山西省大同市浑源县2020-2021学年七年级下学期期中质量检测数学（扫描版含答案）

这是一份山西省大同市浑源县2020-2021学年七年级下学期期中质量检测数学（扫描版含答案），共11页。试卷主要包含了选择题，填空题，解答题等内容，欢迎下载使用。

2021年七年级下册期中数学试题参考答案及评分标准
一、选择题（本大题共10个小题，每小题3分，共30分）
二、填空题（本大题共5个小题，每空3分，共18分）
11. . 12. 如果两个角相等，那么它们的余角也相等.
13. 4. 14. 5，75° 15. 52.5°
三、解答题（本大题含8个小题，共72分）
16.（本题8分）
（1）解：原式 ··· ·········· ············ ···································（2分）
= ····· ·························································（4分）
解：原式 = ··· ·········· ············ ···································（2分）
= ····· ·························································（4分）
y
x
17. （本题8分）
解：
如图所示 ··· ·········· ············ ···································（2分）
左，4； 上，2； ··· ·········· ············ ···································（6分）
如图所示 ··· ·········· ············ ···································（8分）
18.（本题6分）
解：依据1：对顶角相等 ··· ·········· ············ ···································（1分）
依据2：同旁内角互补，两直线平行 ······· ············ ···································（2分）
∠C , 依据3：两直线平行，同旁内角互补 ········································（4分）
依据4：内错角相等，两直线平行 ··········· ············ ···································（5分）
依据5：两直线平行，内错角相等 ··········· ············ ···································（6分）
19.（本题8分）


如图所示 ········ ··································（3分）
（3,0）; （0，-1）；（4，-2） ··········································（6分）
··········································（8分）
20.（本题10分）
解： （1）3； ····································································（4分）
（2） 即
∴的整数部分为4，
小数部分为（），即a=; ························································（6分）
，即
∴的整数部分为5，即b=5 ······················································（7分）
∴==1 ······················································（8分）
（3）则的相反数是 ． ··············································（10分）
21.（本题8分）
解：（1）当h＝1.7时，S2＝1.7×1.7， ··· ·········· ············ ···································（2分）
∴S＝﹣1.7（舍）或S＝1.7， ··· ·········· ············ ···································（3分）
答：当眼睛离海平面的高度是1.7m时，能看到约1.7km远； ····························（4分）
（2）当S＝1.7×3＝5.1时，可得5.12＝1.7h， ······ ···································（5分）
解得h＝15.3， ························································（6分）
15.3﹣1.7＝13.6（米）， ························································（7分）
答：观望台离海平面的高度约为13.6米． ························································（8分）
（本题12分）
（1）过点C作CG∥DF
∴∠DFE=∠FCG ························································（1分）
∵BC⊥MN
∴∠BCF=90°
∴∠BCG+∠FCG=90° ························································（2分）
∴∠BCG+∠DFE=90°
∵∠ABC+∠DFE=90°
∴∠ABC=∠BCG ························································（3分）
∴CG∥AB
∴DF∥AB ························································（4分）
（2）解：∵∠ABC=∠ACB，∠DEF=∠DFE，
又∵∠ABC+∠DFE=90°
∴∠ACB+∠DEF=90° ························································（6分）
∵BC⊥MN
∴∠ACE=90°
∴∠ACB+∠ACE=90°
∴∠DEF=∠ACE ························································（7分）
∴DE∥AC ························································（8分）
如图三角形DEF即为所求
························································（10分）
2α ·······················································（12分）
23.（本题12分）
（1）（﹣8，）； ························································（2分）
平行 ························································（4分）
（2）①t= ························································（6分）
② ························································（10分）
③-4 ························································（12分）
题号
1
2
3
4
5
6
7
8
9
10
答案
B
C
A
D
D
B
C
B
D
D