2014年西工大附中中考模拟5

2014年西工大附中中考模拟5考试数学试卷一、选择题：本大题共10小题，每小题3分，共30分． 1. －2的绝对值是A ．2B ．－2C ．-0.5D ．2± 2. 下列计算正确的是 A ．3x 2·4x 2＝12x 2 B ．x 3·x 5＝x 15 C ．x 4÷x ＝x 3 D ．(x 5)2＝x 73. 某同学在“百度”搜索引擎中输入“魅力南通”，能搜索到与之相关的结果个数约为3930000，这个数用科学记数法表示为 A ．0.393×107 B ．393×104 C ．39.3×105 D ．3.93×1064. 若一个多边形的内角和是900°，则这个多边形的边数是A ．5B ．6C ．7D ．8 5. 如图所示，△ABC 的顶点是正方形网格的格点，则sin A 的值为 A ．12BCD6. 如图，点A 、C 、B 、D 分别是⊙O 上四点，OA ⊥BC ，∠AOB =50°则∠ADC 的度数为 A ．20° B ．25° C ．40° D ．50°7. 如图所示的工件的主视图是8. 某鞋店一天中卖出运动鞋11双，其中各种尺码的鞋的销售量如下表：则这11双鞋的尺码组成的一组数据中，众数和中位数分别是A ．24.5，24.5B ．24.5，25C ．25，24.5D ．25，259. 下列轴对称图形中，只用一把无刻度的直尺不能．．画出对称轴的是 A ．菱形 B ．矩形 C ．等腰梯形 D ．正五边形10. 如图已知在Rt △ABC 中,AB =AC =2,在△ABC 内作第一个内接正方形DEFG ;然后取GF的中点P ,连接PD 、PE ,在△PDE 内作第二个内接正方形HIKJ ；再取线段KJ 的中点Q ,在△QHI 内作第三个内接正方形……依次进行下去,则第n 个内接正方形的边长为 A ．21()32n ⋅ B1()2n C ．121()32n -⋅ D ．11()2n -二、填空题：本大题共8小题，每小题3分，共24分． 11.＝ ．12. 将一直角三角板与两边平行的纸条如图所示放置，若∠1=53°则∠2＝ °．24.5 A . B . C . D .12（第12题）13. 已知分式21x x -+的值为0，那么x 的值为 ．14. 一个圆锥的母线长为4，侧面积为12π，则这个圆锥的底面圆的半径是 ． 15. 如图函数2y x =和5y ax =+的图象相交于A (m ,3),则不等式25x ax <+的解集为 __ 16. 设m ，n 是方程220120x x --=的两个实数根，则2m n +的值为 ． 17. 如图，已知正方形ABCD 的边长为2，对角线AC 、BD 相交于点O ，AE 平分∠BAC 交BD 于点E ， 则BE 的长为 ．18. 如图，点A 是双曲线4y x =在第一象限上的一动点，连接AO 并延长交另一分支于点B ，以AB 为斜边作等腰Rt △ABC ，点C 在第二象限，随着点A 的运动，点C 的位置也不断的．三、解答题:本大题共10小题，共计96分．19．（10分） （1）计算：0(cos30°－11()5- （2）解方程组：38 53 4 x y x y +=⎧⎨-=⎩①②20．（8分）化简分式222421444a aa a a -÷--++，并选取一个你认为合适的整数a 代入求值.21（9分）小敏为了解本市的空气质量情况，从环境监测网随机抽取了若干天的空气质量情况作为样本进行统计，绘制了如图所示的条形统计图和扇形统计图（部分信息未给出）．请你根据图中提供的信息，解答下列问题：（1）计算被抽取的天数；（2）请补全条形统计图，并求扇形统计图中表示优的扇形的圆心角度数；（3）请估计该市这一年（365天）达到优和良的总天数．（第15题）本市若干天空气质量情况扇形统计图优良 64%轻微污染轻度污染 中度污染 重度污染污染 污染 15105污染 污染（第18题）（第17题）DC22．（8分） 如图，AB 与⊙O 相切于点C ，OA ＝OB .（1）如图①，若⊙O 的直径为8cm ，AB ＝10cm ，求OA 的长（结果保留根号）； （2）如图②，OA 、OB 与⊙O 分别交于点D 、E ，连接CD 、CE ，若四边形ODCE 为菱形，求ODOA的值．23．（8分）如图，在边长为1的正方形组成的网格中，△ABC 的顶点和O 点均在格点上．（1）以点O 为位似中心，在网格中将△ABC 放大为原来的2倍，得到△A ′B ′C ′；（2）△A ′B ′C ′绕点B ′顺时针旋转90°，画出旋转后得到的△A ″B ′C ″，并求边A ′B ′在旋转过程中扫过的图形面积．24．（8分）如图，某飞机于空中探测某座山的高度，在点A 处飞机的飞行高度是AF ＝3700米，从飞机上观测山顶目标C 的俯角是45°，飞机继续以相同的高度飞行300米到B 处，此时观测目标C 的俯角是50°，求这座山的高度CD ． （参考数据：sin50°≈0.77，cos50°≈0.64，tan50°≈1.20）．25．（9分）甲、乙两个袋中均装有三张除所标数值外完全相同的卡片，甲袋中的三张卡片上所标有的三个数值为－7，－1，3，乙袋中的三张卡片所标的数值为－2，1，6，先从甲袋中随机取出一张卡片，用x 表示取出的卡片上的数值，再从乙袋中随机取出一张卡片，用y 表示取出卡片上的数值．把x 、y 分别作为点A 的横坐标和纵坐标．（1）用列表或画树形图的方法写出点A （x ，y ）的所有情况； （2）求点A 落在直线2y x 上的概率．D F图 ①图 ②26．（10分）甲、乙两组同时加工某种零件，乙组工作中有一次停产更换设备，更换设备后，乙组的工作效率是原来的2倍．两组各自加工零件的数量y (件)与时间x (时)的函数图象如图所示．（1）直接写出甲组加工零件的数量y 与时间x 之间的函数关系式 ；（2）求乙组加工零件总量a 的值；（3）甲、乙两组加工出的零件合在一起装箱，每满300件装一箱，零件装箱的时间忽略不计，求经过多长时间恰好装满第1箱？27. （12分）如图，在矩形ABCD 中，AB =3，BC =4．动点P 从点A出发沿AC 向终点C 运动，同时动点Q 从点B 出发沿BA 向点A 运动，到达A 点后立刻以原来的速度沿AB 返回．点P ，Q 运动速度均为每秒1个单位长度，当点P 到达点C 时停止运动，点Q 也同时停止．连结PQ ，设运动时间为t （t >0）秒．（1）当点Q 从B 点向A 点运动时（未到达A 点），若△APQ ∽△ABC ，求t 的值；（2）伴随着P ，Q 两点的运动，线段PQ 的垂直平分线为直线l ．①当直线l 经过点A 时，射线QP 交AD 边于点E ，求AE 的长；②是否存在t 的值，使得直线l 经过点B ？若存在，请求出所有t 的值；若不存在，请说明理由．28. （14分）如图，二次函数212y x mx n =-++的图象与y 轴交于点N ，其顶点M 在直线32y x =-上运动，O 为坐标原点. （1）当m ＝－2时，求点N 的坐标；（2）当△MON为直角三角形时，求m 、n 的值；（3）已知△ABC 的三个顶点的坐标分别为A (－4，2)，B (－4，－3)，C (－2，2)，当抛物线212y x mx n =-++在对称轴左侧的部分与△ABC的三边有公共点时，求m 的取值范围.（第2问图）（第3问图）2014年西工大附中中考模拟5答案及评分标准说明：本评分标准每题一般只提供一种解法，如有其他解法，请参照本标准的精神给分． 一、选择题：本大题共10小题，每小题3分，共30分．1．A 2．C 3．D 4．C 5．B 6．B 7．A 8．D 9．B 10．D 二、填空题：本大题共8小题，每小题3分，共24分． 11．－3 12．37 13．2 14. 315．32x < 16．2013 17．2- 18. 4(0)y x x =-<（不注明范围不扣分）三、解答题：本大题共10小题，共96分．19. （1） 解：原式＝15+-················································································· 3分 ＝－1 ····································································································· 5分（2） 解：①+②得，612x =，∴2x =. ····································································· 2分把2x =代入①得，238y +=，∴2y =. ························································· 4分∴方程组的解是：22x y =⎧⎨=⎩． ················································································ 5分20. 解：原式＝22(2)(2)1(2)(2)2a a a a a-+-+- ·············································································· 3分 ＝21a a+- ········································································································ 4分 ＝2a················································································································· 5分 由题意可知，a 不能取0，±2，不妨取a =1（答案不唯一）原式＝2 ··················································································································· 8分21. 解：（1）32÷64％＝50（天）所以被抽取的天数为50天．··········································································· 2分 （2）图略，(轻微污染5天)； ················································································ 4分850×360°＝57.6° ··························································································· 6分 （3）达到优和良的总天数为83236529250+⨯=（天） ········································· 9分 22. （1）如图①，连接OC ，则4OC =． ········································································ 1分AB 与O ⊙相切于点C ，∴OC AB ⊥． ································································································ 2分 在OAB △中，由10OA OB AB ==，，得152AC AB ==． ····························································································· 3分 在Rt AOC △中，由勾股定理，得OA ==． ···························································· 4分 （2）如图②，连接OC ，则OC OD =．四边形ODCE 是菱形，∴OD DC =． ·································································································· 5分 ∴ODC △为等边三角形， 60AOC ∠=︒．·················································· 6分 由（1）知，90OCA ∠=︒， ∴30A ∠=︒，∴OC =12OA ． ············································································ 7分 ∴12OD OA =．····································································································· 8分23. 解：（1）见图中△A ′B ′C ′ ····························································································· 3分（直接画出图形，不画辅助线不扣分）（2）见图中△A ″B ′C ″（直接画出图形，不画辅助线不扣分） 6分 S=90360 π ( 22+42)=14···························· 8分24. 解：设EC ＝x 米．在Rt △BCE 中，tan ∠EBC ＝EC BE ，∴BE ＝EC tan ∠EBC ＝56x ． 2分在Rt △ACE 中，tan ∠EAC ＝EC AE ，∴AE ＝ECtan ∠EAC ＝x ． ··································· 4分∴300＋56x ＝x ，∴x ＝1800 ························································································ 6分∴CD ＝DE －EC ＝3700－1800＝1900（米） ··························································· 7分图 ①图 ②答：这座山的高度是1900米． ················································································ 8分25. 解：（1）列表如下：···································································································································· 6分 （2）点A （x ，y ）共9种情况，其中落在直线y =2x 上的有2种，所以P （点A 落在直线y =2x 上）=29． ··········································································· 9分26. 解：（1）60y x = ·········································································································· 2分 （2）当2x =时，100y =．因为更换设备后，乙组工作效率是原来的2倍，所以，10010024.82.82a -=⨯- ················································································ 4分 解得300a = ··································································································· 5分 （3）乙组更换设备后，乙组加工的零件的个数y 与时间x 的函数关系式为100100(2.8)100180y x x =+-=-． ································································ 6分 当0≤x ≤2时，6050300x x +=．解得3011x =．舍去． ···································· 7分 当2<x ≤2.8时，10060300x +=．解得103x =．舍去． ·································· 8分 当2.8<x ≤4.8时，60100180300x x +-=．解得3x =． ····················································· 9分 所以，经过3小时恰好装满第1箱． ····························································· 10分27. 解：（1）∵△APQ ∽△ABC ∴AP AQ AB AC =，即335t t -= 2分 解得98t = 3分 （2）①如图①，线段PQ 的垂直平分线为l 经过点A ，则AP=AQ ，即3－t=t ，∴t=1.5，∴AP=AQ=1.5， ······························································ 4分过点Q 作QO ∥AD 交AC 于点O ，则,BCQO AB AQ AC AO ==∴52AQ AO AC AB =⋅=，5分2=⋅=BC ABAQOQ ，∴PO=AO －AP=1． ····················································· 6分 由△APE ∽△OPQ ，得3,=⋅=∴=OQ OPAPAE OP AP OQ AE ． ···························· 7分 ②（ⅰ）如图②，当点Q 从B 向A 运动时l 经过点B ， BQ ＝BP ＝AP ＝t ，∠QBP ＝∠QAP∵∠QBP ＋∠PBC ＝90°，∠QAP ＋∠PCB ＝90° ∴∠PBC ＝∠PCB CP ＝BP ＝AP ＝t ∴CP ＝AP ＝21AC ＝21×5＝2.5 ∴t ＝2.5． ······················································ 9分 （ⅱ）如图③，当点Q 从A 向B 运动时l 经过点B ， BP ＝BQ ＝3－（t －3）＝6－t ，AP ＝t ，PC ＝5－t ， 过点P 作PG ⊥CB 于点G ，由△PGC ∽△ABC ， 得()t AB AC PC PG BC GC AB PG AC PC -=⋅=∴==553, ()t BC AC PC CG -=⋅=554，BG ＝4－()t -554=t 54··································· 10分 由勾股定理得222PG BG BP +=，即 ()222553)54()6(⎥⎦⎤⎢⎣⎡-+=-t t t ，解得4514t =． ············································· 12分28. 解：（1）∵2211()22y x m m n =--++∴抛物线顶点M 坐标为（m ，212m n +）1分∵顶点在直线32y x =-上∴212m n +＝32m -，2分当m ＝－2时，n ＝1∴点N 的坐标为（0，1） 4分（2）若点M 在第二象限时，△MON 不可能为直角三角形，当点M 在坐标原点时，△MON 不存在，若点M 在第四象限，当△MON 为直角三角形时，显然只有∠OMN ＝90°． · 5分 过点M 作x 轴的垂线，垂足为H ，则△OMN ∽△MHO ，∴2OM MH ON =⋅………6分 设M （m ，32m -），则MH ＝32m ，22134OM m =，而ON ＝－n ，∴2134m =32m ×（-n ），即136n m =- ①·················································· 7分 又212m n +＝32m -，②由①、②解得43m =，269n =- 9分（3）由（1）可知，22113222y x mx m m =-+--，当点A (－4，2)在该抛物线上时，22113(4)42222m m m -----=，整理，得211200m m ++=，Q POEDCBAQPDCBA GQPDC BA （图①）（图②）（图③）解得m =························································································ 11分 ∵在对称轴的左侧，∴m················································· 12分 ∵B (－4，－3)，C (－2，2)， ∴直线BC 的解析式为572y x =+，代入抛物线解析式得22(52)3140x m x m m +-+++=， 令△=0得，22(52)4(314)0m m m --++=，解得3132m =-·························· 13分3132m -≤ ······················分。