山东省济南市高新区学年七年级下学期期末考试数学试题Word版含答案.docx
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山东省济南市高新区学年七年级下学期期末考试数学试题Word版含答案
绝密★启用前
2020至2021学年第二学期期末学业水平测试
高新初中数学七年级试题
本试题分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分．第Ⅰ卷共2页，满分为48分；第Ⅱ卷共5页，满分为102分．本试题共6页，满分为150分．考试时间为120分钟．答卷前，考生务必用0.5毫米黑色墨水签字笔将自己的考点、姓名、准考证号、座号填写在答题卡上和试卷规定的位置上．考试结束后，将本试卷和答题卡一并交回．本考试不允许使用计算器．
第I卷（选择题共48分）
注意事项：
第Ⅰ卷为选择题，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．答案写在试卷上无效．
一、选择题（本大题共12个小题，每小题4分，共48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．）
1．计算
所得结果是（ ）
A．2021B．
C．﹣
D．﹣2021
2．下面四个图形分别是绿色食品、低碳、节能和节水标志，是轴对称图形的是（ ）
A．
B．
C．
D．
3．下列4个袋子中，装有除颜色外完全相同的10个小球，任意摸出一个球，摸到红球可能性最大的是（ ）
A．
B．
C．
D．
4．如图，沿笔直小路DE的一侧栽植两棵小树B，C，小明在A处测得AB＝5米，AC＝7米，则点A到DE的距离可能为（ ）
A．4米B．5米
C．6米D．7米
5．在行进路程s、速度v和时间t的相关计算中，若保持行驶的路程不变，则下列说法正确的是（ ）
A．变量只有速度vB．变量只有时间t
C．速度v和时间t都是变量D．速度v、时间t、路程s都是常量
6．现有两根长度分别3cm和7cm的木棒，若要钉成一个三角形木架，则应选取的第三根木棒长为（ ）
A．4cmB．7cmC．10cmD．13cm
7．如图，一只电子蚂蚁从正方体的顶点A处沿着表面爬到顶点C处，电子蚂蚁的部分爬行路线在平面展开图中的表示如图的虚线，其中能说明爬行路线最短的是（ ）
A．
B．
C．
D．
8．等腰三角形的一个内角为50°，它的顶角的度数是（ ）
A．65°B．80°C．65°或80°D．50°或80°
9．若m，n为常数，等式（x+2）（x﹣1）＝x2+mx+n恒成立，则mn的值为（ ）
A．1B．﹣1C．2D．﹣2
10．如图，将一个长方形纸条折成如图的形状，若已知∠1＝140°，则∠2为（ ）
A．50°B．60°
C．70°D．80°
11．设一个直角三角形的两直角边分别是a，b，斜边是c．若用一把最大刻度是20cm的直尺，可一次直接测得c的长度，则a，b的长可能是（ ）
A．a＝12，b＝16B．a＝11，b＝17C．a＝10，b＝18D．a＝9，b＝19
12．如图有两张正方形纸片A和B，图1将B放置在A内部，测得阴影部分面积为2，图2将正方形AB并列放置后构造新正方形，测得阴影部分面积为20，若将3个正方形A和2个正方形B并列放置后构造新正方形如图3，（图2，图3中正方形AB纸片均无重叠部分）则图3阴影部分面积（ ）
A．22B．24C．42D．44
第Ⅱ卷（非选择题共102分）
注意事项：
1.第II卷必须用0.5毫米黑色签字笔作答，答案必须写在答题卡各题目指定区域内相应的位置，不能写在试卷上；如需改动，先划掉原来的答案，然后再写上新的答案；不能使用涂改液、胶带纸、修正带．不按以上要求作答的答案无效．
2.填空题请直接填写答案，解答题应写出文字说明、证明过程或演算步骤．
二、填空题:
（本大题共6个小题，每小题4分，共24分．）
13．计算（y+2）（y﹣2）的结果等于 ．
14．某人连续抛掷一枚质地均匀的硬币3次，结果都是正面朝上，则他第四次抛掷这枚硬币，正面朝上的概率为 ．
15．如图，在△ABC中，AD平分∠BAC，∠BAC＝80°，
∠B＝35°，则∠ADC的度数为 °．
16．某工程队承建30km的管道铺设，工期60天，施工x天后剩余管道ykm，则y与x的关系式为 ．
17．如图，在△ABC中，AB的垂直平分线分别交AB，AC于D，E两点，且AC＝10，BC＝4，则△BCE的周长为 ．
第17题图第18题图
18．在直线上依次摆着七个正方形（如图），已知斜放置的三个正方形的面积分别为1，2，3，正放置的四个正方形的面积是S1，S2，S3，S4，则S1+S2﹣S3﹣S4＝ ．
三、解答题:
（本大题共12个小题，共78分．解答应写出文字说明、证明过程或演算步骤．）
19．（本题满分4分）计算：
a3•a2•a+（a2）3．
20．（本题满分4分）计算：
（x﹣3）（x+6）．
21．（本题满分4分）如图，在边长为1的小正方形网格中，点A，B，C均落在格点上．
（1）画出△ABC关于直线l的轴对称图形△A1B1C1．
（2）△A1B1C1的形状是．
22．（本题满分5分）填写下列空格：
已知：
如图，CE平分∠ACD，∠AEC＝∠ACE．
求证：
AB∥CD．
证明：
∵CE平分∠ACD（已知），
∴∠ACE＝∠ （ ）．
∵∠AEC＝∠ACE（已知），
∴∠AEC＝∠ （ ）．
∴AB∥CD（ ）．
23．（本题满分5分）已知：
如图，在△ABC中，BC⊥AC，若AC＝8，BC＝6，求AB的长．
24．（本题满分6分）如图是一位病人的体温记录图，看图回答下列问题：
（1）自变量是 ，因变量是 ；
（2）护士每隔 小时给病人量一次体温；
（3）这位病人的最高体温是 摄氏度，最低体温是 摄氏度；
（4）他在4月8日12时的体温是 摄氏度.
25．（本题满分6分）先化简，再求值：
（2x+3y）2﹣（2x+y）（2x﹣y），其中x＝1，y＝﹣1．
26．（本题满分6分）如图，AD是等边△ABC的中线，AE＝AD，求∠AED的度数．
27．（本题满分8分）完成下列推理过程：
如图所示，点E在△ABC外部，点D在BC边上，DE交AC于F，若∠1＝∠2＝∠3，AD＝AB．猜想AC与AE之间的数量关系，并说明理由.
答：
ACAE．
解：
∵∠2＝ ，∠AFE＝∠DFC，
∴180°﹣∠2﹣∠AFE＝180°﹣∠3﹣∠DFC
∴∠E＝ ．
又∵∠1＝∠2，
∴ +∠DAC＝ +∠DAC．
∴∠BAC＝∠DAE（）．
在△ABC和△ADE中，
∴△ABC≌△ADE（）．
∴AC＝AE．
28．（本题满分8分）一圆盘被平均分成10等份，分别标有1，2，3，4，5，6，7，8，9，10这10个数字，转盘上有指针，转动转盘，当转盘停止，指针指向的数字即为转出的数字，现有两人参与游戏，一人转动转盘另一人猜数，若猜的数与转盘转出的数相符，则猜数的获胜，否则转动转盘的人获胜，猜数的方法从下面三种中选一种：
（1）猜“是奇数”或“是偶数”；
（2）猜“是3的倍数”或“不是3的倍数”；
（3）猜“是大于4的数”或“是不大于4的数”．若你是猜数的游戏者，为了尽可能获胜，应选第几种猜数方法？
并请你用数学知识说明理由．
29．（本题满分10分）如图，△ABC与△ADE是以点A为公共顶点的两个三角形，且AD＝AE，AB＝AC，∠DAE＝∠CAB＝90°，且线段BD、CE交于F．
（1）求证：
△AEC≌△ADB．
（2）猜想CE与DB之间的关系，并说明理由．
30．（本题满分12分）“一带一路”让中国和世界更紧密，“中欧铁路”为了安全起见在某段铁路两旁安置了A，D两座可旋转探照灯．假定主道路是平行的，即PQ∥CN，A，B为PQ上两点，AD平分∠CAB交CN于点D，E为AD上一点，连接BE，AF平分∠BAD交BE于点F．
（1）若∠C＝20°，则∠EAP＝ ；
（2）作AG交CD于点G，且满足∠1=
∠ADC，当∠2+
∠GAF＝180°时，试说明：
AC∥BE；
（3）在
（1）问的条件下，探照灯A、D照出的光线在铁路所在平面旋转，探照灯射出的光线AC以每秒5度的速度逆时针转动，探照灯D射出的光线DN以每秒15度的速度逆时针转动，DN转至射线DC后立即以相同速度回转，若它们同时开始转动，设转动时间为t秒，当DN回到出发时的位置时同时停止转动，则在转动过程中，当AC与DN互相平行或垂直时，请直接写出此时t的值．
备用图
2020至2021学年第二学期期末学业水平测试
高新初中数学七年级参考答案及评分标准
一、选择题
题号
1
2
3
4
5
6
7
8
9
10
11
12
答案
A
A
D
A
C
B
A
D
A
C
A
C
二、填空题:
（本大题共6个小题，每小题4分，共24分．）
13．y2﹣4．14．
．15．75．16．y＝30﹣0.5x17．14．18．﹣2．
三、解答题:
（本大题共12个小题，共78分．解答应写出文字说明、证明过程或演算步骤．）
19．（本题4分）解：
原式＝a6+a6·····················································································2分
＝2a6·······················································································4分
20．（本题4分）解：
原式＝x2+6x﹣3x﹣18·············································································2分
＝x2+3x﹣18·················································································4分
21．（本题4分）解：
（1）如图，△A1B1C1为所求；
·······································································································3分
（2）△A1B1C1是等腰直角三角形····················································································4分
22．（本题5分）
DCE；角平分线的定义；DCE；等量代换；内错角相等，两直线平行
23．（本题5分）
解：
∵BC⊥AC
∴∠C＝90°··············································································································1分
∵Rt△ABC中，∠C＝90°，AC＝8，BC＝6·····································································3分
∴BC2+AC2=AB2·······································································································4分
AB=10··········································································································5分
24．（本题6分）
解：
（1）时间，体温··········································································································2分
（2）6························································································································3分
（3）39.5，36.8············································································································5分
（4）37.5·····················································································································6分
25．（本题6分）
解：
原式＝4x2+12xy+9y2﹣（4x2﹣y2）···················································································2分
＝4x2+12xy+9y2﹣4x2+y2＝12xy+10y2··················································································4分
当x＝1，y＝﹣1时，
原式＝﹣12+10＝﹣2·····································································································6分
26．（本题6分）
解：
∵AD是等边△ABC的中线，
∴∠BAC=60°，AD平分∠BAC·····················································································2分
∴∠CAD
∠BAC=30°································································································3分
∵AD＝AE，
∴∠ADE＝∠AED·······································································································5分
∴∠AED=75°·············································································································6分
27．（本题8分）每空1分
答：
＝
解：
∠3，∠C，∠1，∠2，等式性质，∠E＝∠C，AAS
28．（本题8分）
解：
选第2种猜数方法··································································································1分
理由：
P（是奇数）＝0.5，P（是偶数）＝0.5；
P（是3的倍数）＝0.3，P（不是3的倍数）＝0.7；
P（是大于4的数）＝0.6，P（不是大于4的数）＝0.4·········································································7分
∵P（不是3的倍数）最大，
∴选第2种猜数方法，并猜转盘转得的结果不是3的倍数······················································8分
29．（本题10分）
（1）证明：
∵∠BAC＝∠DAE，
∴∠BAC+∠CAD＝∠DAE+∠CAD，
∴∠BAD＝∠CAE·····························································································1分
在△BAD与△CAE中，
···························································································3分
∴△BAD≌△CAE（SAS）···················································································4分
（2）答：
=，⊥············································································································6分
解：
由
（1）知，△BAD≌△CAE，
∴∠ABD＝∠ACE，BD＝CE··············································································7分
∵∠BAC＝90°，
∴∠CBF+∠BCF＝∠ABC+∠ACB＝90°································································9分
∴∠BFC＝90°·······························································································10分
30．（本题12分）
解：
（1）100°···················································································································2分
（2）∵∠1
∠ADC，
∴令∠1
a，则∠ADC=3a························································································3分
∵PQ∥CN，
∴∠ADC＝∠BAD=3a
∵AD平分∠BAC，
∴∠CAD＝∠ADC＝∠BAD=3a················································································4分
∵AF平分∠BAD，
∴∠BAD＝2∠EAF．
∴∠EAF=1.5a
∴∠GAF＝∠1+∠EAF
∴
∠GAF＝3a······································································································5分
∵∠2
∠GAF＝180°，
∴∠2+3a＝180°．
∴∠2+∠CAD＝180°．
∵∠2+∠AEB＝180°，
∴∠CAD＝∠AEB·································································································6分
∴AC∥BE············································································································7分
（3）t的值为2s或11s或12.5s或17s或21.5s···································································12分
升级会员 