2009年山西省太原市中考數(shù)學(xué)試卷及答案

2009年山西省太原市初中畢業(yè)學(xué)業(yè)考試試卷數(shù)學(xué)一、選擇題(本大題含10個(gè)小題,只有一項(xiàng)符合題目要求1．在數(shù)軸上表示的點(diǎn)離開原點(diǎn)的距離等于,每小題3分,共30分)在每小題給出的四個(gè)選項(xiàng)中2,請(qǐng)選出并在答題卡上將該項(xiàng)涂黑.()2B．2C．4D．A．22．下列計(jì)算中,結(jié)果正確的是2a·3a6aB．()3aa6C．2a·aa6aaa3D．62A．233．學(xué)業(yè)考試體育測(cè)試結(jié)束后,某班體育委員將本班50名學(xué)生的測(cè)試成績(jī)制成如下的統(tǒng)計(jì)表．這個(gè)班學(xué)生體育測(cè)試成績(jī)的眾數(shù)是()成績(jī)(分)(人)A．30分20211222234242562652782810296302人數(shù)15B．28分C．25分D．10人3x9x3x4x14．已知一個(gè)多項(xiàng)式與的和等于,則這個(gè)多項(xiàng)式是()22A．5x15x1B．C．13x113x1D．x2x505．用配方法解方程時(shí),原方程應(yīng)變形為()2x126A．x126B．AAx229C．x229D．B△ACB≌△ACB,BCBACA=30°,則的度數(shù)為()6．如圖,A．20°7．如圖,在Rt△ABC中,C=90°,B．30°C．35°D．40°BCABC=10,若以點(diǎn)為圓心,CBABDAC長(zhǎng)為半徑的圓恰好經(jīng)過的中點(diǎn),則的長(zhǎng)等于()53A．52C．B．5D．6C8．如果所得的三角形的A．4B．4.59．如圖三角形的兩邊分別為3和5,那么連接這個(gè)三角形三邊中點(diǎn)周長(zhǎng)可能是()ABDC．5D．5.5,AB是半圓O的直徑,點(diǎn)P從點(diǎn)O出發(fā),沿OAABBO的路徑運(yùn)動(dòng)一周．設(shè)為OPstst之間關(guān)系的是,運(yùn)動(dòng)時(shí)間為,則下列圖形能大致地刻畫與()ssssPABOOOtOtOttA．10．在二行三列的方格棋盤上沿骰子的某和5點(diǎn),3點(diǎn)和4點(diǎn)),在每一種翻動(dòng)方式中,骰子不能后退．開B．C．D．條棱翻動(dòng)骰子(相對(duì)面上分別標(biāo)有1點(diǎn)和始時(shí)骰子如圖(1)那樣擺放,朝6點(diǎn),2點(diǎn)上的點(diǎn)數(shù)是2；最后翻動(dòng)到如圖(2)所示的位置,此時(shí)骰子朝上的點(diǎn)數(shù)不可能是下列數(shù)中的．．．()圖（1）B．4圖（2）A．5C．3D．1二、選擇題(本大題含10個(gè)小題,每小題2分,共20分)把答案填在題中的橫線上或按要求作答．2的結(jié)果等于．211．計(jì)算12．若反比例函數(shù)的圖象經(jīng)過點(diǎn)A2，1,則它的表達(dá)式是．13．自2005年以來,太原市城市綠化走上了快車道．目前我市園林綠化總面積達(dá)到了7101.5萬平方米．這個(gè)數(shù)據(jù)用科學(xué)記數(shù)法表示為萬平方米．5214．方程x12x的解是．CAB15．如圖是一種貝殼的俯視圖,點(diǎn)分線段近似于黃金分割．已知AB=100.1cm)cmAC,則的長(zhǎng)約為cm．(結(jié)果精確到16．甲、乙兩盞路燈底部間的距離是30米,一天晚上,當(dāng)小華走到距路燈乙底部5米處時(shí),發(fā)現(xiàn)自己的身影頂部正好接觸路燈乙的底部．已知小華的身高為1.5米,那么路燈甲的高為米．甲小華乙,每部售價(jià)由17．某種品牌的手機(jī)經(jīng)過四、五月份連續(xù)兩次降價(jià)x3200元降到了2500元．設(shè)平均每月降價(jià)的百分率為,根據(jù)題意列出的方程是．ABAC⊙OCAC=30°,過點(diǎn)的切18．如圖、是的兩條弦,OBDD線與的延長(zhǎng)線交于點(diǎn),則的度數(shù)為．D19．有兩把不同的鎖和三把鑰匙,其中兩把鑰匙分別能打開其中一把鎖,第三把鑰匙不能打開這兩把鎖,任意取出一把鑰匙去開任意的一把鎖,B一次打開鎖的概率為．AABCD20．如圖,在等腰梯形中,AD∥BC,BC=4AD=42,B=45°．直角三角板含45°角DA的頂點(diǎn)在邊上移動(dòng),一直角邊始終經(jīng)過點(diǎn),斜邊與CD交于點(diǎn)EBCAFF．若△ABE為等腰三角形,則CF的長(zhǎng)等于．CB三、解答題(本大題含9個(gè)小題,共70分)解答應(yīng)寫出文字說明、證明過程或演算步驟21．(每小題滿分5分)E411x24x2x2化簡(jiǎn):22．(本小題滿分5分)y4x28x．寫出這個(gè)函數(shù)圖象的對(duì)稱軸和頂點(diǎn)坐標(biāo)已知,二次函數(shù)的表達(dá)式為,并求x圖象與軸的交點(diǎn)的坐標(biāo)．23．(本小題滿分6分)ww20件,其總產(chǎn)值(萬元)滿足:1150＜＜1200,相關(guān)數(shù),公司應(yīng)怎樣設(shè)計(jì)這兩種產(chǎn)品的生產(chǎn)方案．某公司計(jì)劃生產(chǎn)甲、乙兩種產(chǎn)品共據(jù)如下表．為此產(chǎn)品名稱每件產(chǎn)品的產(chǎn)值(萬元)甲4575乙24．(本小題滿分8分)CAB60°如圖,從熱氣球上測(cè)得兩建筑物、底部的俯角分別為30°和．如果這時(shí)氣球的高CDADBAB,求建筑物、間的距離．度為90米．且點(diǎn)、、在同一直線上CEF30°60°ABD25．(本小題滿分8分)為了解某校學(xué)生每周購買瓶裝飲料的情況,課外活動(dòng)小組從全校30個(gè)班中采用科學(xué)的方,結(jié)果如法選了5個(gè)班．并隨機(jī)對(duì)這5個(gè)班學(xué)生某一天購買瓶裝飲料的瓶數(shù)進(jìn)行了統(tǒng)計(jì)下圖所示．(1)求該天這5個(gè)班平均每班購買飲料的瓶數(shù)；(2)估計(jì)該校所有班級(jí)每周(以5天計(jì))購買飲料的瓶數(shù)；(3)若每瓶飲料售價(jià)在1.5元至2.5元之間,估計(jì)該校所有學(xué)生一周用于購買瓶裝飲料的費(fèi)用范圍．瓶數(shù)/瓶131211109876543210ABCDE班數(shù)26．(本小題滿分9分)AMONOMAE∥ON．,如圖,是邊上一點(diǎn)MON(1)在圖中作OB的角平分線B于點(diǎn)；,交AE(要求:尺規(guī)作圖,保留作圖痕跡,不寫作法和證明)AOB的垂線,垂足為點(diǎn)D,交ONCCB于點(diǎn),連接(2)在(1)中,過點(diǎn)畫OABC,將圖形補(bǔ)充完整,并證明四邊形是菱形．MEAON27(本小題滿分8分)8個(gè)班,一班必須參加在同一個(gè)品牌的四個(gè)乒乓球上分別標(biāo)上數(shù)字1,2,3,4,并放入一個(gè)不透明的袋中某中學(xué)九年級(jí)有,要從中選出兩個(gè)班代表學(xué)校參加社區(qū)公益活動(dòng)．各班都想?yún)⒓?另外從二至八班中再選一個(gè)班．有人提議用如下的方法,搖勻后從,但由于特定原因:中隨機(jī)摸出兩個(gè)乒乓球,兩個(gè)球上的數(shù)字和是幾就選幾班,你認(rèn)為這種方法公平嗎？請(qǐng)用列表或畫樹狀圖的方法說明理由．123428．(本小題滿分9分)、兩座城市之間有一,并始終在駛．甲車駛往城,乙車駛往城,甲車在行駛過程中AB條高速公路,甲、乙兩輛汽車同時(shí)分別從這條路兩端的入口處BA駛?cè)敫咚俟飞险Ｐ蠦yx速度變．甲車距城高速公路入口處的距離(千米)與行駛時(shí)間(時(shí))之間的關(guān)系如圖．yx(1)求關(guān)于的表達(dá)式；(2)已知乙車以60千米s/時(shí)的速度勻速行駛,設(shè)行駛過程中,兩車相距的路程為(千sx關(guān)于的表達(dá)式；米)．請(qǐng)(3)當(dāng)乙車按(2)中的狀結(jié)果比甲車晚40分鐘到達(dá)終點(diǎn),求乙車變化后的速度．在下圖中畫出乙車離開直接寫出a態(tài)行駛與甲車相遇后,速度隨即改為(千米/時(shí))并保持勻速行駛,aByx城高速公路入口處的距離(千米)與行駛時(shí)間(時(shí))之間的函數(shù)圖象．y/千米36030024018012060O12345x/時(shí)29．(本小題滿分12分)問題解決MFBCDEA折疊,使點(diǎn)落在邊上一點(diǎn)(不ABCDDEC如圖(1),將正方形紙片CE1AMCDMN與點(diǎn),重合),壓平后得到折痕．當(dāng)時(shí),求的CD2BN值．方法指導(dǎo)：AM為了求得的值，可先求、BN圖（1）BNAMAB的長(zhǎng)，不妨設(shè)：=2BN類比歸納CE1AM則的值等于CE1AM，則的值等CD4BN，在圖(1)中,若；若CD3BNCE1AMnn．(用含的式子表示)于；若(為整數(shù)),則的值等于CDnBN聯(lián)系拓廣ABCDBCDEC，D,使點(diǎn)落在邊上一點(diǎn)(不與點(diǎn)重合),壓平如圖(2),將矩形紙片折疊AB1CE1AMMN，m1，，m，n．(用含的后得到折痕設(shè)則的值等于CDnBNBCm式子表示)FMABDECN圖（2）2009年山西省太原市初中畢業(yè)生學(xué)業(yè)考試試卷數(shù)學(xué)試題參考答案一、選擇題題號(hào)1答案A2345678910DCBABBADC二、填空題211．2；12．y；13．7.1015×；14．(或5)；15．6.2；16．910x53x1x2250032x64x70或32(1x)225)17．3200(或21518．30°19．20．,2,423．32三、解答題4x2x2x2121．解:原式=x2x2x2·········································2分x2·x2·································································4分=x2x2=1．··························································································5分22．解:在y4x28x中,a4，b8，c0．8241，4acb2440824．4a44∴b2ax1,頂點(diǎn)坐標(biāo)是:1，4．圖象的對(duì)稱軸是∴這個(gè)函數(shù)···················２分評(píng)分說明:直接寫出正確結(jié)果也得2分．令y＝０,則4x28x0．·······························································3.分解得x0，x2．·······································································4分21∴函數(shù)圖象與x軸的交點(diǎn)的坐標(biāo)為0，0，2，0．································5分23．解:設(shè)計(jì)劃生產(chǎn)甲產(chǎn)品x件,則生產(chǎn)乙產(chǎn)品20x件,········································1分45x7520x1200．················································3分45x7520x1150，根據(jù)題意,得解得10x353．···········································································4分xx11．20x9(件)．···········································5分,為整數(shù),∴此時(shí)答:公司應(yīng)安排生產(chǎn)甲產(chǎn)品11件,乙產(chǎn)品9件．··············································6分ECA30°，F(xiàn)CB60°，CD90，:由已知,得24．解EF∥AB，CDABD于點(diǎn)．AECA30°，BFCB60°．·········································２分CD在Rt△ACD中,CDA90°，tanA=，ADADCD90903903．···············································4分tanA333CD在Rt△BCD中,CDB90°，tanB=，BDCD90DBtanB303．·····························································6分3ABADBD9033031203(米)．答:建筑物A、B1203間的距離為米．······················································8分18912111010(瓶)．25．解:(1)5答:該天這5個(gè)班平均每班購買飲料10瓶．··················································3分(2)105301500(瓶)．答:該校所有班每周購買飲料1500瓶．························································6分(3)1.515002250(元),2.515003750(元)．答:該校所有班級(jí)學(xué)生一周用于購買瓶裝飲料的費(fèi)用為2250元至3750元．·········8分26．解:(1)如圖,射線OB為所求作的圖形．·························································3分MEABDOCNOB平分MON，AOBBOC．(2)方法一:AE∥ON，ABOBOC．AOBABO，AOAB．····························································5分ADOB，BDOD．···································································6分ABDCOD，在ADB和中BDOD，△△CDOADBCDO，△ADB≌△CDO，ABOC．··························································7分AB∥OC，∴四邊形OABC是平行四邊形．AOAB，∴四邊形OABC是菱形．方法二:同方法一,AOBABO，AOAB．········································5分DODDB，ADOCDO90°．·········································8分··················································9分ADOB于點(diǎn),∴······················6分AODCOD，在AOD和中ODOD，△△CODADOCDO，∴AOD∴四邊形OABC是平行四邊形．AOAB(或ACOB),∴四邊形OABC是菱形．CODADCD························································7分△≌△，．···························································8分·······························9分一次摸球可能出現(xiàn)的結(jié)果列表如下:·································4分27．解:這種方法不公平．123.4123.4(1,1)(1,2)(1,3)(1,4)(2,1)(3.,1)(4,1)(2,2)(3.,2)(4,2)(2,3)(3,3)(4,3)(2,4)(3.,4)(4,4)由上表可知,一次摸球出現(xiàn)的結(jié)果共有16種可能的情況,且每種情況出現(xiàn)的可能性為2的一種,和為3的兩種,和為4的三種,和為5的四種,和為6的三種,和為7的兩種,和為8的一種．·····························································6分相同．其中和121P(和為2)=P(和為8)=,P(和為3)=P(和為7)=,16316841P(和為4)=P(和為6)=,P(和為5)=．161641311．所以416816······································································7分因?yàn)槎嘀涟税喔靼啾贿x中的概率不全相等,所以這種方法不公平．············8分評(píng)分說明:只要計(jì)算出二至八班中有兩個(gè)班被選中的概率不相等,就可得分．28．解:(1)方法一:由圖知y是x的一次函數(shù),設(shè)ykxb．·····································1分b300，圖象經(jīng)過點(diǎn)(0,300),(2,120),∴2kb120．····································2分b300k90，············································································3分解得．∴y90x300．即y關(guān)于x的表達(dá)式為y90x300．··················4分x0y300x2y120．方法二:由圖知,這條高速公路長(zhǎng)為300千米．2小時(shí)的行程為300－120=180(千米)．∴甲車的行駛速度為180÷2=90(千米／時(shí))．·······································3分,當(dāng)時(shí),；時(shí),所以甲車yxy30090x(y90x300)．·····················4分∴關(guān)于的表達(dá)式為(2)s150x300．··············································································5分(3)在s150x300s0x2．中．當(dāng)時(shí),即甲乙兩車經(jīng)過2小時(shí)相遇．·························································6分10在y90x300中,當(dāng)y0，x．所以,相遇后乙車到達(dá)終點(diǎn)所用的時(shí)間310222(小時(shí)為)．33y/千米乙車與甲車相遇后的速度360300240180a300260290(千米/時(shí))．a(chǎn)90∴(千米/時(shí))．································7分By乙車離開城高速公路入口處的距離(千米)與行12060x駛時(shí)間(時(shí))之間的29．問題解決解:方法一:如圖(1-1),連接BM，EM，BE．AMF函數(shù)圖象如圖所示．················9分O12345x/時(shí)DECBN圖（1-1）由題設(shè),得四邊形ABNM和四邊形FENMMN關(guān)于直線對(duì)稱．MNBEBMEM，BNEN．····································1分∴垂直平分．∴DC90°,ABBCCDDA2．A∵四邊形ABCD是正方形,∴CE1，CEDE1．設(shè)BNx，則NEx，NC2x．∵CD2在Rt△CNE中,NE2CN2CE2．x2x212．解得x54,即BN5．···········································3分4∴2在Rt△ABM和在Rt△DEM中,AM2AB2BM2,DM2DE2EM2,AM2AB2DM2DE2．·····························································5分設(shè)AMy，則DM2y，y22y212．∴22y，AM1．1解得即4··············································