2022年南京市建鄴區(qū)初三九上數(shù)學(xué)期中試題和答案(蘇科版)

2021－2022學(xué)年度第一學(xué)期期中學(xué)業(yè)質(zhì)量監(jiān)測(cè)九年級(jí)數(shù)學(xué)學(xué)科注意事項(xiàng)：1．本試卷共6頁(yè)．全卷滿分120分．考試時(shí)間為120分鐘．考生答題全部答在答題卡上，答在本試卷上無(wú)效．2．請(qǐng)認(rèn)真核對(duì)監(jiān)考教師在答題卡上所粘貼條形碼的姓名、考試證號(hào)是否與本人相符合，、考試證號(hào)用0.5毫米黑色墨水簽字筆填寫在答題卡3．答選擇題必須用2B鉛筆將答題卡上對(duì)非選擇題必須用0.5毫米黑色墨水簽字筆寫在答題卡上的指定再將自己的姓名及本試卷上．應(yīng)的答案標(biāo)號(hào)涂黑．如需改動(dòng)，請(qǐng)用橡皮擦干凈后，再選涂其他答案．答位置，在其他位置答題一律無(wú)效．4．作圖必須用2B鉛筆作答，并請(qǐng)加黑加粗，（本大題共6小題，2分，共12分．在每小題所給出的四個(gè)選項(xiàng)中，是符合題目要的求，請(qǐng)將正確選項(xiàng)1．下列方程是一元二次方程的是描寫清楚．一、選擇題每小題恰有一項(xiàng)前的字母代號(hào)填涂在答題卡相應(yīng)位置上）．．．．．．．1A．x2＝0B．＋xy2＝1C．x＝2x3－3D．3x＋＝1x2．小明在一次射擊訓(xùn)練時(shí)，連續(xù)10次的成績(jī)?yōu)?次10環(huán)、4次9環(huán)，則小明這10次射擊的平均成績(jī)?yōu)锳．9.3環(huán)B．9.4環(huán)C．9.5環(huán)D．9.6環(huán)3的圓，下列結(jié)3．在△ABC中，AB＝AC＝5，BC＝8，以A為圓心作一個(gè)半徑為論中正確的是A．點(diǎn)B．點(diǎn)A上C．直線BC與⊙D．直線BC與⊙4．小明根據(jù)演講比賽中9位評(píng)委所給的分?jǐn)?shù)制作了如下表格：B在⊙A內(nèi)C在⊙A相切A相離平均數(shù)中位數(shù)眾數(shù)方差8.58.38.10.15如果去掉一個(gè)最高分和一個(gè)最低分，那么表格中數(shù)據(jù)一定不發(fā)生變化的是A．平均數(shù)5．關(guān)于x的一元二次方程ax2－4x－1＝0有兩個(gè)不相等的實(shí)數(shù)根，則A．a(chǎn)≥－B．a(chǎn)＞－C．a(chǎn)≥－4且a≠0D．a(chǎn)＞－4且a≠06．在平面直角坐標(biāo)系中，以若A（2，－1）為圓心，交于C、D，則CD的最小值是A．C．22B．2B．中位數(shù)C．眾數(shù)D．方差a的取值范圍是442為半徑的⊙A與過(guò)點(diǎn)B（1，0）的直線2D．4數(shù)學(xué)試卷第1頁(yè)（共11頁(yè)）二、填空題（本大題共10小題，每小題2分，共20分．請(qǐng)把答案填寫在答題卡相應(yīng)位置）7．南京2021年11月1號(hào)的最高氣溫為22℃，最低氣溫為12℃，該日的氣溫極差為▲．8．某件羊毛衫的售價(jià)為1000元，因換季促銷，在經(jīng)過(guò)連續(xù)兩次降價(jià)后，現(xiàn)售價(jià)為810元，x，根據(jù)題意可列方程為▲．9．如圖，A、B是⊙O上的點(diǎn)，且∠AOB=60°．在這個(gè)圖中，僅用無(wú)刻度的設(shè)平均每次降價(jià)的百分率為直尺能畫出的角的度數(shù)可以是▲．（只要求寫出四個(gè)）BADAOOBC（第9題）（第10題）10．如圖，⊙O是四邊形ABCD的內(nèi)切圓，若∠BOC＝118°，則∠AOD＝▲．11．用一個(gè)半徑為3，圓心角度數(shù)為120°的扇形圍成一個(gè)圓錐的側(cè)面，則該圓錐的底面圓的半徑為▲．12．若代數(shù)式x2＋4x＋6可以表示為(x＋1)＋a(x＋1)＋b，則2a＋b＝▲．213．若點(diǎn)A（1，2），B（3，－3），C（5，n）三點(diǎn)可以確定一個(gè)圓，則n需要滿足的條件為▲．－b＋b2－4ac2a14．當(dāng)a＝1，b＝m，c＝－15時(shí)，若代數(shù)式的值為3，則代數(shù)式－b－b2－4ac的值為▲．2a15．如圖，某酒店有一張桌面邊長(zhǎng)為2米的正六邊形桌子，每邊圍坐兩人（平均每人占據(jù)1米長(zhǎng)的桌沿），可坐下12人.現(xiàn)酒店方想將桌面改成正十二邊形，每邊坐1人，也可坐下12人.改造方案如下：在原正六邊形桌面的頂點(diǎn)處分別截去一個(gè)等腰三角形，則桌面改造后，圍坐的12人每人占據(jù)的桌沿長(zhǎng)度比改造前減少▲米（.精確到0.01米，參考數(shù)據(jù)：3≈1.73）ABDMPC（第15題）（第16題）16．如圖，在矩形ABCD中，AB＝4，AD＝8，M是CD的中點(diǎn)，P是BC上一個(gè)動(dòng)點(diǎn)，若∠DPM的度數(shù)最大，則BP＝▲．三、解答題（本大題共11小題，共88分．請(qǐng)?jiān)诖痤}卡指定區(qū)域內(nèi)作答，解答時(shí)應(yīng)寫出文．．．．．．．字說(shuō)明、證明過(guò)程或演算步驟）17．（8分）解下列方程（1）x(x＋1)－2(x＋1)＝0；（2）3x－5x＋1＝0．218．（7分）如圖，PA，PB是圓O的切線，A，B為切點(diǎn)，Q是圓上一點(diǎn)，且OQ∥PB，∠P＝34°．求∠Q的度數(shù)．AQPOB（第18題）19．（7分）如圖，⊙O的弦AB、CD的延長(zhǎng)線相交于點(diǎn)P，且AB＝CD．求證：PB＝PD．ABOPDC（第19題）20．（7分）如圖，四邊形ABCD是⊙O的內(nèi)接四邊形，AD與BC的延長(zhǎng)線交于點(diǎn)E，∠DCB＝100°，∠B＝50°．求證：△CDE是等腰三角形．ECDOBA（第20題）數(shù)學(xué)試卷第3頁(yè)（共11頁(yè)）21．（7分）學(xué)校舉行廚藝大賽，參賽選手人數(shù)是評(píng)委人數(shù)的5倍少2人，每位參賽者需在規(guī)定時(shí)間內(nèi)，將制作好的菜品分到小盤中給每位評(píng)委一小盤試吃評(píng)分．若本次比賽評(píng)委共試吃168個(gè)小盤菜品，求參賽選手的人數(shù)．22．（8分）如圖，一張正方形紙片的邊長(zhǎng)為2cm，將它剪去4個(gè)全等的直角三角形，四邊形EFGH的面積可能為1cm2嗎？請(qǐng)說(shuō)明理由．GCDFHAEB（第22題）23．（10分）甲乙兩人在相同條件下完成了5次射擊訓(xùn)練，兩人的成績(jī)?nèi)鐖D所示．乙5次射擊訓(xùn)練成績(jī)統(tǒng)計(jì)圖甲5次射擊訓(xùn)練成績(jī)條形統(tǒng)計(jì)圖成績(jī)/環(huán)成績(jī)/環(huán)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）甲射擊成績(jī)的（2）計(jì)算兩人射擊成績(jī)的方差；（3）根據(jù)訓(xùn)練成績(jī)，眾數(shù)為▲環(huán)，乙射擊成績(jī)的中位數(shù)為▲環(huán)；你認(rèn)為選派哪一名隊(duì)員參賽更好，為什么？數(shù)學(xué)試卷第4頁(yè)（共11頁(yè)）24．（6分）如圖，點(diǎn)A在直線l上，點(diǎn)P在直線l外，作⊙O經(jīng)過(guò)P，A兩點(diǎn)且與l相切．PlA（第24題）25．（8分）已知關(guān)于x的一元二次方程(x－5)2＝m＋1有實(shí)數(shù)根．（1）求m的取值范圍；（2）若方程的兩根分別為x、x，且x＋x2－xx12＝3，求m的值．12126．（9分）如圖，D為⊙O上一點(diǎn)，點(diǎn)C是直徑BA延長(zhǎng)線上的一點(diǎn)，且∠CDA＝∠CBD．（1）求證：CD是⊙O的切線；（2）過(guò)點(diǎn)B作⊙O的切線BE交CD的延長(zhǎng)線于點(diǎn)E．若BC＝12，AC＝4，求BE的長(zhǎng)．EDBCOA（第26題）數(shù)學(xué)試卷第5頁(yè)（共11頁(yè)）27．（11分）【問(wèn)題提出】（1）如圖1，在四邊形ABCD中，AB＝AD，∠BCD＝∠BAD＝90°，AC＝4，求BC＋CD的值；小明提供了他研究這個(gè)問(wèn)題的思路：延長(zhǎng)CD至點(diǎn)M，使得DM＝BC，連接AM．可以構(gòu)造三角形全等，結(jié)合勾股定理便可解決這個(gè)問(wèn)題．【問(wèn)題解決】（2）如圖2，有一個(gè)直徑為10cm的圓形配件⊙O，現(xiàn)需在該配件上切割出一個(gè)四邊形孔洞OABC，要求∠O＝60°，∠B＝30°，OA＝OC，求四邊形OABC的面積的最小值．BAABCOCMD（圖1）（圖2）數(shù)學(xué)試卷第6頁(yè)（共11頁(yè)）2021－2022學(xué)年度第一學(xué)期期中學(xué)業(yè)質(zhì)量監(jiān)測(cè)九年級(jí)數(shù)學(xué)試卷參考答案及評(píng)分標(biāo)準(zhǔn)說(shuō)明：本評(píng)分標(biāo)準(zhǔn)每題給出了一種或幾種解法供參考．如果考生的解法與本解答不同，參照本評(píng)分標(biāo)準(zhǔn)的精神給分．一、選擇題（本大題共6小題，每小題2分，共12分）題號(hào)答案123456ADCBDC二、填空題（本大題共10小題，每小題共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題，不出1個(gè)或三、解答題（本大題共11小題，共88分）17．（本題8分）1）x(x＋1)－2(x＋1)＝0，5．15．0.08．16．8－22．寫單位，不扣分．第9題，寫出4個(gè)即可，寫出2個(gè)或者3個(gè)給1分，寫者所寫答案中有錯(cuò)誤，不給分．第15題，答案為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．（本題解：連接OA，OB．∵PA，PB是圓O的切線，∴AO⊥AP，BO⊥BP．7分）A，B為切點(diǎn)，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數(shù)學(xué)試卷第7頁(yè)（共11頁(yè)）∴∠AOQ＝∠AOB－∠QOB＝146°－90°＝56°．··········································5分∵OA＝OQ，∴∠OAQ＝∠OQA．180°－56°∴∠Q＝＝62°．······································································7分219．（本題7分）解：連接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．（本題四邊形ABCD是⊙∴∠CDA＋∠B＝180°．∵∠B＝50°，∴∠CDA＝180°－50°＝130°．·································································7分）解：∵O的內(nèi)接四邊形，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．(本題7分)x人，則參賽選手有x(5x－2)＝168．····································································4分解：設(shè)評(píng)委有(5x－2)人．根據(jù)題意，得28x＝6，x＝－（不合題意，舍）．·······································6分512解這個(gè)方程，得5x－2＝5×6－2＝28．答：參賽選手有28人．···················································································7分22．（本題8分）解：∵在正方形紙上剪去4個(gè)全等的直角三角形，∴HG＝GF＝FE＝EH，∠AHE＝∠DGH．∴四邊形EFGH為菱形．∵四邊形ABCD是正方形，數(shù)學(xué)試卷第8頁(yè)（共11頁(yè)）∴∠D＝90°．∠DGH＋∠DHG＝90°．∠EHG＝180°－∠AHE－∠DHG＝90°．四邊形EFGH為正方形．·································································3分∴∴∴GCD設(shè)AH＝xcm，則AE＝(2－x)cm，x＋(2－x)2＝1．··································································6分F根據(jù)題意，得2整理，得2x－4x＋3＝0．2∵b2－4ac＝(－4)2－4×2×3＝－8＜0，H∴方程沒(méi)有實(shí)數(shù)根．四邊形EFGH的面積不可能為1cm2．·························································8分23．（本題10分）1）7，8．·····························································································4分AEB解：（1－（2）x＝×(7＋7＋10＋9＋7)＝8（環(huán)），甲5＝×[(7－8)2＋(7－8)2＋(10－8)2＋(9－8)2＋(7－8)2)]＝（851環(huán)）．S251甲－x＝×(8＋8＋7＋8＋9)＝8（環(huán)）5乙＝×[(8－8)2＋(8－8)2＋(7－8)2＋(8－8)2＋(9－8)2)]＝（2515環(huán)）．······8分S2乙－－x＝x，S＞S，說(shuō)明兩人實(shí)力相當(dāng)，但乙射擊的成績(jī)比甲穩(wěn)定．22甲乙甲乙（3）因?yàn)樗赃x擇乙參加比賽．（本題答案不惟一，言之有理即可得分）．···············10分24．（本題解：作出AP的垂直平分過(guò)點(diǎn)A作l的垂線．·················································································4分6分）線．············································································2分作⊙O．·································································································6分POlA25．（本題8分）解：（1）解法一：∵(x－5)≥0，···································································2分2∴m＋1≥0．∴m的取值范圍是m≥－解法二：∵關(guān)于x的一元二次方1．·····························································4分程(x－5)2＝m＋1有實(shí)數(shù)根，∴關(guān)于x的一元二次方程x－10x＋(24－m)＝0有實(shí)數(shù)根．2∵b2－4ac＝(－10)2－4×1×(24－m)≥0，············································2分∴m的取值范圍是m≥－（2）根據(jù)題意，得x1＋x2＝10，xx12＝24－m．················································6分∵x1＋x2－xx12＝3，1．·····························································4分?jǐn)?shù)學(xué)試卷第9頁(yè)（共11頁(yè)）∴10－(24－m)＝3．∴m＝17．·····················································································8分26．(本題9分)證明：（1）連接OD．∴∠ADO＝∠OAD，∵AB是⊙O的直徑，∴∠BDA＝90°，∴∠ABD＋∠BAD＝90°，∵∠CDA＝∠CBD，∴∠CDO＝∠CDA＋∠ADO＝90°，即CD⊥OD．················································································3分又OD為半徑，∴CD是⊙O的切線．·····································································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的切線，∴∠CBE＝90°，BE＝DE．設(shè)BE＝x，BCOA在Rt△EBC中，∵BE2＋BC2＝EC2，∴x2＋122＝(x＋43)2．∴x＝43．BE即的長(zhǎng)為43．··········································································9分27．(本題11分)證明：（1）延長(zhǎng)CD至點(diǎn)M，使得DM＝BC，連接AM．∵∠BCD＝∠BAD＝90°，∴∠BCD＋∠