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09 学探诊初三上 期末试题(1)

09 学探诊初三上 期末试题(1)
09 学探诊初三上 期末试题(1)

期末检测题(一)

一、选择题(本题共32分,每小题4分)在下列各题的四个各选答案中,只有一个是正确的.

1.一元二次方程x 2

-9=0的根为( )

A .x =3

B .x =-3

C .x 1=3,x 2=-3

D .x 1=0,x 2=3 2.如图,在△ABC 中,D

E ∥BC ,DE 分别与AB 、AC 相交于

点D 、E ,若AD =4,DB =2,则EC

AE 的值为( )

A .

2

1 B .

2 C .3

2 D .

2

3

3.已知正六边形的外接圆半径为R ,那么这个正六边形的边长为( )

A .R

B .R 2

C .2R

D .R 3

4.已知⊙O 1、⊙O 2的半径分别为2和5,圆心距O 1O 2=7,则⊙O 1和⊙O 2的位置关系

是( ) A .外切 B .内切 C .相交 D .相离

5.盒中装有4只白球5只黑球,从中任取一只球,取出的球是白球的概率是( )

A .

20

5 B .

9

5 C .

20

4 D .

9

4

6.若将抛物线y =3x 2平移,得到抛物线y =3(x -2)2-1可采用的办法是( )

A .向左平移2个单位,再向上平移1个单位

B .向左平移2个单位,再向下平移1个单位

C .向右平移2个单位,再向上平移1个单位

D .向右平移2个单位,再向下平移1个单位

7.北京市为迎接2008年奥运会,决定改善城市面貌,绿化环境,计划经过两年时间,

绿地面积增加69%,则这两年平均每年绿地面积的增长率是( ) A .29% B .30% C .31% D .35% 8.如果一个圆锥的轴截面是等边三角形,它的边长为4cm ,那么圆锥的全面积是( )

A .8πcm 2

B .10πcm 2

C .12πcm 2

D .9πcm 2

二、填空题(本题共16分,每小题4分)

9.学校招收书法班学生,从每5个报名的人中录取3人,如果有200人报名,那么估

计有______人被录取.

10.关于x 的方程0122

=++x k x 有两个不相等的实数根,则k 的取值范围为______.

11.大矩形的周长是与它相似的小矩形周长的2倍,小矩形的面积为5cm 2

,大矩形的面积为______cm 2

12.如图,A 点是半圆上一个三等分点,B 点是的中点,P 点

是直径MN 上一动点,⊙O 的半径为1,则AP +BP 的最小值是______.

三、解答题(本题共30分,每小题5分)

13.解方程:.020522=-+x x

14.从地面竖直向上抛出一个小球.小球的上升高度h (单位:m)与小球运动时间t (单

位:s )的关系式是h =20t -5t 2

.小球运动的时间是多少时,小球最高?小球运动中的最大高度是多少?

15.已知:如图,AB ,CD 是⊙O 的直径,∠C =∠B ,

求证:CF =BE .

16.已知:如图,△ABC 中,DE ∥BC ,EF ∥AB ,

求证:

?=EC

EF AE

AD

17.如图,有一表面凸凹不平的圆盘和一把L 型且带有刻度的直角三角尺,尺的两直

角边的长度大于圆盘的半径,但小于圆盘的直径,请你设计能计算出圆盘直径的测

量方案(请画出图形,并说明测量步骤).

18.小明有红、黄、白、黑四件衬衫,又有米色、蓝色、灰色三条长裤.如果他喜欢穿白色衬衫和米色长裤,那么他在黑暗中随机摸出一套衣裤正是他喜欢的搭配,这种

巧合发生的概率是多少,并用列表或树图说明理由.

四、解答题(本题共20分,每小题5分)

19.如图,⊙O 中,弦AB ,CD 相交于P ,且四边形OEPF 是正方形,连接OP .若⊙

O 的半径为5cm ,cm 23 OP ,求AB 的长.

20.已知二次函数图象的顶点坐标为M (3,-2),且与y 轴交于N (0,2

5).

(1)求该二次函数的解析式,并用列表、描点画出它的图象;

(2)若该图象与x 轴交于A 、B 两点,在对称轴上侧的图象上存在点C ,使得△ABC 的面积等于12,求出C 点的坐标.

21.如图,在△ABC 中,若AB =5,AC =2,∠BAC =120°.以BC 为边作等边三角形BCD ,把△ABD 绕D 点按顺时针方向旋转60°到△ECD 的位置.

(1)求∠BAD 的度数; (2)求AE 的长.

22.某商店销售一批小家电,平均每天可售出20个,每个盈利50元,为扩大销售,增

加盈利,减少库存,商场决定采用适当降价的措施.经调查发现,如果每个小家电每降价1元,商店平均每天可多售出2个,若商场平均每天要盈利1600元,每个小家电应降价多少元商店可达到减少库存的目的.

五、解答题(本题共22分,第23题6分,第24题8分,第25题8分)

23.如图,AB,AC分别是⊙O的直径和弦,D为劣弧上的一点,DE⊥AB于点H,交⊙O于点E,交AC于点F,P为ED的延长线上的一点.

(1)当△PCF满足什么条件时,PC与⊙O相切,用给出的条件证明结论;

(2)当点D在劣弧的什么位置时,才能使AD2=DE·DF,请加以证明.

24.如图,直角坐标系内的梯形AOBC(O为原点)中AC∥OB,AO⊥OB,AC=1,OA =2,OB=5.

(1)求经过O,C,B三点的抛物线的解析式;

(2)延长AC交抛物线于点D,求线段CD的长;

(3)在(2)的条件下,动点P、Q分别从O、D同时出发,都以每秒1个单位的速度

运动,其中点P沿OB由O向B运动,点Q沿DC由D由C运动(其中一个点

运动到终点后,另一个点运动也随之停止),过点Q作QM⊥CD交BC于点M,

连结PM.设动点运动的时间为t秒,请你探索:当时间t为何值时,△PMB中

有一个角是直角.

25.如图1,在等腰梯形ABCD 中AB ∥DC ,已知AB =12,,24 BC ∠DAB =45°,

以AB 所在直线为x 轴,A 为坐标原点建立直角坐标系,将等腰梯形AB -CD 绕A 点按逆时针方向旋转90°得到等腰梯形OEFG (O 、E 、F 、G 分别是A 、B 、C 、D 旋转后的对应点).

图1 图2

(1)写出C 、F 两点的坐标;

(2)将等腰梯形ABCD 沿x 轴的负半轴平行移动,设移动后的OA 的长度是x ,如图2,等腰梯形ABCD 与等腰梯形OEFG 重合部分的面积为y ,当点D 移动到等腰梯形OEFG 的内部时,求y 与x 之间的函数关系式并写出自变量x 的取值范围;

(3)在直线CD 上是否存在点P ,使△EFP 为等腰三角形,若存在,求出P 点的坐标,若不存在,说明理由.

答案与提示

期末检测题(一)

二、填空题(本题共16分,每小题4分)

9.120. 10.k >1. 11.20. 12..2 三、解答题(本题共30分,每小题5分) 13.解:.20,52,1-===c b a

0100)20(14)52(42

2

>=-??-=-ac b

,1

2100

52?±-=

x

.55,5521--=+-=x x

14.解:h =-5t 2+20t =-5(t 2-4t +4)+20=-5(t -2)2+20

所以,t =2时,h =20.

答:当t =2s 时,小球最高,最大高度是20m . 另解:h =-5t 2+20t ,a =-5,b =20,c =0. 所以,a

b t 2-=

时,h 运动到最大高度,即.2)

5(220=-?-=t

.20)

5(4200)5(442

2

=-?-?-?=

-=

b a

c h

答:当t =2s 时,小球最高,最大高度是20m . 15.证明:连结AE ,FD .

∵AB ,CD 是⊙O 的直径.

∴∠AEB =∠DFC =90°,AB =CD . ∵∠C =∠B . ∴△ABE ≌△DCF . ∴FC =BE . 另证:连结FO ,OE ∵∠B =∠C , ∴∠FOD =∠EOA 有

∵AB ,CD 是⊙O 的直径, ∴

=

.∴

.∴FC =BE .

16.解:∵在△ABC 中,DE ∥BC ,EF ∥AB ,

∴∠ADE =∠B =∠EFC .

∵DE ∥BC ,

∴∠AED =∠C , ∴△ADE ∽△EFC .

?

=∴EC

EF AE

AD

17.方案:(1)L 型直角尺两直角边紧靠圆盘,如图所示,图中点A 、B 表示圆盘与直角尺两

直角边的切点.

(2)量出MA 的长度,再乘以2就是圆盘的直径.

18.

?121

四、解答题(本题共20分,每小题5分) 19.解:连结OA .

∵四边形OEPF 是正方形,

∴OE ⊥AB 且平分AB 有AE =EB .

,cm 23=OP

∴OE 2+PE 2=OP 2

有OE =3cm ,

∵OA =5cm ,∴AE 2=OA 2-OE 2有AE =4cm . ∵AB =2AE ,∴AB =8cm .

20.(1)由于二次函数图象的顶点是(3,-2),设所求的二次函数解析式是y =a (x -3)2

-2.由

于所求图象过),25

,0(N

可得

.2)30(2

52

--=a

解得?=2

1a 所以?+

-=

2

532

12

x x y

列表:

(2)当02

5321

2=+

-x x 时,x 1=1,x 2=5.

∴点A (1,0),点B (5,0), 则 AB =4.

∵△ABC 的面积为12.

,12||2

1=?∴h AB

∴|h |=6.

∴抛物线顶点是(3,-2). h 1=6,h 2=-6(舍去).

?+

-=

2

532

162

x x

解出,x 1=7,x 2=-1. 由于抛物线对称轴是x =3, 所以x 2=-1(舍去).有点c (7,6).

21.解:(1)∵把△ABD 绕D 点按顺时针方向旋转60°,到△ECD 位置,

∴∠ADE =60°,AD =DE ,AB =CE .

.602

60

180

=-=

∠=∠∴DEA DAE

∵∠BAC =120°,∴∠BAD =120°-60°=60°.

(2)由(1)知CE =AB =5,AC =2,∠BAD =60°,有∠DCE +∠BCD +∠BAC =180°, ∴AE =7.

22.解:设每个小家电应降价x 元,根据题意,得

(50-x )(20+2x )=1600. 即x 2-40x +300=0.

得,x 1=30,x 2=10.

因为要尽量减少库存,所以x =30. 答:每个小家电应降价30元.

五、解答题(本题共22分,第23题6分,第24题8分,第25题8分)

23.解:(1)当PC =PF (或∠PCF =∠PFC ,或△PCF 为等边三角形)时,

PC 与⊙O 相切,下面对满足条件PC =PF ,进行证明

连结OC ,则∠OCA =∠FAO . ∵DE ⊥AB 于H ,PC =PF , ∴∠AHF =90°, ∠PCF =∠PFC . ∵∠AFH =∠PFC .

∴∠OCA +∠PCF =∠FAH +∠AFH =90°. 即OC ⊥PC ,∴PC 与⊙O 相切. (2)当点D 是劣弧

的中点,AD 2=DE ·DF .

连结AE , ∵D 点是劣弧的中点,

∴∠DAF =∠DEA . ∵∠ADF =∠ADE , ∴△ADF ∽△EDA .

AD

DF DE

AD =

,即AD 2

=DE ·DF .

24.解:(1)由题意知,O (0,0),C (1,2),B (5,0).

设过O 、C 、B 三点的抛物线的解析式为y =ax 2

+bx , 将C 、B 点坐标代入y =ax 2+bx ,得

??

?=+=+.0525,2b a b a 可得???

????=-=.

25,21b a .2

5212

x x y +

-

=∴ (2)当y =2时,则,22

5212

=+

-

x x

解得,x 1=1,x 2=4.

∴CD =4-1=3.

(3)延长QM 交x 轴于点N ,有MN ⊥OB . ①当点P 与点N 重合时,有

MP ⊥OB ,则四边形AOPQ 是矩形.

∴AQ =OP 即4-t =t ∴t =2.

②若MP ⊥BM ,则△PNM ∽△MNB . ∴MN 2=PN ·BN . ∵CQ ∥NB ,

∴△CQM ∽△BNM . ,CQ

BN MQ

MN =∴即

?----=

-t

t MN

MN 14)4(52

则?+=2

1t MN

∵BN =1+t ,PN =5-(1+t )-t =4-2t ,

).1)(24()

21(

2

+-=+∴t t t

解得,t 1=-1(舍去),,3

52=

t 综合①,②知,当t =2或3

5=t 时,△PMB 中有一个角是直角.

25.解:(1)过C 作CH ⊥x 轴于点H .

,24=BC ∠CBA =∠DAB =45°.

∴CH =HB =4.

∴C 点坐标为(8,4).

同理可求得F 点坐标为(-4,8).

(2)设AD 、CD 分别与OG 、OE 交于点M 、N . ∵∠DAB =∠GOA =45°,

.4,2

222===

=∴ON x OA AB OM 连结OD ,则S 四边形MOND =S △DMO +S △DNO , 即ON DN MO DM y ?+

?=

2

12

1

4)4(2

1

22)2224(2

1?-+?-=

x x x ).84(844

12

<<-+-=x x x

(3)设P 点坐标为(a ,4).

①若PE =PF ,

在Rt △PNE 和Rt △PGF 中,

由PE 2=PN 2+NE 2=PG 2+FG 2=PF 2, 得a 2+(12-4)2=(a +4)2+42 解得a =4.

②若PF =EF .

则由PF 2

=PG 2

+FG 2

=EF 2

, 得.)24(4)4(222=++a 解得a 1=0,a 2=-8(舍去). ③若PE =EF ,

则由PE 2=PN 2+NE 2=EF 2, 得.)24()412(222=-+a

化简得a 2+32=0,方程无解,此时P 点不存在. 综合①、②、③知,所求P 点坐标为P 1(4,4),P 2(0,4).

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