济南高新区2021-2022学年第二学期八年级S数学学业质量抽测
2022/7/8 9:56:27 考试真题那些事儿
济南高新区2021-2022学年第二学期八年级学业质量抽测
数 学 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.要使分式
有意义，x的取值应满足( )
A.x≠0 B.x≠1 C.x≠2 D.x为任意实数
2.下列等式从左到右的变形，是因式分解的是( )
A.(x+y)(x﹣y)＝x2﹣y2 B.x2﹣2x+1＝(x﹣1)2
C.x2+2x+2＝(x+1)2+1 D.12xy2＝2x?6y2
3.在平行四边形ABCD中，若∠A+∠C＝80°，则∠B的度数是( )
A.140° B.100° C.40° D.120°
4.已知x＝1是方程x2﹣3x+c＝0的一个根，则实数c的值是( )
A.﹣1 B.0 C.1 D.2
5.如图，在菱形ABCD中，两条对角线长AC＝6，BD＝8，
则此菱形的面积为( )
A.48 B.24
C.20 D.12
6.从口袋中随机摸出一球，再放回口袋中，不断重复上述过程，共摸了150次，其中有50次摸到黑球，已知口袋中有黑球10个和若干个白球，由此估计口袋中大约有多少个白球( )
A.10个 B.20个 C.30个 D.无法确定
7.下列式子中，能运用平方差公式分解因式的是( )
A.﹣4a2+b2 B.x2+4 C.a2+c2﹣2ac D.﹣a2﹣b2
8.化简
的结果是( )
A.m B.
C.﹣m D.
9.已知关于x的一元二次方程x2﹣2x﹣k﹣1＝0有两个实数根，则k的取值范围是( )
A.k>﹣2 B.k≥﹣2 C.k≤﹣2 D.k<﹣2
10.如图，将△ABC沿着它的中位线DE对折，点A落在F处.
若∠C＝120°，∠A＝20°，则∠FEB的度数是( )
A.140° B.120°
C.100° D.80°
11.已知a，b，c，d都是正数，如果M＝(a+b+c)(b+c+d)，N＝(a+b+c+d)(b+c)，那么M，N的大小关系是( )
A.M>N B.M＝N C.M
12.如图，在平面直角坐标系xOy中，P(4，4)，A、B分别是x轴正半轴、y轴正半轴上的动点，且△ABO的周长是8，则P到直线AB的距离是( )
A.4 B.3 C.2.5 D.2
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.因式分解：2ab﹣4a＝ .
14.已知一个正n边形的每个内角都为120°，则n＝ .
15.随机闭合开关S1，S2，S3中的两个，能够让灯泡发亮
的概率为 .
16.若关于x的方程
0产生增根，则m＝ .
17.如图，在一块长11m，宽为7m的矩形空地内修建三条宽
度相等的小路，其余部分种植花草.若花草的种植面
积为60m2，则小路宽为 m.
18.如图，在矩形ABCD中，E，F分别是边AB，AD上的动
点，P是线段EF的中点，PG⊥BC，PH⊥CD，G，H为
垂足，连接GH.若AB＝8，AD＝6，EF＝6，则GH的最
小值是 .
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6分)化简：
.
20.(本题6分)解方程：2x2﹣3x＝1﹣2x.
21.(本题6分)已知：如图，?ABCD的对角线AC，BD相交于点O，点E，F分别在AO，CO上，且AE＝CF，求证：∠EBO＝∠FDO.
22.(本题8分)第24届冬季奥林匹克运动会(简称“冬奥会”)于2022年2月4日在北京开幕，本届冬奥会设7个大项、15个分项、109个小项.某校组织了关于冬奥知识竞答活动，随机抽取了七年级若干名同学的成绩，并整理成如下不完整的频数分布表、频数分布直方图和扇形统计图：
分组
频数
60
4
70
12
80
16
90
请根据图表信息，解答下列问题：
(1)本次知识竞答共抽取七年级同学 名；在扇形统计图中，成绩在“90
(2)该校计划对此次竞答活动成绩最高的小颖同学：奖励两枚“2022?北京冬梦之约”的邮票.现有如图所示“2022?北京冬梦之约”的四枚邮票供小颖选择，依次记为A，B，C，D，背面完全相同.将这四枚邮票背面朝上，洗匀放好，小颖从中随机抽取一枚不放回，再从中随机抽取一枚.请用列表或画树状图的方法，求小颖同学抽到的两枚邮票恰好是B(冰墩墩)和C(雪容融)的概率.
23.(本题8分)
如图，在正方形ABCD中，E为AB的中点，连接CE，将△CBE沿CE对折，得到△CGE，延长EG交CD的延长线于点H.
(1)求证：△HCE是等腰三角形.
(2)若AB＝4，求HD的长度.
24.(本题10分)
某汽车贸易公司销售A，B两种型号的新能源汽车，A型车每台进货价格比B型车每台进货价格少3万元，该公可用24万元购买A型车的数量和用30万元购买B型车的数量相同.
(1)求购买一台A型、一台B型新能源汽车的进货价格各是多少万元？
(2)该公可准备用不超过300万，采购A，B两种新能源汽车共22台，问最少需要采购A型新能源汽车多少台？
25.(本题10分)
先阅读下面的材料，再解决问题：
因式分解多项式：am+an+bm+bn，
先把它的前两项分成一组，并提出a；把它的后两项分成一组，并提出b：
得：a(m+n)+b(m+n)
再提公因式(m+n)，得：(m+n)(a+b).
于是得到：am+an+bm+bn＝a(m+n)+b(m+n)＝(a+b)(m+n).
这种因式分解的方法叫做分组分解法.
请用上面材料中提供的方法解决问题：
(1)将多项式ab﹣ac+b2﹣bc分解因式；
(2)若△ABC的三边a、b、c满足条件：a4﹣b4+a2c2+b2c2＝0，试判断△ABC的形状.
26.(本题12分)
利用完全平方公式(a+b)2＝a2+2ab+b2和(a﹣b)2＝a2﹣2ab+b2的特点可以解决很多数学问题.下面给出两个例子：
例1.分解因式：
x2+2x﹣3＝x2+2x+1﹣4
＝(x+1)2﹣4
＝(x+1+2)(x+1﹣2)
＝(x+3)(x﹣1)
例2.求代数式2x2﹣4x﹣6的最小值：
2x2﹣4x﹣6＝2(x2﹣2x)﹣6
＝2(x2﹣2x+1﹣1)﹣6
＝2[(x﹣1)2﹣1]﹣6
＝2(x﹣1)2﹣8
又∵2(x﹣1)2≥0
∴当x＝1时，代数式2x2﹣4x﹣6有最小值，最小值是﹣8.
仔细阅读上面例题，模仿解决下列问题：
(1)分解因式：m2﹣6m﹣7；
(2)当x、y为何值时，多项式2x2+y2﹣8x+6y+20有最小值？并求出这个最小值；
(3)已知△ABC的三边长a、b、c都是正整数，且满足a2+b2＝8a+6b﹣25，求△ABC周长的最大值.
27.(本题12分)
【问题原型】如图1，在四边形ABCD中，∠ADC＝90°，AB＝AC.点E、F分别为AC、BC的中点，连接EF，DE.试说明：DE＝EF.
【探究】如图2，在问题原型的条件下，当AC平分∠BAD，∠DEF＝90°时，求∠BAD的大小.
【应用】如图3，在问题原型的条件下，当AB＝2，且四边形CDEF是菱形时，直接写出四边形ABCD的面积.
济南高新区2021-2022学年第二学期八年级学业质量抽测
数学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
B
B
A
D
B
B
A
C
B
C
A
A
二、填空题:(本大题共6个小题，每小题4分，共24分.)
13.2a(b﹣2). 14.6. 15.
. 16.﹣5. 17.1. 18.7.
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6分)解：原式
··················································································4分
···························································································4分
20.(本题6分)解：原方程化为2x2﹣x﹣1＝0·······································································2分
∵a＝2，b＝﹣1，c＝﹣1，
∴Δ＝b2﹣4ac＝(﹣1)2﹣4×2×(﹣1)＝9>0····································································4分
∴x
，
∴x1＝1，x2
···········································································································6分
21.(本题6分)证明：连接DE、BF，如图所示：
∵四边形ABCD是平行四边形，
∴OB＝OD，OA＝OC···································································································2分
∵AE＝CF，
∴OE＝OF···················································································································3分
∴四边形BEDF是平行四边形··························································································4分
∴BE∥DF···················································································································5分
∴∠EBO＝∠FDO·········································································································6分
22.(本题8分)解：(1)40，72·························································································2分
(2)画树状图如下：
································································································5分
共有12种等可能的结果··································································································6分
其中小颖同学抽到的两枚邮票恰好是B(冰墩墩)和C(雪容融)的结果有2种·························7分
∴P小颖同学抽到的两枚邮票恰好是B(冰墩墩)和C(雪容融)的概率=
····························································8分
23.(本题8分)(1)证明：在正方形ABCD中，AB∥CD，
∴∠BEC＝∠ECD·········································································································1分
根据翻折，可得∠BEC＝∠GEC························································································2分
∴∠ECD＝∠GEC·········································································································3分
∴HE＝HC，即△HCE是等腰三角形·················································································4分
(2)设HD＝x，
∵AB＝4，
∴BC＝CD＝4，
∵E为AB的中点，
∴EB＝2，
根据翻折，GC＝BC＝4，EG＝EB＝2，
∵HC＝4+x··················································································································5分
∴HE＝4+x，
∴HG＝4+x﹣2＝2+x·····································································································6分
在Rt△HGC中根据勾股定理，
得(x+4)2＝42+(x+2)2································································································7分
解得x＝1，即HD＝1·····································································································8分
24.(本题10分)
解：设一台B型新能源汽车的进货价格是x万元，则一台A型新能源汽车的进货价格是(x-3)万元·····1分
由题意可得：
···································································································3分
解得：x＝15·················································································································4分
经检验，x＝15是原方程的解，且符合题意·········································································5分
∴x﹣3＝12(万元)，
∴购买一台A型新能源汽车的进货价格是12万元，购买一台B型的是15万元··························6分
(2)设需要采购A型新能源汽车a台···············································································7分
由题意可得：12a+15(22﹣a)≤300·················································································8分
∴a≥11·······················································································································9分
∴最少需要采购A型新能源汽车11台··············································································10分
25.(本题10分)解：(1)ab﹣ac+b2﹣bc
＝(ab﹣ac)+(b2﹣bc)·······························································································2分
＝a(b﹣c)+b(b﹣c)
＝(a+b)(b﹣c)·········································································································4分
(2)由已知，得(a2﹣b2)(a2+b2)+c2(a2+b2)＝0····························································6分
即(a2+b2)(a2﹣b2+c2)＝0
∵a2+b2>0
∴a2﹣b2+c2＝0·············································································································8分
即 a2+c2＝b2··············································································································9分
∴△ABC是直角三角形·································································································10分
26.(本题12分)解：(1)m2﹣6m﹣7
＝m2﹣6m+9﹣9﹣7·······································································································1分
＝(m﹣3)2﹣16···········································································································2分
＝(m﹣3+4)(m﹣3﹣4)
＝(m+1)(m﹣7)········································································································4分
(2)2x2+y2﹣8x+6y+20
＝(2x2﹣8x)+y2+6y+9+11
＝2(x2﹣4x+4﹣4)+y2+6y+9+11
＝2(x﹣2)2﹣8+(y+3)2+11
＝2(x﹣2)2+(y+3)2+3································································································6分
∵2(x﹣2)2≥0，(y+3)2≥0，··························································································7分
∴当x＝2，y＝﹣3时，2x2+y2﹣8x+6y+20有最小值，最小值是3············································8分
(3)∵a2+b2＝8a+6b﹣25，
∴a2﹣8a+16+b2﹣6b+9＝0，
∴(a﹣4)2+(b﹣3)2＝0······························································································9分
∴a﹣4＝0，b﹣3＝0，
∴a＝4，b＝3··············································································································10分
∵4﹣3
∴1
∵c为正整数，
∴c最大取6················································································································11分
∴△ABC周长的最大值＝3+4+6＝13，
∴△ABC周长的最大值为13···························································································12分
27.(本题12分)
解：【问题原型】证明：
在△ABC中，点E，F分别为AC，BC的中点
∴EF∥AB，且EF
AB··································································································2分
在Rt△ACD中，点E为AC的中点
∴DE
AC·················································································································4分
∵AB＝AC，
∴DE＝EF···················································································································5分
【探究】解：∵AC平分∠BAD，EF∥AB，DEAC＝AE＝EC
∴∠BAC＝∠DAC，∠CEF＝∠BAC
∠DEC＝2∠DAC＝∠BAD······························································································7分
∵∠DEF＝90°
∴∠CEF+∠DEC＝∠BAC+2∠DAC＝90°
∴∠BAC＝∠DAC＝30°·································································································9分
∴∠BAD＝60°············································································································10分
【应用】四边形ABCD的面积为：
··············································································12
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