初三中考数学模拟试卷及答案

初三中考数学模拟试卷及答案初三中考数学模拟试卷本试卷满分为150分,考试时间为120分钟.一、选择题(每小题只有一个正确选项,将正确选项的序号填入题后的括号内.10小题,每小题3分,共30分)1.2(2)3--的值是()b5E2RGbCAPA.1B.2C.1-D.2-2.下列运算中,不正确的是()A.23ababab+=B.2ababab-=C.22ababab⨯=p1EanqFDPwD.122abab÷=3.如果一个四边形绕对角线的交点旋转90,所得的图形与原来的图形重合,那么这个四边形一定是()A.平行四边形B.矩形C.菱形D.正方形DXDiTa9E3d4.下列物体中,是同一物体的为()A.(1)与(2)B.(1)与(3)C.(1)与(4)D(2)与(3)5.“数轴上的点并不都表示有理数,如图中数轴上的点P,这种利用图形直观说明问题的方式体现的数学思想方法叫()A.代入法B.换元法C.数形结合的思想方法D.分类讨论的思想方法6.在一幅长60cm,宽40cm的矩形风景画的四周镶一条金色纸边,制成一幅矩形挂图,如图所示.如果要使整个挂图的面积是22816cm,设金色纸边的宽为cmx,那么x满足的方程是()RTCrpUDGiTA.(602)(402)2816xx++=B.(60)(40)2816xx++=C.(602)(40)2816xx++=D.(60)(402)2816xx++=7.夏天,一杯开水放在桌子上,杯中水的温度()TC随时间t变化的关系的大致图象是()8.如图,ABC△的边长都大于2,分别以它的顶点为圆心,1为半径画弧(弧的端点分别在三角形的相邻两边上),则这三条弧的长的和是()A.4πB.3πC.6πD.5π5PCzVD7HxA9.一人乘雪橇沿如图所示的斜坡笔直滑下,滑下的距离S(米)与时间t(秒)间的关系式为210Stt=+,若滑到坡底的时间为2秒,则此人下滑的高度为()A.24米jLBHrnAILgB.12米C.D.6米10.某超市(商场)失窃,大量的商品在夜间被罪犯用汽车运走.三个嫌疑犯被警察局传讯,警察局已经掌握了以下事实:(1)罪犯不在甲、乙、丙三人之外;(2)丙作案时总得有甲作从犯;(3)乙不会开车.在此案中,能肯定的作案对象是()A.嫌疑犯乙B.嫌疑犯丙C.嫌疑犯甲D.嫌疑犯甲和丙二、填空题(把答案填在题中的横线上.8小题,每小题4分,共32分)xHAQX74J0X11.2005年10月12日9时15分,我国“神州六号”载人飞船发射成功.飞船在太空飞行了77圈,路程约3300000千米,用科学记数法表示这个路程为千米.12.如图中两圆的位置关系是(相交,外切,外离).LDAYtRyKfE13.反比例函数2kyx=的图象的两个分支分别位于象限.14.下列是三种化合物的结构式及分子式,则按其规律第4个化合物的分子式为.15.小明将一把钥匙放进自己家中的抽屉中,他记不清到底放进三个抽屉中的哪一个了,那么他一次选对抽屉的概率是.Zzz6ZB2Ltk16.如图,ABCD∥,EG平分BEF∠,若260∠=,则1∠=.第17题图ABCDEFAEBCFGD12第16题图17.如图,ADACDF==∠∠,,则需要补充条件:(写出一个即可),才能使ABCDEF△≌△.dvzfvkwMI118.二次函数2yaxbxc=++图象上部分点的对应值如下表:则使的取值范围为.三、解答题(解答时,必须写出必要的解题步骤.5小题,共38分)19.(6分)先化简,再求值:23111aaaaaa-⎛⎫-⎪-+⎝⎭·,其中2a.20.(8分)有7名同学测得某楼房的高度如下(单位:米):29,28.5,30,30,32,31,33.(1)求这组数据的中位数、众数、平均数;(2)你认为此楼房大概有多高?21.(8分)一次函数图象如图所示,求其解析式.22.(8分)如图,已知ABDCACDB==,.求证:12∠=∠.xADO1223.(8分)为节约用电,某学校在本学期初制定了详细的用电计划.如果实际每天比计划多用2度电,那么本学期的用电量将会超过2990度;如果实际每天比计划节约2度电,那么本学期的用电量将不超过2600度.若本学期的在校时间按130天计算,那么学校原计划每天用电量应控制在什么范围内?rqyn14ZNXI四、解答题(解答时,必须写出必要的解题步骤.5小题,共50分)24.(8分)如图是两个半圆,点O为大半圆的圆心,AB是大半圆的弦关与小半圆相切,且24AB.问:能求出阴影部分的面积吗?若能,求出此面积;若不能,试说明理由.25.(10分)一架长5米的梯子AB,斜立在一竖直的墙上,这时梯子底端距墙底3米.如果梯子的顶端沿墙下滑1米,梯子的底端在水平方向沿一条直线也将滑动1米吗?用所学知识,论证你的结论.EmxvxOtOco26.(10分)某公司现有甲、乙两种品牌的计算器,甲品牌计算器有ABC,,三种不同的型号,乙品牌计算器有DE,SixE2yXPq5两种不同的型号,新华中学要从甲、乙两种品牌的计算器中各ABEDC选购一种型号的计算器.(1)写出所有的选购方案(利用树状图或列表方法表示);(2)如果(1)中各种选购方案被选中的可能性相同,那么A型号计算器被选中的概率是多少?(3)现知新华中学购买甲、乙两种品牌计算器共40个(价格如图所示),恰好用了1000元人民币,其中甲品牌计算器为A型号计算器,求购买的A型号计算器有多少个?27.(12分)如图,在M中,AB所对的圆心角为120,已知圆的半径为2cm,并建立如图所示的直角坐标系.(1)求圆心M的坐标;6ewMyirQFL(2)求经过ABC,,三点的抛物线的解析式;(3)点D是弦AB所对的优弧上一动点,求四边形ACBD的最大面积;(4)在(2)中的抛物线上是否存在一点P,使PAB△和ABC△相似?若存在,求出点P的坐标;若不存在,请说明理由.kavU42VRUs张掖市2006年初中毕业学业与升学考试(新课程)x数学试题参考答案与评分标准一、选择题:每小题3分,共30分ACDBCABDBC二、填空题:每小题4分,共32分11.63.310⨯12.外离13.一,三14.410CH15.1316.6017.答案不唯一;如:CF=∠∠,或ABDE=,等18.23x-三、解答题:共38分19.本小题满分6分解:原式24a=+,·························································································4分y6v3ALoS89当2a时,原式=.··········································································6分说明:会运用分配律且对者给2分,会用公式21(1)(1)xxx-=-+且对者给1分,若先通分且对者给2分.M2ub6vSTnP20.本小题满分8分解:(1)在这组数据中,中位数是30,·······························································2分众数是30,···································································································4分平均数是30.5;······························································································6分(2)该楼房大概高30米(未写单位不扣分)······················································8分21.本小题满分8分0YujCfmUCw解:设一次函数解析式为ykxb=+,·································································1分则0120.kbkb=⨯+⎧⎨-=⨯+⎩,····························································································5分eUts8ZQVRd解得22.kb=⎧⎨=-⎩,所以,一次函数解析式为22yx=-.·································································8分说明:只要求出22.kb=⎧⎨=-⎩sQsAEJkW5T,无最后一步不扣分.22.本小题满分8分说明:ABDCACDBBCBC=⎧⎪=⎨⎪=⎩,,,························································································2分GMsIasNXkAABCDCB∴△≌△.·······················································································4分TIrRGchYzgAD∴∠=∠.································································································6分又AOBDOC∠=∠,7EqZcWLZNX12∴∠=∠.··································································································8分23.本小题满分8分lzq7IGf02E解:设学校原计划每天用电量为x度,································································1分依题意得130(2)2990130(2)2600.xx+⎧⎨-⎩zvpgeqJ1hk,≤·············································································5分NrpoJac3v1解得2122x≤.即学校每天的用电量,应控制在21~22度(不包括21度)范围内.························8分说明:只要求出2122x≤,无最后一步不扣分.1nowfTG4KI四、解答题:共50分24.本小题满分8分解法1:能(或能求出阴影部分的面积).·······································································1分设大圆与小圆的半径分别为Rr,,·····································································2分作辅助线如图所示(作对),·············································································4分可得22212Rr-=,·························································································6分fjnFLDa5Zo221(ππ)72π2SRr∴=-=阴影.·······································································8分tfnNhnE6e5解法2:能(或能求出阴影部分的面积).·························