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

2021－2022學年度第一學期期中學業(yè)質量監(jiān)測九年級數(shù)學學科注意事項：1．本試卷共6頁．全卷滿分120分．考試時間為120分鐘．考生答題全部答在答題卡上，答在本試卷上無效．2．請認真核對監(jiān)考教師在答題卡上所粘貼條形碼的姓名、考試證號是否與本人相符合，、考試證號用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．應的答案標號涂黑．如需改動，請用橡皮擦干凈后，再選涂其他答案．答位置，在其他位置答題一律無效．4．作圖必須用2B鉛筆作答，并請加黑加粗，（本大題共6小題，2分，共12分．在每小題所給出的四個選項中，是符合題目要的求，請將正確選項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前的字母代號填涂在答題卡相應位置上）．．．．．．．1A．x2＝0B．＋xy2＝1C．x＝2x3－3D．3x＋＝1x2．小明在一次射擊訓練時，連續(xù)10次的成績?yōu)?次10環(huán)、4次9環(huán)，則小明這10次射擊的平均成績?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3．在△ABC中，AB＝AC＝5，BC＝8，以A為圓心作一個半徑為論中正確的是A．點B．點A上C．直線BC與⊙D．直線BC與⊙4．小明根據(jù)演講比賽中9位評委所給的分數(shù)制作了如下表格：B在⊙A內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如果去掉一個最高分和一個最低分，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．關于x的一元二次方程ax2－4x－1＝0有兩個不相等的實數(shù)根，則A．a≥－B．a＞－C．a≥－4且a≠0D．a＞－4且a≠06．在平面直角坐標系中，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與過點B（1，0）的直線2D．4數(shù)學試卷第1頁（共11頁）二、填空題（本大題共10小題，每小題2分，共20分．請把答案填寫在答題卡相應位置）7．南京2021年11月1號的最高氣溫為22℃，最低氣溫為12℃，該日的氣溫極差為▲．8．某件羊毛衫的售價為1000元，因換季促銷，在經過連續(xù)兩次降價后，現(xiàn)售價為810元，x，根據(jù)題意可列方程為▲．9．如圖，A、B是⊙O上的點，且∠AOB=60°．在這個圖中，僅用無刻度的設平均每次降價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）BADAOOBC（第9題）（第10題）10．如圖，⊙O是四邊形ABCD的內切圓，若∠BOC＝118°，則∠AOD＝▲．11．用一個半徑為3，圓心角度數(shù)為120°的扇形圍成一個圓錐的側面，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A（1，2），B（3，－3），C（5，n）三點可以確定一個圓，則n需要滿足的條件為▲．－b＋b2－4ac2a14．當a＝1，b＝m，c＝－15時，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處分別截去一個等腰三角形，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，P是BC上一個動點，若∠DPM的度數(shù)最大，則BP＝▲．三、解答題（本大題共11小題，共88分．請在答題卡指定區(qū)域內作答，解答時應寫出文．．．．．．．字說明、證明過程或演算步驟）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，Q是圓上一點，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的延長線相交于點P，且AB＝CD．求證：PB＝PD．ABOPDC（第19題）20．（7分）如圖，四邊形ABCD是⊙O的內接四邊形，AD與BC的延長線交于點E，∠DCB＝100°，∠B＝50°．求證：△CDE是等腰三角形．ECDOBA（第20題）數(shù)學試卷第3頁（共11頁）21．（7分）學校舉行廚藝大賽，參賽選手人數(shù)是評委人數(shù)的5倍少2人，每位參賽者需在規(guī)定時間內，將制作好的菜品分到小盤中給每位評委一小盤試吃評分．若本次比賽評委共試吃168個小盤菜品，求參賽選手的人數(shù)．22．（8分）如圖，一張正方形紙片的邊長為2cm，將它剪去4個全等的直角三角形，四邊形EFGH的面積可能為1cm2嗎？請說明理由．GCDFHAEB（第22題）23．（10分）甲乙兩人在相同條件下完成了5次射擊訓練，兩人的成績如圖所示．乙5次射擊訓練成績統(tǒng)計圖甲5次射擊訓練成績條形統(tǒng)計圖成績/環(huán)成績/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）甲射擊成績的（2）計算兩人射擊成績的方差；（3）根據(jù)訓練成績，眾數(shù)為▲環(huán)，乙射擊成績的中位數(shù)為▲環(huán)；你認為選派哪一名隊員參賽更好，為什么？數(shù)學試卷第4頁（共11頁）24．（6分）如圖，點A在直線l上，點P在直線l外，作⊙O經過P，A兩點且與l相切．PlA（第24題）25．（8分）已知關于x的一元二次方程(x－5)2＝m＋1有實數(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，點C是直徑BA延長線上的一點，且∠CDA＝∠CBD．（1）求證：CD是⊙O的切線；（2）過點B作⊙O的切線BE交CD的延長線于點E．若BC＝12，AC＝4，求BE的長．EDBCOA（第26題）數(shù)學試卷第5頁（共11頁）27．（11分）【問題提出】（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問題的思路：延長CD至點M，使得DM＝BC，連接AM．可以構造三角形全等，結合勾股定理便可解決這個問題．【問題解決】（2）如圖2，有一個直徑為10cm的圓形配件⊙O，現(xiàn)需在該配件上切割出一個四邊形孔洞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)學試卷第6頁（共11頁）2021－2022學年度第一學期期中學業(yè)質量監(jiān)測九年級數(shù)學試卷參考答案及評分標準說明：本評分標準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，參照本評分標準的精神給分．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個即可，寫出2個或者3個給1分，寫者所寫答案中有錯誤，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，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)學試卷第7頁（共11頁）∴∠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：設評委有(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)學試卷第8頁（共11頁）∴∠D＝90°．∠DGH＋∠DHG＝90°．∠EHG＝180°－∠AHE－∠DHG＝90°．四邊形EFGH為正方形．·································································3分∴∴∴GCD設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∴方程沒有實數(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，說明兩人實力相當，但乙射擊的成績比甲穩(wěn)定．22甲乙甲乙（3）因為所以選擇乙參加比賽．（本題答案不惟一，言之有理即可得分）．···············10分24．（本題解：作出AP的垂直平分過點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：∵關于x的一元二次方1．·····························································4分程(x－5)2＝m＋1有實數(shù)根，∴關于x的一元二次方程x－10x＋(24－m)＝0有實數(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分數(shù)學試卷第9頁（共11頁）∴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BE＝x，BCOA在Rt△EBC中，∵BE2＋BC2＝EC2，∴x2＋122＝(x＋43)2．∴x＝43．BE即的長為43．··········································································9分27．(本題11分)證明：（1）延長CD至點M，使得DM＝BC，連接AM．∵∠BCD＝∠BAD＝90°，∴∠BCD＋∠