初一数学试题.docx
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初一数学试题
2012~2013学年度下学期期末学情诊断(东平)
数学试题
本试题第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分,共120分,考试时间120分钟。
第Ⅰ卷(选择题共60分)
一、选择题(本大题共20小题,每小题3分,共60分,在每小题给出的四个选项中,只有一项是正确的,请把正确选项的序号填到下面的答题栏里.)
1.下列说法正确的是
A.由同一端点出发的两条直线组成的图形叫做角
B.大于直角的角都是钝角
C.同角或等角的补角相等
D.120.5°=120°50′
2.用一副三角板的内角(其中一个三角板的内角是45°,45°,90°,另一个的是30°,60°,90°)可以画出大于0°且小于176°的不同度数的角共有
A.8种B.9种C.10种D.11种
3.两条直线被第三条直线所截,下列条件中不能判断这两条直线平行的的是
A.同位角相等B.内错角相等
C.同旁内角互补D.同旁内角相等
4.如图,AB∥CD,DB⊥BC,∠1=40°,则∠2的度数是
A.40°
B.50°
C.60°
D.140°
5.如图,有一块含有45°角的直角三角板的两个顶点放在直尺的对边上.如果∠1=20°,那么∠2的度数是
A.30°
B.25°
C.20°
D.15°
6.二元一次方程
有无数多个解,下列四组值中不是该方程的解的是
A.
B.
C.
D.
7.若│x-2│+(3y+3)2=0,则
的值是
A.-1B.-2C.2D.
8.已知
是二元一次方程组
的解,则
的值为
A.±2B.-2C.2D.4
9.下列运算正确的是
A.x2
x3=x6B.
C.
D.40=1
10.下面的多项式中,能因式分解的是
A.
B.
C.
D.
11.把多项式
因式分解结果正确的是
A.
B.
C.x(x-1)(x+1)D.x2(x-2)
12.已知三角形两边的长分别是5和10,则此三角形第三边的长可能是
A.4B.5C.11D.16
13.一个多边形的内角和是720°,这个多边形的边数是
A.4B.5C.6D.7
14.多边形的边数增加一条,则它的内角和
A.增加180°B.增加360°C.不变D.减少180°
15.如图所示,一个60°角的三角形纸片,剪去这个60°角后,得到一个四边形,则∠1+∠2的度数为
A.120°
B.180°
C.240°
D.300°
16.下列说法正确的是
A.直径是弦,弦是直径.B.半圆是弧,劣弧不大于半圆.
C.半径不相等的两圆是同心圆.D.面积相等的圆是等圆.
17.三角形的一个外角小于与它相邻的内角,则这个三角形是
A.直角三角形B.锐角三角形
C.钝角三角形 D.不能确定
18.一个正多边形的内角是外角的3倍,则这个正多边形是
A.正方形B.正六边形C.正八边形D.正十边形
19.在平面直角坐标系中,已知点P(-3,
),则点P在
A.第一象限B.第二象限C.第三象限D.第四象限
20.在平面直角坐标系xoy中,若A点坐标为(-3,3),B点
坐标为(2,0),则△ABO的面积为
A.3B.6
C.7.5D.15
选择题答题栏
题号
1
2
3
4
5
6
7
8
9
10
答案
题号
11
12
13
14
15
16
17
18
19
20
答案
第Ⅱ卷(非选择题共60分)
二、填空题(每小题3分,满分12分)
21.计算a2÷a-6=.
22.分解因式:
-16=.
23.如图,在△ABC中,∠A=80°,点D是BC延长线上一点,∠ACD=150°,则∠B=.
24.如图,已知棋子“车”的坐标为(-2,3),棋子“马”的坐标为(1,3),则棋子“炮”的坐标为.
三、解答题(本大题共5小题,满分48分,写出必要的文字
说明,证明过程或演算步骤)
25.(每小题4分,满分8分)
(1)计算:
30+2
×
-(-3)2÷(-3)-
(2)因式分解:
26.(本小题满分9分)
已知
,求代数式
的值.
27.(本小题满分9分)
甲、乙两件服装的成本共500元,商店老板为获取利润,决定将甲服装按50%的利润定价,乙服装按40%的利润定价.在实际出售时,应顾客要求,两件服装均按9折出售,这样商店共获利157元,求甲、乙两件服装的成本各是多少元?
28.(本小题满分10分)
如图,在Rt△ABC中,∠B=90°,∠BAC,∠BCA的外角平分线交于点P,求∠P的度数.
29.(本小题满分12分)
在直角坐标系中,第一次将△OAB变换成OA1B1,第二次将△OA1B1变换成△OA2B2,第三次将△OA2B2变换成△OA3B3,已知A(1,3),A1(2,3),A2(4,3),A3(8,3),B(2,0),B1(4,0),B2(8,0),B3(16,0).
O
(1)观察每次变换前后的三角形有何变化,找出规律,按此规律再将△OA3B3变换成△OA4B4,则A4的坐标为,B4的坐标为.
(2)按以上规律将△OAB进行n次变换得到△OAnBn,则可知An的坐标为,
Bn的坐标为.
(3)可发现变换的过程中A、A1、A2…An纵坐标均为.
2012~2013学年度下学期期末学情诊断(东平)
数学试题参考答案
一、选择题
题号
1
2
3
4
5
6
7
8
9
10
答案
C
D
D
B
B
B
D
D
D
C
题号
11
12
13
14
15
16
17
18
19
20
答案
B
C
C
A
C
D
C
C
B
A
二、填空题
21.
22.(x2+4)(x+2)(x-2)23.70°24.(3,2)
三、解答题
25.解:
(1)
··································································2分
·············································································3分
·························································································4分
(2)
···········································································2分
·································································4分
26.解:
a(a+4b)-(a+2b)(a-2b)=a2+4ab-(a2-4b2)·············································3分
=4ab+4b2·····················································5分
由a2+2ab+b2=0,得:
a+b=0·························································7分
所以:
原式=4b(a+b)=0.····································································9分
27.解:
设甲服装的成本是x元,乙服装的成本是y元,依题意得:
······················1分
·················································5分
解得x=300,y=200.·····································································8分
答:
甲、乙两件服装的成本分别为300元、200元.·······························9分
28.解:
因为∠BAC,∠BCA的外角平分线交于点P.
所以∠PAC+∠PCA=
(∠DAC+∠ECA)
又因为∠DAC=∠B+∠BCA,∠ECA=∠B+∠BAC,·································2分
所以∠DAC+∠ECA=2∠B+(∠BCA+∠BAC),·········································4分
又因为∠B=90°,∠BAC+∠BCA=90°,
所以
3×90°=270°,··················································6分
°=135°,························································8分
∠P=180°-(∠PAC+∠PCA)=180°-135°=45°.········································10分
29.解:
(1)(16,3)(32,0)······································································4分
(2)(2n,3)(2n+1,0)···································································10分
(3)3························································································12分