[初二数学试题] 初二数学综合题
初二数学试题
（总分：120分 考试时间：120分钟） 一、选择题（每小题4分，共32分）
1． 在平面直角坐标系中，点P （2，– 3）在第（ ）象限．
A ．一
B ．二
C ．三
D ．四
2． 下列命题中，真命题是（ ）
A ．有两个角相等的梯形是等腰梯形 C ．梯形的两条对角线相等
B ．对角线互相平分的梯形是等腰梯形 D ．对角线相等的梯形是等腰梯形
3． 若一个图形绕着一个定点旋转一个角α（0︒
图形叫做旋转对称图形．例如：等边三角形绕着它的中心旋转120︒（如图），能够与原来的等边三角形重合，因而等边三角形是旋转对称图形．下面①—④四个图形中，旋转对称图形的个数有（ ）
① ② ③ ④
A ．1个 B ．2个 C ．3个 D ．4个
4． 某产品的生产流水线每小时可生产100件产品，生产前没有产品积压，生产3小时后停止生
产并安排工人装箱．若每小时装产品150件，未装箱的产品数值（y ）是时间（t ）的函数，那么这个函数的大致图象是图中（ ）
A
B
C
D
（5题图） B C x A k
6． 若关于x 的方程有增根，则k =（ ） =2+
x -33-x
D
A
A ．1
B ．0
C ．– 2
D ．– 3
B
（7题图）
C
7． 如图，在锐角△ABC 中，∠A =50︒，AB 、AC 两边的中垂线相交于点O ，则∠BOC 的度数为
（ ）
A ．50° C ．100°
B ．75° D ．115°
0
顺时针旋转A 后，再向面对的方向沿直线行走a 个单位．若机器人的位置在原点面向y 轴的负半轴，则它完成一次指令[2，60︒]后，所在位置的坐标为（ ） A
．(-1，
B
．(-1)
C
．-1)
D ．
(-1
二、填空题（每小题3分，共24分）
9．
函数y =x 的取值范围是______________．
⎧y =ax +b
10． 如图，已知函数y =ax +b 和y =kx 的图象交于点P ，则根据图象可得关于⎨的二元
y =kx ⎩
一次方程组的解为______________．
11． 已知菱形的边长为5cm ，一条对角线长为6cm ，则菱形的面积为______________． 12． 已知：
11a -2ab -b -=2，则=______________． a b 2b -7ab -2a
13． 已知，如图，在矩形ABCD 中，AC 、BD 交于点O ，AE ⊥BD 于点E ，且BE ∶DE = 1∶3，AB
= 6cm，则AC = ______________．
14． 已知(a 2+b 2) 2-2(a 2+b 2) -24=0，则a 2+b 2=______________．
D b
（13题图）
C
B
（15题图）
C
（10题图）
15． 如图，梯形ABCD 中，AD ∥BC ，∠B =60︒，∠C =30︒，AD = 2，BC = 8，则CD =
______________．
16． 在平面直角坐标系中，O 是坐标原点，已知点A 的坐标为（2，2），请你在y 轴上找出点B ，
使△AOB 为等腰三角形，则符合条件的点B 共有______________个．
三、解答题（共64分）
17． 分解因式：（每小题5分，共10分）
(1) -2x 3+16x 2-32x
18． 化简求值．（第(1)小题5分，第(2)小题6分，共11分）
(2) (a 2+b 2-c 2) 2-4a 2b 2
(1)
3-x 5
÷(x +2-) x -2x -2
(2)
已知x =1，求(
19． (8分) 作图题
x +1x 1
-2) ÷的值． 2
x -x x -2x +1x
如图，方格纸中的每个小方格都是边长为1个单位的正方形，在建立平面直角坐标系后，△ABC 的顶点均在格点上，点C 的坐标为（4，– 1）．
(1) 把△ABC 向左平移6个单位后得到对应的
△A 1B 1C 1，画出△A 1B 1C 1．
(2) 以原点O 为对称中心，再画出与△ABC 关
于原点O 对称的△A 2B 2C 2．
(3) 图中点B 1的坐标为__________________；
点B 2的坐标为___________________； 点B 1、B 2之间的距离为____________．
20． (8分) 如图，正方形ABCD 的边长为1cm ，AC 为对角线，AE 平分∠BAC ，交BC 于点E ，EF
⊥AC 于点F ，求BE 的长．
D
F
B
C
E
21． (9分) 甲、乙两人骑自行车同时从学校出发，沿同一路线去博物馆。甲行驶20分钟因事耽误
一会儿，事后继续按原速行驶。下图表示甲、乙二人骑自行车行驶的路程y （千米）随时间x （分）变化的图象（全程），根据图象回答下列问题： (1) 求线段OD 、BC 的解析式； (2) 乙比甲晚多长时间到达博物馆？ (3) 甲因事耽误了多长时间？
∠AFD =60︒．22． (9分) 如图，在□ABCD 中，E 、F 分别在AB 、CD 上，且满足AF = AD，CE = CB，
(1) 证明：四边形AECF 是平行四边形．
(2) 若去掉已知条件中的“∠AFD =60︒”，而加上“EF 平分∠AFC ”，试判断四边形AECF
的形状，并加以证明．
23． (9分) 已知：一次函数y =kx +b 的图象与正比例函数y =mx 的图象相交于点P （a ，4），且这
两个函数的图象与y 轴所围成的三角形面积为6．正比例函数y =mx 的图象与直线4x +2y -5=0平行．
(1) 求a 、m 的值； (2) 求一次函数的解析式．
初二数学试题参考答案
一、选择题（每小题4分，共32分）
二、填空题（每小题3分，共24分） 9．x ≥2
⎧x =-310．⎨
y =2⎩
11．24 cm2
12．
4
3
13．12 cm
三、解答题（共64分）
14．6 15
．16．4
17．(1) 解：原式=-2x (x 2-8x +16) ···································································· 2分
··········································································· 5分 =-2x (x -4) 2 ·
(2) 解：原式=(a 2+b 2-c 2+2ab )(a 2+b 2-c 2-2ab ) ·········································· 2分
······················································· 3分 =[(a +b ) 2-c 2][(a -b ) 2-c 2] ·
=(a +b +c )(a +b -c )(a -b +c )(a -b -c ) ········································· 5分
18．(1) 解：原式=
3-x (x +2)(x -2) -5
··························································· 1分 ÷
x -2x -2
=
3-x (x +3)(x -3)
································································ 3分 ÷
x -2x -23-x x -2= x -2(x +3)(x -3)
1
················································································· 5分 x +3
x +1x 1
-]÷
x (x -1) (x -1) 2x
=-
(2) 解：原式=[
(x +1)(x -1) -x 21=÷ ······························································· 2分
x (x -1) 2x =
-1
x ·········································································· 3分
x (x -1) 2
1
············································································· 4分
(x -1) 2
=-
当x =1时，原式==-
1 ··········································· 6分 2
19．解：(1) 图略 ···························································································· 2分
(2) 图略 ···························································································· 4分 -4） ·(3) 点B 1的坐标为（-1，······························································· 5分
点B 2的坐标为（-5，4） ·································································· 6分 点B 1、B 2
······························· 8分
20．解：∵ 四边形ABCD 为正方形
∴ ∠B =90︒，∠ACB =45︒，AB = BC = 1 cm
∵ EF ⊥AC ，∴ ∠EFA =∠EFC =90︒ ······················································ 1分 ∴ △EFC 是等腰直角三角形
∴ EF = FC ························································································ 2分 在△ABE 和△AFE 中， ⎧∠BAE =∠FAE
⎪
∵ ⎨∠B =∠EFA =90︒
⎪
⎩AE =AE
∴ △ABE ≌△AFE （AAS ） ··································································· 5分 ∴ AB = AF = 1 cm，BE = EF
在Rt △ABC
中，AC ································· 7分 ∴
FC =AC -AF =1) cm
∴
BE =FC =1) cm ······································································ 8分
21．解：(1) 设线段OD 的解析式为y =k 1x
当x = 60时，y = 10 ∴ 10 = 60 k 1 ∴ k 1=∴ y =
1
6
1
···················································································· 1分 x ·6
设线段BC 的解析式为y =k 2x +b ∵ 当x = 60时，y = 10，∴ 10=60k 2+b
∵ 当x = 80时，y = 15，∴ 15=80k 2+b
11
解得：k 2=，b =-5 ∴ y =x -5 ············································ 3分
441
(2) 当y = 15时，x =15，x =90，90-80=10（分钟）
6∴ 乙比甲晚10分钟达到博物馆 ······················································· 6分
15-10
(3) v 甲==0.25（千米/分）
79-60
15÷0.25=60（分钟）
80 – 60 = 20（分钟）
∴ 甲因事耽误了20分钟。 ····························································· 9分
22．证明：(1) ∵ 四边形ABCD 是平行四边形
∴ AB // CD，∠B =∠D
∴ ∠AFD =∠EAF ，∵∠AFD =60︒ ∴ ∠EAF =60︒ ·························· 1分 ∵ AF = AD
∴ △ADF 是等边三角形
∴ ∠D =60︒，∠B =60︒ ······························································· 2分
∵ CE = CB
∴ ∠CEB =∠B =60︒ ∴∠CEB =∠FAE
∴ AF // CE ················································································· 3分 ∵ FC // AE
∴ 四边形AECF 是平行四边形 ······················································· 4分 (2) 四边形AECF 是菱形，理由如下：
∵ 四边形ABCD 是平行四边形
∴ AB // CD，∠B =∠D ， ∴ ∠AFD =∠F AE ∵ AF = AD ，CE = CB
∴ ∠D =∠AFD ，∠B =∠CEB ······················································· 5分 ∴ ∠AFD=∠CEB
∴ ∠F AE=∠CEB ········································································· 6分 ∴ AF // CE ∵ AF // CE
∴ 四边形AECF 是平行四边形 ······················································· 7分 ∵ EF 平分∠AFC ∴ ∠AFE=∠CFE ∵ FC // AE ∴ ∠CFE=∠AEF ∴ ∠AFE=∠AEF ∴ AE = AF
∴ 四边形AECF 是菱形 ································································ 9分
5
23．解：(1) 由4x +2y -5=0得y =-2x +，∴ m =-2 ········································· 2分
2
∵ 点P （a ，4）在直线y =-2x 上，
∴ 4=-2a ，∴ a =-2 ··································································· 3分 (2) ∵ a =-2，∴ P （– 2，4）
∵ 点P （– 2，4）在直线y =kx +b 上
∴ 4=-2k +b ··········································································· 4分 直线y =kx +b 与y 轴交于点A （0，b ）
直线y =-2x 与y 轴交于点（0，0） ···················································· 5分 ∴ AO =|b |
1
∴ S ∆AOP = |b | |-2|=6
2
∴ b =±6 ····················································································· 7分 ①当b = 6时，4=-2k +6，∴ k =1，∴ y =x +6······························· 8分 ②当b = – 6时，4=-2k -6，∴ k =-5，∴ y =-5x -6 ······················· 9分