2023年山東省濟南市東南片區(qū)中考一模數(shù)學試題（含答案）

2023年九年級學業(yè)水平模擬測試一數(shù)學試題一、選擇題，本大題共10個小題，每小題4分，共40分．在每小題給出的四個選項中，只有一項是符合題目要求的．）1．的相反數(shù)是（）A． B．2 C． D．2．如圖所示的幾何體，其俯視圖是（）正面A． B． C． D．3．為完善城市軌道交通建設，提升城市公共交通服務水平，濟南市城市軌道交通2020~2025年第二期建設規(guī)劃地鐵總里程約為159600米．把數(shù)字“159600”用科學記數(shù)法表示為（）A． B． C． D．4．如圖，平行線，被直線所截，平分，若，則的度數(shù)是（）A．39° B．51° C．78° D．102°5．下列圖案中，既是中心對稱圖形又是軸對稱圖形的是（）A． B． C． D．6．已知實數(shù)，在數(shù)軸上對應點的位置如圖所示，則下列判斷正確的是（）A． B． C． D．7．“二十四節(jié)氣”是中華農耕文明與天文學智慧的結晶，被國際氣象界譽為“中國第五大發(fā)明”．小明購買了“二十四節(jié)氣”主題郵票，他要將“立春”“立夏”“秋分”三張郵票中的兩張送給好朋友小亮．小明將它們背面朝上放在桌面上（郵票背面完全相同），讓小亮從中隨機抽取一張（不放回），再從中隨機抽取一張，則小亮抽到的兩張郵票恰好是“立春”和“秋分”的概率是（）A． B． C． D．8．函數(shù)與在同一坐標系中的圖象如圖所示，則函數(shù)的大致圖象為（）A． B． C． D．9．如圖，已知銳角，按如下步驟作圖：（1）在射線上取一點，以點為圓心，長為半徑作，交射線于點，連接；（2）分別以點，為圓心，長為半徑作弧，交于點，；③連接，，．根據(jù)以上作圖過程及所作圖形，下列結論中錯誤的是（）A． B．若，則C． D．10．已知二次函數(shù)，將其圖象在直線左側部分沿軸翻折，其余部分保持不變，組成圖形．在圖形上任取一點，點的縱坐標的取值滿足或，其中．令，則的取值范圍是（）A． B． C． D．二、填空題（本大題共6個小題，每小題4分，共24分．）11．因式分解：______．12．如圖，一個可以自由轉動的轉盤，被分成了9個相同的扇形，轉動轉盤，轉盤停止時，指針落在陰影區(qū)域的概率等于______．13．比大的最小整數(shù)是______．14．如圖，扇形紙片的半徑為4，沿折疊扇形紙片，點恰好落在上的點處，圖中陰影部分的面積為______．15．如圖（1），已知小正方形的面積為1，把它的各邊延長一倍得到新正方形；把正方形邊長按原法延長一倍得到正方形（如圖（2））…；以此下去，則正方形的面積為______．16．正方形的邊長為8，點、分別在邊、上，將四邊形沿折疊，使點落在處，點落在點處，交于．以下結論：①當為中點時，三邊之比為；②連接，則；③當三邊之比為時，為中點；④當在上移動時，周長不變．其中正確的有______（寫出所有正確結論的序號）．三、解答題（本大題共10個小題，共86分．解答應寫出文字說明、證明過程或演算步驟．）17．（6分）計算：18．（6分）解不等式組：，并寫出它的所有非負整數(shù)解．19．（6分）在中，點，在對角線上，且，連接，．求證：．20．（8分）為深入學習貫徹黨的二十大精神，某校開展了以“學習二十大，永遠跟黨走，奮進新征程”為主題的知識競賽．發(fā)現(xiàn)該校全體學生的競賽成績（百分制）均不低于60分，現(xiàn)從中隨機抽取名學生的競賽成績進行整理和分析（成績得分用表示，共分成四組），并繪制成如下的競賽成績分組統(tǒng)計表和扇形統(tǒng)計圖，其中“”這組的數(shù)據(jù)如下：82，83，83，84，84，85，85，86，86，86，87，89．競賽成績分組統(tǒng)計表組別競賽成績分組頻數(shù)平均分18652a763b854c94請根據(jù)以上信息，解答下列問題：（1）______．（2）“”這組數(shù)據(jù)的眾數(shù)是______分，方差是______；（3）隨機抽取的這名學生競賽成績的中位數(shù)是______分，平均分是______分；（4）若學生競賽成績達到85分以上（含85分）為優(yōu)秀，請你估計全校1200名學生中優(yōu)秀學生的人數(shù)．21．（8分）如圖，一艘游輪在處測得北偏東45°的方向上有一燈塔B，游輪以海里/時的速度向正東方向航行2小時到達處，此時測得燈塔在處北偏東15°的方向上．（1）求到直線的距離；（2）求游輪繼續(xù)向正東方向航行過程中與燈塔的最小距離是多少海里?（結果精確到1海里，參考數(shù)據(jù)：，，，，）22．（8分）如圖，是的直徑，，是上兩點，且，過點的切線交的延長線于點，交的延長線于點，連結，交于點．（1）求證：；（2）若，的半徑為2，求的長．23．山地自行車越來越受到中學生的喜愛，各品牌相繼投放市場．某車行經營的A型車去年銷售總額為50000元，今年每輛銷售價比去年降低400元，若賣出的數(shù)量相同，銷售總額將比去年減少20%．（1）今年A型車每輛售價多少元?（2）該車行計劃新進一批A型車和新款B型車共60輛，且B型車的進貨數(shù)量不超過A型車數(shù)量的兩倍，應如何進貨才能使這批車獲利最多?A，B兩種型號車的進貨和銷售價格如下表：A型車B型車進貨價格（元）11001400銷售價格（元）今年的銷售價格200024．如圖，在矩形中，，，分別以，所在的直線為軸和軸建立平面直角坐標系．反比例函數(shù)的圖象交于點，交于點，．（1）求的值與點的坐標；（2）在軸上找一點，使的周長最小，請求出點的坐標；（3）在（2）的條件下，若點是軸上的一個動點，點是平面內的任意一點，試判斷是否存在這樣的點，，使得以點，，，為頂點的四邊形是菱形．若存在，請直接寫出符合條件的點坐標；若不存在，請說明理由．25．（12分）某校數(shù)學興趣學習小組在一次活動中，對一些特殊幾何圖形具有的性質進行了如下探究：（1）發(fā)現(xiàn)問題：如圖1，在等腰中，，點是邊上任意一點，連接，以為腰作等腰，使，，連接．求證：．（2）類比探究：如圖2，在等腰中，，，，點是邊上任意一點，以為腰作等腰，使，．在點運動過程中，是否存在最小值？若存在，求出最小值，若不存在，請說明理由．（3）拓展應用：如圖3，在正方形中，點是邊上一點，以為邊作正方形，是正方形的中心，連接．若正方形的邊長為8，，求的面積．26．（12分）拋物線過點，點，頂點為，與軸相交于點．點是該拋物線上一動點，設點的橫坐標為．（1）求拋物線的表達式及點的坐標；（2）如圖1，連接，，，若的面積為3，求的值；（3）連接，過點作于點，是否存在點，使得．如果存在，請求出點的坐標；如果不存在，請說明理由．九年級數(shù)學試卷（2023.4）參考答案及評分標準一、選擇題題號12345678910答案BCCABDCADD二、填空題11．4(a+1)(a-1)12．4913．314．163π-8三、解答題17．原式=3-22-2+2×22······························································································4=1-18．解：解不等式①，得：x＜5，··················································································2分解不等式②，得：x＜4，······························································································4分原不等式組的解集是x＜4．··························································································5分非負整數(shù)解為0，1，2，3·························································································6分19．證明：∵四邊形ABCD是平行四邊形，∴AD＝BC，AD∥BC，····································································································2分∴∠DAE＝∠BCF，··································································································3分∵AE=FC，∠DAE＝∠BCF，AD＝BC，∴△ADE≌△CBF（SAS），·······························································································4分∴∠DEA=∠BFC············································································································5分∴∠DEC=∠BFA∴DE∥BF·····················································································································6分20．解：（1）20························································································2分（2）86，3.5································································································4分（3）85.583.6··························································································6分（4）1200×2750=648答：獲獎的人數(shù)是648人．································································································8分21．解：（1）如圖，由題意可得，∠CAB＝45°，過點C作CE⊥AB于點E，·······································1分在△ABC中，∠BAC＝45°，∴△ACE是等腰直角三角形，·································2分由題意得：AC＝2×202=402，∴CE=22即點C到線段AB的距離為40海里；····················································································4分（2）由題意可得，∠DCB＝15°，則∠ACB＝105°，∵∠ACE＝45°，∴∠CBE＝30°，·································································································5分∵在Rt△BEC中，AE＝CE＝40，∴BE=3CE＝403∴AB＝AE+BE＝40+403··························································································7分作BF⊥AC于點F，則∠AFB=90°在Rt△BEC中，cos∠BAC＝BFAB=∴BF=202+206≈77答：與燈塔B的最小距離是77海里．········································································8分22．解：（1）證明：如圖，連接OD，∵DE是⊙O的切線，∴DE⊥OD，∴∠ODF=90°···········································································1分∵BD=CD，∴∠CAD=∠DAB，·····················································································2分∵OA=OD，∴∠DAB=∠ODA，∴∠CAD=∠ODA，∴OD∥AE，························································································3分∴∠AEF=∠ODF=90°∴AE⊥EF························································································4分（2）解：∵∠CAD=∠ODA，∠AGE=∠OGD，∴△OGD∽△EGA，∴，················································································5分∵∠AEF=∠ODF，∠F=∠F∴△ODF∽△AEF∴·································································································6分∴·····································································································7分∴BF=2························································································8分23．解：（1）設今年A型車每輛售價x元，則去年售價每輛為（x+400）元，····················1分由題意，得50000x+400=解得：x＝1600．························································································3分經檢驗，x＝1600是原方程的根，且符合題意，······························································4分答：今年A型車每輛售價1600元；··········································································5分（2）設今年新進A型車a輛，則B型車（60﹣a）輛，獲利y元，由題意，得················6分y＝（1600﹣1100）a+（2000﹣1400）（60﹣a），y＝﹣100a+36000．························································································7分∵B型車的進貨數(shù)量不超過A型車數(shù)量的兩倍，∴60﹣a≤2a，∴a≥20．························································································8分∵y＝﹣100a+36000．∴k＝﹣100＜0，∴y隨a的增大而減?。ぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁ?分∴a＝20時，y最大＝34000元．∴B型車的數(shù)量為：60﹣20＝40輛．∴當新進A型車20輛，B型車40輛時，這批車獲利最大．···········································10分24．解：（1）∵在矩形OABC中，OA＝6，OC＝4，∴AB=4，BC=6∵BE=4∴點E（2，4）························································································1分把E（2，4）代入y=kx中，得：∴k＝8．························································································2分當x＝6時，y=4∴F(6，4（2）作點F關于x軸的對稱點G(6，-43)，則連接GE與x軸交于點M，連接EF，此時△EMF的周長最?。ぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁ?分設EG的函數(shù)關系式為y＝ax+b把E（2，4），G(6，-43)代入y＝ax得：2a+b=46a+b=-43∴y=-4當y＝0時，x＝5，∴M（5，0）．························································································6分（3）點P的坐標為（0，0）或(-1，0)或(10，0)或·································10分25．(1)證明：∵∠BAC=∠MAN，∴∠BAC－∠CAM=∠MAN－∠CAM，即∠BAM=∠CAN，········································1分∵AB=AC，AM=AN，∴△ABM≌△ACN，························································································3分∴∠ACN=∠ABM．························································································4分(2)解：AN存在最小值，理由如下：∵AM=MN，AB=BC，∴又∵∠AMN=∠B，∴△ABC∽△AMN，∴AMAB=ANAC，∠BAC∴∠BAC-∠MAC=∠MAN-∠MAC即∠BAM=∠CAN∴△ABM∽△ACN∴∠ACN=∠B=30°························································································7分過點A作AH⊥CN交CN延長線于點H，此時AN最小，最小值為AH，Rt△ACH中，∠ACN=30°AH=12AC=1故AN存在最小值，最小值為4···················································································8分(3)解：連接BD，EH，過H作HQ⊥CD于Q，如圖所示，∵H為正方形DEFG的中心，∴DH=EH，∠DHE=90°，∵四邊形ABCD為正方形，∴BC=CD，∠BCD=90°，∴∠BDE+∠CDE=∠CDH+∠CDE=45°，∴∠BDE=∠CDH，∵BDCD∴△BDE∽△CDH，························································································9分∴∠DCH=∠DBC=45°，BE=2CH=6設CE=x，則CD=x+6，∵DE=8，∴由勾股定理得：x2解得：x=23-3或x=-∴CD=23+3，在Rt△CDH中，CQ=QH=3，∴△CDH的面積為1226．解：（1）將點A（﹣1，0），點B（3，0）代入y＝ax2+bx+3得：a-b+3=0解得：a=-1b=2．·····················································································∴拋物線的表達式為y＝﹣x2+2x+3．···································································3分∵y＝﹣x2+2x+3＝﹣（x﹣1）2+4，∴頂點C（1，4）．·····································································4分(2)∵點D（0，3），點B（3，0），∴直線BC解析式為y＝﹣x+3，··········································································5分過點P作PQ∥y軸交BD于點Q，設點P（m，﹣m2+2m+3），點Q（m，﹣m+3），∴S△PBD＝12×PQ×OB＝12×3（﹣m2+2m+3+m-3）＝-32∵△PBD的面積為3，