广东省深圳市龙华区2022-2023学年九年级上学期期中考试数学试卷(含答案)

这是一份广东省深圳市龙华区2022-2023学年九年级上学期期中考试数学试卷(含答案)，共15页。试卷主要包含了11等内容，欢迎下载使用。
2022-2023学年第一学期期中学情调查九年级  数学2022.11本试卷分两部分，试卷共6页，满分100分，考试时间90分钟。注意事项：1．答卷前，考生务必在答题卡上用黑色字迹的钢笔或签字笔填写自己的学校、班级、姓名和考生号；用2B铅笔将对应的考生号码涂黑。2．选择题每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑；如需改动，用橡皮擦干净后，再选涂其他答案标号；不能答在试卷上。3．非选择题必须用黑色字迹的钢笔或签字笔作答，答案必须写在答题卡各题目指定区域内的相应位置上；如需改动，先划掉原来的答案，然后再写上新的答案；不准使用铅笔、圆珠笔和涂改液。不按以上要求作答的答案无效。一、选择题：本大题共10小题，每小题3分，共30分。每小题只有一个选项符合题目要求。1．关于x的一元二次方程的二次项系数、一次项系数、常数项分别是（    ）A．5，， B．5，2， C．，2，1 D．，，2．一元二次方程，用配方法解该方程，配方后的方程为（    ）A． B． C． D．3．若，则（    ）A． B． C． D．4．如图，，且，，则AE的长为（    ）A．6 B．9 C．3 D．45．在今年“十一”期间，小康和小明两家准备进行徒步活动，从塘朗山、阳台山，梧桐山三个地点中分别选择一个地点，他们两家去同一地点徒步的概率是（    ）A． B． C． D．6．如图，在矩形ABCD中，，对角线AC与BD相交于点O，，垂足为E，，则BC的长为（    ）A． B．6cm C． D．7．一花户，有26 m长的篱笆，要围成一边靠住房墙（墙长12 m）的面积为的长方形花园，且垂直于住房墙的一边留一个1 m的门，设垂直于住房墙的其中一边长为x，则可列方程为（    ）A．  B．C．  D．8．下列说法中，正确的是（    ）A．顺次连接对角线互相垂直的四边形各边中点所组成的图形是菱形B．关于x的方程有两个不相等实根，则k的取值范围且C．正方形的对角线所在的直线是它的对称轴它有2条对称轴D．点P是线段AB的一个黄金分割点（），若，则9．如图，在正方形ABCD中，对角线AC、BD相交于点O．E、F分别为AC、BD上一点，且，连接AF，BE，EF．若，则的度数为（    ）A．55° B．65° C．45° D．70°10．如图，矩形ABCD中，点E在BC边上，且，作于点F，连接DE，BF，BF的延长线交DE于点O，交CD于点G．以下结论：①，②DE为的角平分线③若，则，④若AE平分，，则矩形ABCD的面积为则正确结论的个数是（    ）A．①②③ B．①②④ C．①③④ D．①②③④二、填空题（本大题共5小题，每小题3分，共15分．）11．关于x的一元二次方程的一个根为0，则a的值为________．12．在一个不透明的箱子里装有m个球，其中红球6个，这些球除颜色外都相同，每次将球搅拌均匀后，任意摸出一个球记下颜色后再放回，大量重复试验后发现，摸到红球的频率在0.3，那么可以估算出m的值为________．13．如图，点E是正方形ABCD中CD边上的中点，对角线交点为O，连接BE交AC于F点，则________．14．如图，四边形ABCD是菱形，对角线AC，BD相交于点O，，，点P是AC上一动点，点E是AB的中点，则的最小值为________．15．如图，P是边长为6的正方形的边AD上的一个动点（P与B、C不重合）连接CP，过点B作，将沿CP所在直线翻折得到，延长交CB的延长线于点G。当时，PG的长为________．三、解答题（本大题共7题，共55分。本大题有7题，其中16题9分，17题6分，18题6分，19题7分，20题8分，21题9分，22题10分，共55分）16．解方程（1）  （2）（用配方法）（3）（用公式法）17．定义：在边长为1的小正方形方格纸中，把顶点落在方格交点上的线段、三角形、四边形分别称为格点线段、格点三角形、格点四边形，在的正方形网格中，若每一个小正方形的边长均为1，请仅用无刻度直尺按要求画图．（1）在图①中画一个以AB为边画一个格点正方形．（2）在图②中画一个格点平行四边形，使平行四边形面积为6（3）在图③中画一个格点菱形．不是正方形（温馨提示：请画在答题卷相对应的图上）18．现有甲、乙两个不透明袋子，分别装有3个除颜色外完全相同的小球，其中甲袋装有2个白球、1个红球；乙袋装有2个红球、1个白球．（1）将甲袋摇匀，然后从甲袋中随机摸出一个小球，摸出的小球是白球的概率________．（2）A、B两人商定一个游戏规则：摇匀后，从甲、乙两袋中随机各摸出一球，若两球颜色相同，则A获胜；若颜色不同，则B获胜．请用列表法或树状图法说明这个游戏规则对双方是否公平．19．如图，四边形为菱形，点E在AC的延长线上，．（1）求证：；（2）当，时，求AE的长．20．2022北京冬奥会期间，冰墩墩和雪容融受到人们的广泛喜爱．某网店以每套96元的价格购进了一批冰墩墩和雪容融，由于销售火爆，销售单价经过两次的调整，从每套150元上涨到每套216元，此时每天可售出16套冰墩墩和雪容融．（1）若销售价格每次上涨的百分率相同，求每次上涨的百分率；（2）冬奥会闭幕后需求有所下降，需尽快将这批冰墩墩和雪容融售出，决定降价出售．经过市场调查发现：销售单价每降价15元，每天多卖出3套，商店想使每天利润达到2000元，每套价格应为多少元？21．矩形中，，点E是边BC的中点，连接AE，过点E作AE的垂线EF，与矩形的外角平分线CF交于点F．【特例证明】（1）如图，当时，求证：；【类比探究】（2）如图，当时，求的值（用含k的式子表示）；【拓展运用】如图（3），当时，P为边CD上一点，连接AP，PF，，，则BC的长为________．22．已知：在平面直角坐标系中，直线：与x轴，y轴分别交于A、B两点，直线经过点A，与y轴交于点．（1）求直线的解析式；（2）如图1，点P为直线一个动点，是否存在以点P、C、A为顶点的三角形与相似，若存在请求出点P的坐标及此时的面积。（3）如图2，将沿着x轴平移，平移过程中的记为，请问在平面内是否存在点D，使得以、、C、D为顶点的四边形是菱形？若存在，直接写出点D的坐标． 2022-2023学年度第一学期初三年级期中考试数学参考与评分标准一、选择题：本题共10小题，每小题3分，共30分．在每小题给出的四个选项中，只有一项是符合题目要求的．题号12345678910答案BDDABADBBD二、填空题：本题共5小题，每小题3分，共15分．题号1112131415答案4202：1三、解答题：本题共7题，共55分．解答应写出文字说明、证明过程或演算步骤．16．（1）解：·····································································1分·················································································2分或或···············································································3分（2），解：，，，···············································································1分，···············································································2分，或，∴，·············································································3分（3）解：    ∵，·············································································1分∴，·············································································2分解得：，··········································································3分17．（1）（2）（3）  每个图符合条件即可，每个图2分18．解：（1）共有3种等可能结果，而摸出白球的结果有2种∴；·············································································2分（2）根据题意，列表如下： 红1红2白白1（白1，红1）（白1，红2）（白1，白）白2（白2，红1）（白2，红2）（白2，白）红（红，红1）（红，红2）（红，白）由上表可知，共有9种等可能结果，其中颜色不相同的结果有5种，颜色相同的结果有4种∴，，···········································································5分∵，∴这个游戏规则对双方不公平．·····················································6分19．（1）证明：∵四边形ABCD为菱形，∴，∵，∴，·············································································2分∵，∴；·············································································3分（2）解：∵，∴···············································································5分∵菱形，∴，，∴···············································································6分∴．·············································································7分20．解：（1）设每次上涨的百分率为x，根据题意得：，·····································3分解得：，（不合题意，舍去），························································4分答：每次上涨的百分率为20%；（2）设每套价格降价为a元根据题意得：，····································································6分·················································································7分售价：元答：商店使每天利润达到2000元，每套价格应为196元．······································8分21．证明：如图，在BA上截取，连接EH．∵，∴．∵，，∴，·············································································1分∴．．∵CF平分，，∴．∴．·············································································2分∵，∴，∵，∴，∵，，∴，∴，∴；··········································································3分（2）解：在BA上截取，连接EH．∵，，∴，∴，∵CF平分，，∴．∴，∴，∴，·············································································5分∴，∵，E是BC边的中点，∴设    ∴，∴；·············································································6分（3）············································································9分∵∴为等腰直角三角形为等腰直角三角形易证∴∵由（2）得∴  易证点P为AQ中点∴易知又∵由（2）知  ∴  易证四边形为正方形，由（2）得在中，，由勾股定理得  22．解：（1）设直线的解析式，∵直线：与x轴，y轴分别交于A、B两点，∴，，∵直线经过点A，与y轴交于点，························································1分∴，·············································································2分∴  ∴直线的解析式：；······························································3分（2）设当时，∴当时，∴∴即在中，∴在中，∴存在以点P、C、A为顶点的三角形与相似，理由如下：∵∴当时，解得：（舍），P点坐标··········································································4分过点P作轴于点D，·················································································5分若P与B重合，   此时三角形ABC的面积为6·················································6分综上所述存在点  面积为15或6（3）设  当沿x轴向左平移时设，则，，①当时，，，∴即∴②当时∴，，，∴即∴③当时（舍），∴，，，∴即∴当沿x轴向右平移时设，则，，∵∴∴，，，∴即∴综上所述：符合条件的点D的坐标为或或·················································10分