原创2023学年中考数学预测适应性考试卷(附答案)

原创2023学年胡文1．截止2023学年年6月7日12时，全国各地支援四川地震灾区的临时安置房已经安装了40600套．这个数用科学记数法表示为（）A．C．套B．套套0.4061054.06104406102套D．40.61032．如图1，直线，l分别与）相交，如果，l∥l1l，l2120l1212l1l2那么1的度数是（2A．B．C．D．75304560图13．下列事件中是必然事件的是（A．阴天一定下雨）B．随机掷一枚质地均匀的硬币，正面朝上C．男生的身高一定比女生高D．将油滴在水中，油会浮在水面上4．图2是由几个相同的小正方体搭成的一个几何体，它的俯视图是（）A．B．C．D．图25．下列命题中正确的是（）A．两条对角线互相平分的四边形是平行四边形B．两条对角线相等的四边形是矩形C．两条对角线互相垂直的四边形是菱形D．两条对角线互相垂直且平分的四边形是正方形6．若反比例函数yk(k0)x的图象经过点(2，1)，则这个函数的图象一定经过点（）1/21原创2023学年胡文A．211，2(1，2)C．B．D．1，(1，2)2≤的解集在数轴上表示正确的是（7．不等式组213x）x3-301A．-301B．-301C．-301D．8．图3是对称中心为点O的正八边形．如果用一个含45角的直角三角板的角，借助点O（使角的顶点落在点O处）把这个正八边形的面积n等分．那么n的所有可能的值有（A．2个B．3个）C．4个D．5个图3二、填空题（每小题3分，共24分）9．分解因式：．x3y4xy10．体育老师对甲、乙两名同学分别进行了8次跳高测试，经计算这两名同学成绩的平均数相同，甲同学的方差是，乙同学的方差是28.2，那么这S26.4甲S乙两名同学跳高成绩比较稳定的是同学．11．一元二次方程x22x10的解是．12．如图4，分别是△ABC的边D，E上的点，DE∥BC，AB，ACAD2，则DBS:S．△ADE△ABCADEBC图5图413．如图5，假设可以在图中每个小正方形内任意取点（每个小正方形除颜色外完全相同），那么这个点取在阴影部分的概率是．14．一个圆锥底面周长为4cm，母长线为5cm，则这个圆锥的侧面积是．2/21原创2023学年胡文15．如图6，观察下列图案，它们都是由边长为1cm的小正方形按一定规律拼接而成的，依此规律，则第16个图案中的小正方形有个．……图案1图案2图案3图案4图616.如图7，直线y3x3x与轴、y轴分别相交于A，By3B两点，圆心P的坐标为(10，)，P与y轴相切于点O．若将PxAOP沿x轴向左移动，当P与该直线相交时，横坐标为整数的点P图7有个．三、（每小题8分，共16分）17．先化简，再求值：3a，其中a2．2aa1a1a1a18．如图8所示，在网格中建立了平面直角坐标系，每个小正方形的边长均为1个单位长度，将四边形ABCD绕坐标原点O按顺时针方向旋转180后得到四边形．ABCD1111（1）直接写出D点的坐标；1（2）将四边形平移，得到四边形，若，画出平移后的图D(45，)ABCDABCD1111222223/21原创2023学年胡文形．（友情提示：画图时请不要涂错阴影的位置哦！）四、（每小题10分，共20分）19．如图9，有四张背面相同的纸牌A，B，C，D图形，小明将这四张纸牌背面朝上洗匀后随机摸出一张，放回后洗匀再随机摸出一张．，其正面分别画有四个不同的（1）用树状图（或列表法）表示两次摸牌所有可能出现的结果（纸牌用A，B，C，D表示）；（2）求两次摸牌的牌面图形既是中心对称图形又是轴对称图形的概率．图94/21原创2023学年胡文20.如图10，AB为O的直径，D为弦BE的中点，连接OD并延长交O于点F，与过B点的切线相交于点C．若点E为AF的中点，连接AE．CFE求证：．△ABE≌△OCBDABO图10五、（每小题10分，共20分）21．某中学开展以“我最喜欢的职业”为主题的调查活动．通过对学生的随机抽样调查得到一组数据，下面两图（如图11、图12）是根据这组数据绘制的两幅不完整的统计图．请你根据图中所提供的信息解答下列问题：5/21原创2023学年胡文（1）求在这次活动中一共调查了多少名学生？（2）在扇形统计图中，求“教师”所在扇形的圆心角的度数．（3）补全两幅统计图．人数80其它教师604020医生15%军人公务员10%职业0教师医生公务员军人其它20%图11图1222．在“汶川地震”捐款活动中，某同学对甲、乙两班捐款情况进行了统计：甲班捐款人数比乙班捐款人数多3人，甲班共捐款2400元，乙班共捐款1800元，乙班平均每人捐款的钱数是甲班平均每人捐款钱数的4倍．求甲、乙两班各5有多少人捐款？6/21原创2023学年胡文六、（每小题10分，共20分）23．如图13，某数学兴趣小组在活动课上测量学校旗杆高度．已知小明的眼睛与地面的距离(AB)是1.7m，看旗杆顶部M的仰角为45；小红的眼睛与地面的距离(CD)是1.5m，看旗杆顶部M的仰角为30．两人相距28米且位于旗杆两侧（点B，N，D在同一条直线上）．请求出旗杆MN的高度．（参考数据：，，结果保留整数）2≈1.43≈1.7M45°A30°CBND图1324．2023学年胡文年6月1日起，我国实施“限塑令”，开始有偿使用环保购物袋．为了满足市场需求，某厂家生产A，B两种款式的布质环保购物袋，每天共生产4500个，两种购物袋的成本和售价如下表，设每天生产A种购物袋x个，每天共获利y元．成本（元/售价（元/个）个）7/21原创2023学年胡文232.33.5AB（1）求出y与x的函数关系式；（2）如果该厂每天最多投入成本10000元，那么每天最多获利多少元？七、（本题12分）25.如图14，在Rt△ABC中，，，，另有一等腰梯形DEFGA90ABACBC42（GF∥DE）的底边DE与BC重合，两腰分别落在AB，AC上，且G，F分别是AAB，AC的中点．（1）求等腰梯形DEFG的面积；GF(D)BC(E)图148/21原创2023学年胡文（2）操作：固定△ABC，将等腰梯形DEFG以每秒1个单位的速度沿BC方向向右运动，直到点D与点C重合时停止．设运动时间为x秒，运动后的等腰梯形为DEFG（如图15）．探究1：在运动过程中，四边形BDGG能否是菱形？若能，请求出此时x的值；若不能，请说明理由．AGFGFEBDC图15探究2：设在运动过程中△ABC与等腰梯形DEFG重叠部分的面积为y，求与yx的函数关系式．9/21原创2023学年胡文八、（本题14分）26．如图16，在平面直角坐标系中，直线y3x3x与轴交于点A，与y轴交于点C，抛物线23yax2xc(a0)经过A，B，C三点．3（1）求过三点抛物线的解析式并求出顶点F的坐标；A，B，C（2）在抛物线上是否存在点P，使△ABP为直角三角形，若存在，直接写出P点坐标；若不存在，请说明理由；（3）试探究在直线AC上是否存在一点M，使得△MBF的周长最小，若存在，求出M点的坐标；若不存在，请说明理由．yxOBACF图1610/21原创2023学年胡文11/21原创2023学年胡文答案一、选择题（每小题3分，共24分）题号答案1B2345678BCDDADA二、填空题（每小题3分，共24分）10．甲9．11．712．13．xy(x2)(x2)xx114:925214．10cm2（丢单位扣1分）三、（每小题8分，共16分）15．13616．317．解法一：原式3a(a1)a(a1)a1·········································2分2a1a22a4········································································································6分当a2时，原式2248···································································8分解法二：原式3a(a1)(a1)a(a1)(a1)······························2分a1aa1a2a4········································································································6分当a2时，原式2248···································································8分18．解：（1）（2）····························································································2分描对一个点给1分．·········································6分B，C，D222D(3，1)1，A2画出正确图形（见图1）········································································8分12/21原创2023学年胡文图1四、（每小题10分，共20分）19．（1）解法一：第二次第一次ABCDAB（A，A）（A，B）（A，C）（A，D）（B，A）（B，B）（B，C）（B，D）（C，A）（C，B）（C，C）（C，D）（D，A）（D，B）（D，C）（D，D）···················6分CD（2）从表中可以得到，两次摸牌所有可能出现的结果共有16种，其中既是中心对称图形又是轴对称图形的有9种．·············································8分故所求概率是9．················································································10分1619．（1）解法二：开始第一次牌面的字母第二次牌面的字母ABCDABCDABCDABCDABCD所以可能出现的结果：（A，A），（A，B），（A，C），（A，D），（B，A），（B，B），（B，C），（B，D），（C，A），（C，B），（C，C），（C，D），（D，A），（D，B），13/21原创2023学年胡文（D，C），（D，D）．························································6分（2）以下同解法1．20.解：（1）证明：如图2．CEFAB是O的直径．DE90·······························································1分ABO又BC是O的切线，OBC90图2EOBC·························································3分OD过圆心，BDDE，EFFBBOCA．··························································································6分E为AF中点，EFBFAEABE30·······························································································8分E90AE1ABOB························································································9分2△ABE≌△OCB．·················································································10分五、（每小题10分，共20分）21．（1）被调查的学生数为40200（人）···········································2分20％（2）“教师”所在扇形的圆心角的度数为115％20％10％70100％36072··············································5分20014/21原创2023学年胡文（3）如图3，补全图·············································································8分如图4，补全图·····················································································10分人数80教师其它20%60402035%医生15%公务员军人10%职业0教师医生公务员军人其它20%图3图422．解法一：设乙班有人捐款，则甲班有(x3)人捐款．···········1分x根据题意得：····························································································5分240041800x35x解这个方程得x45．·············································································8分经检验x45是所列方程的根．·····························································9分x348（人）答：甲班有48人捐款，乙班有45人捐款．·································10分解法二：设甲班有人捐款，则乙班有(x3)人捐款．····················1分x根据题意得：····························································································5分240041800x5x3解这个方程得x48．·············································································8分经检验x48是所列方程的根．·····························································9分x345（人）答：甲班有48人捐款，乙班有45人捐款．·································10分六、（每小题10分，共20分）23．解法一：解：过点作于，过点作于，·····················1分AAEMNECCFMNF15/21原创2023学年胡文则在EFABCD1.71.50.2···································································2分Rt△AEM中，AEM90，MAE45AEME···································································································3分设AEMEx（不设参数也可）MMFx0.2，5412-0·································5分45°AE30°C在Rt△MFC中，，FMFC90MCF30BND图5MFCFtanMCF3(28x)·········································7分x0.23x≈10.0···································································································9分MN≈12答：旗杆高约为12米．·····································································10分解法二：解：过点作于，过点作于，····1分AAEMNECCFMNF则在EFABCD1.71.50.2···································································2分Rt△AEM中，，AEM90MAE45AEME设在AEx，则MFx0.2·········································································3分Rt△MFC中，，MFC90MCF30CFMFtan603(x0.2)·······································································5分BNNDBDx3(x0.2)28·····················································································7分解得x≈10.2···································································································9分MN≈12答：旗杆高约为12米．·····································································10分16/21原创2023学年胡文（注：其他方法参照给分）24．解：（1）根据题意得：y(2.32)x(3.53)(4500x)0.2x2250·········2分（2）根据题意得：2x3(4500x)≤10000············································5分解得x≥3500元··························································································6分，随增大而减小·························································8分k0.20yx当x3500时y0.2350022501550·········································································9分答：该厂每天至多获利1550元．····················································10分七、（本题12分）25．解：如图6，（1）过点作于．GGMBCM，，，为中点ABACBAC90BC42GABAGM2．······································1分G又分别为G，F的中点AB，ACFGF1BC22·······························2分(D)BC(E)M2图6S1(2242)262梯形DEFG等腰梯形DEFG的面积为6．·····························································3分（2）能为菱形·························································································4分如图7，由BG∥DG，GG∥BC四边形BDGG是平行四边形········6分A当BDBG1AB2时，四边形BDGG为菱形，GFGF2此时可求得x2EBDMC当秒时，四边形BDGG为菱形．8分x2图717/21原创2023学年胡文（3）分两种情况：0≤x22时，方法一：①当，GM2S2xBDGG重叠部分的面积为：y62x当0≤x22时，与的函数关系式为y62x······················10分yx方法二：当0≤x22时，A，，FG22xDC42xGM2FGPF重叠部分的面积为：Gy(22x)(42x)262xBEDQC2图8当0≤x22时，与的函数关系式为y62x···················10分=yx②当22≤x≤42时，设FC与DG交于点P，则PDCPCD45，CPD90PCPD作PQDC于Q，则PQDQQC1(42x)2重叠部分的面积为：y1(42x)1(42x)1(42x)21x222x8························12分2244八、（本题14分）26．解：（1）直线y3x3与x轴交于点A，与y轴交于点C．，A(10，)C(0，3)··················································································1分点A，C都在抛物线上，23330aca3c33c18/21原创2023学年胡文323x3·············································3分3抛物线的解析式为yx23······················································································4分43顶点F1，3（2）存在··································································································5分····································································································7分P(0，3)1····································································································9分P(2，3)2（3）存在·······························································································10分理由：解法一：延长BC到点B，使BCBC，连接交直线AC于点M，则点M就是所求的点．BF··································································11分过点作于点H．BBHABy323x23B点在抛物线x3上，