2022年南京市建邺区初三九上数学期中试题和答案(苏科版)

2021－2022学年度第一学期期中学业质量监测九年级数学学科注意事项：1．本试卷共6页．全卷满分120分．考试时间为120分钟．考生答题全部答在答题卡上，答在本试卷上无效．2．请认真核对监考教师在答题卡上所粘贴条形码的姓名、考试证号是否与本人相符合，、考试证号用0.5毫米黑色墨水签字笔填写在答题卡3．答选择题必须用2B铅笔将答题卡上对非选择题必须用0.5毫米黑色墨水签字笔写在答题卡上的指定再将自己的姓名及本试卷上．应的答案标号涂黑．如需改动，请用橡皮擦干净后，再选涂其他答案．答位置，在其他位置答题一律无效．4．作图必须用2B铅笔作答，并请加黑加粗，（本大题共6小题，2分，共12分．在每小题所给出的四个选项中，是符合题目要的求，请将正确选项1．下列方程是一元二次方程的是描写清楚．一、选择题每小题恰有一项前的字母代号填涂在答题卡相应位置上）．．．．．．．1A．x2＝0B．＋xy2＝1C．x＝2x3－3D．3x＋＝1x2．小明在一次射击训练时，连续10次的成绩为6次10环、4次9环，则小明这10次射击的平均成绩为A．9.3环B．9.4环C．9.5环D．9.6环3的圆，下列结3．在△ABC中，AB＝AC＝5，BC＝8，以A为圆心作一个半径为论中正确的是A．点B．点A上C．直线BC与⊙D．直线BC与⊙4．小明根据演讲比赛中9位评委所给的分数制作了如下表格：B在⊙A内C在⊙A相切A相离平均数中位数众数方差8.58.38.10.15如果去掉一个最高分和一个最低分，那么表格中数据一定不发生变化的是A．平均数5．关于x的一元二次方程ax2－4x－1＝0有两个不相等的实数根，则A．a≥－B．a＞－C．a≥－4且a≠0D．a＞－4且a≠06．在平面直角坐标系中，以若A（2，－1）为圆心，交于C、D，则CD的最小值是A．C．22B．2B．中位数C．众数D．方差a的取值范围是442为半径的⊙A与过点B（1，0）的直线2D．4数学试卷第1页（共11页）二、填空题（本大题共10小题，每小题2分，共20分．请把答案填写在答题卡相应位置）7．南京2021年11月1号的最高气温为22℃，最低气温为12℃，该日的气温极差为▲．8．某件羊毛衫的售价为1000元，因换季促销，在经过连续两次降价后，现售价为810元，x，根据题意可列方程为▲．9．如图，A、B是⊙O上的点，且∠AOB=60°．在这个图中，仅用无刻度的设平均每次降价的百分率为直尺能画出的角的度数可以是▲．（只要求写出四个）BADAOOBC（第9题）（第10题）10．如图，⊙O是四边形ABCD的内切圆，若∠BOC＝118°，则∠AOD＝▲．11．用一个半径为3，圆心角度数为120°的扇形围成一个圆锥的侧面，则该圆锥的底面圆的半径为▲．12．若代数式x2＋4x＋6可以表示为(x＋1)＋a(x＋1)＋b，则2a＋b＝▲．213．若点A（1，2），B（3，－3），C（5，n）三点可以确定一个圆，则n需要满足的条件为▲．－b＋b2－4ac2a14．当a＝1，b＝m，c＝－15时，若代数式的值为3，则代数式－b－b2－4ac的值为▲．2a15．如图，某酒店有一张桌面边长为2米的正六边形桌子，每边围坐两人（平均每人占据1米长的桌沿），可坐下12人.现酒店方想将桌面改成正十二边形，每边坐1人，也可坐下12人.改造方案如下：在原正六边形桌面的顶点处分别截去一个等腰三角形，则桌面改造后，围坐的12人每人占据的桌沿长度比改造前减少▲米（.精确到0.01米，参考数据：3≈1.73）ABDMPC（第15题）（第16题）16．如图，在矩形ABCD中，AB＝4，AD＝8，M是CD的中点，P是BC上一个动点，若∠DPM的度数最大，则BP＝▲．三、解答题（本大题共11小题，共88分．请在答题卡指定区域内作答，解答时应写出文．．．．．．．字说明、证明过程或演算步骤）17．（8分）解下列方程（1）x(x＋1)－2(x＋1)＝0；（2）3x－5x＋1＝0．218．（7分）如图，PA，PB是圆O的切线，A，B为切点，Q是圆上一点，且OQ∥PB，∠P＝34°．求∠Q的度数．AQPOB（第18题）19．（7分）如图，⊙O的弦AB、CD的延长线相交于点P，且AB＝CD．求证：PB＝PD．ABOPDC（第19题）20．（7分）如图，四边形ABCD是⊙O的内接四边形，AD与BC的延长线交于点E，∠DCB＝100°，∠B＝50°．求证：△CDE是等腰三角形．ECDOBA（第20题）数学试卷第3页（共11页）21．（7分）学校举行厨艺大赛，参赛选手人数是评委人数的5倍少2人，每位参赛者需在规定时间内，将制作好的菜品分到小盘中给每位评委一小盘试吃评分．若本次比赛评委共试吃168个小盘菜品，求参赛选手的人数．22．（8分）如图，一张正方形纸片的边长为2cm，将它剪去4个全等的直角三角形，四边形EFGH的面积可能为1cm2吗？请说明理由．GCDFHAEB（第22题）23．（10分）甲乙两人在相同条件下完成了5次射击训练，两人的成绩如图所示．乙5次射击训练成绩统计图甲5次射击训练成绩条形统计图成绩/环成绩/环108106498720第1次第2次第3次第4次第5次0第1次第2次第3次第4次第5次（1）甲射击成绩的（2）计算两人射击成绩的方差；（3）根据训练成绩，众数为▲环，乙射击成绩的中位数为▲环；你认为选派哪一名队员参赛更好，为什么？数学试卷第4页（共11页）24．（6分）如图，点A在直线l上，点P在直线l外，作⊙O经过P，A两点且与l相切．PlA（第24题）25．（8分）已知关于x的一元二次方程(x－5)2＝m＋1有实数根．（1）求m的取值范围；（2）若方程的两根分别为x、x，且x＋x2－xx12＝3，求m的值．12126．（9分）如图，D为⊙O上一点，点C是直径BA延长线上的一点，且∠CDA＝∠CBD．（1）求证：CD是⊙O的切线；（2）过点B作⊙O的切线BE交CD的延长线于点E．若BC＝12，AC＝4，求BE的长．EDBCOA（第26题）数学试卷第5页（共11页）27．（11分）【问题提出】（1）如图1，在四边形ABCD中，AB＝AD，∠BCD＝∠BAD＝90°，AC＝4，求BC＋CD的值；小明提供了他研究这个问题的思路：延长CD至点M，使得DM＝BC，连接AM．可以构造三角形全等，结合勾股定理便可解决这个问题．【问题解决】（2）如图2，有一个直径为10cm的圆形配件⊙O，现需在该配件上切割出一个四边形孔洞OABC，要求∠O＝60°，∠B＝30°，OA＝OC，求四边形OABC的面积的最小值．BAABCOCMD（图1）（图2）数学试卷第6页（共11页）2021－2022学年度第一学期期中学业质量监测九年级数学试卷参考答案及评分标准说明：本评分标准每题给出了一种或几种解法供参考．如果考生的解法与本解答不同，参照本评分标准的精神给分．一、选择题（本大题共6小题，每小题2分，共12分）题号答案123456ADCBDC二、填空题（本大题共10小题，每小题共20分）2分，7．10℃．10．62°．8．1000(1－x)2＝810．9．60°，90°，30°，150，120°．11．1．12．7．13．n≠－8．14．－注：第7题，不出1个或三、解答题（本大题共11小题，共88分）17．（本题8分）1）x(x＋1)－2(x＋1)＝0，5．15．0.08．16．8－22．写单位，不扣分．第9题，写出4个即可，写出2个或者3个给1分，写者所写答案中有错误，不给分．第15题，答案为0.07不扣分．解：（(x＋1)(x－2)＝0．···········································································2分x＋1＝0或x－2＝0．x＝－1，x＝2．················································································4分12（2）3x2－5x＋1＝0，∵a＝3，b＝－5，c＝1，b2－4ac＝(－5)2－4×3×1＝25－12＝13．·············································6分5±135±13∴x＝2×3＝．65＋1365－136x＝，x＝．·································································8分1218．（本题解：连接OA，OB．∵PA，PB是圆O的切线，∴AO⊥AP，BO⊥BP．7分）A，B为切点，A∴∠PAO＝∠PBO＝90°．·······································································1分∵∠P＝34°，Q∴∠AOB＝360°－∠PAO－∠PBO－∠P＝146°．································O·········3P分∵OQ∥PB，∴∠QOB＝180°－∠PBO＝90°．B数学试卷第7页（共11页）∴∠AOQ＝∠AOB－∠QOB＝146°－90°＝56°．··········································5分∵OA＝OQ，∴∠OAQ＝∠OQA．180°－56°∴∠Q＝＝62°．······································································7分219．（本题7分）解：连接AC，∵AB＝CD，⌒⌒∴AB＝CD．························································································2分⌒⌒⌒⌒A∴AB＋BD＝CD＋BD．⌒⌒即AD＝BC．························································································4分BO∴∠A＝∠C．∴PA＝PC．·························································································6分∵AB＝CD，PDC∴PA－AB＝PC－CD．即PB＝PD．························································································7分20．（本题四边形ABCD是⊙∴∠CDA＋∠B＝180°．∵∠B＝50°，∴∠CDA＝180°－50°＝130°．·································································7分）解：∵O的内接四边形，EC2分∴∠CDE＝180°－∠CDA＝180°－130°＝50°．·············································D3分∵∠DCB＝100°，∴∠CDE＋∠E＝100°．O∴∠E＝50°．·······················································································5分BA∴∠E＝∠CDE．∴CD＝CE．∴△CDE是等腰三角形．·······································································7分21．(本题7分)x人，则参赛选手有x(5x－2)＝168．····································································4分解：设评委有(5x－2)人．根据题意，得28x＝6，x＝－（不合题意，舍）．·······································6分512解这个方程，得5x－2＝5×6－2＝28．答：参赛选手有28人．···················································································7分22．（本题8分）解：∵在正方形纸上剪去4个全等的直角三角形，∴HG＝GF＝FE＝EH，∠AHE＝∠DGH．∴四边形EFGH为菱形．∵四边形ABCD是正方形，数学试卷第8页（共11页）∴∠D＝90°．∠DGH＋∠DHG＝90°．∠EHG＝180°－∠AHE－∠DHG＝90°．四边形EFGH为正方形．·································································3分∴∴∴GCD设AH＝xcm，则AE＝(2－x)cm，x＋(2－x)2＝1．··································································6分F根据题意，得2整理，得2x－4x＋3＝0．2∵b2－4ac＝(－4)2－4×2×3＝－8＜0，H∴方程没有实数根．四边形EFGH的面积不可能为1cm2．·························································8分23．（本题10分）1）7，8．·····························································································4分AEB解：（1－（2）x＝×(7＋7＋10＋9＋7)＝8（环），甲5＝×[(7－8)2＋(7－8)2＋(10－8)2＋(9－8)2＋(7－8)2)]＝（851环）．S251甲－x＝×(8＋8＋7＋8＋9)＝8（环）5乙＝×[(8－8)2＋(8－8)2＋(7－8)2＋(8－8)2＋(9－8)2)]＝（2515环）．······8分S2乙－－x＝x，S＞S，说明两人实力相当，但乙射击的成绩比甲稳定．22甲乙甲乙（3）因为所以选择乙参加比赛．（本题答案不惟一，言之有理即可得分）．···············10分24．（本题解：作出AP的垂直平分过点A作l的垂线．·················································································4分6分）线．············································································2分作⊙O．·································································································6分POlA25．（本题8分）解：（1）解法一：∵(x－5)≥0，···································································2分2∴m＋1≥0．∴m的取值范围是m≥－解法二：∵关于x的一元二次方1．·····························································4分程(x－5)2＝m＋1有实数根，∴关于x的一元二次方程x－10x＋(24－m)＝0有实数根．2∵b2－4ac＝(－10)2－4×1×(24－m)≥0，············································2分∴m的取值范围是m≥－（2）根据题意，得x1＋x2＝10，xx12＝24－m．················································6分∵x1＋x2－xx12＝3，1．·····························································4分数学试卷第9页（共11页）∴10－(24－m)＝3．∴m＝17．·····················································································8分26．(本题9分)证明：（1）连接OD．∴∠ADO＝∠OAD，∵AB是⊙O的直径，∴∠BDA＝90°，∴∠ABD＋∠BAD＝90°，∵∠CDA＝∠CBD，∴∠CDO＝∠CDA＋∠ADO＝90°，即CD⊥OD．················································································3分又OD为半径，∴CD是⊙O的切线．·····································································4分（2）∵BC＝12，CA＝4，∴OC＝8，OD＝4．∵∠CDO＝90°，E∴CD＝OC2－OD2＝8－42＝43．2D∵BE、CD是⊙O的切线，∴∠CBE＝90°，BE＝DE．设BE＝x，BCOA在Rt△EBC中，∵BE2＋BC2＝EC2，∴x2＋122＝(x＋43)2．∴x＝43．BE即的长为43．··········································································9分27．(本题11分)证明：（1）延长CD至点M，使得DM＝BC，连接AM．∵∠BCD＝∠BAD＝90°，∴∠BCD＋∠