2023年12月2日发(作者：湖南对口2020数学试卷)
2020-2021学年度第一学期期中学情分析样题
八年级数学
（考试时间100分钟，试卷总分100分）
一、选择题（每小题2分，共16分）
1．下列图案中，不是轴对称图形的是
．．
A． B． C． D．
2．下列说法正确的是
A．全等三角形是指形状相同的两个三角形
B．全等三角形的周长和面积分别相等
C．全等三角形是指面积相等的两个三角形
D．所有的等边三角形都是全等三角形
3．下列各组线段能构成直角三角形的一组是
A．1，2，3 B．2，3，4
C．3，4，5 D．4，5，6
4．如图，已知点A、D、C、F在同一条直线上，AB＝DE，∠A＝∠EDF，再添加一个条件，可使△ABC
≌
△DEF，下列条件不符合的是
．．．
A
． ∠B＝∠E B． BC∥EF
C．
5．如图，用直尺和圆规作一个角的平分线，该作法的依据是
A．SSS
B．SAS
C．ASA
D．AAS
AD＝CF D． AD＝DC
6．在如图所示的正方形网格中，△ABC的顶点A、B、C都是网格线的交点，则△ABC的外角∠ACD的度数等于
A．130° B．135° C．140° D．145°
B E
B
C
D
F
A
A
D C
（第6题）
（第5题）
（第4题）
7．如图，AB⊥CD，且AB＝CD．E、F是AD上两点，CE⊥AD，BF⊥AD．若CE＝a，BF＝b，EF＝c，则AD的长为
A．a＋c B．b＋c C．a－b＋c D．a＋b－c
8．如图，在△ABC中，∠BAC＝90°，AD是高，BE是中线，CF是角平分线，CF交AD于G，交BE于H．下列结论：①S△ABE＝S△BCE；②∠AFG＝∠AGF；③∠FAG＝2∠ACF；④BH＝CH．其中所有正确结论的序号是
A．①②③④ B．①②③ C． ②④ D．①③
C
A
E
B
F
A
G
H
E
F
D
B
D
C
（第7题）
（第8题）
二、填空题（每小题2分，共20分）
9．等腰三角形的底角是顶角的2倍，则顶角的度数是 ▲ °．
10．等边三角形的两条中线相交所形成的锐角等于 ▲ °．
11． 如图，△ABC≌△DEC， CA和CD， CB和CE是对应边，∠ACD＝28°， 则∠BCE＝ ▲ °．
12．如图，在△ABC中，AB＝AC，AD是BC边上的高，点E、F是AD的三等分点，若AD＝6cm，CD＝3cm，则图中阴影部分的面积是 ▲ cm2；
13．如图，在△ABC中，AC的垂直平分线分别交BC、AC于点D、E，若AB＝10cm，BC＝18cm，则△ABD的周长为
A▲ cm．
A
E
（第11题）
B
C
D
F
E
A
E
C
B（第12题）
DCB
D
（第13题）
14．如图，点P为等边三角形ABC的边BC上一点，且∠APD＝80°， AD＝AP，则∠DPC ＝ ▲ °．
15．在△ABC中，将∠B、∠C按如图所示方式折叠，点B、C均落于边BC上一点G处，线段MN、EF为折痕．若∠A＝82°，则∠MGE＝ ▲ °．
16．如图，将△ABC绕点C逆时针旋转得到△A′B′C，其中点A′与点A是对应点，点B′与点A
B是对应点，点B′ 落在边AC上，连接A′B，若∠ACB＝45°，AC＝3，BC＝2，则A′B＝ ▲ ．
D
B
P
（第14题）
C
B
G
F
N
（第15题）
C
B
2A
M
E
C
A′
B′
（第16题）
A
17．如图，在△ABC中，∠ACB＝90°，∠CAB＝30°，以AB长为一边作△ABD，且AD＝BD，
∠ADB＝90°，取AB中点E，连DE、CE、CD．则∠EDC＝ ▲ °．
18．如图，在等腰△ABC中，AB＝AC＝10，高BD＝8，AE平分∠BAC，则△ABE的面积为
▲ ．
三、解答题(本大题共8小题，共64分)
19．（6分）如图，AD、BC交于点O，AC＝BD，BC＝AD．
求证：∠C＝∠D．
20．(7分) 如图，AD是△ABC的角平分线， DE、DF分别是△ABD和△ACD的高．
求证：AD垂直平分EF．
B
D
（第20题）
E
F
C
A
A
（第19题）
B
C
O
D
E
A
E
（第17题）
B
B
（第18题）
C
D
D
C
A
21．(6分)如图，已知△ABC，请用直尺和圆规以C为一个公共顶点作△CDE，使△CDE与△ABC全等，则全等的依据是 ▲ ．（不写作法，保留作图痕迹）
B
C
A
22．（6分）如图，在△ABC中，AB＝AC，点E在CA的延长线上，EP⊥BC，垂足为P，
EP交AB于点F．求证：△AEF是等腰三角形．
23．（9分）如图，一架2.5米长的梯子AB斜靠在竖直的墙AC上，这时B到墙底端C的距离
为0.7米．如果梯子的顶端沿墙面下滑0.4米，那么点B将向左滑动多少米？
24．（9分）如图，求证：有两条高相等的三角形是等腰三角形．
B
C
E D
A
B1
B
（第23题）
A
A1
B
C
P
（第22题）
F
E
A
C
25．（10分）已知：如图，AB＝AC，AD＝AE，BE与CD相交于点P．
（1）求证：PC＝PB；
（2）求证：∠CAP＝∠BAP；
（3）利用（2）的结论，用直尺和圆规作∠MON的平分线．
C
E
P
A
D
B
（第25题）M
O
N
26．（11分）在Rt△ABC中，∠ACB＝90°，BC＝a，AC＝b，AB＝c．将Rt△ABC绕点O依次旋转90°、180°和270°，构成的图形如图所示．该图是我国古代数学家赵爽制作的“勾股圆方图”，也被称作“赵爽弦图”，它是我国最早对勾股定理证明的记载，也成为了2002年在北京召开的国际数学家大会的会标设计的主要依据．
（1）请利用这个图形证明勾股定理；
（2）请利用这个图形说明a2＋b2≥2ab，并说明等号成立的条件；
（3）请根据（2）的结论解决下面的问题：长为x，宽为y的长方形，其周长为8，求当x，y取何值时，该长方形的面积最大？最大面积是多少？
C
A
b
c
a
O
B
（第26题）
2020-2021学年度第一学期期中学情分析样题(2)
八年级数学评分标准
一、选择（每题2分，共16分）
题号
答案
1
D
2
B
3
C
4
D
5
A
6
B
7
D
8
B
二、填空题（每题2分，共20分）
9．36 10．60 11．28 12．9 13．28
14．20 15．82 16．13 17．75 18．15．
三、解答题
19．(6分) 证明：在△ABC和△BAD中，
∵AC＝BD，BC＝AD，AB＝BA，
B
A
∴△ABC≌△BAD． ······································································ 4分
（第19题）
C
O
D
∴∠C＝∠D． ··············································································· 6分
·20．(7分) 证明：∵AD是△ABC的角平分线，
∴∠BAD＝∠CAD， ······································································ 1分
∵ DE、DF分别是△ABD和△ACD的高，
E
F
A
∴∠DEA＝∠DFA＝90°， ····························································· 2分
∵AD＝AD，
B
D
（第20题）
C
∴△AED≌△AFD． ······································································ 4分
∴DE＝DF，AE＝AF， ································································· 5分
∴A、D在EF的垂直平分线上， ···················································· 6分
E
∴AD垂直平分EF． ····································································· 7分
A
21．(6分)可分别利用平移、翻折、旋转作图． ······································ 4分
理由． ··························································································· 6分
F
22．(6分) 证明：∵AB＝AC，
∴∠B＝∠C ···················································································· 1分
∵EP⊥BC，
B
C
P
G
（第22题）
∴∠B＋∠BFP＝∠C＋∠E＝90°， ···················································· 3分
∵∠BFP＝∠AFE
∴∠AFE＝∠E ················································································ 5分
∴AE＝AF，
即△AEF是等腰三角形． ································································· 6分
证法二：过A作AG⊥BC，································································ 1分
∵AB＝AC，∴∠BAG＝∠CAG， ······················································· 2分
∵EP⊥BC，∴∠AGC＝∠EPC＝90°，
∴AG∥EP， ··················································································· 3分
∴∠BAG＝∠AFE，∠CAG＝∠E， ····················································· 4分
∴∠AFE＝∠E ················································································ 5分
∴AE＝AF，
即△AEF是等腰三角形． ································································· 6分
23．（9分）
A
A1
·解：在△ABC中，∠C＝90°，∴AC2＋BC2＝AB2， ······························ 2分
即AC2＋0.72＝2.52，∴AC＝2.4． ························································ 4分
在△A1B1C中，∠C＝90°，∴A1C2＋B1C2＝A1B12， ······························ 6分
B1
B
即(2.4–0.4)＋B1C＝2.5，∴B1C＝1.5． ············································ 8分
（第23题）
·∴B1B＝1.5–0.7＝0.8，即点B将向左移动0.8米．
································· 9分
24．(9分)
已知： 在△ABC中，BD⊥AC于点D，CE⊥AB于点E，且BD＝CE， ····· 2分
求证：△ABC是等腰三角形．（或AB＝AC） ········································· 3分
A
证明：∵BD⊥AC于点D，CE⊥AB于点E，
∴∠BDC＝∠CEB＝90°， ································································ 4分
在△BDC和△CEB中，
∵BD＝CE，BC＝CB，
C
B
∴△BDC≌△CEB（HL）． ································································ 7分
E D
2 22C
∴∠DCB＝∠EBC． ········································································· 8分
∴AB＝AC，
即△ABC是等腰三角形． ·································································· 9分
25．(10分)证明：（1）
∵AB＝AC，AD＝AE，∠BAE＝∠CAD．
C
M
∴△BAE≌△CAD（SAS）， ································································ 2分
∴∠C＝∠B，
∵AB＝AC，AD＝AE，
∴CE＝BD，
∵∠CPE＝∠BPD，
∴△CPE≌△BPD（AAS）， ······························································· 4分
∴PC＝PB． ··················································································· 5分
（2）∵AB＝AC，∠C＝∠B， PC＝PB，
∵△ACP≌△ABP（SAS）， ································································ 7分
∴∠CAP＝∠BAP． ········································································· 8分
（3）如图． ················································································· 10分
26．(11分)解：（1）因为边长为c的正方形面积为c2，···························· 1分
它也可以看成是由4个直角三角形与1个边长为（a– b）的小正方形组成的，
1它的面积为4×ab＋(a– b)2＝a2＋b2， ·············································· 3分
2所以c2＝a2＋b2． ········································································· 4分
（2）∵(a– b)2≥0， ······································································ 5分
∴a2＋b2–2ab≥0，∴a2＋b2≥2ab， ················································· 6分
当且仅当a＝b时，等号成立． ························································ 7分
（3）依题意得2(x＋y)＝8，∴x＋y＝4，长方形的面积为xy，
由（2）的结论知2xy≤x2＋y2＝(x＋y)2–2xy， ···································· 9分
∴4xy≤(x＋y)2，∴xy≤4， ··························································· 10分
当且仅当x＝y=2时，长方形的面积最大，最大面积是4． ··················· 11分
A
D
E
P
B
O
N
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