2024—2025学年度第一学期教学质量监测
八年级数学科参考答案及评分标准
一、选择题（共 10题，每题 3分，共 30分）
1．C 2．B 3．D 4．A 5．D 6．A 7．D 8．C 9．C 10．B
二、填空题（共 5题，每题 3分，共 15分）
x＝1 12
11．5 12．＞ 13．1或 3 14． 15．
y＝4 5
三、解答题（一）（共 3题，每题 7分，共 21分）
16．（1）解：原式＝4 3÷ 3― 6＋2 6 ······································································ 2分
＝4＋ 6．···································································································3分
（2）解：方程组整理得
6x―5y＝30①
，···························································································4分
5x―6y＝14②
①×6―②×5得：36x―25x＝180―70······························································ 5分
解得：x＝10，
将 x＝10代入①得：y＝6，·············································································6分
x＝10
∴方程组的解为： ．·············································································7分
y＝6
17．解：设修建一个 A种光伏车棚需投资 x万元，修建一个 B种光伏车棚需投资 y万元，······ 1分
2x＋y＝8
根据题意得： ，················································································· 4分
5x＋3y＝21
x＝3
解得： ．····································································································6分
y＝2
答：修建一个 A种光伏车棚需投资 3万元，修建一个 B种光伏车棚需投资 2万元．········· 7分
18．（1）解：如图所示； ·························································································· 2分
（2）如上图，△ABC即为所求，其中 C(―1，3)；···················································· 4分
（3）如上图，△A1B1C1即为所求．········································································· 7分
八年级数学科参考答案及评分标准 第 1 页 (共 4 页)
四、解答题（二）（共 3题，每题 9分，共 27分）
19．（1）证明：∵CD平分∠ACB，
∴∠DCB＝∠1，·························································································· 1分
∵∠1＝∠D，
∴∠DCB＝∠D，··························································································2分
∴DF//BC；································································································ 4分
（2）解：∵DF//BC，∠DFE＝34°，
∴∠B＝∠DFE＝34°，···················································································5分
在△ABC中，∠A＝36°，∠B＝34°，
∴∠ACB＝180°―36°―34°＝110°，·································································· 6分
∵CD平分∠ACB，
1
∴∠1＝ 2∠ACB＝55°，·················································································7分
∴∠2＝180°―36°―55°＝89°．········································································9分
20．（1）解：补全图形如下：······················································································2分
m＝20；······································································································3分
11＋11
（2）抽取的黄瓜根数的众数为 10根，中位数为 ＝11(根)；································· 6分
2
1
（3） ×(10×14＋11×8＋12×10＋13×4＋14×4)≈11(根)，····································8分
40
答：这个品种的黄瓜平均每株结 11根．····························································9分
21．（1）证明：∵AB＝AC，
∴∠B＝∠C，······························································································ 1分
∵EF⊥BD，
∴∠AEF＋∠AED＝90°，··············································································· 2分
∵∠AEF＝∠B，∠B＝∠C，
∴∠AEF＝∠C，·························································································· 3分
∴∠C＋∠AED＝90°，
∴∠EAC＝90°，
∴AE⊥AC；································································································ 4分
八年级数学科参考答案及评分标准 第 2 页 (共 4 页)
（2）解：∵AB＝AC＝10，BC＝16，点 D是 BC的中点，
1
∴BD＝DC＝ ×16＝8，AD⊥BC，·································································· 5分
2
∴AD＝ AC2―DC2＝ 102―82＝6，·································································6分
∵∠EAC＝90°，
∴AE2＋AC2＝CE2，
∵CE＝CD＋DE＝DE＋8，
∴AE2＝CE2―AC2＝(DE＋8)2―102，································································7分
又∵在 Rt△ADE中，AE2＝AD2＋DE2＝62＋DE2，
∴(DE＋8)2―102＝62＋DE2，··········································································8分
解得：DE＝4.5．·························································································· 9分
五、解答题（三）（共 2题，22题 13分，23题 14分，共 27分）
22．（1）y1＝―2x＋48 y2＝x＋3················································································· 4分
（2）由(1)得，y1＝―2x＋48，y2＝x＋3，
当 x＝13时，y1＝22，y2＝16，········································································ 6分
∵可变车道为自东向西方向，
∴自东向西方向的车道数为 3，自西向东方向的车道数为 2，
u y1 22 y∴ ＝ ＝ ，u ＝ 2 161 3 3 2 2 ＝ 2 ＝8，··································································· 7分
∵u1＜u2，
∴自西向东方向更拥堵．················································································8分
y y
（3）在没有可变车道的情况下，两个方向的车道数均为 2，即u1＝ 1 22，u2＝ 2，·············· 9分
当 u1＝u2时，y1＝y2，·················································································· 10分
∴―2x＋48＝x＋3，解得 x＝15，··································································· 11分
经判断，在 8时至 15时，可变车道设置为自东向西方向；在 15时至 20时，可变车道设置为
自西向东方向．·························································································· 13分
23．（1）解：如图，过点 A作 AD⊥OC于点 D， ···························································1分
∵等边△AOB的边长为 2，
∴AB＝AO＝OB＝2，OD＝DB＝1，································································· 2分
在 Rt△AOD中，由勾股定理得：AD＝ AO2―OD2＝ 3，···································· 3分
∴A(1， 3)；······························································································ 4分
八年级数学科参考答案及评分标准 第 3 页 (共 4 页)
（2）将点 A 3的坐标代入直线 y＝― 3 x＋b得：
3 3＝― 3 ＋b，····························································································5分
解得：b 4 3＝ 3 ，··························································································· 6分
3 4 3
故该直线的表达式为 y＝― 3 x＋ 3 ；·····························································7分
（3）在 x轴上存在点 Q，使得△ACQ是以 AC为腰的等腰三角形，理由如下：
在 y 3 x 4 3＝― 3 ＋ 3 中，令 y＝0时，x＝4，即点 C(4，0)；··································· 8分
设点 Q(x，0)，
由点 A、C、Q的坐标得：AC2＝12，AQ2＝(x―1)2＋3，CQ2＝(x―4)2，················· 9分
当 AC＝AQ时，即 12＝(x―1)2＋3，······························································ 10分
解得：x＝4(舍去)或 x＝―2，
∴点 Q(―2，0)；······················································································· 11分
当 AC＝CQ时，则 12＝(x―4)2，···································································12分
解得：x＝4±2 3，
∴点 Q(4＋2 3，0)或 Q(4―2 3，0)，···························································13分
综上，在 x轴上存在点 Q，使得△ACQ是以 AC为腰的等腰三角形；点 Q的坐标为(―2，0)
或(4＋2 3，0)或(4―2 3，0)．···································································· 14分
八年级数学科参考答案及评分标准 第 4 页 (共 4 页)2024一2025学年度第一学期教学质量监测
八年级数学科试卷
温馨提示：请将答案写在答题卡上；考试时间为120分钟，满分120分。
一、选择题（共10题，每题3分，共30分)
1.下列实数中，是无理数的是（)
A.3.14
B.V⑧
C.5
D.
3
2.如图，直线AB,CD相交于点E,AB/DF,若∠BEC=125°，则∠D等于（)
A.45°
B.55°
C.65°
D.125°
3.下列计算正确的是()
A.疙V2
1
B.V2+V3=√5
C.V6÷2=3
D.(2W3)2=12
4.如图是甲、乙两名同学6次射击成绩的折线统计图，甲、乙两人射击成绩的方差分别记作S，$足，
下列结论正确的是()
成绩/环
10
◆一甲
8F
◆-乙
7
0123456次数
A.S>2
B.2=
C.D,无法确定
5.下列条件中，不能判断△ABC是直角三角形的是（)
A.AB BC AC=3:4:5
B.AB BC AC=1:2:3
C.∠A-∠B=∠C
D.∠A:∠B:∠C=3:4:5
八年级数学科试卷第1页(（共6页)
6.在量子物理的研究中，科学家需要精确计算微观粒子的能量.已知某微观粒子的能量E可以用公式
E=,a2十b表示.当a=2,b=9时，该微观粒子的能量E的值在（)
A.3和4之间
B.4和5之间
C.5和6之间
D.6和7之间
7.一次函数y=a一b与正比例函数y=bx(k,b为常数，且kb≠0)在同一直角坐标系内的大致图象不
可能的是(
8.我国古代数学著作《孙子算经》有“多人共车”问题：“今有三人共车，二车空；二人共车，九人步.问：
人与车各几何？”其大意如下：有若干人要坐车，如果每3人坐一辆车，那么有2辆空车；如果每2人坐
一辆车，那么有9人需要步行，问人与车各多少？设共有x人，y辆车，则可列方程组为（)
3(y-2）=x
3(y+2)=x
3(y-2)=x
[3(y-2)=x
A.
B.
C.
D
2y-9=x
2y+9=x
2y+9=x
2y+x=9
9.如题9图，AD是△ABC的角平分线，点B、C、E共线，则a、B、y之间的数量关系是()
A.a+B=y
B.2a-B=y
C.28-a=y
D.
2y-a=B
B
D
题9图
题10图
10.某校“灯谜节”的奖品是一个底面为等边三角形的灯笼（如题10图），在灯笼的侧面上，从顶点A
到顶点A'缠绕一圈彩带.已知此灯笼的高为50cm,底面边长为40cm,则这圈彩带的长度至少为
(
)
A.50cm
B.130cm
C.120cm
D.150cm
八年级数学科试卷第2页（共6页）
