2026年南京市联合体八上数学期中试卷含答案

2025～2026学年度第一学期期中学情分析样题

八年级数学

注意事项：

1．本试卷共6页．全卷满分100分．考试时间为100分钟．考生答题全部答在答题卡上，

答在本试卷上无效．

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在其他位置答题一律无效．

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一、选择题（本大题共8小题，每小题2分，共16分．在每小题所给出的四个选项中，恰有

一项是符合题目要求的，请将正确选项前的字母代号填涂在答题卡．．．相应位置．．．．上）

1．下列各数中，是无理数的是

1

A．2．．D．

B2C42

2．下列长度的三条线段能组成三角形的是

A．2，3，5B．4，4，8C．5，6，12D．9，9，16

3．下列说法正确的是

A．形状相同的两个三角形全等B．面积相等的两个三角形全等

C．周长相等的两个三角形全等D．全等三角形的对应边相等

4．一个正方形的面积是12，估计它的边长大小在

A．2与3之间B．3与4之间C．4与5之间D．5与6之间

5．如图，在△ABC中，∠C＝90°，BD平分∠ABC，AC＝8，AD＝5，则点D到AB的距离为

A．2B．3C．4D．5

A

A

CP

D

ODB

BC

（第5题）（第6题）

6．用直尺和圆规作一个角的平分线的示意图如图所示，则能说明∠AOP＝∠BOP的依据是

A．SSSB．ASA

C．SASD．AAS

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7．如图，∠AOB＝60°，OA＝OB，动点C从点O出发，沿射线OB方向移动，以AC为边在

右侧作等边△ACD，连接BD，则BD所在直线与OA所在直线的位置关系是

A．相交B．垂直C．平行D．无法确定

8．如图，CD是△ABC的外角∠BCE的平分线，DA＝DB，DE⊥AC，垂足为E．若BC＝10，

AC＝4，则CE的长为

A．1.5B．2C．2.5AD．3

A

BC

DE

OCB

D

（第7题）（第8题）

二、填空题（本大题共10小题，每小题2分，共20分．请把答案填写在答题卡相应位置．．．．．．．上）

9．25的算术平方根是▲．

10．小明用天平称得一个罐头的质量为2.163kg，将2.163精确到0.1是▲.

11．等腰三角形的一个内角为100°，则它的底角的度数为▲°．

5－11

12．比较大小：▲．（填“＞”“＝”或“＜”号）

22

13．若a是无理数，且2＜a＜3，写出一个a的值：▲.

14．如图，在Rt△ABC中，∠C＝90°，D为AB的中点，CD＝2，则AB的长为▲．

A

BB

E

DE

A

D

CACBDC

（第题）

（第14题）15（第16题）

15．如图，AB＝AC，要证明△ABD≌△ACE，还需添加一个条件：▲．

16．如图，在△ABC中，AC的垂直平分线分别交BC，AC于点D，E，连接AD，若△ABC

的周长为22，△ADB的周长为15，则CE的长为▲．

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17．如图，在△ABC中，点D在AC上，BD平分∠ABC，AB＋AD＝BC．若∠A＝88°，则

∠ABD＝▲°．

A

A

C

DE

D

BCOB

（第17题）（第18题）

18．已知等腰直角三角形的底边长是其腰长的2倍．如图，△OAB和△OCD是等腰直角三

角形，∠AOB＝∠COD＝90°，点D在AB上，E是CD的中点，连接AE．若OB＝2，

则AE的长的最小值为▲．

三、解答题（本大题共8小题，共64分．请在答题卡指定区域．．．．．．．内作答，解答时应写出文字说

明、证明过程或演算步骤）

19．（8分）计算：

3

（1）9＋－27；（2）(－2)2－364＋(3)2．

20．（8分）求下列各式中的x．

（1）3x2＝48；（2）8(x＋1)3－27＝0．

21．（8分）如图，点A，F，C，D在同一直线上，点B，E分别在直线AD的两侧，且AB＝

DE，∠A＝∠D，AF＝DC．

B

求证：△ABC≌△DEF．

D

AFC

E

（第21题）

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22．（8分）证明：等腰三角形两腰上的中线相等．

A

已知：如图，在△ABC中，AB＝AC，▲．

求证：▲．

FE

证明：

BC

（第22题）

23．（6分）已知a＞b＞0，比较a与b的大小，说明理由．

24．（8分）如图，在△ABC中，AD是边BC上的高，BE是边AC上的中线，BD＝CE，DF⊥BE

于点F．

（1）求证：BF＝EF；

A

（2）若∠AEB＝72°，求∠EBC的度数．

E

F

BDC

（第24题）

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25．（9分）尺规作图：根据要求补全图形．（保留作图痕迹，写出必要的文字说明）

（1）在图①中，作△DEF，使△DEF≌△ABC．

A

BCEF

①

（2）在图②中，作等腰三角形△ABC，使AB＝2BC．

AB

②

（3）在图③中，作直角三角形△ABC，使∠C＝90°，且AB＝2BC．

AB

③

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26．（9分）

（1）如图①，在等边三角形ABC中，点D在BC上，CD的垂直平分线交BA的延长线于

E

点E，连接ED，EC，DE交AC于点F．

E

A

FA

BC

DBC(D)

①②

【特殊化】

(Ⅰ)当点D与点C重合时，如图②，直接写出AE与BD的数量关系．

【一般化】

(Ⅱ)当点D与点C不重合时，如图①，判断AE与BD的数量关系，并说明理由．

【应用】

（2）如图③，ED＝EC，点A在△DEC外，∠DAE＝120°，∠ADC＝60°，AD交CE于点F，

若AE＝CD，直接写出AF与DF的数量关系．

E

A

F

DC

③

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2025～2026学年度第一学期期末学情分析样题

八年级数学参考答案

说明：本评分标准每题给出了一种或几种解法供参考．如果考生的解法与本解答不同，参照本

评分标准的精神给分．

一、选择题（本大题共8小题，每小题2分，共16分）

题号12345678

答案BDDBBACD

二、填空题（每小题2分，共20分）

9．510．2.211．4012．＞13．5(答案不唯一)

2

18．

14．415．AD＝AE16．3.517．242

三、解答题（本大题共8小题，共64分）

19．（8分）

（1）解原式＝3＋(－3)··························································································2分

＝0··········································································································4分

（2）解原式＝2－4＋3···························································································7分

＝1··········································································································8分

20．（8分）

（1）解：x2＝16，·······································································································2分

x＝±4．······································································································4分

27

（2）解：(x＋1)3＝，······························································································5分

8

3

(x＋1)＝．··································································································7分

2

1

x＝．··········································································································8分

2

21．（8分）证明：

∵AF＝DC，

∴AF＋FC＝DC＋FC，即AC＝DF．··································································3分

在△ABC和△DEF中，··························································································4分

AC＝DF，

∠A＝∠D，········································································································7分

AB＝DE．

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∴△ABC≌△DEF(SAS)························································································8分

22．（8分）

E、F分别是AC、AB的中点，···················································································1分

BE＝CF．·····················································································································2分

证法1：∵E、F分别是AC、AB的中点，

11

∴BF＝AB，CE＝AC．···························································································3分

22

∵AB＝AC，

∴∠FBC＝∠ECB，·································································································4分

BF＝CE．··············································································································5分

在△FBC和△ECB中，

BF＝CE，

∠FBC＝∠ECB，

BC＝BC．

∴△FBC≌△ECB(SAS)·························································································7分

∴BE＝CF．··········································································································8分

证法2：∵E、F分别是AC、AB的中点，

11

∴AF＝AB，AE＝AC．···························································································3分

22

∵AB＝AC，

∴AF＝AE．················································································································5分

在△AEB和△AFC中，

AF＝AE，

∠A＝∠A，

AB＝AC．

∴△AEB≌△AFC(SAS)·························································································7分

∴BE＝CF．··········································································································8分

23．（6分）

解：a＞b；理由如下·························································································2分

方法1说理：将a理解成面积为a的正方形边长，b理解成面积为b的正方形边长．

················································································································4分

∵a＞b＞0，

∴面积为a的正方形可以放置在面积为b的正方形内，如图所示．················5分

∴由图可得：a＞b．········································································6分

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方法2说理：∵(a)2＝a，(b)2＝b，a＞b＞0

∴(a)2＞(b)2······················································································3分

∴(a)2－(b)2＞0

∵根据乘法公式：(a＋b)(a－b)＝(a)2－(b)2

∴(a＋b)(a－b)＞0·····································································5分

又(a＋b)＞0

∴(a－b)＞0，即a＞b；····························································6分

24．（8分）

（1）∵AD是边BC上的高，BE是边AC上的中线，

∴∠ADC＝90°，AE＝CE．··············································································1分

1

∴DE＝AC＝CE．······························································································2分

2

又BD＝CE，

∴BD＝DE．········································································································3分

又DF⊥BE，

∴BF＝EF．·········································································································4分

（2）设∠EBC＝α．

∵BD＝DE，

∴∠EBC＝∠DEB＝α．······················································································5分

∴∠EDC＝∠EBC＋∠DEB＝2α．

∵DE＝EC，

∴∠EDC＝∠C＝2α．·························································································6分

∴∠AEB＝∠EBC＋∠C＝3α．···········································································7分

又∠AEB＝72°，即3α＝72°，

∴α＝24°，即∠EBC＝24°．···············································································8分

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25．（9分）

（1）

AD

B

CEF

①

········································································································································3分

（2）C

AB

②

········································································································································6分

（3）

AB

③

········································································································································9分

26．（9分）

（1）

(Ⅰ)AE＝BD．···············································································································2分

(Ⅱ)AE＝BD．···············································································································3分

证法1：在BE上截取BG＝BD．

∵△ABC是等边三角形，E

∴∠B＝∠BAC＝∠ACB＝60°，

A

∴△BGD是等边三角形．F

数学试卷第10页（共6页）

G

BDC