m>
7.若用半径为9,圆心角为120°的扇形围成一个圆锥的侧面（接缝忽略不计,则这个圆锥的底面半径是（.A.1.5B.2C.3D.6
8.化简2
ba
aaab⎛⎫-⎪-⎝⎭
的结果是（.A.ab-B.ab+C.
1
ab-D.
1
ab
+
9.如图,9030AOBB∠=∠=°,°,AOB''△可以看作是由AOB△绕点O顺时针旋转α角度得到的.若点A'在AB上,
则旋转角α的大小可以是（.
A.30°B.45°C.60°D.90°
10.根据下表中的二次函数2
yaxbxc=++的自变量x与函数
y的对应值,可判断该二次函数的图象与x轴（.
（第3题图
120°
（第7题图
A
O
B
A'B'
（第9题图
AC.有两个交点,且它们均在y轴同侧D.无交点
第Ⅱ卷（非选择题共90分
二、填空题（共6小题,每小题3分,计18分11.0
31--=__________.
12.如图,ABCD∥,直线EF分别交ABCD、于点EF、,147∠=°,则2∠的大小是__________.13.若1122（（AxyBxy,,,是双曲线3y
x
=
上的两点,
且120xx>>,则12_______yy{填“>”、“=”、“<”}.
14.如图,在梯形ABCD中,DCAB∥,DACB=.若104ABDC==,,tan2A=,则这个梯形的面积是__________.
15.一家商店将某种商品按成本价提高50%后,标价为450元,又以8折出售,则售出这件商品可获利润__________元.
16.如图,在锐角ABC△中,45ABBAC=∠=°,
BAC∠的平分线交BC于点DMN,、分别是AD和AB上
的动点,则BMMN+的最小值是___________.
三、解答题（共9小题,计72分17.（本题满分5分解方程:
2
2312
4
xxx--=
+-.
18.（本题满分6分
如图,在ABCD中,点E是AD的中点,连接CE并延长,交BA的延长线于点F.求证:
FAAB=.
A
BD
CF
1
（第12题图
A
B
C
D
（第14题图
A
B
C
D
N
M
（第16题图
ABCD
EF（第18题图
19.（本题满分7分
某校为了组织一项球类对抗赛,在本校随机调查了若干名学生,对他们每人最喜欢的一项球类运动进行了统计,并绘制成如图①、②所示的条形和扇形统计图.
根据统计图中的信息,解答下列问题:
（1求本次被调查的学生人数,并补全条形统计图;
（2若全校有1500名学生,请你估计该校最喜欢篮球运动的学生人数;
（3根据调查结果,请你为学校即将组织的一项球类对抗赛提出一条合理化建议.20.（本题满分8分
小明想利用太阳光测量楼高.他带着皮尺来到一栋楼下,发现对面墙上有这栋楼的影子,针对这种情况,他设计了一种测量方案,具体测量情况如下:
如示意图,小明边移动边观察,发现站到点E处时,可以使自己落在墙上的影子与这栋楼落在墙上的影子重叠,且高度恰好相同.此时,测得小明落在墙上的影子高度1.2CD=m,0.8CE=m,30CA=m（点AEC、、在同一直线上.
已知小明的身高EF是1.7m,请你帮小明求出楼高AB（结果精确到0.1m.
项目①
足球20%
篮球26%乒乓球32%
他②
（第19题图
（第20题图
21.（本题满分8分
在一次运输任务中,一辆汽车将一批货物从甲地运往乙地,到达乙地卸货后返回.设汽车从甲地出发x（h时,汽车与甲地的距离为y（km,y与x的函数关系如图所示.根据图象信息,解答下列问题:
（1这辆汽车的往、返速度是否相同?
请说明理由;（2求返程中y与x之间的函数表达式;
（3求这辆汽车从甲地出发4h时与甲地的距离.
22.（本题满分8分
甲、乙两同学用一副扑克牌中牌面数字分别是3、4、5、6的4张牌做抽数学游戏.游戏规则是:
将这4张牌的正面全部朝下,洗匀,从中随机抽取一张,抽得的数作为十位上的数字,然后,将所抽的牌放回,正面全部朝下、洗匀,再从中随机抽取一张,抽得的数作为个位上的数字,这样就得到一个两位数.若这个两位数小于45,则甲获胜,否则乙获胜.你认为这个游戏公平吗?
请运用概率知识说明理由.23.（本题满分8分
如图,O⊙是ABC△的外接圆,ABAC=,过点A作APBC∥,交BO的延长线于点P.（1求证:
AP是O⊙的切线;
（2若O⊙的半径58RBC==,,求线段AP的长.
（第21题图
（第23题图
24.（本题满分10分
如图,在平面直角坐标系中,OBOA⊥,且2OBOA=,点A的坐标是（12-,.（1求点B的坐标;
（2求过点AOB、、的抛物线的表达式;
（3连接AB,在（2中的抛物线上求出点P,使得ABPABOSS=△△.25.（本题满分12分问题探究
（1请在图①的正方形ABCD内,画出使90APB∠=°的一个..点P,并说明理由.（2请在图②的正方形ABCD内（含边,画出使60APB∠=°的所有..的点P,并说明理由.
问题解决
（3如图③,现在一块矩形钢板43ABCDABBC==,,.工人师傅想用它裁出两块全等的、面积最大的APB△和CPD'△钢板,且60APBCPD'∠=∠=°.请你在图③中画出符合要求的点P和P',并求出APB△的面积（结果保留根号.
DCBA①DCB
A③DCBA②（第25题图
2009年陕西省初中毕业学业考试
数学试题参考答案
A卷
一、选择题（共10小题,每小题3分,计30分
二、填空题（共6小题,每小题3分,计18分
11.212.133°13.<14.4215.6016.4三、解答题（共9小题,计72分17.（本题满分5分解:
2
2
（2（43xx---=.························································································（2分
45x-=-.
54
x=
.·······················································································（4分
经检验,54
x=
是原方程的解.····················································································（5分
18.（本题满分6分
证明:
四边形ABCD是平行四边形,ABDCABDC∴=,∥.
FAEDFECD∴∠=∠∠=∠,.············（3分又EAED=,
AFEDCE∴△≌△.·······························（5分AFDC∴=.AFAB∴=.·············································（6分
19.（本题满分7分解:
（11326%50÷=,
∴本次被调查的人数是50.·
·········（2分补全的条形统计图如图所示.·······（4分
（第19题答案图
项目
A
B
C
D
E
F
（2150026%390⨯=,
∴该校最喜欢篮球运动的学生约为390人.·
·······························································（6分（3如“由于最喜欢乒乓球运动的人数最多,因此,学校应组织乒乓球对抗赛”等.（只要根据调查结果提出合理、健康、积极的建议即可给分··············································（7分20.（本题满分8分
解:
过点D作DGAB⊥,分别交ABEF、于点GH、,则1.2EHAGCD===,
0.830DHCEDGCA====,.························（2分EFAB∥,FHDHBG
DG
∴=.·························································（5分
由题意,知1.71.20.5FHEFEH=-=-=.0.50.8
30
BG∴
=,解之,得18.75BG=.···················（7分18.751.219.9520.0ABBGAG∴=+=+=≈.
∴楼高AB约为20.0米.·
·····························································································（8分21.（本题满分8分解:
（1不同.理由如下:
往、返距离相等,去时用了2小时,而返回时用了2.5小时,
∴往、返速度不同.·
·····································································································（2分（2设返程中y与x之间的表达式为ykxb=+,
则1202.505.
kbkb=+⎧⎨
=+⎩,
解之,得48240.kb=-⎧⎨=⎩,
········································································································（5分
∴48240yx=-+.
（2.55xx≤≤
（评卷时,自变量的取值范围不作要求·····（6分
（3当4x=时,汽车在返程中,48424048y∴=-⨯+=.
∴这辆汽车从甲地出发4h时与甲地的距离为48km.·················································（8分22.（本题满分8分
解:
这个游戏不公平,游戏所有可能出现的结果如下表:
表中共有16种等可能结果,小于45的两位数共有6种.··········································（5分
（第20题答案图
（（6310516
8
16
8
PP∴==
=
=
甲获胜乙获胜,.·····································································（7分
3588
≠,
∴这个游戏不公平.·
·····································································································（8分23.（本题满分8分解:
（1证明:
过点A作AEBC⊥,交BC于点E.ABAC=,AE∴平分BC.
∴点O在AE上.·
······································（2分又APBC∥,AEAP∴⊥.
AP∴为O⊙的切线.································（4分（2142
BEBC==,
3OE∴=
=.
又AOPBOE∠=∠,OBEOPA∴△∽△.···································································································（6分
BEOEAP
OA
∴=.即
435
AP
=
.
203
AP∴=.·················································································································（8分
24.（本题满分10分
解:
（1过点A作AFx⊥轴,垂足为点F,过点B作BEx⊥轴,垂足为点E,
则21AFOF==,.
OAOB⊥,
90AOFBOE∴∠+∠=°.又90BOEOBE∠+∠=°,AOFOBE∴∠=∠.
RtRtAFOOEB∴△∽△.
2BE
OE
OB
OFAFOA
∴===.
24BEOE∴==,.（42B∴,.·····················································································································（2分（2设过点（12A-,,（42B,,（00O,的抛物线为2
yaxbxc=++.
216420.
abcabcc-+=⎧⎪
∴++=⎨⎪=⎩,,解之,得12
320abc⎧
=⎪⎪
⎪=-⎨⎪=⎪⎪⎩
,.
P
（第23题答案图
∴所求抛物线的表达式为2
132
2
yxx=
-
.··································································（5分
（3由题意,知ABx∥轴.
设抛物线上符合条件的点P到AB的距离为d,则112
2
ABPSABdABAF=
=
△.
2d∴=.
∴点P的纵坐标只能是0,或4.·
··············································································（7分令0y=,得213
022
xx-=.解之,得0x=,或3x=.
∴符合条件的点1（00P,
2（30P,.令4y=,得2
1342
2
xx-
=.解之,得32
x±
=
.
∴
符合条件的点33（
42
P-
43（
42
P+
.
∴综上,符合题意的点有四个:
1（00P,,2（30P,
33（
42
P-
43（
42
P+
.·
·········································（10分（评卷时,无1（00P,
不扣分25.（本题满分12分
解:
（1如图①,
连接ACBD、交于点P,则90APB∠=°.
∴点P为所求.·
······················································（3分（2如图②,画法如下:
1以AB为边在正方形内作等边ABP△;
2作ABP△的外接圆O⊙,分别与ADBC、交于点EF、.在O⊙中,弦AB所对的
APB上的圆周角均为60°,EF
∴上的所有点均为所求的点P.···················（7分（3如图③,画法如下:
1连接AC;
2以AB为边作等边ABE△;
3作等边ABE△的外接圆O⊙,交AC于点P;4在AC上截取APCP'=.则点PP'、为所求.·············································（9分（评卷时,作图准确,无画法的不扣分过点B作BGAC⊥,交AC于点G.在RtABC△中,43ABBC==,.
D
C
B
A
①
②
③
（第25题答案图
5AC∴=
=.
12
5
ABBC
BGAC∴=
=
.·····························································································（10分
在RtABG△中,4AB=,
16
5
AG∴=
=
.
在RtBPG△中,60BPA∠=°,
12tan605
3
5
BGPG∴=
=
⨯
=
°
.
∴165
5
APAGPG=+=
+
.
11
16122
255
525APBSAPBG⎛∴=
=
⨯+⨯=⎝⎭
△.·································（12分
升级会员 