2023 年春季学期阶段性自主评估训练（一）
七年级 数学参考答案
一、选择题（本大题共 12小题，每小题 3分，共 36分．）
题号 1 2 3 4 5 6 7 8 9 10 11 12
答案 C D D B A A C B C B D B
二、填空题（本大题共 6小题，每小题 2分，共 12分．）
13．＜ 14．55° 15．如果等式两边加同一个数，那么结果仍是等式
16． 3
1
V 17．144 18． (α＋β)－90°2
三、解答题（本大题共 8小题，共 72分．）
19．（本题满分 6分）
1
解：（1 1）－ 3 ＝－ ；···············································································2分
8 2
（2） 0.64 ＝0.8；·······················································································4分
（3） ( 3)2 ＝ 9 ························································································5分
＝3．······················································································· 6分
20．（本题满分 6分）
解：（1）x3＋27＝0；
x3＝－27．················································································ 1分
x＝－3．···················································································3分
（2）(x－1)2＝4．
x－1＝±2．······················································································4分
所以 x－1＝2或 x－1＝－2．······································································5分
所以 x1＝3，x2＝－1．················································································ 6分
21．（本题满分 10分）
解：（1）三角形 A′B′C′如图所示．·························· 4分
（2）过点 A的 BC的垂线和点 P如图所示．············· 7分
（3）过点 A的 BC的平行线和点 Q如图所示．··········10分
22．（本题满分 10分）
解：∵E是直线 CA上一点，∠FEA＝40°，
七年级数学参考答案（第 1页 共 3页）
∴∠CEF＝180°－∠FEA＝180－40°＝140°．··························2分
∵EB平分∠CEF，
1 1
∴∠BEF＝∠CEB＝ ∠CEF＝ ×140°＝70°．······················ 5分
2 2
∵GE⊥EF，
∴∠GEF＝90°．···························································································7分
∴∠GEB＝∠GEF－∠BEF＝90°－70°＝20°．···················································10分
23．（本题满分 10分）
证明：∵DE∥AB（已知），
∴∠A＝ ∠CED （ 两直线平行，同位角相等 ）．····················4分
又∵∠BFD＝∠CED（已知），
∴∠A＝∠BFD（ 等量代换 ）．······················· 6分
∴DF∥ AC （ 同位角相等，两直线平行 ）．·······················8分
∴∠EGF＋∠AEG＝180°（ 两直线平行，同旁内角互补 ）．····· 10分
24．（本题满分 10分）
解：不能．理由如下：·················································································· 1分
设：裁出的纸片的长为 5xcm，宽为 4xcm．························································2分
由题意，得 5x 4x＝800．··············································································· 3分
解得 x＝ 40．···························································································· 5分
所以长方形纸片的长为 5 40．·······································································6分
因为 40＞36，
所以 40＞6．····························································································· 7分
所以 5 40＞30，即长方形纸片的长大于 30cm．················································ 8分
因为 900＝30，
所以正方形纸片边长为 30cm．这样，长方形纸片的长大于正方形纸片的边长．········ 9分
所以小李不能裁出符合要求的纸片．·······························································10分
25．（本题满分 10分）
解：（1）因为 x的平方根是(a＋3)与(2a－15)，
七年级数学参考答案（第 2页 共 3页）
所以 a＋3＋2a－15＝0．················································································ 1分
解得 a＝4．································································································· 3分
所以 x a 3 2＝ 4 3 2 49．····································································· 5分
因为 2b 1＝3，
所以 2b－1＝9，即 b＝5．·············································································· 7分
（2）由（1）得 3 a b 1 3 4 5 1 2．······················································ 10分
26．（本题满分 10分）
证明：（1）如图 1，过点 G作 GH∥AB．······················· 1分
∵AB∥CD，
∴AB∥GH∥CD．
∴∠BEG＝∠EGH，∠DFG＝∠FGH．
∴∠BEF＋∠DFE＝180°．············································································· 2分
∵EG平分∠BEF，FG平分∠DFE，
1 1
∴∠GEB＝ ∠BEF，∠GFD＝ ∠DFE．
2 2
1 1 1
∴∠GEB＋∠GFD＝ ∠BEF＋ ∠DFE＝ (∠BEF＋∠DFE)＝90°．····················3分
2 2 2
∴在三角形 EFG中，∠EGF＝∠GEB＋∠GFD＝90°．
∴EG⊥FG．································································································4分
（2）如图 2，由（1）得∠BEG＋∠DFG＝90°．·················· 5分
∵EM平分∠BEG，FM平分∠DFG，
1
∴∠BEM＋∠MFD＝ (∠BEG＋∠DFG)＝45°．··················6分
2
∴∠EMF＝∠BEM＋∠MFD＝45°．································································· 7分
（3）如图 3，由题意得∠EOF＝∠BEO＋∠DFO，∠EPF＝∠BEP＋∠DFP．·········· 8分
∵EP平分∠BEO，FP平分∠DFO，
∴∠BEO＝2∠BEP，∠DFO＝2∠DFP．···························· 9分
∴∠EOF＝2∠EPF．·····················································10分
七年级数学参考答案（第 3页 共 3页）浦北三中2023年春季学期阶段性自主评估训练（一）
7.估计√27的值在
A.3和4之间
B.4和5之间
七年级数学
C.5和6之间
D.6和7之间
(考试时间：120分钟满分：120分)
8.下列说法错误的是
注意事项：
A.0的算术平方根是0
B.一1的平方根是一1
1.答题前，考生务必将姓名、准考证号、座位号填写在试卷和答题卡上。
C.0的立方根是0
D.式子3有意义
2.考生作答时，请在答题卡上作答（答题注意事项见答题卡），在本试卷上作答无效。
9.如图，下列条件中不能判定AB∥CD的是
A.∠1=∠5
B.∠3=∠4
(第9题图)
第I卷
C.∠3=∠5
D.∠1+∠4=180°
10.若25.36=2.938,25360=29.38，则325360000等于
一、选择题（共12小题，每小题3分，共36分，在每小题给出的四个选项中只有一项是符合要求的，
A.632.9B.293.8C.2938D.6329
用2B铅笔把答题卡上对应题目的答案标号涂黑.)
11.如图，两个直角三角形重叠在一起，将其中一个三角形沿着点B到
1.3的算术平方根是
点C的方向平移到三角形DEF的位置，∠B=90°，AB=8,DH=3,
A.-9B.±5C.3D.9
B E
平移距离为4，则阴影部分的面积为
(第11题图)
2.下列图形中，∠1和∠2互为对顶角的是
A.20B.24C.25D.26
12.将一条两边沿互相平行的纸带如图折叠，已知∠a比∠1大30°，则
2g一
∠a的度数为
A
B
3.下列图形中，不能通过其中一个四边形平移得到的是
A.69°B.70°C.71°D.72°
(第12题图)
00W今安
第Ⅱ卷
A
B
D
二、填空题（本大题共6小题，每小题2分，共12分.)
4.经过一点（已知直线上或直线外），能画出已知直线的垂线数为
13.比较大小：√8▲一√0（填写“>”，“<”或“=”）·
(第14题图)》
A.0条B.1条C.2条D.无数条
14.如图，两条直线相交于点0，若∠1+∠2=250°，则∠3=▲一
5.下列命题中，是真命题的是
15.将命题“等式两边加同一个数，结果仍是等式.”改写成“如果…那
A,对顶角相等
B.同旁内角互补
么…”的形式：▲一·
20米
(第17题图)
C.互补的角是邻补角
D.相等的角是对顶角
16.如果一个正方体的体积为”，那么这个正方体的棱长为▲
6,如图，过直线外一点作已知直线的平行线，其依据是
17.如图，在长方形草地内修建了宽为2米的道路，则草地面积为▲平方米.
A.同位角相等，两直线平行B.内错角相等，两直线平行
18.如图，AC∥BD,EP,FP分别平分∠AEF,∠EFB,若∠A=a,∠B=B,
(第6题图)
R
C.两直线平行，同位角相等D.两直线平行，内错角相等
则∠P的大小为▲（用含a,的式子表示），
(第18题图)
七年级数学试卷第1页（共4页）
七年级数学试卷第2页（共4页）
