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2025—2026 学年度第二学期期中适应性练习
八年级数学参考答案
一、 选择题（共 10小题，每小题 4分）
1. A 2. C 3. B 4. C 5. C 6. B 7. A 8. D 9. D 10. B
二、 填空题（共 6小题，每小题 4分，满分 24分）
11. 2 12. 540 13. 13
14. 8 15. 12 16. 5 3 5
5
三、解答题（共 9小题，满分 86分）
17.(8 分)
解：(1) 原式= 2 2 3 2 2 ................................................................................2‘
=0........................................................................................................ 4‘
(2) 原式= 18 5 2 ................................................................................................. 6‘
3
= ...............................................................................................................8‘
5
18.(8 分)
解：∵ 四边形 ABCD是平行四边形
∴ AB∥CD，AO=CO ............................................................................................. 2'
A E B
∴ ∠EAO=∠FCO.................................................................................................... 3'
∵ ∠AOE=∠COF..........................................................................O.......................... 4'
∴ △AEO≌△CFO（ASA）.....................................D................F......C........................6'
∴ AE=CF...................................................................................................................8'
19.(8 分)
解：原式= m 2 2(m 1) .......................................................................................2 4'm 1 (m 2)
= 2 ........................................................................................................ 6'
m 2
当m 2 2时
原式= 2 2 .............................................................................................. 7'
2 2 2 2
2 ...........................................................................................................................8'
20.(8 分)
A
解：依题意，得： ACB 90 ,CD 1.5米
∴ 在 Rt ACB中, AC AB2 BC 2 172 152 8 ............................................6'
B
∴ 在 AD=AC+CD=8+1.5=9.5 米 C
∴ 风筝距离地面的垂直高度为 9.5 米.........................................................D......... 8'
21.(8 分)
(1) 1 1 1 1 1 1 2 2 1 1 . . . . . . . . . . . . 3'4 5 4 5 20
(2)解：原式=1 1 1 1 1 1 1 1 1 1 1 1 .................................... 5'
2 2 3 3 4 9 10
10 1 .................................................................................................7'
10
9 9 ......................................................................................................8'
10
22.(10 分)
解：ADCF为矩形，理由如下：
∵ AF ∥ BC
∴ ∠AFE = ∠DBE ..................................................A..............F.................................1'
∵ E 是 AD中点
∴ AE = DE ..................................................................E............................................2'
∵ ∠AEF = ∠DEB .................................................................................................3'
B C
∴ △AEF ≌ △DEB (AAS) .....................................
D..............................................4'
∴ AF = DB ..............................................................................................................5'
∵ D是 BC中点
∴ BD = CD ..............................................................................................................6'
∴ AF∥ CD ............................................................................................................. 7'
∴ ADCF为平行四边形 ..........................................................................................8'
∵ AB = AC
∴∠ADC=90°.........................................................................................................9'
∴ 平行四边形 ADCF为矩形..................................................................................10'
23. (10 分)
解：(1)作出菱形 ABCD如图所示........................................................................... 5'
（做图略）
（作垂直平分线得 2分，找出点B、D各得一分，回答得一分）
(2) 设 AC、BD交于点O
∵ ABCD是菱形
∴ AO
1
AC 3 3 , AOB 90 ......................................................................6'
2
∵ ∠MAN=30°
∴ AB = 2OB
在 △ AOB中，AB2=OB2+AO2
∴ BO=3，AB=6
∴ BD=2OB=6 .........................................................................................................8'
∴ S
1 1
AC BD 6 3 6 18 3 ................................................................10'
2 2
24. (12 分)
(1)∵ AB2+BC2=92+122=225................................................................................1'
AC2=152=225........................................................................................................... 2'
∴AB2+BC2=AC2
∴∠ABC=90°..........................................................................................................3'
(2)∵ AD2+AC2=82+152=289
CD2=172=289
∴AD2+AC2=CD2
∴∠DAC=90°.........................................................................................................4'
街 D
∴ S阴 S S
E
ABC DAC 道
G
1
AB BC 1 AD AC
2 2 A H
1 9 12 1 8 15
2 2 F
114m2 ....................................................B...............................C.....街......道........................ 6'
∴费用为114 100 11400（元）............................................................................7'
(3) 方案一：15+13=28m
28×50=1400（元）...................................................................................................8'
方案二：设 HF=x，则 EH=14-x
∴152- (14-x)2=132-x2................................................................................................9'
解得 x=5
∴GH EG 2 EH 2 12m .....................................................................................10'
∴费用为50 (12 14) 1300（元）.......................................................................11'
∴选择方案二.............................................................................................................12'
25. (14 分)
(1)证：∵ ABCD是正方形
∴ AD=DC,∠ADE = ∠C=90° ............................................................................1'
∴ ∠DAE + ∠AED=90°
∵MN AE
∴ ∠CDN + ∠AED=90°
∴ ∠DAE = ∠CDN ................................................................................................ 2'
∴ △ADE≌△DCN（ASA）.................................................................................. 3'
∴ DE=CN...................................................................................................................4'
(2)连接 AG、CG、EG
∵ MN AE，F为 AE中点
∴ MN为 AE的垂直平分线
∴ AG = EG ...........................................................A....................M...............D.................5'
∵AD=CD F
E
∴ ∠ADG = ∠CDG=45°，DG=DG
G
∴ △ADG≌△CDG（SAS）
∴ AG=CG,∠DAG = ∠DCG...............................B.........N............................C................6'
∴ CG = EG
图 2
∴ ∠GEC = ∠DCG = ∠DAG
∵ ∠DEG + ∠GEC=180°
∴ ∠DAG + ∠DEG=180°....................................................................................7'
∴ 在四边形AGED中∠AGE=90°..................................................................... 8'
∴ FG 1 AE 1 3 AD2 DE 2 ..........................................................................10'
2 2 2
（单独计算出AE=3 得 1分）
(3)在 BC上截取BH=BN，连接EH
∵ P为 NE中点
BP 1∴ HE .........................................................................................................12'
2 A（M） D
同理（1）可得BN=DE
E
∴ BH = DE
P
∴ CE = CH
∴ EH 2CE ,即CE 2 HE ....................N..........B..........H......................
C................13'
2 图 3
1
BP HE∴ 2 2 .............................................................................................. 14'
CE 2 2HE
2
法二：HC=EC= 2 2 1
∴ EH 2 2 2 1 4 2
∴ BP 1 2 EH 2
2 2
2 2
∴ BP 2 4 2 14 2 2
EC 2 2 1 4 2 2 28 2

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