辽宁省鞍山市2024—2025学年度第一学期期末质量检测八年级数学试卷(图片版，含答案)

2024—2025 学年度第一学期期末质量检测
八年级数学参考答案及评分标准
一、 选择题：（每题 3分，共 24分）
题号 1 2 3 4 5 6 7 8
答案 D A B C B C D C
二、填空题：（每题 3分，共 15分）
9. 10. 34°或 20° 11. 4 12. 13. (5，0)
三、解答题：（本题共 8小题，共 61分）
14. (每题 5 分，共 10 分)
解：（1）原式= ············································································· (5 分)
（2）原式= ·································································· (5 分)
15. （本题 6 分）
解：(1) ① ···························································································· （1 分）
(2)原式= ·································································· （3 分）
∵ ············································································ （4 分）
解得 x=5 ···················································································· （6 分）
16. （本题 6 分）
解：（1）原式=（ ··········································· （2 分）
= ······································································ （4 分）
当 x= ，y= 时，原式=3 ···························································· （6 分）
17. （本题 6 分）
解：（1）答案不唯一，例如：a=1，b=2，c=3，d=6 ································ (1 分)
（2）当 a=1，b=2，c=3，d=6 时
，
∴ ················································································ (3 分)
证明： ···································· (4 分)
∵
∴ad=bc，即 ad－bc=0
∴ ，即 ··········································· (6 分)
18. （本题 7 分）
解：（1）如图，△A1B1C1即为所求 ······························································ (2 分)
A1(1，－2)；B1(5，－5)；C1(6，－2) ··········································· （5 分）
（2）如图，△ACD 与△CAB 成轴对称 ················································· （6 分）
直线 l 即为所求 ······································································ （7 分）
l
y
D B
M
C
A C
O x F H
A1 C1
A D E B
B1 N
18 题图 19 题图
19. （本题 8 分）
解： （1）如图，直线 MN 即为所求 ······················································ （2 分）
（2）AF⊥CE·························································································· (3 分)
理由是： 延长 AF 交 CE 与点 H
∵AC=BC
∴点 C 在直线 MN 上 ···························································· （4 分）
又∵∠ACB=90°
∴∠B=∠CAB=45°,∠ACD=∠DCB=45°
∴∠ACD=∠B
∵CF=BE
∴△ACF≌△CBE ································································ （6 分）
∴∠CAF=∠BCE
∵∠ACH＋∠BCE =90°
∴∠ACH＋∠CAF =90°
∴∠CHA =90°
即 AF⊥CE ··········································································· (8 分)
20. （本题 8 分）
解：（1）设两个班花费的费用均为 m 元，则一班的单位面积费用为
元，二班的
单位面积费用为 元，∵
∴八年二班绿化的单位面积费用高 ················································ （3 分）
（2）设学校计划区域 B 的费用为 n 元，根据题意得
················································································ (5 分)
解得 a=3，经检验 a=3 是原方程的解 ··················································· (7 分)
则区域 A 得面积为： 23 －1=8， 区域 的面积为（ － ）2B 3 1 =4
答：区域 的面积为 2A 8m ，区域 B 的面积为 24m . ································ （8 分）
21. （本题 10 分）
解：(1)答案不唯一，例如： ·································································· （2 分）
①关于某一个顶点对称的顶点对称四边形的对称轴必经过与它相对的顶点.
②关于某一个顶点对称的顶点对称四边形在对称顶点两侧的邻边(角)相等.
(2)∵△ACD 是等边三角形
∴∠ADC=∠ACD =∠CAD= 60°
∵AB⊥BC，AB⊥AD
∴∠ABC=∠BAD=90°
A
∴BC∥AD
∴∠BCD＋∠ADC =180° B E
∴∠BCD=120°
C D
∵∠BAC＋∠CAD =90°
∴∠BAC=30°
根据对称性，∠DAE=30°,∠E=90°
∴五边形的内角分别为 90°，120°，120°，90°120° ······················ （6 分）
(3)连接 OB，OE
根据对称性，OB=OF，OC=OE
∵CF= OF＋OC=5
∴OB＋OE=5
即点 O 与点 B，E 的距离之和为 5 ················································· （8 分）
在△BOC 与△EOF 中 l
∵OB=OF，OC=OE，BC=EF A
∴△BOC≌△EOF B F
∴∠BOC=∠EOF O
∵∠COD＋∠DOE＋∠EOF =180° C E
∴∠COD＋∠DOE＋∠BOC=180° D
即∠BOE=180°
∴对角线 BE 经过点 O ····························································· （10 分）