南京市建邺区2021-2021学年度第|一学期期(中|考)试八年级|数学试卷
(考试时间100分钟, 试卷总分100分 )
一、选择题 (本大题共6小题, 每题2分, 共12分 ) 1.在下面的四个京剧脸谱中, 不是轴对称图形的是 ( ▲ )
A.
B.
C.
D.
2.以下长度的三条线段能组成直角三角形的是 ( ▲ ) A.1, 2, 3
B.2, 3, 4
C.3, 4, 5
D.5, 6, 7
3.等腰三角形两边长分别为2和4, 那么这个等腰三角形的周长为 ( ▲ ) A.6
B.8
C.10
D.8或10
4.如图, 在数轴上表示实数7+1的点可能是 ( ▲ ) A.P
B.Q
A'
O
C.R
B
D.S
D C
H P Q R S A B'
0 1 2 3 4 5 A C B
(第4题 ) (第5题 ) (第6题 )
5.如图是跷跷板的示意图, 支柱OC与地面垂直, 点O是AB的中点, AB绕着点O上下转动.当A端落地时, ∠OAC=20°, 跷跷板上下可转动的最|大角度 (即∠A′OA )是 ( ▲ ) A.20°
B.40°
C.60°
D.80°
6.如图, 在四边形ABCD中, AB=AC=BD, AC与BD相交于H, 且AC⊥BD.
①AB∥CD;②△ABD≌△BAC;③AB2+CD2=AD2+CB2;④∠ACB+∠BDA=135°.其中真命题的个数是 ( ▲ ) A.1
B.2
C.3
D.4
二、填空题 (本大题共10小题, 每空2分, 共20分 ) 7.5的相反数是 ▲ .
8.一个罐头的质量约 kg, kg可得近似值 ▲ kg.
9.如图, 点A, D, C, F在同一条直线上, AB=DE, ∠B=∠E, 要使△ABC≌△DEF, 还需要添加一个条件是 ▲ .
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B 10.如图, 在Rt△ABC中, CD是斜边AB上的中线, 假设AB=2, 那么CD= ▲ . B
E
B
A D (第9题 )
C F C D D
A A (第10题 )
E F C (第11题 )
11.如图, 在△ABC中, AB=AC, ∠B=66°, D, E分别为AB, BC上一点, AF∥DE, 假设∠BDE
=30°, 那么∠FAC的度数为 ▲ .
12.如图, 一块形如 \"Z〞字形的铁皮, 每个角都是直角, 且AB=BC=EF=GF
=1, CD=DE=GH=AH=3, 现将铁片裁剪并拼接成一个和它等面积的正方形, 那么正方形的边长是 ▲ .
13.如图, △ABC, △ADE均是等腰直角三角形, BC与DE相交于F点, 假设AC
=AE=1, 那么四边形AEFC的周长为 ▲
D
(第12题 )
A
A B C G F
E
B
(第13题 ) E
F C D
B
D (第14题 )
C
H E
A
14.如图, △ABC是边长为6的等边三角形, D是BC上一点, BD=2, DE⊥BC交AB于点E,
那么AE= ▲ .
A ABC的角平分线A 15.如图, 在△ABC中, AB=4, AC=3, BC=5, AD是△, DE⊥AB于点E, 那
么DE长是 ▲ . A
B
D (第15题 )
C C
B
E E P
D
C B
(第16题 )
16.如图, 在△ABC中, ∠C=90°, ∠A=34°, D, E分别为AB, AC上一点, 将△BCD, △
ADE沿CD, DE翻折, 点A, B恰好重合于点P处, 那么∠ACP= ▲ .
三、解答题 (本大题共10题, 共68分 )
17. (6分 )计算
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(1 )(-2)+64-4; (2 )
2
3
9
1+(π-3)0-|1-3|. 16
18. (6分 )求以下各式中的x
(1 )(x+2)2=4; (2 )1+(x-1)3=-7.
19. (6分 )请在以下图中画出三个以AB为腰的等腰△ABC.
(要求:1.锐角三角形, 直角三角形, 钝角三角形各画一个;2.点C在格点上.)
A B
A B
A B
(锐角三角形 ) (直角三角形 )
(钝角三角形 )
20. (6分 )如图, AC⊥BC, BD⊥AD, 垂足分别为C, D, AC=BD.求证BC=AD. D C
A B
(第20题 )
21. (6分 )如图, 在△ABC中, 边AB, AC的垂直平分线相交于点P.求证PB=PC.
P
B
(第21题 )
C
A
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22. (6分 )如图, 点P为△ABC边BC上一点.请用直尺和圆规作一条直线EF, 使得
A关于EF的对称点为P. (保存作图痕迹, 不写作法 )
23. (7分 )如图, 在长方形ABCD中, AB=8, AD=10, 点E为BC上一点, 将△ABE沿AE
折叠, 使点B落在长方形内点F处, 且DF=6, 求BE的长.
A 24. (8分 )如图, 在△ABC中, AB=AC, ∠A=48°, 点D、E、F分别在BC、AB、AC边上,
且BE=CF, BD=CE, 求∠EDF的度数.
B
E
(第24题 )
B
E (第23题 )
F
A
D
A (第22题 )
B
P C C
D F C
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25. (8分 )阅读理解:求107的近似值.
解:设107=10+x, 其中0<x<1, 那么107=(10+x)2, 即107=100+20x+x2.
因为0<x<1, 所以0<x2<1, 所以107≈100+20x, 解之得x≈, 即107的近似值为10.35.
理解应用:利用上面的方法求97的近似值 (结果精确到0.01 ).
26. (9分 )如图, 在四边形ABCD中, AB∥CD, ∠D=90°, 假设AD=3, AB=4, CD=8, 点P
为线段CD上的一动点, 假设△ABP为等腰三角形, 求DP的长.
A B
D (第26题 )
C
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D (备用图3 )
C
A
B
D (备用图2 )
C
A
B
D (备用图1 )
C
A
B
南京市建邺区2021-2021学年度第|一学期期中学情试卷
八年级|数学参考答案及评分标准
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 (每题2分, 共计12分 )
题号 答案 1 D 2 C 3 C 4 B 5 B 6 B
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二、填空题 (每题2分, 共计20分 ) 7.-5. 12.10.
三、解答题 (本大题共10小题, 共计68分 ) 17.(此题6分)
解: (1 )(-2)+64-4
=4+4-2
=6 ······························································································· 3分 (2 )
91+(π-3)0-|1-3| 16
5
= +1-(3-1) 4
13
=-3. ·················································································· 6分
4
18.(此题6分)
解: (1 )x-2=±2 ··················································································· 1分 x=±2+2
x=0, x2=4. ··············································································· 3分
(2 )(x-1)3=-8 ··················································································· 4分 x-1=-2 ··················································································· 5分
x=-1. ····················································································· 6分
19.(此题6分)图略.
20.(此题6分)
证明:∵ AC⊥BC, BD⊥AD, ∴ ∠C=∠D=90°.
在Rt△ABC和Rt△BAD中,
AB=BA
AC=BD.
2
8.. 13.22.
9.BC=EF (答案不惟一 ). 14.2.
10.1. 12
15..
7
11.18. 16.22.
3
∴ Rt△ABC≌Rt△BAD(HL).
∴ BC=AD. ························································································· 6分
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21.(此题6分)
证明:∵ 边AB, AC的垂直平分线相交于点P, ∴ PA=PB, PA=PC.
∴ PB=PC. ················································································ 6分
22.(此题6分)图略.
23.(此题7分)
解:∵ 将△ABE沿AE折叠, 使点B落在长方形内点F处,
∴ ∠AFE=∠B=90°, AB=AF=8, BE=FE. 在△ADF中, ∵ AF2+DF2 =62+82 =100 =102=AD2,
B
E
F
C
A D
∴ △ADF是直角三角形, ∠AFD=90°. ····················································· 3分 ∴ D, F, E在一条直线上. ····································································· 4分 设BE=x, 那么EF=x, DE=6+x, EC=10-x, 在Rt△DCE中, ∠C=90°, ∴ CE2+CD2=DE2, 即 (10-x) 2+82=(6+x) 2. ∴ x=4.
∴ BE=4. ··························································································· 7分
24.(此题8分)
(1 )证明:∵ AB=AC, ∠A=48°,
∴ ∠B=∠C=(180°-48°)÷2=66°. ················································ 2分 在△DBE和△ECF中,
A D BD=CE
∠B=∠C BE=CF.
∴ ∠FEC=∠BDE,
∴ ∠DEF=180°-∠BED-∠FEC
F ∴ △DBE≌△ECF(SAS). ································································· 4分
B
E
C
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=180°-∠DEB-∠EDB=∠B=66°. ············································· 6分 ∵ △DBE≌△ECF(SAS),
∴ DE=FE.∴△DEF是等腰三角形.
∴ ∠EDF =(180°-66°)÷2=57°. ···················································· 8分
25.(此题8分)
解:设97=10-x, 其中0<x<1, 那么97=(10-x)2, 即97=100-20x+x2.
因为0<x<1, 所以0<x2<1,
所以97≈100-20x, 解之得x≈0.15, 即97的近似值为9.85. ··························· 8分 (设97=9+x, 求出97的近似值为9.89也给总分值.)
26.(此题9分)
解:①AB=AP时, DP1=4-3=7; ·························································· 2分
11②BP=AP时, DP2=AB=×4=2; ························································· 4分
22③BA=BP时, 过点B作BH⊥CD于H, 那么BH=AD=3, 由勾股定理得, FP=4-3=7, DP3=4-7, 或者DP4=4+7.
综上所述, DP的值为7, 2, 4-7, 或者4+7. ········································ 9分
A
B
2222D P3 P2 P1 H P4 C
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