根据题意列表如下： 黑1 黑2红 黑1 / （黑1，黑2） （黑1，红）黑2 （黑2，黑1） /（黑2，红）红（红，黑1） （红，黑2）/ 列表如图，一共有6种等可能事件，其中颜色不同的等可能事件有4种，∴颜色不同的概率为4263=， 答：颜色不同的概率为23； 【小问2详解】由图表可得摸到红球概率为34， 设加入了x 个红球，则1334x x +=+， 解得：5x =， 经检验：5x =是原方程的解且符合题意，答：加入了5个红球．22. 如图，在Rt ACB △中，90C ∠=°，30cm AC =，25cm BC =．动点P 从点C 出发，沿CA 方向运动，速度是2cm/s ；同时，动点Q 从点B 出发，沿BC 方向运动，速度是1cm/s ．（1）几秒后P 、Q 两点相距25cm ？（2）几秒后PCQ △与ABC 相似？【答案】（1）10秒后P 、Q 两点相距25cm（2）758秒或12517秒后PCQ △与ABC 相似 【解析】【分析】本题考查了相似三角形的判定与性质、一元二次方程的应用、勾股定理，熟练掌握以上知识点并灵活运用，采用分类讨论的思想是解此题的关键．（1）设x 秒后P 、Q 两点相距25cm ，用x 表示出CP 、CQ ，根据勾股定理列出方程，解方程即可； （2）分PCQ ACB ∽和PCQ BCA ∽两种情况，利用相似三角形的性质，列出关系式，求解即可．【小问1详解】解：设x 秒后P 、Q 两点相距25cm ，则2cm CP x =，cm BQ x =，()25cm CQ x ∴=−，由题意得：()()22222525x x +−=，解得：110x =，20x =（不符合题意，舍去）， ∴10秒后P 、Q 两点相距25cm ；【小问2详解】解：设y 秒后PCQ △与ABC 相似，则2cm CP y =，cm BQ y =，()25cm CQ y ∴=−，当PCQ ACB ∽时，CP CQ CA CB =，即2253025y y −=， 解得：758y =； 当PCQ BCA ∽时，CP CQ CB CA =，即2252530y y −=， 解得：12517y =， ∴758秒或12517秒后PCQ △与ABC 相似．23. 如图，直线2y x =与双曲线(0,0)k y x k x=>>相交于点A ，AB x ⊥轴于点B ，以AB 为边在右侧作正方形ABCD ，CD 与双曲线相交于点E ，连结AE 、OE ．（1）当4BC =时，求点E 的坐标；（2）当24AOE S =△时，求k 的值；（3）是否存在实数k ，满足AE OA ⊥，若存在，求出k 的值；若不存在，请说明理由．【答案】（1）46,3（2）18k =（3）不存在，理由见解析【解析】【分析】（1）根据正方形的性质，得到A 点的纵坐标为4，点A 在直线2y x =上，求出A 点坐标，进而求出反比例函数的解析式，求出OC E 在反比例函数上，进行求解即可；（2）设()(),20A a a a >，同法（1）求出E 点坐标，利用24AOE ABCES S ==梯形△，列式计算即可； （3）假设存在，推出()ASA OAB EAD ≌△△，得到OB DE =，推出0k =，与0k <矛盾，即可得出结论．【小问1详解】解：∵四边形ABCD 为正方形，4AB BC ==， ∴A 点的纵坐标为4，∵A 在直线2y x =上，∴42A x =，∴2A x =，∴()2,4A ，∴2OB =， ∴42k=，∴8k ， ∴反比例函数解析式为8y x =，∵6OC OB BC =+=， ∴6E x =， ∴8463E y ==，∴点E 的坐标为46,3；【小问2详解】 设()(),20A a a a >， ∴2ka a =，OB a =，2AB BC a ==， ∴22k a =，3OC OB BC a =+=， ∴3E x a =， ∴22233E E ka y a x a===， ∴23,3a E a∵2=2AOB kS a =△，22COE kS a ==△，∴24AOE ABCE S S ==梯形△， ∴12222423a a a×+×= ，解得29a =，∴2218k a ==；【小问3详解】 不存在．理由如下： ∵四边形ABCD 是正方形，∴AB AD =，90ABO BAD D ∠=∠=∠=°，要使AE OA ⊥，则90OAB BAE ∠+∠=°，∵90DAE BAE ∠+∠=°，∴OAB DAE =∠∠，∴()ASA OAB EAD ≌△△，∴OB DE =，由（2）可知，()(),20A a a a >，则点23,3a E a， ∴OB a =，24233a a DE a =−=， ∴43a a =，得0a =， ∴0k =，∵0k >，∴不符合题意，不存在．【点睛】本题考查反比例函数与一次函数的综合应用，反比例函数与几何的综合应用．熟练掌握k 值的几何意义，利用数形结合的思想进行求解，是解题的关键．24. （1）问题如图1，在四边形ABCD 中，点P 为AB 上一点，当90DPC A B ∠=∠=∠=°时，求证：AD BC AP BP ⋅=⋅．（2）探究若将90°角改为锐角（如图2），其他条件不变，上述结论还成立吗？说明理由． （3）应用如图3，在ABC 中，AB =，45B ∠=°，以点A 为直角顶点作等腰Rt ADE △．点D 在BC 上，点E 在AC 上，点F 在BC 上，且45EFD ∠=°，若CE =，求CD 的长．【答案】（1）见解析；（2）成立；理由见解析；（3）5【解析】【分析】（1）由90DPC A B ∠=∠=∠=°可得ADP BPC ∠=∠,即可证到ADP BPC ∽，然后运用相似三角形的性质即可解决问题；（2）由DPC A B α∠=∠=∠=可得ADP BPC ∠=∠,即可证到ADP BPC ∽，然后运用相似三角形的性质即可解决问题；（3）证明ABD DFE ∽△△,求出4DF =，再证EFC DEC ∽△△,可求1FC =,进而解答即可．【详解】解：（1）证明：如图1，90DPC ∠=°90BPC APD ∴∠+∠=°，90A ∠=° ，90ADP APD ∴∠+∠=°APD BPC ∴∠=∠，又90A B ∠=∠=°ADP BPC ∴∽△△，::AD BP AP BC ∴=AD BC AP BP ∴⋅=⋅；（2）结论AD BC AP BP ⋅=⋅仍成立；理由：如图2，BPD DPC BPC ∠=∠+∠ ，又BPD A APD ∠=∠+∠ ，DPC BPC A APD ∴∠+∠=∠+∠，DPC A α∠=∠= ，BPC APD ∴∠=∠，又A B α∠=∠= ，ADP BPC ∴∽△△，::AD BP AP BC ∴=AD BC AP BP ∴⋅=⋅；（3）45EFD ∠=° ,45B ADE ∴∠=∠=°,BAD EDF ∴∠=∠,ABD DFE ∴ ∽::AB DF AD DE ∴=Rt ADE △是等腰直角三角形:1:AD DE ∴:AB DF ∴AB =4DF ∴=Rt ADE △等腰直角三角形45AED ∴∠=°45EFD ∠=°18045135DEC EFC ∴∠=∠=°−°=°又C C ∠=∠DEC EFC ∴ ∽::DC EC EC CF ∴=即2(4)EC FC FC ⋅+EC =∴54()FC FC +1FC ∴=解得5CD =．【点睛】本题考查相似三角形的综合题，三角形的相似，正切值的求法，能够通过构造45°角将问题转化为一线三角是解题的关键．是。
2019-2020学年佛山市南海区南海中学高三英语三模试题及答案解析
2019-2020学年佛山市南海区南海中学高三英语三模试题及答案解析第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项APlan Your Visit to the British MuseumWe've made a few changes to ensure you can safely enjoy your visit:How to book tickets·Tickets for the permanent collection can only be hooked online in advance.·To book tickets for exhibitions, visit our exhibition pages. If you want to guarantee your visit to the permanent collection as well as the exhibition on your visit, you will need to book a separate ticket.·We are limiting numbers of people in the museum to ensure there is room for you to safely enjoy your visit. Everyone, including members and volunteers, will therefore need to book a free, timed ticket in advance.Important information about your ticket booking·You can book up to six tickets for people who live in the same family.·We will email yourticket to you, and you will need to show it either printed out or on your device (设备) when you arrive.·Please arrive at the time stated on your ticket-we cannot guarantee admission before or after it.Following the routeWe have created a new one-way route through the museum which will allow you to safely explore world cultures from the Mediterranean to the Middle East and from the Americas to Africa.·View a map of the route (Open in a new window).·Our audio guides (音频指南) are not available. However, youcan pay to download the audio introductions from iTunes (Open in a new window) or Google Play (Open in a new window) to each of the galleries and listen on your own device. Please remember to bring along your own headphones.1.Why does the museum offer timed tickets?A.To earn more money.B.To hire more volunteers.C.To save room for collections.D.To control the number of visitors.2.How can a visitor learn about the new route in advance?A.View an online map.B.Buy a new headphone.C.Buy audio guides from the museum.D.Download video introductions from iTunes.3.Where can you probably find the text?A.In a newspaper.B.In a magazine.C.On a website.D.In a report.BThe Rise of Voice TechnologyVoice technology has come a long way. Just a few years ago, it would have been unusable. But now, those who follow the technology know that it has got considerably better.Writing with your voice raises several interesting questions. How difficult is it actually? Human speech involves a lot more starting and stopping with errors and the need for repairing broken sentences than you may think. Even gifted speakers make mistakes. To turn the spoken word into reasonable writing requires lots of planning. You’ll need some kinds of notes or other organisers to make it work.Another question turning speech into writing raises is the style. How would writing make the change that people speak their writing rather than type? Chances are that it would come up with many more short sentences and more concrete language, which is good. It would probably also rely on prepared phrases a lot more often, which is not available when you are speaking quickly.To confirm this, a column was not written, but dictated (听写). It was composed from brief notes written down for structure only, and it was edited for length, with all of the original errors kept in. Here were the results. The first was that the literal accuracy was extremely high. There weren’t many cases wherethe software had heard one word incorrectly and written down another. But the other result was that the readability of this column was rather bad. Obviously, the blame is not with the technology at all, which turns out to be rather good. Speaking into writing relies on a better human brain than the one we currently possess. Writing is hard. There’s a reason it can’t be done at the speed of speech, in real time.To clarify the matter, this time paragraph breaks were added after the whole writing. Punctuations (标点) had to be spoken aloud, and after a full stop, the first word in the new sentence was capitalized automatically. Some minor punctuation marks were added to make it clearer. To improve accuracy, people “trained” the software beforehand, reading a prewritten passage aloud. Actually it turns out to be more effective. All of these ensure the satisfactory completion of turning speech into writing.Language is the most important tool for human interaction, and voice is one beautiful part of language. With the maturity of modern technology, it has given birth to a great change in the human-computer voice interaction.4. According to the passage, which helps to turn speech into writing in terms of style?A. There is careful planning in advance.B. Errors and broken sentences are avoided.C. People type words as fast as they say them.D. The writing contains more prepared phrases.5. To achieve better results, the author mentions some changes for ______.A. processing errors in a column.B. adding minor punctuation marks.C. increasing the number of brief notes.D. integrating short paragraphs in writing.6. The author suggests that ______.A. human brains are responsible for poor dictated writing.B. writing with voice promises to improve the quality of writing.C. writing is an unnatural act that can hardly be learned and improved.D. technology has a long way to go in the human-computer voice interaction.7. What is the passage mainly about?A. Why people fully intend to turn speech into writing.B. What role voice technology plays in improving readability.C. Where the human-computer voice interaction is at an advantage.D. How voice technology enables the change from speech into writing.CWhat a day! I started at my new school this morning and had the best time. I made lots of new friends and really liked my teachers. I was nervous the night before, but I had no reason to be. Everyone was so friendly and polite. They made me feel at ease. It was like I'd been at the school for a hundred years!The day started very early at 7:00 am. I had my breakfast downstairs with my mom. She could tell that I was very nervous. Mom kept asking me what was wrong. She told me I had nothing to worry about and that everyone was going to love me. If they didn't love me, Mom said to send them her way for a good talking to. I couldn't stop laughing.My mom dropped me off at the school gates about five minutes before the bell. A little blonde girl got dropped off at the same time and started waving at me. She ran over and told me her name was Abigail. She wasvery nice and we became close straight away. We spent all morning together and began to talk to another girl called Stacey. The three of us sat together in class all day and we even made our way home together! It went so quickly. Our teacher told us that tomorrow we would really start learning and developing new skills.I cannot wait until tomorrow and feel as though I am really going to enjoy my time at my new school. I only hope that my new friends feel the same way too.8. How did the author feel the night before her new school?A. Tired.B. ConfidentC. Worried.D. homesick9. What did the author think of her mother’s advice?A. Clear.B. Funny.C. OptionalD. Respectable10. What happened on the author's first day of school?A. She met many nice people.B. She had a hurried breakfast.C. She learned tome new skills.D. She arrived at school very early.11. What can we infer about Abigail?A. She disliked Stacey.B. She was shy and quiet.C. She got on well with the author.D. She was an old friend of the author.DPlastic is piling up in ecosystems all over the world. Although its harmful impacts on both species and ecosystems have been documented, a few animals—like bowerbirds and hermit crabs—are doing what they can to recycle it. And according to a recent study, wild bees in Canada have joined the effort, which is a rare observation of behavioral flexibility in species especially insects, in increasingly plastic-rich environments.The researchers found two species of leafcutter bees putting plastic into their nests. One of the bees they studied, the alfalfa leafcutter bee, normally bites off pieces of leaves and flowers while the second bee gathers sticky substances from trees. Leafcutter bees don't build big nests or store honey like honeybees, choosing instead small nests in underground holes, tree holes or cracks（裂缝）in buildings. But the researchers found that three of eight brood cells（育雏巢室）contained pieces of plastic bags, replacing 23 percent of the cut leaves in each cell on average.While they don't make honey,alfalfa leafcutter bees still make money for theU. S. and Canadian farmers by pollinating（给......传授花粉）crops including alfalfa , carrots and melons. The European insects were introduced to North America in the 1930s for that purpose, and they've since become wild, joining the continent's many native species of leafcutter bees.In a separate study conducted in Argentina between 2017 and 2018, researchers found a bee nest made entirely of plastic, which consisted of three separate cells. It's the first known example of such construction worldwide. Compared to the other nests the researchers examined, which were made of natural materials, this one had a pretty lower success rate of the bees' survival. One of the cells had a dead baby bee , another seemed to have housed an adult that had left the nest, and the third was unfinished.12. What does the animals' use of plastic show according to the study?A. How widely plastic is used.B. How strange the behavior of wildlife is.C. How some wildlife is adapting to plastic.D. How plastic pollution has harmed them.13. What do leaves mean to alfalfa leafcutter bees?A. Food.B. Shelter.C A plastic substitute. D. Traditional nest materials.14. Which is one characteristic of leafcutter bees?A. They have great economic value.B. They store honey like honeybees.C. They prefer to live in tree holes.D. They have evolved into a new species.15. What was the nest made entirely of plastic like?A. It might be warmer.B. It might be unhealthy.C. It might be easy to finish.D. It might be recyclable.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
2019-2020学年广东省实验中学南海学校九年级（上）月考数学试卷（10月份）
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广东省佛山市南海区南海外国语学校2019-2020学年九年级上学期第三次月测数学试卷
2021-2021学年度上学期九年级第三次月测数学试卷、选择题〔本大题共 10题,每题3分,共30分.每题给出的四个选项中只有一个是正 确的,请将所选选项的字母写在题目后面的括号内〕4.以下命题正确的选项是〔 A. 一组邻边相等的四边形是菱形 B.有三个角是直角的四边形是矩形 C.对角线相等的四边形是矩形D.有一个角是直角的平行四边形是正方形6 .如图,假设 4ABC 与△A 1B 1C 1是位似图形,那么位似中央的坐标为〔1.某几何体的三视图如下图,该几何体是〔 ).B.圆柱C.三棱锥D.球2.假设△ABC S /XABC ,相似比为 1:2, 那么 AABC 与 AABC的周长的比为〔〕.A. 2:1B. 1:2C. 4:1D.1:43.x 1 , *2是一元二次方程2_X 22x 0的两个实数根,以下结论错误的选项是〔A. x 1x 2B.2 - X 12x1 C. x 1x 2 2D. x 1 x 2 25.如图,在网格中,小正方形的边长为1, Z XABC 的顶点都是格点,那么cos BAC 的值为、5 A . -5 2、5B. -----5C. 5D.7 .如图,电灯P 在横杆AB 的正上方,AB 在灯光下白影子为 CD , AB// CD , AB 1m,CD 3m ,点P 到CD 的距离为9m,那么AB 与CD 间的距离是〔8 .如图,在正方形ABCD 中,点E, F 分别在BC, CD 上,BE CF ,那么图中与 AEB 相等的角的个数是〔〕.A. 1B. 2C. 3D. 49 .如图, AB// CD//EF, AD : AF 3:5, BC 6, CE 的长为().A. 2B. 4C. 3D. 510 .如图,在平面直角坐标系中,菱形ABCD 的顶点A, B 在反比例函数ky -〔k 0,x 0〕的图象上,横坐标分别为2, 6,对角线BD//x 轴.假设菱形ABCD 的面 xA. (1,0)B. (0,1)C. ( 1,0)D. (0, 1)A. 6B. 5C. 4D. 3积为16,那么k 的值为〔).、.一 一 2 一 11 .计算：2cos 45 12 . 4a 3b(ab值范围是.16 .在一个不透明的布袋中,有黄色、白色的玻璃球共有质地等完全相同,小刚通过屡次摸球试验后发现摸到黄色球的频率在 球的个数有可能是个.17 .如图,CE 是YABCD 的边AB 的垂直平分线,垂足为点 O, CE 与DA 的延长线交于3二、填空题〔本大题共15B.—47题,每题4分,共C. 6D. 728分.请把答案填写在横线上〕13.：如图,ZXABC 的面积为12,点D 、 E 分别是边AB 、AC 的中点,那么四边形BCED120元降至到现在80元,设平均每次降价的k 1 ..........15.反比例函数 y —^一 〔k 是常数,k 1〕的图象有一支在第二象限,那么 k 的取60个,除颜色外,形状、大小、20%,那么布袋中白色 A.的面积百分率为x ,那么列方程为.药价从原来每盒点E .连接AC , BE , DO , DO与AC交于点F ,那么以下结论:①四边形ACBE是菱形;② ACD ABE;③ AF:BE 1:3；④ S 四边形AFOE : S*A COD 2 : 3；其中正确的结论有.〔填写所有正确结论的序号〕三、解做题〔一〕〔本大题共3题,每题6分,共18分〕218.解万程：〔x 3〕 x 3.19.如图,四边形ABCD中,对角线AC、BD相交于点O , AO OC , BO OD ,且AOB 2 OAD .〔1〕求证,四边形ABCD是矩形;一4 3⑵假设tan ADO 一,AC 10 .求AABO的面积. 420.为了促进“足球进校园〞活动的开展,我校举行了中学生足球比赛活动. 现从A, B,C三支获胜足球队中,随机抽取两支球队分别到两所遥远地区学校进行交流.〔1〕请用列表或画树状图的方法,表示出抽到的两支球队的所有可能结果；〔2〕求出抽到B队和C队参加交流活动的概率.四、解做题〔二〕〔本大题共3,每8分,共24分〕21.如图,某测量小组为了测量山BC的高度,在地面A处测得山顶B的仰角45 ,然后沿着坡度为i 1:点的坡面AD走了300米到达D处,此时在D处测得山顶B的仰角为60 ,求山高BC 〔结果保存根号〕.22. 一商店销售某种商品,平均每天可售出20件,每件盈利40元.为了扩大销售、增加盈利,该店采取了降价举措,在每件盈利不少于25元的前提下,经过一段时间销售,发现销1200 元?k 223 .如图,一次函数 y k i x b 的图象与反比例函数 y 二■的图象相交于 A, B 两点,其中点A 的坐标为(1,3),点B 的坐标为(3,n).(1)根据函数图象,直接写出满足k 1x b 殳的x 的取值范围是；(2)求这两个函数的表达式；(3)点P 在线段AB 上,且S/x A0P : S A BOP 2:3,求点P 的坐标.2题,每题10分,共20分)△ABC 是等腰直角三角形,过斜边 AB 的中点D 作正方形DECF ,分别交 BC , AC 于点 E , F ,求证：AF BE —AB ;2(2)问题解决：如图②,在任意直角 ZXABC 内,找一点D,过点D 作正方形DECF ,分别交BC , AC 于点E , F ,假设AB BE AF ,求 DABDBA 的度数；(3)联系拓广；如图③,在(2)的条件下,分别延长 ED , FD ,交AB 于点M , N ,假设AM 2, BN 3,求MN 的长.25 .如图,在矩形 ABCD 中,AB 8, AD 10, E 是CD 边上一点,连接 AE ,将矩形ABCD 沿AE 折叠,顶点D 恰好落在BC 边上点F 处,延长AE 交BC 的延长线于点 G ,售单价每降低1元,平均每天可多售出 2件. (1)假设降价4元,那么平均每天销售数量为件； (2)当每作商品降价多少元时,该商店每天销售利润为 24. (1)数学理解：如图①,五、解做题(三)(本大题共连接DG .(1)求tan EFC 的值；(2)求证：四边形AFGD是菱形；(3)如图2, M , N分别是线段AG, DG上的动点(与端点不重合) 设AM x, DN y,请解决以下相关问题：①写出y关于x的函数解析式；②是否存在这样的点M ,使4DMN是等腰三角形？假设存在,请求出请说明理由.且DMN DAM ,x的值；假设不存在,。
2019-2020学年佛山外国语学校高三英语第三次联考试题及答案
2019-2020学年佛山外国语学校高三英语第三次联考试题及答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AContact CCA's Special Programs to learn more about our wide range of credit and noncredit opportunities for academic and personal development.Summer ProgramAn exceptional 4-week program for students currently completing their study of high school.Earn 3 college credits!Our setting is CCA's Oakland campus, where studio-focused, college-level courses help you push beyond concepts to portfolio(作品集)－ready pieces. On-campus Housing & Scholarships available!Summer CoursesCCA offers about 35 for-creditdegree courses each summer, scheduled on theSan FranciscoandOaklandcampuses.Undergraduate and graduate students can work with a visiting artist, develop skills and stay on track for graduation by earning credits...all while still allowing time to visit family, work, travel, etc.For information about summer housing, please contact the Office of Residential Life at 510594722 or **************.Summer StartSummer Start is a six-week program designed specifically for international undergraduate and graduate students. This is a for-credit program; students receive six credits towards their degree. During the program, you must live on CCA'sSan Franciscocampus.Tuition includes six units of discounted graduate college credit, which will include books, field trip expenses and one-on-one tutoring. Housing fees include six weeks at CCA's San Francisco Panoramic Residence.Company-Customized ProgramsCCA's Office of Special Programs can design customized workshops to develop and strengthen employee skills within your company. Customized programs have included:An Ideation Sketching class, for the design team of a national clothing retailer, scheduled Friday afternoons onSan Franciscocampus.Software-specific workshops, for designers from a national paper products and gifts retailer, scheduled in a two-day workshop format during the week on ourOaklandcampus.1.Who would most potentially apply for Summer Program?A.Employees in companies.B.International undergraduate and graduate students.C.High school graduates.D.Undergraduate and graduate students.2.What can students do if they take Summer Courses?A.Go on a field trip.B.Visit famous artists.C.Make some friends.D.Earn some credits.3.What are students required to do if they take Summer Start?A.Live on one of the campuses.B.Pay school fees in advance.C.Read as many books as possible.plete the courses on time.BJack was born without eyes. He was very lucky as he grew up having other kittens (young cats) to socialize (交往)with, and was used to people from the moment he was born. However, when it came time to find the kittens homes, no one knew where Jack would end up.That’s when I got an e-mail from my friend. All she asked was “Do you still want one of the kittens? There’s one here with no eyes and no one would like to take him”. Without thinking I told her that I did want the kitten.When we first brought him home, Jack stayed mostly in my room. After about a day he had noissues running around and climbing on everything. At times he gets lost in the house, he’ll stop. But we just call his name and talk to him and it isn’t long before he finds his way back to us.A few weeks after getting Jack, we got a new cat named Bear. Jack and Bear have become best friends. It doesn’t matter that he can’t see. He always knows when Bear is around. He’ll run across the yard straight to Bear and wrap his front legs around his neck in a big hug. They run after each other around and wrestle (摔跤). They’ll lie down in the grass together when tired.Jack is truly an inspiration. I’ve owned lots of kittens in my life, but Jack is the happiest and most playful. He doesn’t feel sorry for himself. He doesn’t need pity. I think Jean, owner of Gumbo, another eyeless cat, said it best when she told me that cats don’t have disabilities; they have adaptability.4. Why did Jack come to the author’s home?A. The author cared for an eyeless cat.B. The author didn’t mind whether he was blind.C. No other young cats kept him company.D. The author’s friend begged the author to take him home.5. Which of the following statements is TRUE?A. Jack often wrestles with Bear indoors.B. Jack likes to play with a new eyeless cat.C. Jack quickly adapts to the new environment.D. Jack is good at talking and playing with people.6. What does the underlined word “issue” in Paragraph 3 mean?A. Trouble.B. Fun.C. Luck.D. Business.7. What does the passage mainly tell us?A. A cat has nine lives.B. All is well that ends well.C. God helps those who help themselves.D. A good beginning makes a good ending.CIn order to develop the pandemic-stricken economy, China recovered the street vendors（商贩）in a new nation-wide method known as “street stall economy'', allowing residents to set up open-air shops on the sidewalks or other available public spaces.Street vendors were once an important part ofChina's economy and urban landscape. However, sinceChina's economy took off in the last decade, street vendors have gradually disappeared from the streets and many of them opened shops of their own.Fast forward to today, street vendors have come into our sight again after cities such asChengduand Yantai succeeded in creating hundreds of thousands of jobs by giving street stalls permission to operate.China's tech industry was quick to jump on the street vendor trend, with tech giants including Alibaba, Tencent, Meituan and flocking todish outcheap loans, offer support and payment solutions to millions of owners of newly established small businesses.Ant Financial, Alibaba's fintech arm, promised its mobile wallet app Alipay will give interest-free loans to 30 million vendors, and 70 billion RMB of interest-free credit line to consumers to make purchases everywhere, including street vendors. provided 50 billion RMB worth of quality goods for street vendors, and provided each one of them with a maximum 100,000 RMB interest-free loan to stock up.Tencent's WeChat announced to offer payment solutions, marketing supports and even training for up to 50 million street vendors, with the end goal of digitally transforming them to increase their income.Guangzhoucity partnered with WeChat this month to hold a live streaming shopping festival to improve sales of local produce. In attempts to promote various live streaming platforms, many tech CEOs also made their own live streamed e-commerce debuts（首次亮相）selling goods coming from all over the country.8. What does the underlined part “dish out" in Paragraph 3 mean?A. turn down.B. provide with.C. pay off.D. apply for.9. Compared with and Tencent, what unique measure did Ant Financial take?A. It provides interest-free loans for vendors.B. It offers interest-free credit line to customers.C. It provides quality goods for street vendors.D. It offers marketing support to businessmen.10. What's the purpose of the cooperation between WeChat andGuangzhoucity?A. To volunteer to train street vendors.B. To give away free goods to the poor.C. To promote to develop the local economy.D. To help CEOs make their own streaming platforms.11. What can be the best title for the text?A. Chengdu and Yantai Succeeded in Creating Job OpportunitiesB. "Street Vendor Economy” Greatly Increases People's IncomeC. The Whole Nation Are Involved in a New Economy ModelD.China's Major Tech Companies Are Helping With "Street Vendor Economy”DPlease take a few seconds and think of your personal biggest goal. Imagine telling someone you meet todaywhat you’re going to do. Imagine their congratulations and their high image of you. Doesn’t it feel good to say it out loud? Don’t you feel one step closer already? Well, bad news: you should have kept your mouth shut, because that good feeling will make you less likely to do it.Any time you have a goal, there is some work that needs to be done to achieve it. Ideally, you would not be satisfied until you’d actually done the work. But when you tell someone your goal and he acknowledges(认可) it, psychologists have found it’s called a “social reality”. The mind is kind of tricked into feeling that it’s already done. And then, because you’ve felt that satisfaction, you’re less motivated to do the actual hard work necessary. This goes againstthe traditional wisdom that we should tell our friends our goals, right?In 1982, Peter Gollwitzer, a Professor of Psychology, wrote a whole book about this. And in 2009, he did some new tests that were published. It goes like this: 163 people across four separate tests—everyone wrote down their personal goal. Then half of them announced their commitment(许诺) to this goal to the room, and half didn’t. Then everyone was given 45 minutes of work that would directly lead them towards their goal, but they were told that they could stop at any time. Now those who kept their mouths shut worked the entire 45 minutes on average, and when asked afterwards, said they felt they had a long way to go to achieve their goal. But those who had announced it quit after only 33 minutes on average, and when asked afterwards, said that they felt much closer to achieving their goal.12. What do the words “social reality” in Paragraph 2 mean?A. Completion of the goal.B. Necessary hard work.C. People's acknowledgement.D. A sense of satisfaction.13. What does Peter Gollwitzer try to tell us?A. Writing down the goal is very helpful.B. Achieving personal goal needs more time.C. Keeping the goal secret makes people work harder.D. Making the goal public makes people less satisfied.14. How did Peter Gollwitzer prove his idea about people’s goal?A. By giving figures.B. By giving examples.C. By making a survey.D. By making comparison tests.15. What will probably happen if you tell your friends your goal?A. You will be more confident.B. You will not gain satisfaction.C. You are less likely to realize it.D. You’ll be much moremotivated.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
2019-2020学年佛山外国语学校高三英语第三次联考试卷及参考答案
2019-2020学年佛山外国语学校高三英语第三次联考试卷及参考答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项ABritain's brilliant bridges have aided trade and brought communities together and are always the most exciting part of the journey. These must-see bridges are now tourist attractions in their own right.Clifton Suspension Bridge, BristolDescribed byits legendary engineer Isambard Kingdom Brunel as “my first love, my darling”, in was originally designed for horse-drawn traffic. Now, more than four million vehicles a year cross the 1,352ft-Jong toll (收费) bridge over the Avon Gorge. The £ l toll for every journey pays for its repair. The history of the bridge, dating back to 1864, is kept alive through a programme of tours, events and exhibitions.Infinity Bridge, Stockton-On-TeesA pedestrian and cycle footbridge across the River Tees, its working title was the North Shore Footbridge, before it was given its grander name when opened in 2009. It is particularly spectacular (壮观) at night. The arches of the bridge are also lit white and, on calm nights, their reflection in the water appears as an infinity symbol, thereby inspiring the name which was chosen by the public.Tower Bridge, LondonAn engineering wonder built from thousands of tons of Cornish granite, Portland stone and steel, it took construction workers eight years to complete. More than 120 years old, it's a popular tourist attraction, as well as a functional bridge. Visitors can take in the views over the capital and experience seeing London life through the Glass Floor.Iron Bridge, ShropshireOpened in 1781, this is the first arch bridge in the world to be made out of cast iron. Recognised as one of the great symbols of the industrial revolution, it transformed the cart of bridge building and was a crucial factor in the development of the iron trade in Shropshire.1.Which bridge has the longest history?A.Clifton Suspension Bridge.B.Infinity Bridge.C.Tower Bridge.D.Iron Bridge.2.What can we know about Infinity Bridge?A.People can just walk on it.B.The public give it two names.C.It's well worth visiting at night.D.It's arch is the biggest on the earth.3.What do Clifton Suspension Bridge and Tower Bridge have in common?A.They are both over 120 years old.B.Visitors should pay for passing them.C.They have the same original design.D.Visitors can have a good view of London on them.BMy first week working in a restaurant, one of the servers said something that stuck with me: Everyone should work in a restaurant for at least a year. At the time, I didn't get it, but I took the advice to heart and worked in restaurants on and off for the next eight years. Before realizing it, I mastered many important skills, one of which is communication skill.When I was little, I was so shy that I used to hide behind my mom whenever someone spoke to me. And when I first started in restaurants, I had two personalities: Restaurant Lizzy and Home Lizzy. It was easier to pretend to be a different person while at work, since it was so different from who I actually was. But gradually, the skills I learned working in restaurants helped Home Lizzy come out of her shell in the real world.When you work in a restaurant, you don't have the luxury of hiding behind your parents to avoid talking to people. I'm still 110% an introvert, but restaurant work helped me communicate. Working in a restaurant not only helped me speak clearly, deliberately and directly but also taught me how to talk about almost everything. Some guests don't want their servers to interact too much with them, and that's fine. But some sit at the bar simply to chat with you. You learn how to judge your guests' level of interest in communicating with you, and how to exit a conversation at the appropriate time.My restaurant work is something that I'm most proud of and I know I wouldn't be the person I am today without those eight years of experience. If you're still on the fence about working in a restaurant for that long, start with one year. I doubt that you'll look back.4. What did the writer think of the server's words?A. Impressive.B. Ridiculous.C. Amusing.D. Logical.5. What do we infer from Paragraph 2?A. The writer tried different jobs.B. The writer became more sociable.C. The writer used an invented name.D. The writer had a hard time at work.6. Which of the following best describes the writer's restaurant work?A. Boring.B. Relaxing.C. Worthwhile.D. Unchallenging.7. What message does the writer try to convey in this passage?A. A strong-willed soul can reach his goal.B. Things are difficult before they are easy.C. Communication skills advance your career.D. Restaurantwork helps to achieve a better self.CRichard Campbell is a secondary school student. He is15 years old. He lives in a small town in the north ofEngland. Every morning, he gets up at eight o’clock, puts on his uniform and walks to school.One hour later, the lessons start. The students usually study maths, English, history and geography in the morning. They usually study music and drawing and they play sports after lunch. They have a ten-minute break between classes. They also spend a long time in the school library, reading books and doing their homework.Richard likes his school very much. His favourite subjects are English and geography, but he doesn’t like maths because he is not good at it.Richard and all his friends spend the whole day at school. Lunch is at one o’clock. He doesn’t like the food that the school serves. This is why he often brings a packed lunch from home. He always has his lunch in the dining hall, but some of his friends sometimes eat in the courtyard or outside the school gate.At weekends, he always goes out with his friends because he doesn’t go to school. On Saturday, he always goes to the cinema or to the sports centre. On Sunday, he just goes for a walk with his dog.8. What time do Richard’s lessons begin in the morning?A. At 8:00.B. At 8:30.C. At 9:00.D. At 9:10.9. What does Richard like best?A. Maths and English.B. English and geography.C. History and maths.D. Geography and history.10. Where does Richard have lunch on weekdays?A. In the dining hall.B. At home.C. In the courtyard.D. Outside his school gate.11. What does Richard always do on Saturday?A. He reads and does homework.B. He goes to see films or does sports.C. He walks his dog or does exercise.D. He spends a long time in the library.DIf you’ve ever had a dog, you know just how deep a connection you can develop with “man’s best friend”. But a dog has a much shorter life span — about 12 to 15 years long — than humans, which means every dog owner has to go through the heartbreaking moment when their loving pet passes away.Why not make a clone of that dog then? This is the solution offered by a South Korean company, Sooam Biotech Research Foundation. The company has successfully cloned at least 400 dogs, mostly for US customers, ever since it pioneered the technique in 2005. Now, Sooam Biotech is planning to introduce their business toUKdog owners, offering them dogs that look just like their lost ones.Meanwhile, another dog is selected to supply an egg.Researchers then replace the DNA in the egg with that from the skin cell and implant the egg into the womb (子宫) of a female dog. The egg grows into a puppy over the following two months. To clone a dog, researchers first need to take a skin cell from a living dog or one that has just died.The whole process takes less than a day, but it comes at a shockingly high price — around £63,000 (614,000 yuan). But if you can’t afford it now, you can also save the cells in a laboratory and access them at a later date. Just like identical twins of humans, they share the exact same DNA but there will still be small differences between them. “The spots on a Dalmatian clone will be different, for example,” Insung Hwang, head of Sooam Biotech, toldThe Guardian. However, as magical as cloning might sound, there is no guarantee that the cloned dog will be a perfect replica of the original one.Dog owners will also have to accept the fact that personality is not “clone-able”. Apart from genes, personality is also determined by upbringing and environment, which are both “random elements [that] cloning technologies simply cannot overcome”, Professor Tom Kirkwood atNewcastle University,UK, toldThe Telegraph.Perhaps bringing our dogs back with cloning is not the best way to remember them after all.Kirkwood, a dog owner himself, pointed out: “An important aspect of our relationship with them is coming to terms with the pain of letting go.”12. According to the article, Sooam Biotech Research Foundation is ______.A. working on plans to help dog owners enjoy their pets longerB. offering a way to help dogs give birth to more puppiesC. providing a service that will make copies of pet dogsD. introducing a completely new technique to clone dogs13. Which of the following statements about dog cloning is TRUE according to the article?A. Dog cloning technology hadn’t been put into practice until recently.B. Dog cloning is very expensive and usually takes several months to complete.C. Dog cloning is very popular among US andUKpet owners.D. Cloned dogs might develop different habits and characteristics even though they look very similar.14. Which of the following shows the correct order of the dog cloning process?a. an egg is taken from another dogb. a skin cell is taken from the pet dog and saved in a laboratoryc. the egg is placed in the womb of a female dogd. the DNA of the egg is replaced by the DNA from the skin celle. the egg grows into a puppy in two monthsA. acbde.B. adbce.C. bacde.D. badce.15. We can learn from the article thatKirkwood______ dog cloning.A. disapproves ofB. supportsC. is afraid ofD. is curious about第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
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2019-2020学年度上学期九年级第三次月测
数学试卷
一、选择题（本大题共10题，每题3分，共30分．每题给出的四个选项中只有一个是正确的，请将所选选项的字母写在题目后面的括号内）
1．某几何体的三视图如图所示，该几何体是（ ）．
A ．圆锥
B ．圆柱
C ．三棱锥
D ．球
2．若ABC A B C '''△∽△，相似比为1:2，则ABC △与A B C '''△的周长的比为（ ）．
A ．2:1
B ．1:2
C ．4:1
D ．1:4
3．已知1x ，2x 是一元二次方程220x x -=的两个实数根，下列结论错误的是（ ）．
A ．12x x ≠
B ．2
1120x x -= C ．122x x += D ．122x x ⋅=
4．下列命题正确的是（ ）．
A ．一组邻边相等的四边形是菱形
B ．有三个角是直角的四边形是矩形
C ．对角线相等的四边形是矩形
D ．有一个角是直角的平行四边形是正方形
5．如图，在网格中，小正方形的边长为1，ABC △的顶点都是格点，则cos BAC ∠的值为（
）
．
A ．5
B ．5
C ．5
D 6．如图，若ABC △与111A B C △是位似图形，则位似中心的坐标为（ ）．
A ．(1,0)
B ．(0,1)
C ．(1,0)-
D ．(0,1)-
7．如图，电灯P 在横杆AB 的正上方，AB 在灯光下的影子为CD ，AB CD ∥，1AB m =，3CD m =，点P 到CD 的距离为9m ，则AB 与CD 间的距离是（ ）m ．
A ．6
B ．5
C ．4
D ．3
8．如图，在正方形ABCD 中，点E ，F 分别在BC ，CD 上，BE CF =，则图中与AEB ∠相等的角的个数是（ ）．
A ．1
B ．2
C ．3
D ．4
9．如图，已知AB CD EF ∥∥，:3:5AD AF =，6BC =，CE 的长为（ ）．
A ．2
B ．4
C ．3
D ．5
10．如图，在平面直角坐标系中，菱形ABCD 的顶点A ，B 在反比例函数(0,0)k y k x x
=>>的图象上，横坐标分别为2，6，对角线BD x ∥轴．若菱形ABCD 的面积为16，则k 的值为（ ）．
A ．53
B ．154
C ．6
D ．7
二、填空题（本大题共7题，每题4分，共28分．请把答案填写在横线上）
11．计算：22cos 45︒= ．
12．已知43(0)a b ab =≠，则:a b = ．
13．已知：如图，ABC △的面积为12，点D 、E 分别是边AB 、AC 的中点，则四边形BCED 的面积为 ．
14．一种药品经过两次降价，药价从原来每盒120元降至到现在80元，设平均每次降价的百分率为x ，则列方程为 ．
15．已知反比例函数1y k x -=
（k 是常数，1k ≠）的图象有一支在第二象限，那么k 的取值范围是 ．
16．在一个不透明的布袋中，有黄色、白色的玻璃球共有60个，除颜色外，形状、大小、质地等完全相同，小刚通过多次摸球试验后发现摸到黄色球的频率在20%，则布袋中白色球的个数有可能是 个．
17．如图，CE 是ABCD Y 的边AB 的垂直平分线，垂足为点O ，CE 与DA 的延长线交于点E ．连接AC ，BE ，DO ，DO 与AC 交于点F ，则下列结论：
①四边形ACBE 是菱形；
②ACD ABE ∠=∠；
③:1:3AF BE =；
④S 四边形:2:3AFOE COD S =△；
其中正确的结论有 ．（填写所有正确结论的序号）
三、解答题（一）（本大题共3题，每题6分，共18分）
18．解方程：2
(3)3x x -=-．
19．如图，四边形ABCD 中，对角线AC 、BD 相交于点O ，AO OC =，
BO OD =，且2AOB OAD ∠=∠． （1）求证，四边形ABCD 是矩形；
（2）若3tan 4
ADO ∠=，10AC =．求ABO △的面积．
20．为了促进“足球进校园”活动的开展，我校举行了中学生足球比赛活动．现从A ，B ，C 三支获胜足球队中，随机抽取两支球队分别到两所边远地区学校进行交流．
（1）请用列表或画树状图的方法，表示出抽到的两支球队的所有可能结果；
（2）求出抽到B 队和C 队参加交流活动的概率．
四、解答题（二）（本大题共3，每8分，共24分）
21．如图，某测量小组为了测量山BC 的高度，在地面A 处测得山顶B 的仰角45︒，然后沿着坡度为
1:i =AD 走了300米达到D 处，此时在D 处测得山顶B 的仰角为60︒，求山高BC （结果保留根号）．
22．一商店销售某种商品，平均每天可售出20件，每件盈利40元．为了扩大销售、增加盈利，该店采取了降价措施，在每件盈利不少于25元的前提下，经过一段时间销售，发现销售单价每降低1元，平均每天可多售出2件．
（1）若降价4元，则平均每天销售数量为 件；
（2）当每作商品降价多少元时，该商店每天销售利润为1200元？
23．如图，一次函数1y k x b =+的图象与反比例函数2k y x =
的图象相交于A ，B 两点，其中点A 的坐标为(1,3)-，点B 的坐标为(3,)n ．
（1）根据函数图象，直接写出满足21k k x b x
+≤
的x 的取值范围是 ； （2）求这两个函数的表达式；
（3）点P 在线段AB 上，且:2:3AOP BOP S S =△△，求点P 的坐标．
五、解答题（三）（本大题共2题，每题10分，共20分）
24．（1）数学理解：如图①，ABC △是等腰直角三角形，
过斜边AB 的中点D 作正方形DECF ，分别交BC ，
AC 于点E ，F ，求证：2
AF BE AB +=； （2）问题解决：如图②，在任意直角ABC △内，找一点D ，过点D 作正方形DECF ，分别交BC ，AC 于点E ，F ，若AB BE AF =+，求DAB DBA ∠+∠的度数；
（3）联系拓广；如图③，在（2）的条件下，分别延长ED ，FD ，交AB 于点M ，N ，若2AM =，3BN =，求MN 的长．
25．如图，在矩形ABCD 中，8AB =，10AD =，E 是CD 边上一点，连接AE ，将矩形ABCD 沿AE 折叠，顶点D 恰好落在BC 边上点F 处，延长AE 交BC 的延长线于点G ，连接DG ．
（1）求tan EFC ∠的值；
（2）求证：四边形AFGD 是菱形；
（3）如图2，M ，N 分别是线段AG ，DG 上的动点
（与端点不重合），且DMN DAM ∠=∠，设AM x =，DN y =，请解决以下相关问题：
①写出y 关于x 的函数解析式；
②是否存在这样的点M ，使DMN △是等腰三角形？若存在，请求出x 的值；若不存在，请说明理由．。