陕西中考数学试题及答案.docx
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陕西中考数学试题及答案.docx
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陕西中考数学试题及答案
2009年陕西省初中毕业学业考试
数学
第Ⅰ卷(选择题共30分
A卷
一、选择题(共10小题,每小题3分,计30分.每小题只有一个选项是符合题意的1.12
-
的倒数是(.
A.2B.2-C.12
D.12
-
2.1978年,我国国内生产总值是3645亿元,2007年升至249530亿元.将249530亿元
用科学记数表示为(.A.13
24.95310⨯元B.12
24.95310⨯元C.13
2.495310⨯元D.14
2.495310⨯元
3.图中圆与圆之间不同的位置关系有(.A.2种B.3种C.4种D.5种4.王老师为了了解本班学生课业负担情况,在班中随机调查了10名学生,他们每人上周平均每天完成家庭作业所用的时间分别是(单位:
小时:
1.5,2,2,2,2.5,2.5,2.5,2.5,3,3.5.则这10个数据的平均数和众数分别是(.A.2.4,2.5B.2.4,2C.2.5,2.5D.2.5,25.若正比例函数的图象经过点(1-,2,则这个图象必经过点(.A.(1,2B.(1-,2-C.(2,1-D.(1,2-6.如果点(12Pmm-,在第四象限,那么m的取值范围是(.A.102
m<<
B.102
m-
< 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
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