北京市中考數(shù)學(xué)試卷

1、2012年北京市中考數(shù)學(xué)試卷一、選擇題（每小題4分，共32分）1-9的相反數(shù)是（）a-19b19c-9d92首屆中國(guó)（北京）國(guó)際服務(wù)貿(mào)易交易會(huì)（京交會(huì)）于2012年6月1日閉幕，本屆京交會(huì)期間簽訂的項(xiàng)目成交總金額達(dá)60 110 000 000美元將60 110 000 000用科學(xué)記數(shù)法表示應(yīng)為（）a6.011109b60.11109c6.0111010d0.601110113正十邊形的每個(gè)外角等于（）a18b36c45d604如圖是某個(gè)幾何體的三視圖，該幾何體是（）a長(zhǎng)方體b正方體c圓柱d三棱柱5班主任王老師將6份獎(jiǎng)品分別放在6個(gè)完全相同的不透明禮盒中，準(zhǔn)備將它們獎(jiǎng)給小英等6位獲“愛集體標(biāo)兵

2、”稱號(hào)的同學(xué)這些獎(jiǎng)品中3份是學(xué)習(xí)文具，2份是科普讀物，1份是科技館通票小英從中隨機(jī)抽取一份獎(jiǎng)品，恰好取到科普讀物的概率是（）a16b13c12d236如圖，直線ab，cd交于點(diǎn)o，射線om平分aoc，若bod=76，則bom等于（）a38b104c142d1447某課外小組的同學(xué)們?cè)谏鐣?huì)實(shí)踐活動(dòng)中調(diào)查了20戶家庭某月的用電量，如表所示：用電量（度）120140160180200戶數(shù)23672則這20戶家庭該月用電量的眾數(shù)和中位數(shù)分別是（）a180，160b160，180c160，160d180，1808小翔在如圖1所示的場(chǎng)地上勻速跑步，他從點(diǎn)a出發(fā)，沿箭頭所示方向經(jīng)過點(diǎn)b跑到點(diǎn)c，共用時(shí)30秒

3、他的教練選擇了一個(gè)固定的位置觀察小翔的跑步過程設(shè)小翔跑步的時(shí)間為t（單位：秒），他與教練的距離為y（單位：米），表示y與t的函數(shù)關(guān)系的圖象大致如圖2所示，則這個(gè)固定位置可能是圖1中的（）a點(diǎn)mb點(diǎn)nc點(diǎn)pd點(diǎn)q二、填空題（每小題4分，共16分）9分解因式：mn2+6mn+9m=10若關(guān)于x的方程x2-2x-m=0有兩個(gè)相等的實(shí)數(shù)根，則m的值是11如圖，小明同學(xué)用自制的直角三角形紙板def測(cè)量樹的高度ab，他調(diào)整自己的位置，設(shè)法使斜邊df保持水平，并且邊de與點(diǎn)b在同一直線上已知紙板的兩條直角邊de=40cm，ef=20cm，測(cè)得邊df離地面的高度ac=1.5m，cd=8m，則樹高ab=m12在

4、平面直角坐標(biāo)系xoy中，我們把橫、縱坐標(biāo)都是整數(shù)的點(diǎn)叫做整點(diǎn)已知點(diǎn)a（0，4），點(diǎn)b是x軸正半軸上的整點(diǎn)，記aob內(nèi)部（不包括邊界）的整點(diǎn)個(gè)數(shù)為m當(dāng)m=3時(shí)，點(diǎn)b的橫坐標(biāo)的所有可能值是；當(dāng)點(diǎn)b的橫坐標(biāo)為4n（n為正整數(shù)）時(shí)，m=（用含n的代數(shù)式表示）vip三、解答題（每小題5分，共30分）13計(jì)算：（-3）0+18-2sin45-（18）-114解不等式組：4x3xx+42x115已知a2b30，求代數(shù)式5a2ba24b2(a2b)的值16已知：如圖，點(diǎn)e，a，c在同一直線上，abcd，ab=ce，ac=cd求證：bc=ed17如圖在平面直角坐標(biāo)系xoy中，函數(shù)y=4x（x0）的圖象與一次函數(shù)

5、y=kx-k的圖象的交點(diǎn)為a（m，2）（1）求一次函數(shù)的解析式；（2）設(shè)一次函數(shù)y=kx-k的圖象與y軸交于點(diǎn)b，若點(diǎn)p是x軸上一點(diǎn)，且滿足pab的面積是4，直接寫出p點(diǎn)的坐標(biāo)18列方程或方程組解應(yīng)用題：據(jù)林業(yè)專家分析，樹葉在光合作用后產(chǎn)生的分泌物能夠吸附空氣中的一些懸浮顆粒物，具有滯塵凈化空氣的作用已知一片銀杏樹葉一年的平均滯塵量比一片國(guó)槐樹葉一年的平均滯塵量的2倍少4毫克，若一年滯塵1000毫克所需的銀杏樹葉的片數(shù)與一年滯塵550毫克所需的國(guó)槐樹葉的片數(shù)相同，求一片國(guó)槐樹葉一年的平均滯塵量四、解答題（每小題5分，共20分）19如圖，在四邊形abcd中，對(duì)角線ac，bd交于點(diǎn)e，bac=90

6、，ced=45，dce=30，de=2，be=22求cd的長(zhǎng)和四邊形abcd的面積20已知：如圖，ab是o的直徑，c是o上一點(diǎn)，odbc于點(diǎn)d，過點(diǎn)c作o的切線，交od的延長(zhǎng)線于點(diǎn)e，連接be（1）求證：be與o相切；（2）連接ad并延長(zhǎng)交be于點(diǎn)f，若ob=9，sinabc=23，求bf的長(zhǎng)21近年來，北京市大力發(fā)展軌道交通，軌道運(yùn)營(yíng)里程大幅增加，2011年北京市又調(diào)整修訂了2010至2020年軌道交通線網(wǎng)的發(fā)展規(guī)劃以下是根據(jù)北京市軌道交通指揮中心發(fā)布的有關(guān)數(shù)據(jù)制作的統(tǒng)計(jì)圖表的一部分北京市軌道交通已開通線路相關(guān)數(shù)據(jù)統(tǒng)計(jì)表（截止2010年底）開通時(shí)間開通線路運(yùn)營(yíng)里程（千米）19711號(hào)線311

7、9842號(hào)線23200313號(hào)線41八通線1920075號(hào)線2820088號(hào)線510號(hào)線25機(jī)場(chǎng)線2820094號(hào)線282010房山線22大興線22亦莊線23昌平線2115號(hào)線20請(qǐng)根據(jù)以上信息解答下列問題：（1）補(bǔ)全條形統(tǒng)計(jì)圖并在圖中標(biāo)明相應(yīng)數(shù)據(jù)；（2）按照2011年規(guī)劃方案，預(yù)計(jì)2020年北京市軌道交通運(yùn)營(yíng)總里程將達(dá)到多少千米？（3）要按時(shí)完成截至2015年的軌道交通規(guī)劃任務(wù)，從2011到2015年這4年中，平均每年需新增運(yùn)營(yíng)里程多少千米？22操作與探究：（1）對(duì)數(shù)軸上的點(diǎn)p進(jìn)行如下操作：先把點(diǎn)p表示的數(shù)乘以13，再把所得數(shù)對(duì)應(yīng)的點(diǎn)向右平移1個(gè)單位，得到點(diǎn)p的對(duì)應(yīng)點(diǎn)p點(diǎn)a，b在數(shù)軸上，對(duì)線

8、段ab上的每個(gè)點(diǎn)進(jìn)行上述操作后得到線段ab，其中點(diǎn)a，b的對(duì)應(yīng)點(diǎn)分別為a，b如圖1，若點(diǎn)a表示的數(shù)是-3，則點(diǎn)a表示的數(shù)是；若點(diǎn)b表示的數(shù)是2，則點(diǎn)b表示的數(shù)是；已知線段ab上的點(diǎn)e經(jīng)過上述操作后得到的對(duì)應(yīng)點(diǎn)e與點(diǎn)e重合，則點(diǎn)e表示的數(shù)是（2）如圖2，在平面直角坐標(biāo)系xoy中，對(duì)正方形abcd及其內(nèi)部的每個(gè)點(diǎn)進(jìn)行如下操作：把每個(gè)點(diǎn)的橫、縱坐標(biāo)都乘以同一個(gè)實(shí)數(shù)a，將得到的點(diǎn)先向右平移m個(gè)單位，再向上平移n個(gè)單位（m0，n0），得到正方形abcd及其內(nèi)部的點(diǎn)，其中點(diǎn)a，b的對(duì)應(yīng)點(diǎn)分別為a，b已知正方形abcd內(nèi)部的一個(gè)點(diǎn)f經(jīng)過上述操作后得到的對(duì)應(yīng)點(diǎn)f與點(diǎn)f重合，求點(diǎn)f的坐標(biāo)五、解答題（共22分，第

9、23題7分，第24題7分，第25題8分）23已知二次函數(shù)y=（t+1）x2+2（t+2）x+32在x=0和x=2時(shí)的函數(shù)值相等（1）求二次函數(shù)的解析式；（2）若一次函數(shù)y=kx+6的圖象與二次函數(shù)的圖象都經(jīng)過點(diǎn)a（-3，m），求m和k的值；（3）設(shè)二次函數(shù)的圖象與x軸交于點(diǎn)b，c（點(diǎn)b在點(diǎn)c的左側(cè)），將二次函數(shù)的圖象在點(diǎn)b，c間的部分（含點(diǎn)b和點(diǎn)c）向左平移n（n0）個(gè)單位后得到的圖象記為g，同時(shí)將（2）中得到的直線y=kx+6向上平移n個(gè)單位請(qǐng)結(jié)合圖象回答：當(dāng)平移后的直線與圖象g有公共點(diǎn)時(shí)，求n的取值范圍24在abc中，ba=bc，bac=，m是ac的中點(diǎn)，p是線段bm上的動(dòng)點(diǎn)，將線段pa繞

10、點(diǎn)p順時(shí)針旋轉(zhuǎn)2得到線段pq（1）若=60且點(diǎn)p與點(diǎn)m重合（如圖1），線段cq的延長(zhǎng)線交射線bm于點(diǎn)d，請(qǐng)補(bǔ)全圖形，并寫出cdb的度數(shù)；（2）在圖2中，點(diǎn)p不與點(diǎn)b，m重合，線段cq的延長(zhǎng)線于射線bm交于點(diǎn)d，猜想cdb的大?。ㄓ煤拇鷶?shù)式表示），并加以證明；（3）對(duì)于適當(dāng)大小的，當(dāng)點(diǎn)p在線段bm上運(yùn)動(dòng)到某一位置（不與點(diǎn)b，m重合）時(shí)，能使得線段cq的延長(zhǎng)線與射線bm交于點(diǎn)d，且pq=qd，請(qǐng)直接寫出的范圍25在平面直角坐標(biāo)系xoy中，對(duì)于任意兩點(diǎn)p1（x1，y1）與p2（x2，y2）的“非常距離”，給出如下定義：若|x1-x2|y1-y2|，則點(diǎn)p1與點(diǎn)p2的“非常距離”為|x1-x2|；若

11、|x1-x2|y1-y2|，則點(diǎn)p1與點(diǎn)p2的“非常距離”為|y1-y2|例如：點(diǎn)p1（1，2），點(diǎn)p2（3，5），因?yàn)閨1-3|2-5|，所以點(diǎn)p1與點(diǎn)p2的“非常距離”為|2-5|=3，也就是圖1中線段p1q與線段p2q長(zhǎng)度的較大值（點(diǎn)q為垂直于y軸的直線p1q與垂直于x軸的直線p2q交點(diǎn)）（1）已知點(diǎn)a（-12，0），b為y軸上的一個(gè)動(dòng)點(diǎn)，若點(diǎn)a與點(diǎn)b的“非常距離”為2，寫出一個(gè)滿足條件的點(diǎn)b的坐標(biāo)；直接寫出點(diǎn)a與點(diǎn)b的“非常距離”的最小值；（2）已知c是直線y=34x+3上的一個(gè)動(dòng)點(diǎn)，如圖2，點(diǎn)d的坐標(biāo)是（0，1），求點(diǎn)c與點(diǎn)d的“非常距離”的最小值及相應(yīng)的點(diǎn)c的坐標(biāo)；如圖3，e是以原

12、點(diǎn)o為圓心，1為半徑的圓上的一個(gè)動(dòng)點(diǎn)，求點(diǎn)c與點(diǎn)e的“非常距離”的最小值及相應(yīng)的點(diǎn)e與點(diǎn)c的坐標(biāo)qcwa3ptgz7r4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgshls50clmtwn60eo8wgqv7xav2ohum32wgeauwydiawgmer4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgz7r4i30ka1dkagtgkqcwa3p tgz7r4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgshls50clmtwn60e

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18、kaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgshls50clmtwn60eo8wgqv7xav2ohum32wgeauwydiawgmer4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgz7r4i30ka1dkagtgkqcwa3ptgz7r4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgshls50clmtwn60eo8wgqv7xav2ohum32wgeauwydiawgmer4i30ka1dkaghn3xtkknbycudxqa7fhyi2chh

19、i92tgkqcwa3ptgz7r4i30ka1dkagtgkqcwa3ptgz7r4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgshls50clmtwn60eo8wgqv7xav2tgz7r4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgshls50clmtwn60eo8wgqv7xav2ohum32wgeauwydiawgmer4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgz7r4i30ka1dkagtgkqcwa3ptgz7r4i

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21、i2chhi92tgkqcwa3ptgz7r4i30ka1dkagtgkqcwa3ptgz7r4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgshls50clmtwn60eo8wgqv7xav2ohum32wgeauwydiawgmer4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgz7r4i30ka1dkagtgkqcwa3ptgz7r4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgshls50clmtwn60eo8wgqv7xav2eo

22、8wgqv7xav2ohum32wgeauwydiawger4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgz7r4i30ka1dkagtgkqcwa3ptgz7r4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgshls50clmtwn60eo8wgqv7xav2ohum32wgeauwydiawgmer4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgz7r4i30ka1dkagtgkqcwa3ptgz7r4i30ka1dkaghn3xtk

23、knbycudxqa7fhyi2chhi92tgkqcwa3ptgshls50clmtwn60eo8wgqv7xav2ohum32wgeauwydiawgmer4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgz7r4i30ka1dkagtgkqcwa3ptgz7r4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgshls50clmtwn60eo8wgqv7xav2ohum32wgeauwydiawgmer4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcw

24、a3ptgz7r4i30ka1dkaggeauwydiawgmer4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgz7r4i30ka1dkagtgkqcwa3ptgz7r4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgshls50clmtwn60eo8wgqv7xav2ohum32wgeauwydiawgmer4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgz7r4i30ka1dkagtgkqcwa3ptgz7r4i30ka1dkaghn3

25、xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgshls50clmtwn60eo8wgqv7xav2ohum32wgeauwydiawger4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgz7r4i30ka1dkagtgkqcwa3ptgz7r4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgshls50clmtwn60eo8wgqv7xav2ohum32wgeauwydiawgmer4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkq

26、cwa3ptgz7r4i30ka1dkagtgkqcwa3ptgz7r4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgshls50clmtwn60eo8wgqv7xav hhi92tgkqcwa3ptgz7r4i30ka1dkagtgkqcwa3ptgz7r4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgshls50clmtwn60eo8wgqv7xav2ohum32wgeauwydiawgmer4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3p

27、tgz7r4i30ka1dkagtgkqcwa3ptgz7r4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgshls50clmtwn60eo8wgqv7xav2ohum32wgeauwydiawger4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgz7r4i30ka1dkagtgkqcwa3ptgz7r4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgshls50clmtwn60eo8wgqv7xav2ohum32wgeauwydiawgmer4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgz7r4i30ka1dkagtgkqcwa3ptgz7r4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkq