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4.数学活动课上，小明将一根14厘米长的木棒截，成三段，首尾相连围成一个等腰三角
怀仁市2024一2025学年度第一学期八年级期中学业质量监测
形.如图，如果第一次在4厘米处（剪刀处）截断，那么第二次可以在(
)处戳断
A.①或②
B.②或③
数学
C.①或
D.③或④
①
②③
④
注意事项：
1.试卷分第I卷和第Ⅱ卷两部分.全卷共8页，满分120分，考试时间120分钟
2答案全部在答题卡上完成，答在本试卷上无效、
第4题图
第5通图
5,如图，将透明直尺叠放在正五边形上，若正五边形有两个顶点恰好落在直尺的边
第I卷选择题（共30分）
上，且L2=52°，则∠1的度数为
A.35
B.30
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，
C.20°
D.10°
只有一项符合题目要求，请选出并在答题卡上将该项涂黑】
6.放风筝在中国有者悠久的历史，是中国文化、传统习俗的重要组成部分.在校内劳
1.在2024年巴黎奥运会上，中国代表团取得了优异成绩，下列巴黎奥运会项目的图标
动课上，小明所在小组的同学们设计了如图所示的风筝框架.已知∠B=∠E,AB=
中，在文字上方的图标是轴对称图形的是
DE,BF=EC,△ABC的周长为24cm,FC=3cm,则制作该风筝框架所需材料的总长
度至少为
A.44 cm
B.45 cm
C.46cm
D.48 cm
羽毛球
田径
跳水
足球
&
C
D
2.小林在家打扫卫生，为方便拆取窗帘拿来一个人字模，在打开梯
子时发现中间有一根拉杆（如图），这样设计所蕴含的数学道理是
A.两点之间，线段最短
D
B.两点确定一条直线
第6题图
第7题园
第8照图
7.如图，在△ABC中，AB=AC,D,E分别是AB,AG上的点，要使△ABE=△ACD,下列
C.三角形具有稳定性
第2随图
条件能作为补充条件的是
D.三角形两边之和大于第三边
A.∠ABE=∠ACD
B.∠A=∠A
3.下列四位同学所画的△ABC中BC边上的高AE,正确的是
C.∠1BC=∠ACB
D.BE =CD
8.如图，在△ABC中，AB=AG,∠BAC=108°，在BC上分别取点D,E,使∠BAD=∠B,
LCAE=∠C,图中等腰三角形共有
A.3个
B.4个
C.5个
D.6个
八年级数学第1页（共8页）
八年级数学第2页（共8页）2024—2025 学年度第一学期八年级期中学业质量监测
数学试题参考答案及评分建议
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 D C C B C B A D A B
二、填空题（本大题共 5 个小题，每小题 3 分，共 15 分）
9
11. 55° 12. 4 13. 240 14. 3 15.
2
三、解答题（本大题共 8 个小题，共 75 分）
16.（本题 8 分）
解：AC=EF.证明如下： ···································································································· 2 分
∵AB∥DE，∴∠B=∠D. ····································································································· 3 分
∵BF=CD，∴ BF-CF=CD-CF，即 BC=DF. ············································································· 4 分
在△ABC 和△EDF 中，AB=ED，∠B=∠D，BC=DF，∴△ABC≌△EDF. ····································· 6 分
∴AC=EF. ························································································································ 8 分
17. （本题 7 分）
解：∵∠ABM 是△BCM 的外角，∴∠ABM=∠M+∠ACM. ························································ 2 分
∵∠ABM=2∠ACM，∴∠M+∠ACM=2∠ACM.
∴∠M=∠ACM. ·················································································································4 分
∴BM=BC. ·······················································································································5 分
∴只要测得∠ABM=2∠ACM 及 BC 的长，就可得到 BM 的长.····················································· 7 分
18.（本题 9 分）
解：由题可知 CD⊥DB，AB⊥DB，∴∠CDP=90°，∠ABP=90°. ················································· 1 分
∵∠APB＝72°，∴∠BAP=90°-∠APB=90°-72°=18°. ································································· 2 分
∵∠DPC＝18°，∴∠BAP=∠DPC. ························································································ 3 分
∵PB=9 m，CD=9 m，∴PB=CD. ·························································································· 4 分
在△PBA 和△CDP 中，∠BAP=∠DPC，∠PBA=∠CDP，PB=CD，∴△PBA≌△CDP. ···················· 6 分
∴AB=PD. ······················································································································· 7 分
∵DB=36 m，PB=9 m，∴PD=DB-PB=36-9=27 m. ····································································· 8 分
∴AB=27 m.
答：楼 AB 高为 27 m. ········································································································· 9 分
19.（本题 8 分）
解：∵AD，CE 为△ABC 的两条高，∴AD⊥BC，CE⊥AB.
∴∠ADC=90°，∠AEC=90°. ································································································· 1 分
在 Rt△CDA 和 Rt△AEC 中，AC=CA，AD=CE，∴Rt△CDA≌Rt△AEC. ······································· 4 分
∴∠ACD=∠EAC. ············································································································· 5 分
∵∠ACD=65°，∴∠EAC=65°. ······························································································ 6 分
∴∠ACE=90°-∠EAC=90°-65°=25°. ························································································ 7 分
第 1 页（共 4 页）
{#{QQABZYqAogCAABAAAQgCQQVgCkGQkgGACQgOBBAIIAABCQFABAA=}#}
∴∠OCD=∠ACD-∠ACE=65°-25°=40°. ·················································································· 8 分
20.（本题 10 分）
1
解：BE= CD ·················································································································· 2 分
2
∵CD 平分∠ACB，∴∠BCE=∠ACE. ···················································································· 3 分
∵BE⊥CD，∴∠BEC=∠FEC=90°. ······················································································ 4 分
在△BEC 和△FEC 中，∠BCE=∠ACE，CE=CE，∠BEC=∠FEC，∴△BEC≌△FEC.
1
∴BE=EF= BF. ··············································································································· 5 分
2
∵∠BAC=90°，∴∠BAF=90°. ∴∠FBA+∠F=90°.
∵∠BEC=90°，∴∠ECF+∠F=90°.
∴∠FBA=∠ECF，即∠FBA=∠DCA. ····················································································7 分
在△ABF 和△ACD 中，∠FBA=∠DCA，AB=AC，∠BAF=∠CAD，∴△ABF≌△ACD. ················· 8 分
∴BF=CD. ······················································································································ 9 分
1
∴BE= CD. ·················································································································· 10 分
2
21.（本题 8 分）
解：（1）如图：
线段 A1B1即为所求； ········································································································· 2 分
（说明：画对线段得 1 分，正确标记两个顶点字母再得 1 分，不写结论不扣分）
(2) (-m，n)； ···················································································································· 4 分
(3) (4，6) ， (-2，2) ，(0，-1)，(6，3) ················································································· 8 分
(每写对 1 个得 1 分，有错写的一共扣 1 分）
解析：满足条件的点 C 如图中点 C1，C2，C3，C4.
第 2 页（共 4 页）
{#{QQABZYqAogCAABAAAQgCQQVgCkGQkgGACQgOBBAIIAABCQFABAA=}#}
22.（本题 14 分）
解：（1）D ······················································································································· 3 分
（2）理由如下：
AB AD，
在△ABC 和△ADC 中， BC DC，
AC AC，
∴△ABC≌△ADC. ············································································································ 5 分
∴∠BAC=∠DAC.
∴沿 AC 画一条射线 AE，则 AE 就是∠PRQ 的平分线. ····························································· 6 分
（3）①角的内部到角的两边的距离相等的点在角的平分线上 ····················································· 9 分
②∵点 A，B，C 在一条直线上，∠CBM=90°，
∴∠ABM=180°-∠CBM=90°，∴∠ABM=∠CBM. ·································································· 10 分
∵BM 所在直线过∠EOF 的顶点 O，∴∠ABO=∠CBO.
在△ABO 和△CBO 中，
AB CB，
ABO CBO，
OB OB，
∴△ABO≌△CBO（SAS）. ······························································································ 12 分
∴∠COB=∠AOB.
又∵点 C 在 OE 上，∴∠EOB=∠AOB.
∴∠EOB=∠AOB=∠AOD .
∴射线 OA 和射线 OB 将∠EOF 三等分. ··············································································· 14 分
23.（本题 11 分）
解：（1）16 ······················································································································· 2 分
（2）如图，连接 AD. ··········································································································· 3 分
第 3 页（共 4 页）
{#{QQABZYqAogCAABAAAQgCQQVgCkGQkgGACQgOBBAIIAABCQFABAA=}#}
∵△ABC 是等腰直角三角形，AB=AC，∠BAC=90°，∴∠B=∠C=45°. ·········································· 4 分
∵D 是 BC 中点，即 AD 是 BC 边上的中线，∴AD⊥BC，BD=CD.
∴∠ADB=∠ADC=90°.
∴∠B=∠BAD=∠C=∠CAD=45°.
∴AD=BD=CD. ····················································································································· 5 分
∵∠FDE=∠ADC=90°，∴∠FDE-∠ADE=∠ADC-∠ADE，即∠ADM=∠CDN.································· 6 分
∴△ADM≌△CDN（ASA）. ································································································ 7 分
1 1 1
∴ 2S 重叠=S△ADC= S△ABC= × ×8×8=16 cm . ·········································································· 8 分
2 2 2
（3）BF+AF=EF. ················································································································ 11 分
第 4 页（共 4 页）
{#{QQABZYqAogCAABAAAQgCQQVgCkGQkgGACQgOBBAIIAABCQFABAA=}#}
