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2023-2024学年上学期期末学业水平调研测试
八年级数学 参考答案与评分标准
一、选择题（本大题共 10小题，每小题 3分，共 30分。每小题有四个选项，其中只有一项
是正确的）
题号 1 2 3 4 5 6 7 8 9 10
答案 A C A C B A D B D B
二、填空题（本大题共 5小题，每小题 3分，共 15分）
x 2 15
11．2 12．48 13． 14．36（ 写36 不扣分） 15．
y 3 2
三、解答题（本大题共 7小题，共 55分）
16．（共8分）
（1） 45 125 20 ；
解：原式=3 5 5 5 2 5 ······································································· 3分
（说明：第一步计算中，每化简正确一个给 1分）
=6 5 .························································································4分
2 50 32（ ） 3 2
8
5 2 4 2
解：原式= 3 2 50 32（或者 3 2 ）··································· 3分
2 2 8
10 2 3 2
7 2．·····················································································4分
（其他算法酌情给分）
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17.（共 5 分）
3m 2n 7 ①

3m n 5 ②
解法一
解：由②得： n 3m 5 . ③ ···················································· 1分
将③ 代入①得：3m 2(3m 5) 7 .····················································· 2分
m 1．···································································· 3分
将m 1代入③ ,得： n 2．····························································· 4分
m 1,
所以原方程组的解是 ·································································5分
n 2.
解法二
解：①－②得： n 2 . ···································································· 2分
n 2． ···································································· 3分
将 n 2代入① ,得：3m 2 ( 2) 7，
m 1.···································································4分
m 1,
所以原方程组的解是 ·································································5分
n 2.
（其他算法酌情给分）
18.（共 8 分）
解：（1）设胸口朥售价是每份 x元,肥胼售价是每份 y元.······························ 1分
2x 3y 196,
根据题意，可得 ·····················································2分
4x y 192.
x 38,
解这个方程组，得 ···························································3分
y 40.
答：胸口朥售价是每份 38 元,肥胼售价是每份 40 元.·························4分
（2）胸口朥的进价是 38-6=32（元）. ·················································1分
肥胼的进价是 40-8=32（元）. ················································· 2分
180 32 5760（元）. ······················································ 3分
答：火锅店老板实际进货用了 5760 元．········································· 4分
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19. （共 8分）
（1）证明： BD DE，
DEB DBE .································································1分
DE // BC，
DEB CBE .································································ 2分
DBE CBE .································································3分
BE平分 ABC .··································································· 4分
A
D E
B C
（2）解： EBC 30 , （第 19题图）
DBE CBE 30 .······························································1分
EBC 60 .········································································· 2分
A 50 , A ABC C 180 ,
ACB 180 50 60 70 .·················································· 4分
20.（共 8分）
（1） 3 ，·························································································2分
（2）(共 2分) y y
3
32
21 1
-5 -4 -3 -2 -1 O 1 2 3 x -5 -4 -3 -2 -1 O 1 2 3 x-1
-1
-2
-22
2-32
2-3
22
22
（说明：描点正确给 1分,图象需画出“v”字型的两条射线,否则不给画图这一分）
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（3） ① 函数的最小值是0;
② x 2 时,y随x的增大而增大（减小而减小）;
③ x 2 时,y随 x的增大而减小（减小而增大）；
④函数不经过第三象限 或 函数不经过第四象限或函数经过第一、二象限;
⑤函数与 x轴只有一个交点或函数与 y轴只有一个交点.
（只要以上性质中的 2个就应该满分，如有其他性质正确的也可以参照给分）
（4） 4 .····················································································· 2分
21.（共9分）
（1）解： a 5 ,b 6 , c 7 ,
p 5 6 7 9.··································································· 2分
2
s 9(9 5)(9 6)(9 7) 6 6 .················································ 3分
B
（2）解法一
解： a 7 ,b 8 , c 9 , ┐
C D A
7 8 9
p 12 . （第 21题图）
2
s 12(12 7)(12 8)(12 9) 12 5 .··········································· 1 分
AC BD
12 5 8 BD, 即 12 5 .
2 2
BD 3 5 .··········································································· 2分
在 Rt△ BDC中，CD BC 2 BD2 72 (3 5)2 49 45 2 .· 3分
解法二：设CD为x,则AD为（8 x）.由题意可得：······························· 1分
72 x2 92 (8 x)2 .························································· 2分
解得： x 2 .
CD 2 .··············································································· 3分
（3） s p 15 , a 10 ,
15 15(15 10)(15 b)(15 c) .
化简得：bc 15(b c) 225 3.············································· 1分
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15 10 b c ,
2
b c 20 . ···································································· 2分
bc 15 20 225 3 .
bc 78.············································································ 3分
22.（共9分）
（1）解：令 y 0 4,即 x 4 0，得：
3
x 3.························································································ 1分
令 x 0 ,得： y 4 .··································································· 2分
A（3,0）,B（0,4）.
AB 32 42 5.··································································· 3分
y
B
A C
O x
E
D
（第 22 题图）
（2）解： AB AC 5 ,
OC 8 ,C(8,0) .································································ 1分
设OD a ,那么DC DB 4 a，
a2 82 (a 4)2 .
解得： a 6 ,即OD 6 .
D（0, 6）.········································································· 2分
设直线CD的表达式为 y kx b ,则
8k b 0,

0 b 6.

k
3
,
解得: 4
b 6.
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直线CD 3的表达式为 y x 6 .···················································· 3分
4
（3）当点 P在第一象限时,
2 x 3 x 6 14 .
4
52
解得： x .
7
52 8 ,此时点 P不可能在第一象限.············································· 1分
7
（说明：语言表述C的横坐标为8,比长方形长加宽的和大,不可能在第一象限也可给分.）
当点 P在第四象限时,
2 x 3 x 6 14 .
4
解得： x 4 .
把 x 4代入 y 3 x 6得,
4
y 3 .
即 PM 3,PN 4 .
S长方形OMPN 4 3 12 .································································· 2分
当点 P在第三象限时,
2 x 3 x 6 14.
4
x 4解得： .
7
x 4把 代入 y 3 x 6得,
7 4
y 45 .
7
即 PM 45 ,PN 4 .
7 7
S 45 4 180 长方形OMPN .····························································· 3分7 7 49
{#{QQABCYQEoggoAAJAABgCEQGaCkGQkBECAKoGQBAEMAAAABFABCA=}#}2023-2024学年第一学期学科素养调研测试
八年级
数学
注意事项：
1.本场考试时间为90分钟，总分100分。请考生务必把姓名、班级等信息写在答题卷
指定的位置上，并将考生信息条形码贴在指定方框内。
2第一部分选择题的答案务必用2B铅笔填涂在答题卡上；如需改动，用橡皮棕千净后，
再填涂其他答案。如有作图题，先用2B铅笔作图，最后用黑色签字笔描一次。
3.第二部分非选择题的答案务必用黑色字迹的钢笔或签字笔作答。答案必须写在答题
卷的指定位置。
4.考试结束后，请将试卷、答题卷、草稿纸一并交回。
第一部分
选择题
一、选择题（本大题10个小题，每小题3分，共30分）
1.下列各数中为无理数的是（▲）
A.g
B.0.4
C.4
D.4
2.平面直角坐标系中，点(a,4)在一次函数y=3x+1的图象上，则a的值是（▲）
A.1
B.2
C.3
D.4
3.下列条件中，可以判断△ABC是直角三角形的是（▲）
A.AB+BC>AC
B.AB:BC:AC=3:4:5
C.∠A=75°，∠B=30°
D.∠A:∠B:∠C=3:4:5
4.如右表记录了甲、乙、丙、丁四名跳高运动员最
甲
乙
丙
近几次选拔赛成绩的平均数与方差。根据表中数
平均(cm)
165
160
165
160
据，要从中选择一名成绩好且发挥稳定的运动员
方差
2.5
2.5
6.4
7.1
参加比赛，应该选择（▲）
A.甲
B.乙
●“
C.丙
D.丁
图书馆
教学楼
5.如图所示的是一所学校的平面示意图，若用(2,2)表示教学
楼的位置，(3,0)表示旗杆的位置，则实验楼的位置可表示
校门
旗杆
成（▲）
A.(3,-1)
B.（1,-3)
●-
C.（1,3)
D.(-1,3)
实验楼
第5题图
八年级数学第1页共7页
6.如图，网格中每个小正方形的边长均为1，点A,B,C都在格点上，
以A为圆心，AB为半径画弧，交最上方的网格线于点D,则CD
的长为（▲）
A.√13
B.2.2
c.5
D.3-√5
第6题图
7太阳灶、卫星信号接收锅、探照灯及其他很多灯具都与抛物线有关.如图，从点0照射
到抛物线上的光线OB,0C反射后沿着与P0平行的方向射出，已知图中LAB0=44°，
∠BOC=133°，则∠OCD的度数为（▲）
A.88
B.89
C.90°
D.91°
8.如图，在平面直角坐标系中，A、B两点在一次函数的图象上，其坐标分别为A(x,y),
B(x+a,y+b),下列结论正确的是（▲）
A.a<0,b=0
B.a>0,b>0
C.a<0,b<0
D.ab<0
9.如图，在△ABC中，AE是角平分线，AD⊥BC,垂足为D,点D在点E的左侧，
∠B=60°，∠C=40°，则LDAE的度数为（▲）
A.10°
B.15
C.30°
D.45°
A
B
DE
第7题图
第8题图
第9题图
10.如图，平面直角坐标系中，点A、C的坐标分别为(0，-4)、（9，
0),点B(b,4)在第一象限内，连接AB交x轴于点D,连接
B
AC,∠CBA=2LBAO,则△ABC的面积为（▲）.
A.12
B.20
C.24
D.25
第10题图
第二部分
非选择题
二、填空题（本大题5个小题，每小题3分，共15分）
1.比较大小▲5（填“=”皮>”或“<”）
2
八年级数学第2页共7页

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