2023年山東省濟(jì)南市東南片區(qū)中考一模數(shù)學(xué)試題（含答案）

2023年九年級(jí)學(xué)業(yè)水平模擬測(cè)試一數(shù)學(xué)試題一、選擇題，本大題共10個(gè)小題，每小題4分，共40分．在每小題給出的四個(gè)選項(xiàng)中，只有一項(xiàng)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(shè)，提升城市公共交通服務(wù)水平，濟(jì)南市城市軌道交通2020~2025年第二期建設(shè)規(guī)劃地鐵總里程約為159600米．把數(shù)字“159600”用科學(xué)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(duì)稱圖形又是軸對(duì)稱圖形的是（）A． B． C． D．6．已知實(shí)數(shù)，在數(shù)軸上對(duì)應(yīng)點(diǎn)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(nóng)耕文明與天文學(xué)智慧的結(jié)晶，被國(guó)際氣象界譽(yù)為“中國(guó)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(jī)抽取一張（不放回），再從中隨機(jī)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(biāo)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(diǎn)，以點(diǎn)為圓心，長(zhǎng)為半徑作，交射線于點(diǎn)，連接；（2）分別以點(diǎn)，為圓心，長(zhǎng)為半徑作弧，交于點(diǎn)，；③連接，，．根據(jù)以上作圖過程及所作圖形，下列結(jié)論中錯(cuò)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(cè)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(diǎn)，點(diǎn)的縱坐標(biāo)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(gè)小題，每小題4分，共24分．）11．因式分解：______．12．如圖，一個(gè)可以自由轉(zhuǎn)動(dòng)的轉(zhuǎn)盤，被分成了9個(gè)相同的扇形，轉(zhuǎn)動(dòng)轉(zhuǎn)盤，轉(zhuǎn)盤停止時(shí)，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(diǎn)恰好落在上的點(diǎn)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(zhǎng)一倍得到新正方形；把正方形邊長(zhǎng)按原法延長(zhǎng)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(zhǎng)為8，點(diǎn)、分別在邊、上，將四邊形沿折疊，使點(diǎn)落在處，點(diǎn)落在點(diǎn)處，交于．以下結(jié)論：①當(dāng)為中點(diǎn)時(shí)，三邊之比為；②連接，則；③當(dāng)三邊之比為時(shí)，為中點(diǎn)；④當(dāng)在上移動(dòng)時(shí)，周長(zhǎng)不變．其中正確的有______（寫出所有正確結(jié)論的序號(hào)）．三、解答題（本大題共10個(gè)小題，共86分．解答應(yīng)寫出文字說明、證明過程或演算步驟．）17．（6分）計(jì)算：18．（6分）解不等式組：，并寫出它的所有非負(fù)整數(shù)解．19．（6分）在中，點(diǎn)，在對(duì)角線上，且，連接，．求證：．20．（8分）為深入學(xué)習(xí)貫徹黨的二十大精神，某校開展了以“學(xué)習(xí)二十大，永遠(yuǎn)跟黨走，奮進(jìn)新征程”為主題的知識(shí)競(jìng)賽．發(fā)現(xiàn)該校全體學(xué)生的競(jìng)賽成績(jī)（百分制）均不低于60分，現(xiàn)從中隨機(jī)抽取名學(xué)生的競(jìng)賽成績(jī)進(jìn)行整理和分析（成績(jī)得分用表示，共分成四組），并繪制成如下的競(jìng)賽成績(jī)分組統(tǒng)計(jì)表和扇形統(tǒng)計(jì)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(jìng)賽成績(jī)分組統(tǒng)計(jì)表組別競(jìng)賽成績(jī)分組頻數(shù)平均分18652a763b854c94請(qǐng)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(jī)抽取的這名學(xué)生競(jìng)賽成績(jī)的中位數(shù)是______分，平均分是______分；（4）若學(xué)生競(jìng)賽成績(jī)達(dá)到85分以上（含85分）為優(yōu)秀，請(qǐng)你估計(jì)全校1200名學(xué)生中優(yōu)秀學(xué)生的人數(shù)．21．（8分）如圖，一艘游輪在處測(cè)得北偏東45°的方向上有一燈塔B，游輪以海里/時(shí)的速度向正東方向航行2小時(shí)到達(dá)處，此時(shí)測(cè)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(jié)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(diǎn)，且，過點(diǎn)的切線交的延長(zhǎng)線于點(diǎn)，交的延長(zhǎng)線于點(diǎn)，連結(jié)，交于點(diǎn)．（1）求證：；（2）若，的半徑為2，求的長(zhǎng)．23．山地自行車越來越受到中學(xué)生的喜愛，各品牌相繼投放市場(chǎng)．某車行經(jīng)營(yíng)的A型車去年銷售總額為50000元，今年每輛銷售價(jià)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(jià)多少元?（2）該車行計(jì)劃新進(jìn)一批A型車和新款B型車共60輛，且B型車的進(jìn)貨數(shù)量不超過A型車數(shù)量的兩倍，應(yīng)如何進(jìn)貨才能使這批車獲利最多?A，B兩種型號(hào)車的進(jìn)貨和銷售價(jià)格如下表：A型車B型車進(jìn)貨價(jià)格（元）11001400銷售價(jià)格（元）今年的銷售價(jià)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(biāo)系．反比例函數(shù)的圖象交于點(diǎn)，交于點(diǎn)，．（1）求的值與點(diǎn)的坐標(biāo)；（2）在軸上找一點(diǎn)，使的周長(zhǎng)最小，請(qǐng)求出點(diǎn)的坐標(biāo)；（3）在（2）的條件下，若點(diǎn)是軸上的一個(gè)動(dòng)點(diǎn)，點(diǎn)是平面內(nèi)的任意一點(diǎn)，試判斷是否存在這樣的點(diǎn)，，使得以點(diǎn)，，，為頂點(diǎn)的四邊形是菱形．若存在，請(qǐng)直接寫出符合條件的點(diǎn)坐標(biāo)；若不存在，請(qǐng)說明理由．25．（12分）某校數(shù)學(xué)興趣學(xué)習(xí)小組在一次活動(dòng)中，對(duì)一些特殊幾何圖形具有的性質(zhì)進(jìn)行了如下探究：（1）發(fā)現(xiàn)問題：如圖1，在等腰中，，點(diǎn)是邊上任意一點(diǎn)，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(diǎn)是邊上任意一點(diǎn)，以為腰作等腰，使，．在點(diǎn)運(yùn)動(dòng)過程中，是否存在最小值？若存在，求出最小值，若不存在，請(qǐng)說明理由．（3）拓展應(yīng)用：如圖3，在正方形中，點(diǎn)是邊上一點(diǎn)，以為邊作正方形，是正方形的中心，連接．若正方形的邊長(zhǎng)為8，，求的面積．26．（12分）拋物線過點(diǎn)，點(diǎn)，頂點(diǎn)為，與軸相交于點(diǎn)．點(diǎn)是該拋物線上一動(dòng)點(diǎn)，設(shè)點(diǎn)的橫坐標(biāo)為．（1）求拋物線的表達(dá)式及點(diǎn)的坐標(biāo)；（2）如圖1，連接，，，若的面積為3，求的值；（3）連接，過點(diǎn)作于點(diǎn)，是否存在點(diǎn)，使得．如果存在，請(qǐng)求出點(diǎn)的坐標(biāo)；如果不存在，請(qǐng)說明理由．九年級(jí)數(shù)學(xué)試卷（2023.4）參考答案及評(píng)分標(biāo)準(zhǔn)一、選擇題題號(hào)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(fù)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(jiǎng)的人數(shù)是648人．································································································8分21．解：（1）如圖，由題意可得，∠CAB＝45°，過點(diǎn)C作CE⊥AB于點(diǎn)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(diǎn)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(diǎn)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(shè)今年A型車每輛售價(jià)x元，則去年售價(jià)每輛為（x+400）元，····················1分由題意，得50000x+400=解得：x＝1600．························································································3分經(jīng)檢驗(yàn)，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(jià)1600元；··········································································5分（2）設(shè)今年新進(jìn)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(jìn)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(shí)，y最大＝34000元．∴B型車的數(shù)量為：60﹣20＝40輛．∴當(dāng)新進(jìn)A型車20輛，B型車40輛時(shí)，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(diǎn)E（2，4）························································································1分把E（2，4）代入y=kx中，得：∴k＝8．························································································2分當(dāng)x＝6時(shí)，y=4∴F(6，4（2）作點(diǎn)F關(guān)于x軸的對(duì)稱點(diǎn)G(6，-43)，則連接GE與x軸交于點(diǎn)M，連接EF，此時(shí)△EMF的周長(zhǎng)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(shè)EG的函數(shù)關(guān)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(dāng)y＝0時(shí)，x＝5，∴M（5，0）．························································································6分（3）點(diǎn)P的坐標(biāo)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(diǎn)A作AH⊥CN交CN延長(zhǎng)線于點(diǎn)H，此時(shí)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(shè)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(diǎn)A（﹣1，0），點(diǎn)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(dá)式為y＝﹣x2+2x+3．···································································3分∵y＝﹣x2+2x+3＝﹣（x﹣1）2+4，∴頂點(diǎn)C（1，4）．·····································································4分(2)∵點(diǎn)D（0，3），點(diǎn)B（3，0），∴直線BC解析式為y＝﹣x+3，··········································································5分過點(diǎn)P作PQ∥y軸交BD于點(diǎn)Q，設(shè)點(diǎn)P（m，﹣m2+2m+3），點(diǎn)Q（m，﹣m+3），∴S△PBD＝12×PQ×OB＝12×3（﹣m2+2m+3+m-3）＝-32∵△PBD的面積為3，