2010年中考数学模拟试题及答案(7)

2010年中考模拟题数 学 试 卷（七）*考试时间120分钟 试卷满分150分一、选择题（本大题共7小题，每小题4分，共28分）每题所给的四个选项中只有一项是符合题目要求的，请将所选项的代号字母填在答卷的相应位置处．1 ）A ．BC ．2-D ．22．反比例函数23m y x--=的图象位于（ ）A ．第一、三象限B ．第二、四象限C ．第二、三象限D ．第一、二象限3．从2、3、4、5这四个数中，任取两个数()p q p q ≠和，构成函数2y px y x q =-=+和，并使这两个函数图象的交点在直线2x =的右侧，则这样的有序数对()p q ，共有（ ） A ．12对 B ．6对 C ．5对 D ．3对4．把多项式2288x x -+分解因式，结果正确的是（ ） A ．()224x -B ．()224x -C ．()222x -D ．()222x +5．某等腰三角形的两条边长分别为3cm 和6cm ，则它的周长为（ ） A ．9cm B ．12cm C ．15cm D ．12cm 或15cm6．一次函数y kx b =+（k b ，是常数，0k ≠）的图象如图所示，则不等式0kx b +>的解集是A ．2x >-；B ．0x >；C ．2x <-；D ．0x <7．若0a >且2xa =，3ya =，则x ya-的值为（ ） A ．1- B ．1 C ．23 D ．32二、填空题（本大题共6小题，每小题4分，共24分）把答案直接填在答卷的相应位置处． 8．将点(12)，向左平移1个单位，再向下平移2个单位后得到对应点的坐标是 ．9．幼儿园把新购进的一批玩具分给小朋友．若每人3件，那么还剩余59件；若每人5件，那么最后一个小朋友分到玩具，但不足4件，这批玩具共有 件．10．李师傅随机抽查了本单位今年四月份里6天的日用水量（单位：吨）结果如下：7，8，8，7，6，6，根据这些数据，估计四月份本单位用水总量为 吨．11．我们知道利用相似三角形可以计算不能直接测量的物体的高度，阳阳的身高是1.6m ，他在阳光下的影长是1.2m ，在同一时刻测得某棵树的影长为3.6m ，则这棵树的高度约为 m ．xb +CB12．如图所示的半圆中，AD 是直径，且3AD =，2AC =， 则sin B 的值是 ．13．某个圆锥的侧面展开图形是一个半径为6cm ，圆心角为︒120的扇形，则这个圆锥的底面半径为______________cm ．三、解答题（本大题Ⅰ—Ⅴ题，共10小题，共98分）解答时应在答卷的相应位置处写出文字说明、证明过程或演算过程．Ⅰ．（本题满分12分，第14题6分，第15题6分）14．计算：230116(2)(πtan60)3-⎛⎫--÷-+-- ⎪⎝⎭．15．先化简，再求值：221111121x x x x x +-÷+--+，其中1x =．Ⅱ．（本题满分28分，第16题7分，第17题10分，第18题11分）16．如图，线段AB 与⊙O 相切于点C ，连结OA ，OB ，OB 交⊙O 于点D ，已知6OA OB ==，AB =（1）求⊙O 的半径； （2）求图中阴影部分的面积．17．响应“家电下乡”的惠农政策，某商场决定从厂家购进甲、乙、丙三种不同型号的电冰箱80台，其中甲种电冰箱的台数是乙种电冰箱台数的2倍，购买三种电冰箱的总金额不超过．．．132 000元．已知甲、乙、丙三种电冰箱的出厂价格分别为：1 200元/台、1 600元/台、2 000元/台．(1)至少购进乙种电冰箱多少台？(2)若要求甲种电冰箱的台数不超过丙种电冰箱的台数，则有哪些购买方案？C OABD18．甲、乙两名运动员进行长跑训练，两人距终点的路程y（米）与跑步时间x（分）之间的函数图象如图所示，根据图象所提供的信息解答问题：(1) 他们在进行米的长跑训练，在0＜x＜15的时段内，速度较快的人是；(2) 求甲距终点的路程y（米）和跑步时间x（分）之间的函数关系式；(3) 当x=15时，两人相距多少米？在15＜x＜20的时段内，求两人速度之差．Ⅲ．（本题满分36分，第19题12分，第20题12分，第21题12分）19．把一副扑克牌中的3张黑桃牌（它们的正面牌面数字分别是3、4、5）洗匀后正面朝下放在桌面上．（1）如果从中随机抽取一张牌，那么牌面数字是4的概率是多少？（2）小王和小李玩摸牌游戏，游戏规则如下：先由小王随机抽出一张牌，记下牌面数字后放回，洗匀后正面朝下，再由小李随机抽出一张牌，记下牌面数字．当2张牌面数字相同时，小王赢；当2张牌面数字不相同时，小李赢．现请你利用树状图或列表法分析游戏规则对双方是否公平？并说明理由．分)20．如图，河流两岸a b ，互相平行，C D ，是河岸a 上间隔50m 的两个电线杆．某人在河岸b 上的A 处测得30DAB ∠=，然后沿河岸走了100m 到达B 处，测得60CBF ∠=，求河流的宽度CF 的值（结果精确到个位）．BEDCFab A21．三个生产日光灯管的厂家在广告中宣称，他们生产的日光灯管在正常情况下，灯管的使用寿命为12个月．工商部门为了检查他们宣传的真实性，从三个厂家各抽取11只日光灯管进行检测，灯管的使用寿命（单位：月）如下：试问：（1）这三个厂家的广告，分别利用了统计中的哪一个特征数（平均数、中位数、众数）进行宣传？（2）如果三种产品的售价一样，作为顾客的你选购哪个厂家的产品？请说明理由．Ⅳ（本题满分8分）22．如图， 已知等边三角形ABC 中，点D ，E ，F 分别为边AB ，AC ，BC 的中点，M 为直线BC 上一动点，△DMN 为等边三角形（点M 的位置改变时， △DMN 也随之整体移动） ．（1）如图①，当点M 在点B 左侧时，请你判断EN 与MF 有怎样的数量关系？点F 是否在直线NE 上？都请直接．．．．写出结论，不必证明或说明理由； （2）如图②，当点M 在BC 上时，其它条件不变，（1）的结论中EN 与MF 的数量关系是否仍然成立?若成立，请利用图②证明；若不成立，请说明理由；（3）若点M 在点C 右侧时，请你在图③中画出相应的图形，并判断（1）的结论中EN 与MF 的数量关系是否仍然成立?若成立?请直接写出结论，不必证明或说明理由．图① 图② 图③A·BCD EF··N MFEDCB ANMF EDCBA·Ⅴ（本题满分14分）23．如图，在平面直角坐标系中，以点(11)C ，为圆心，2为半径作圆，交x 轴于A B ，两点，开口向下的抛物线经过点A B ，，且其顶点P 在C 上． （1）求ACB 的大小； （2）写出A B ，两点的坐标； （3）试确定此抛物线的解析式；（4）在该抛物线上是否存在一点D ，使线段OP 与CD 互相平分？若存在，求出点D 的坐标；若不存在，请说明理由．2010年中考模拟题（七） 数学试题参考答案及评分标准一、选择题（本大题共7小题，每小题4分，共28分）二、填空题（本大题共6小题，每小题4分，共24分）8．(00)，；9．152；10．210；11．4.8；12．23；13．4三、解答题（本大题Ⅰ—Ⅴ题，共10小题，共98分） Ⅰ．（本题满分12分，第14题6分，第15题6分） 14．解：原式＝9－16÷（－8）＋1－23×23……………………2分 ＝9＋2＋1－3.……………………………………4分 ＝9 ………………………………6分15．解：原式211(1)1(1)(1)1x x x x x -=-++-+··········································································· 2分 2211(1)(1)1(1)(1)x x x x x x -+--=-=+++ ···································································· 4分 22(1)x =+ ················································································································· 5分当1x =时，原式23== ········································································· 6分 Ⅱ．（本题满分28分，第16题7分，第17题10分，第18题11分）16．（1）连结OC ，则 OC AB ⊥． …………………………………………………1分∵OA OB =，∴1122AC BC AB ===⨯= ………………………………………2分在Rt AOC △中，3OC ===．∴ ⊙O 的半径为3． …………………………………………………………3分 （2）∵ OC =12OB , ∴ ∠B =30o , ∠COD =60o ． ……………………………………5分 ∴扇形OCD 的面积为OCD S 扇形=260π3360⨯⨯=32π． …………………………………5分 阴影部分的面积为：Rt Δ=OBC OCD S S S -阴影扇形=12OC CB ⋅－3π2－3π2．…………………………7分 17．解：（1）设购买乙种电冰箱x 台，则购买甲种电冰箱2x 台，丙种电冰箱(803)x -台，根据题意，列不等式： ····································································· 1分120021600(803)2000132000x x x ⨯++-⨯≤． ································································· 3分解这个不等式，得14x ≥． ·········································································································· 4分∴至少购进乙种电冰箱14台．····································································································· 5分 （2）根据题意，得2803x x -≤． ···························································································· 6分 解这个不等式，得16x ≤． ·········································································································· 7分 由（1）知14x ≥．1416x ∴≤≤．又x 为正整数，141516x ∴=，，． ······························································································································ 8分所以，有三种购买方案：方案一：甲种电冰箱为28台，乙种电冰箱为14台，丙种电冰箱为38台； 方案二：甲种电冰箱为30台，乙种电冰箱为15台，丙种电冰箱为35台；方案三：甲种电冰箱为32台，乙种电冰箱为16台，丙种电冰箱为32台． ·················· 10分 18．解：（1）5000…………………………………2分甲 ………………………………4分（2）设所求直线的解析式为：y =kx +b (0≤x ≤20)， ………5分由图象可知：b =5000，当x =20时，y =0， ∴0=20k +5000，解得k = -250. …7分 即y = -250x +5000 (0≤x ≤20) ……………7分（3）当x =15时，y = -250x +5000= -250×15+5000=5000-3750=1250. ………8分 两人相距：(5000 -1250)-(5000-2000)=750（米）………………9分 两人速度之差:750÷(20-15)=150（米/分）……………11分Ⅲ．（本题满分36分，第19题12分，第20题12分，第21题12分） 19解：（1）P （抽到牌面数字是4）13；············································································· 2分 （2）游戏规则对双方不公平． ······················································································· 5分 理由如下：由上述树状图或表格知：所有可能出现的结果共有9种．分…………………………8分P （抽到牌面数字相同）=3193=， P （抽到牌面数字不相同）=6293=．∵1233<，∴此游戏不公平，小李赢的可能性大． ·················································· 12分 （说明：答题时只需用树状图或列表法进行分析即可）20．解：过点C 作CE AD ∥，交AB 于ECD AE ∥，CE AD ∥ ·············································································································· 2分∴四边形AECD 是平行四边形······································································································ 4分 50AE CD ∴==m ，50EB AB AE =-=m ，30CEB DAB ∠=∠= ···························· 6分又60CBF ∠=，故30ECB ∠=，50CB EB ∴==m ······················································· 8分∴在Rt CFB △中，sin 50sin 6043CF CB CBF =∠=≈m ··········································· 11分 答：河流的宽度CF 的值为43m ． ···························································································· 12分21．答：（1）甲厂的广告利用了统计中的平均数． ································································· 2分 乙厂的广告利用了统计中的众数． ··················································································· 4分 丙厂的广告利用了统计中的中位数． ··················································································· 7分 （2） 选用甲厂的产品． 因为它的平均数较真实地反映灯管的使用寿命 ························ 10分 或选用丙厂的产品．因为丙厂有一半以上的灯管使用寿命超过12个月 ···························· 10分Ⅳ．（本题满分8分）22．（1）判断：EN 与MF 相等 （或EN=MF ），点F 在直线NE 上， ························ 2分 （2）成立． ································································································································ 3分 证明：法一：连结DE ，DF ．∵△ABC 是等边三角形， ∴AB =AC =BC ． 又∵D ，E ，F 是三边的中点，∴DE ，DF ，EF 为三角形的中位线．∴DE =DF =EF ，∠FDE =60°． 又∠MDF +∠FDN =60°， ∠NDE +∠FDN =60°， ∴∠MDF =∠NDE ．在△DMF 和△DNE 中，DF =DE ，DM =DN ， ∠MDF =∠NDE ， ∴△DMF ≌△DNE ． 8∴MF =NE ． ············································································································· 6分 法二：延长EN ，则EN 过点F ．∵△ABC 是等边三角形， ∴AB =AC =BC ． 又∵D ，E ，F 是三边的中点， ∴EF =DF =BF ． ∵∠BDM +∠MDF =60°， ∠FDN +∠MDF =60°，NCA BFMDENCA BFMDEF BC∴∠BDM =∠FDN ．又∵DM =DN ， ∠ABM =∠DFN =60°， ∴△DBM ≌△DFN ．∴BM =FN ．∵BF =EF ， ∴MF =EN ． ·········································································································· 6分 （3）画出图形（连出线段NE ）， 6MF 与EN 相等的结论仍然成立（或MF =NE 成立）． ····················································· 8分Ⅴ．（本题满分14分）23．解：（1）作CH x ⊥轴，H 为垂足，1CH = ，半径2CB = ································································ 1分60BCH ∠=，120ACB ∴∠=············································· 3分（2）1CH = ，半径2CB =HB ∴=(1A ， ···················································· 5分(1B ·······················································································6分 （3）由圆与抛物线的对称性可知抛物线的顶点P 的坐标为(13)， ········································· 7分 设抛物线解析式2(1)3y a x =-+ ································································································· 8分把点(1B 代入上式，解得1a =- ···················································································· 9分222y x x ∴=-++ ························································································································ 10分（4）假设存在点D 使线段OP 与CD 互相平分，则四边形OCPD 是平行四边形 ·········· 11分PC OD ∴∥且PC OD =．PC y ∥轴，∴点D 在y 轴上． ···························································································· 12分 又2PC = ，2OD ∴=，即(02)D ，． 又(02)D ，满足222y x x =-++，∴点D 在抛物线上 ························································································································· 13分所以存在(02)D ，使线段OP 与CD 互相平分． ······································································· 14分。