姓名
准考证号
4.已知2024-号0>2024-？6，则一定有口6.“口中应填的符号是
2023一2024学年第二学期期中学业水平质量监测
A.≤
B.>
C.>
D.<
八年级数学
5.不等式组：+3>0，的解集在数轴上表示正确的是
2-x≥1
注意事项：
-3-2-101
-3-2-10
3-2-10
-201
1.本试卷共8页，满分100分，考这时间120分钟
A
B
D
2答卷前，考生务必将自己的址名、准考证号填写在本试春相应的位置
3.签案全邮在答题卡上定成，答在本议卷上无效」
6.已知△ABC中，AB=AC,求证：∠B<90°，下面写出了运用反证法证明这个命题的四
4.考试结桌后，将本卷和答避一并文。
个步深：
①所以LA+∠B+∠C>180°，这与三角形内角和等于180°矛盾：
第1卷选择题（共30分）
②因此假设不成立，所以∠B<90°；
一、选择题（本大题共10个小题，每小题3分，共30分，在每个小题给出的四个选项中，
③假设在△ABC中，∠B>90°：
只有一项符合题目要求，请选出并在答题卡上将该项涂黑)
④由AB=AC,得∠B=∠C≥90°，即∠B+∠C,180
1.下列运动现象属于平移的是
()
这四个步骤正确的顺序应是
A.0①2③④
B.③④2D
C.3③④12
D.④8D2
7.如图2，一次函数y,=红+4与=x+b的图象相交于点P(1,3),魂察可得关于x的
不等式x+b>x+4的解集是
(
A.x≤1
B.x<1
2.下列感冒胶囊的标识图中，属于中心对称图形的是
C.x≥3
D.x21
,=x+6
3.山西万荣东岳庙的飞云楼的建筑风格是典型的重檐歇
-201
y wkx+4
山式，如图1-①所示，飞云楼的顶端可以近似看作是
2
因3
等腰三角形ABC(如图I-②)，其中AB=AC,AD是BC
8.如图3，△DEF是由△ABC绕点O顺时针旋转得到的，以下说法不一定正确的是
(
边上的中线，已知∠B=61°，则∠BAD的度数为()
A.61
B.58
⊙
A.∠COF=∠BOE
B.OA=OD
D.AB=DE
C.39
D.29
C.AC=EF
八年级数学(BS)第I页（共8页）
八年级数学(BS)第2页（共8页）
9.解决好老百姓的操心事、须心事，是政府一定要办好的实事.在2024年太原市政府工
作报告中，提出今年太原市在承接好省民生实事的基础上，再全力办好10件民生实
三、解答题（本大题共8个小题，共55分.解答应写出文字说明，证明过程或演算步骤】
事，其中将新建二类以上公厕100座（合一类公周和二类公厕）若新建的一类公厕的
16.(5分)解不等式组：
3x-2<7,
数量不低于二类公厕的，则一类公期最少要建的数量（座）满足的不等式为(）
2(x-1)63x+1
A.2>100-x
B.x≥号(100-)
c>100-x
D.x>3(100-x)
10.如图4，在R1△ABC中，∠C=90°，∠BAC的平分线交BC于
点D,DE∥AB,交AC于点E,DF⊥AB于点F,DE=1O,DF=
6,则AC的长为
(
A.13
B.14
C.16
D.18
第Ⅱ卷非选择题（共70分）
4
二、填空题（本大题共5个小题，每小题3分，共15分）
11.命题“直角三角形的两个锐角互余”的逆命题是
12.如图5，把一个三角形纸板的一边紧靠数轴平移，点P平移的距离PP为
17.(5分)如图9.用三角板可按下面的方法同角平分线，在已知的∠AOB的两边上分别
取OM=ON,再分别过点M,N作OA,OB的垂线，交点为P,画射线OP,则OP平分
∠AOB,为什么？
-3-2-0123
图5
日6
13.如图6，将一块含45"角的三角板ABC绕点A按逆时针方向旋转到△AB'C的位置
若∠CAB=15°，则旋转的角度为
14.如图7，在△ABC中，∠C=90°，AB的垂直平分线分别交AB,AC于点D,E,AE=13,
CE=5,则线段CB的长为
15.如图8-①，是一块光学直角棱镜，其截面为图
8-②所示的Rt△ABC,AB所在的面为不透光
的磨砂面，∠ACB=90°，∠A=30°，现有一束单
色光从CB边的点E处垂直射入，到达AB边的
点D,恰有CD⊥AB,经过反射后（即∠CDE=
∠FDC)从AC边的点F处射出.若光线在棱镜内
部经过的路径ED+DF=12 m,则这块棱镜的
高度AC为
cm.
八年级数学(BS)第3页（共8页）
八年级数学(BS)第4页（共8页）2023—2024 学年第二学期期中学业水平质量监测
八年级数学（BS）参考答案及评分标准
一、选择题（本大题共 10 个小题，每小题 3 分，共 30 分）
题 号 1 2 3 4 5 6 7 8 9 10
答 案 B C D D A C D C B D
二、填空题（本大题共 5 个小题，每小题 3 分，共 15 分）
11. 如果一个三角形的两个角互余，那么这个三角形是直角三角形
12. 5 13. 30°
14. 12 15. 16
三、解答题（本大题共 8 个小题，共 55 分）
3x 2 7, ①16.（5 分）解：
2 x 1 3x 1.②
解不等式①，得 x＜3. ··············································································································2 分
解不等式②，得 x≥-3. ············································································································· 4 分
所以原不等式组的解集是-3≤x＜3. ······························································································ 5 分
17. （5 分）解：因为 OM⊥MP，ON⊥NP，所以∠OMP=∠ONP=90°.················································ 1 分
在 Rt△OMP 和 Rt△ONP 中，OM=ON，OP=OP，所以 Rt△OMP≌Rt△ONP（HL）.·································· 3 分
所以∠MOP=∠NOP.················································································································· 4 分
所以 OP 平分∠AOB．·············································································································· 5 分
18.（5 分） 解：（1）三 ········································································································ 1 分
移项没有改变符号 ············································································································ 2 分
（2）不等式的基本性质 3（或填不等式的两边都乘（或除以）同一个负数，不等号的方向改变） ········· 3 分
17
（3）x＜- ·························································································································· 5 分
5
19.（5 分）解：（1）①如图 1，△A 1B 1C1 为所求作 .································································· 1 分
图 1
②减 2 ································································································································· 2 分
加 6······································································································································ 3 分
（2）①如图 1，△A 2B2C2 为所求作 .······················································································· 4 分
②（ -1，3） ··························································································································5 分
20.（6 分）证明：因为 BC=BD，所以∠BCD=∠BDC.·····································································1 分
1
因为∠DBC=30°，所以∠BDC=∠BCD= ×（180°-30°）=75°.························································· 2 分
2
因为∠ACB=45°，所以∠DOC=30°+45°=75°．.……………………………………………………………………4 分
所以∠DOC=∠BDC.所以 CD=CO.··························································································· 5 分
所以△CDO 是等腰三角形．····································································································· 6 分
21.（8 分）解：（1）由题意，得 y1=100+（x-100）×90%=0.9x+10. ················································· 2 分
y2=50+（x-50）×95%=0.95x+2.5.································································································· 4 分
（2）因为小红选择方案二花钱少，所以 y1＞y2.
所以 0.9x+10＞0.95x+2.5.··········································································································· 6 分
解得 x＜150.····························································································································7 分
所以当 100＜x＜150 时，小红选择方案二花钱少.············································································8 分
22.（9 分）（ 1）证明：因为△ABC 是等边三角形，所以∠A=∠C=∠B=60°. ···························1 分
因为 MN∥AC，所以∠BMN=∠C=60°，∠BNM=∠A=60°.······················································2 分
又因为∠B=60°，所以△BMN 是等边三角形.
所以 BM=BN. ························································································································ 3 分
（2）①15 ····························································································································5 分
②证明：因为△ABC 是等边三角形，所以 AB=BC.又因为 BM=BN，所以 AN=MC．························· 6 分
1
因为 CH 是∠ACD 的平分线，∠ACB=60°，所以∠ACH=∠HCD= （180°-∠ACB）=60°.
2
所以∠MCH=120°. ·············································································································· 7 分
又因为∠ANM=180°-∠BNM=120°，所以∠ANM＝∠MCH.
因为∠AMC=∠AMH+∠HMC=∠MAN+∠B，∠AMH=60°，∠B=60°，所以∠HMC=∠MAN.················· 8 分
在△ANM 和△MCH 中，∠MAN＝∠HMC，AN＝MC，∠ANM＝∠MCH，所以△ANM≌△MCH（ASA）.
············································································································································ 9 分
23.（12 分）证明：（1）因为△ABC 和△DEF 均为等腰直角三角形，所以 AB=AC，AD=AE，∠BAC=∠DAE=90°.
··········································································································································· 1 分
所以∠BAC-∠DAC=∠DAE-∠DAC，即∠BAD=∠CAE．······························································ 2 分
所以△ABD≌△ACE（SAS）. ···································································································3 分
（2）135° ····························································································································· 5 分
（3）证明：同（1）可得 AD=AE，∠DAE=90°，△ABD≌△ACE. ···················································6 分
所以 DE= AD2 AE 2 = 2 AD，BD=CE．·················································································7 分
所以 BD=CE=CD+DE=CD+ 2 AD．··························································································· 8 分
（4）证明：如图 2，过点 A 作 AM⊥BD，AN⊥CE，垂足分别为 M，N. ···········································9 分
因为△ABC 和△DEF 都是等腰直角三角形，所以 AB=AC，AD=AE，∠BAC=∠DAE=90°.
所以∠BAC+∠DAC=∠DAE+∠DAC，即∠BAD=∠CAE．所以△ABD≌△ACE（SAS）．·················· 10 分
所以 BD=CE，S△ABD=S△ACE．····································································································11 分
1 BD AM 1所以 CE AN .所以 AM=AN．又因为 AM⊥BD，AN⊥CE，所以 PA 平分∠EPB．··········12 分
2 2
图 2
