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第一学期四校期末联考试卷及答案

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联考试卷九年级数学试题

一 填空题(每小题3分,共30分)

1.当x 时,二次根式x3在实数范围内有意义 2.32_____________。

223.方程x2= x 的根是_______________。 4.方程4x23(4x3)的根的情况是 5.点(4,-3)关于原点对称的点的坐标是 _____________.

6.已知x1是关于x的方程2x2axa20的一个根,则a_____.

7.如图1,AB是⊙O的直径,D是AC的中点,OD∥BC,若BC=8,则OD=_________.

C D O C B A

O A

B

图1

(第8题图)

8.如图,点A、B、C在⊙O上,AO∥BC,∠AOB = 50°. 则∠OAC的度数

是 . 9.如图,△ABC是等边三角形,点P是△ABC内一点。△APC沿逆时针方向 旋转后与△AP'B 重合,则旋转中心是___,最小旋转角等于___°.

AP 'PB第5题C10.观察下列各式:113213,214314,315415……,请你将

发现的规律用含自然数n(n≥1)的等式表示出来__________________________ 二选择题(每小题4分,共28分)在每小题给出的四个选项中,只有一项是符合题目要求的,请把正确选项的代号填在各题后的括号中.

11. 下列计算中,正确的是( )

A、235 B、2222 C、32222 D、182894321

12. 下面的图形中,是中心对称图形的是( )

A. B. C. D.

13.将一元二次方程x22x20配方后所得的方程是 ( )

A.(x2)22 B.(x1)22 C.(x1)23 D.(x2)23 14. m是方程x+x-1=0的根,则式子m+2m+2008的值为( )

A.2007 B.2008 C.2009 D.2010

15. ⊙O的半径为5,圆心O到直线l的距离为3,则直线l与⊙O的位置 关系是( )

A. 相交 B. 相切 C. 相离 D. 无法确定 16.如图2所示,圆O的弦AB垂直平分半径OC.则四边形OACB( ) A. 是正方形 B. 是长方形 C. 是菱形 D.以上答案都不对

图2

17. 如图,把边长为3的正三角形绕着它的中心旋转180°后,重叠部分的面积为( )

A.

9432

3

2

B.

323 C.

343 D.

32

(第17题) 三、解答题(共8小题,计92分)

18.(8分)

312-31+3(31)-20080

-32

19(10分)先化简,再求值 b131ab2b2aba2abb1,其中a13,

20(10分) 已知关于x的一元二次方程x2-mx-2=0. ……①

(1) 若x=-1是方程①的一个根,求m的值和方程①的另一根; (2) 对于任意实数m,判断方程①的根的情况,并说明理由.

21(12分).)某居民小区一处圆柱形的输水管道破裂,维修人员为更换管道,需确定管道圆形截面的半径,下图是水平放置的破裂管道有水部分的截面. (1)请你补全这个输水管道的圆形截面;

(2)若这个输水管道有水部分的水面宽AB=16cm,水面最深地方的高度为4cm,求这个圆形截面的半径.

A B

22(12分)。如图,在平面直角坐标系中,将四边形ABCD称为“基本图形”,

且各点的坐标分别为A(4,4),B(1,3),C(3,3),D(3,1).

(1)画出“基本图形”关于原点O对称的四边形A1B1C1D1,并求出A1,B1,C1,

D1的坐标.

A1( , ),B1( , ),C1( , ),D1( , ) ; (2)画出“基本图形”关于x轴的对称图形A2B2C2D2 ;

(3)画出四边形A3B3C3D3,使之与前面三个图形组成的图形既是中心对称图形

又是轴对称图形.

23(13分)顾客李某于今年“五·一”期间到电器商场购买空调,与营业员有如下的一段对话:

顾客李某:A品牌的空调去年“国庆”期间价格还挺高,这次便宜多了,一次降价幅度就达到19%,是不是质量有问题?

营业员:不是一次降价,这是第二次降价,今年春节期间已经降了一次价,两次降价的幅度相同.我们所销售的空调质量都是很好的,尤其是A品牌系列空调的质量是一流的.

顾客李某:我们单位的同事也想买A品牌的空调,有优惠吗? 营业员:有,请看《购买A品牌系列空调的优惠办法》. 购买A品牌系列空调的优惠办法: 方案一:各种型号的空调每台价格优惠5%,送货上门,负责安装,每台空调另加运输费和安装费共90元. 方案二:各种型号的空调每台价格优惠2%,送货上门,负责安装,免运输费和安装费. 根据以上对话和A品牌系列空调销售的优惠办法,请你回答下列问题: (1)求A品牌系列空调平均每次降价的百分率?

(2)请你为顾客李某决策,选择哪种优惠更合算,并说明为什么?

24(13分)已知:如图,△ABC中,ABAC,以AB为直径的⊙O交BC于点P,PDAC于点D. (1)求证:PD是⊙O的切线;

(2)若CAB120,AB2,求BC的值.

C P D A O B (第24题)

25(14分)已知:正方形ABCD中,MAN45,MAN绕点A顺时针旋转,它的两边分别交CB,DC(或它们的延长线)于点M,N. 当MAN绕点A旋转到BMDN时(如图1),易证BMDNMN. (1)当MAN绕点A旋转到BMDN时(如图2),线段BM,DN和MN之间有怎样的数量关系?写出猜想,并加以证明.

(2)当MAN绕点A旋转到如图3的位置时,线段BM,DN和MN之间又有怎样的数量关系?请直接写出你的猜想.

A D A D A D

N

N

M B

C B B C C

M M

图1 图2

N 图3

参 一

1 .≥3 2 . 1 , 3 . x1=0 , x2=1, 4.两个相等的实数根 ,5 (-4,3) 6.1或-2 7:4 8:25° 9 :A 300° 10 二

11 C 12 D 13 C 14 C 15 A 16 C 17 B 三

18 解:原式=23-3+3-3-1-2+3 (6分)

=3 (8分)

19 解:原式=

abab(ab)(ab)(ab)2b=

abab

当a13,b13时.:原式=3 20 解(1)把x=-1代入得 1+m-2=0 解得m=1

∴x2-x-2=0. ∴x1=2 x2=-1 另一根是2

(2)∵b2-4ac=m2-4 (-2)=m2+8>0 ∴方程①有两个实数根 21解 (2)半径为10

22(1)A1(-4,-4 ),B1(-1,-3),C1(-3,-3),D1(-3,-1) .

正确写出每个点的坐标得4分;正确画出四边形A1B1C1D1给2分. (2)正确画出图形A2B2C2D2给3分. (3)正确画出图形A3B3C3D3给3分.

23.(1)A品牌系列空调平均每次降价的百分率为x

根据题意,得 (1-x)2=1-19%

解得x1=0.1=10﹪ x2=1.9(舍去)

(2)当A品牌系列空调的某一型号的价格为每台小于3000元时,应选方案二;当A品牌系列空调的某一型号的价格为每台3000元时,两种方案都可以选;当A品牌系列空调的某一型号的价格为每台大于3000元时,应选方案一. 24.(1)证明:ABAC,

······················································································································ 2分 CB. ·

又OPOB,

························································································································ 4分 OPBB ·

·················································································································· 5分 COPB. ·························································································································· 6分 OP∥AD ·

又PDAC于D,ADP90,

DPO90. ·················································································································· 7分 PD是⊙O的切线. ··········································································································· 8分

(2)连结AP,AB是直径,

APB90, ······································ 9分

C P D A O B ABAC2,CAB120,

BAP60. ················································································································· 10分

BP3,BC23. ································································································ 13分

25.解:(1)BMDNMN成立. ········································································ (2分) 如图,把△AND绕点A顺时针90,得到△ABE,

A 则可证得E,B,M三点共线(图形画正确) ······ (3分) D

证明过程中,

证得:EAMNAM ····································· (5分)

N 证得:△AEM≌△ANM ······························· (7分)

C MEMN E B M

MEBEBMDNBM

···································································································· (10分) DNBMMN ·

(2)DNBMMN ······························································································· (14分)

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