曲靖市2010年高中（中专）招生统一考试数 学一、选择题（本大题共8个小题，每小题只有一个符合条件的选项，每小题3分，满分24分）1.从3时到6时，钟表的时针旋转角的度数是（ ） Ａ．30︒ Ｂ．60︒ Ｃ．90︒ Ｄ．120︒2.下列各式中，运算正确的是（ ）Ａ．437()x x = Ｂ．842a a a ÷= Ｃ．325385+= Ｄ．315335÷=3.分式方程33122x x x-+=--的解是（ ） Ａ．2 Ｂ．1 Ｃ．－1 Ｄ．－24.下列事件属于必然事件的是（ ） Ａ．367人中至少有两人的生日相同 Ｂ．某种彩票的中奖率为1100，购买100张彩票一定中奖 Ｃ．掷一次骰子，向上的一面是6点 Ｄ．某射击运动员射击一次，命中靶心5.练习本比水性笔的单价少2元，小刚买了5本练习本和3支水性笔正好用去14元.如果设水性笔的单价为x 元，那么下列所列方程正确的是（ ） Ａ．5(2)314x x -+= Ｂ．5(2)314x x ++= Ｃ．53(2)14x x ++= Ｄ．53(2)14x x +-=6.不等式组322(4)1x xx +>⎧⎨--⎩≥的解集在数轴上表示正确的是（ ）7.如图摆放的正六棱柱的俯视图是（ ）8.函数y kx k =-与(0)ky k x=≠在同一坐标系中的大致图象是（ ）二、填空题（本大题共8个小题，每小题3分，满分24分） 9.12-的倒数是___________. 10.在你认识的图形中，写出一个既是轴对称又是中心对称的图形名称：________. 11.如图，AB CD ∥，AC BC ⊥，垂足为C .若40A ∠=︒，则BCD ∠=_______度.12.若2(1)2x -=，则代数式225x x -+的值为________.13.在Rt ABC △中，90C ∠=︒，若10BC AD =，平分BAC ∠交BC 于点D ，且32BD CD =∶∶，则点D 到线段AB 的距离为_______.14.如图，活动衣帽架由三个菱形组成，利用四边形的不稳定性，调整菱形的内角α，使衣帽架拉伸或收缩.当菱形的边长为18cm α=120︒，时，A B 、两点的距离为_______cm.15.在分别写有数字1012-，，，的四张卡片中，随机抽取一张后放回，再随机抽取一张.以第一次抽取的数字作为横坐标，第二次抽取的数字作为纵坐标的点落在第一象限的概率是_____. 16.把一个正三角形分成四个全等的三角形，第一次挖去中间一个小三角形，对剩下的三个A B C D 第11题图 第13题图 DC B A A B小正三角形再重复以上做法……一直到第n 次挖去后剩下的三角形有________个.三、解答题（本大题共8个小题，满分72分） 17.（6分）计算：119(2)(1)3-⎛⎫--+-- ⎪⎝⎭18.（7分）先化简，再求值.2216636x x x x x x x++-÷---，其中3x =19.（8分）如图，小明家所住楼房的高度10AB =米，到对面较高楼房的距离20BD =米，当阳光刚好从两楼房的顶部射入时，测得光线与水平线的夹角为40︒.据此，小明便知楼房CD 的高度.请你写出计算过程（结果精确到0.1米.参考数据：sin 400.6400.77tan 400.84︒≈︒≈︒≈，cos4，）.第二次 第一次 第三次 第四次…BPAC D20.（9分）如图，E F 、是ABCD Y对角线AC 上的两点，且BE DF ∥. 求证：（1）ABE CDF △≌△； （2）12∠=∠.21.（10分）某校对中考前一次数学模拟考试进行抽样分析，把样本成绩按分数段分成A B C D E 、、、、五组（每组成绩含最低分，不含最高分）进行统计，并将结果绘制成下面两幅统计图.请根据图中信息，解答下列问题： （1）求A 组人数在扇形图中所占圆心角的度数； （2）求D 组人数；（3）判断考试成绩的中位数落在哪个组？（直接写出结果，不需要说明理由）22.（10分）如图，O ⊙的直径»12AB BC =，的长为2π，D 在OC 的延长线上，且CD OC =. （1）求A ∠的度数；（2）求证：DB 是O ⊙的切线； （参考公式：弧长公式π180n rl =，其中l 是弧长，r 是半径，n 是圆心角度数）ABC D EFB AAD E A 组 B 组 C 组 D 组 E 组 B DOAC23.（10分）如图，有一块等腰梯形的草坪，草坪上底长48米，下底长108米，上下底相距40米，现要在草坪中修建一条横、纵向的“H ”型甬道，甬道宽度相等，甬道面积是整个梯形面积的213.设甬道的宽为x 米. （1）求梯形ABCD 的周长；（2）用含x 的式子表示甬道的总长；（3）求甬道的宽是多少米？24.（12分）如图，在平面直角坐标系xoy 中，抛物线2y x =向左平移1个单位，再向下平移4个单位，得到抛物线2()y x h k =-+.所得抛物线与x 轴交于A B 、两点（点A 在点B 的左边），与y 轴交于点C ，顶点为D .（1）求h k 、的值；（2）判断ACD △的形状，并说明理由； （3）在线段AC 上是否存在点M ，使AOM △与ABC △相似.若存在，求出点M 的坐标；若不存在，说明理由.A DCFEBA y xBF D C O曲靖市2010年高中（中专）招生统一考试数学参考答案一、选择题1.Ｃ2.Ｄ3.Ｂ4.Ａ5.Ａ6.Ｂ7.Ｄ8.Ｃ 二、填空题9. 2 10.圆（答案不唯一） 11.50 12. 6 13. 4 14. 54 15.1416.3n三、解答题17.解：原式=3213++- ················································································· 4分3=. ····························································································· ６分 18.解：原式=1(6)(6)66(1)x x x x x x x x++--⨯--+ ·························································· 3分 66x x x x +-=-··············································································· 4分 12x =. ··························································································· 5分当x =原式==························································································· 7分 19.解：在Rt ABP △中，10tan 40AB BP BP︒==， 1011.90tan 40BP =︒≈ ···················································································· 4分 在Rt CDP △中，tan 4011.9020CD CDPD ︒==+， ·········································································· 6分 31.900.8426.8CD =⨯≈（米）. 答：楼房CD 的高度为26.8米. ·········································································· 8分 20. 证明：（1）Q 四边形ABCD 是平行四边形，AB CD ∴∥. BAE DCF ∴∠=∠. ························································································ 2分 BE DF Q ∥，BEF DFE ∴∠=∠.AEB CFD ∴∠=∠. ························································································ 4分(AAS)ABE CDF ∴△≌△ ·············································································· 5分 (2)由ABE CDF △≌△得 BE DF =.BE DF Q ∥， ··············································································· 7分 ∴四边形BEDF 是平行四边形. ········································································ 8分 ∴12∠=∠. ·································································································· 9分 21.解：（1）A 组人数所占的百分比：1(26%30%22%12%)10%-+++=， ··············· 2分 A 组人数在扇形图中所占的圆心角的度数：36010%36︒⨯=； ······························· 4分 （2）样本人数：1530%50÷=（人）， ······························································ 6分D 组人数=5022%11⨯=（人）； ······································································ 8分 （3）考试成绩的中位数落在C 组. ···································································· 10分 22.（1）解：设BOC n ∠=︒， 据弧长公式，得π62π180n ⨯=， 60n =︒. ······································································································· 2分 据圆周角定理，得1302A BOC ∠=∠=︒. ···························································· 4分（2）证明：连接BC ，60OB OC BOC =∠=︒Q ，，BOC ∴△是等边三角形. ·················································································· 6分 60OBC OCB OC BC OB ∴∠=∠=︒==，. OC CD =Q ， BC CD ∴=.1302CBD D OCB ∴∠=∠=∠=︒. ···································································· 8分 603090OBD OBC CBD ∴∠=∠+∠=︒+︒=︒. AB BD ∴⊥.DB ∴是O ⊙的切线. ····················································································· 10分 23.解：（1）在等腰梯形ABCD 中， 48AD EF ==，B DCOA()121(10848)23050AE BC DF BC BE CF BC EF AB CD ⊥⊥==-=-=∴===，，，，∴梯形ABCD 的周长=501085048256AB BC CD DA +++=+++=(米). ············· 2分（2）甬道的总长：402482(1282)x x ⨯+-=-米. ·············································· 4分 （3）根据题意，得21(1282)40(48108)132x x -=⨯⨯+. ·································································· 7分 整理，得2642400x x -+=，解之得12460x x ==，.因6048>，不符合题意，舍去.答：甬道的宽为4米. ····················································································· 10分 24.解：（1）2y x =Q 的顶点坐标为（0，0），2()y x h k ∴=-+的顶点坐标(14)D -，，1h k ∴=-，=-4. ··························································································· 3分 （2）由（1）得2(1)4y x =+-. 当0y =时，2(1)40x +-=. 1231x x =-=，.(30)10A B ∴-，，(，). ························································································ 4分当0x =时，22(1)4(01)43y x =+-=+-=-，C ∴点坐标为()03，-.又Q 顶点坐标()14D --，， ·············································································· 5分 作出抛物线的对称轴1x =-交x 轴于点E .作DF y ⊥轴于点F .在Rt AED △中，2222420AD =+=； 在Rt AOC △中，2223318AC =+=； 在Rt CFD △中，222112CD =+=；Q 222AC CD AD +=，ACD ∴△是直角三角形. ·················································································· 7分 （3）存在.由（2）知，AOC △为等腰直角三角形，45BAC ∠=︒， 连接OM ，过M 点作MG AB ⊥于点G ，AC ==①若AOM ABC △∽△，则AO AM AB AC =，即33444AM ⨯===. Q MG AB ⊥，222AG MG AM ∴+=.94AG MG ∴====，93344OG AO AG =-=-=.M Q 点在第三象限，3944M ⎛⎫∴-- ⎪⎝⎭，. ·························································································· 10分②若AOM ACB △∽△，则AO AMAC AB =4AM AM ===，2AG MG ∴====，321OG AO AG =-=-=. M Q 点在第三象限，()12M ∴--，.x综上①、②所述，存在点M 使AOM △与ABC △相似，且这样的点有两个，其坐标分别为()391244⎛⎫---- ⎪⎝⎭，，，. ················································································ 12分。