苏州市立达中学总校胥江部2016-2017学年第二学期初三二
模试卷
数学 2017.5
本试卷由选择题、填空题和解答题三大题组成.共29小题,满分130分.考试时间120分钟. 注意事项:
1.答题前,考生务必将自己的姓名、考点名称、考场号、座位号用0.5毫米黑色墨水签字笔填写在答题卡相应位置上,并认真核对条形码上的准考号、姓名是否与本人的相符; 2.答选择题必须用2B 铅笔把答题卡上对应题目的答案标号涂黑,如需改动,请用橡皮擦干净后,再选涂其他答案;答非选择题必须用0.5毫米黑色墨水签字笔写在答题卡指定的位置上,不在答题区域内的答案一律无效,不得用其他笔答题;
3.考生答题必须答在答题卡上,保持卡面清洁,不要折叠,不要弄破,答在试卷和草稿纸上一律无效.
一、选择题:本大题共10小题,每小题3分,共30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,请将选择题的答案用2B 铅笔涂在答题卡相应位置上. 1
.﹣ 的相反数是 A .3
B .﹣3 C
.
D
.﹣
2.下列运算正确的是( )
A .a 2?a 3=a 6
B .(a 3)
4
=a 12
C .5a ﹣2a =3a 2
D .(x +y )
2
=x 2+y 2 3.如左图是由4个大小相同的正方体组合而成的几何体,其主视图是
A
.
B
.
C
.
D
.
4.函数y=
3
x 中自变量x 的取值范围是
A .x ≥3
B .x ≥﹣3
C .x ≠3
D .x >0且x ≠3
5.如图,直线a ,b 被直线c 所截,若a ∥b ,∠1=110°,则∠2等于
A .70°
B .75°
C .80°
D .85° 6.下列一元二次方程中,有两个相等实数根的是
A .x 2﹣8=0
B .2x 2﹣4x +3=0
C .5x +2=3x 2
D .9x 2+6x +1=0
1
2
b
a
c
)
5(题第
7.抛物线223y x x =++的对称轴是
A .直线x =1
B .直线x = -1
C .直线x =-2
D .直线x =2 8.若x 2﹣3y ﹣5=0,则6y ﹣2x 2﹣6的值为
A .4
B .﹣4
C .16
D .﹣16 9.如图△ABC 中,∠C=90°,AC=4,BC=3,将△ABC 绕点A 逆时针旋转,使点C 落在线段AB 上的点
E 处,点B 落在点D 处,则B 、D 两点间的距离为( ) A .
2
B
.
C .3
D .
2
10.如图点A 、B 在反比例函数y
=(k >0,x >0)图象上,BC ∥x 轴,交y
轴于点C ,动点P 从坐标原点O 出发,沿O →A →B →C (图中“→”所示路线)匀速运动,终点为C ,过P 作PM ⊥x 轴,垂足为M .设三角形OMP 的面积为S ,P 点运动时间为t ,则S 关于x 的函数图象大致为
A
. B
. C
. D
.
二、填空题:本大题共8小题,每小题3分,共24分.把答案直接填在答题卡相应位置上. 11.分解因式:29a -= ▲ .
12.2017年春节期间,在网络上用“百度”搜索引擎搜索“开放二孩”,能搜索到与之相关的结果个数约为45100000,这个数用科学记数法表示为 ▲ .
13.如图,等腰三角形ABC 的顶角为1200,底边BC 上的高AD= 4,则腰长为 ▲ .
第13题 第14题 第15题
O
B
C
D
A
14.小球在如图所示的地板上自由地滚动,并随机地停留在某块方砖上,那么小球最终停留在黑色区域的概率是 ▲ .
15.如图,四边形ABCD 内接于O ,若四边形ABCO 是平行四边形,则ADC ∠的大小为 ▲ .
16.已知扇形的半径为6cm ,面积为10πcm 2,则该扇形的弧长等于▲ . 17.如图,是矗立在高速公路水平地面上的交通警示牌,经测量得到如下数据:AM=4米,AB=8米,∠MAD=45°,∠MBC=30°,则警示牌的高CD 为 ▲ 米(结果保留根号).
第17题 第18题
18.如图,正五边形的边长为2,连接对角线AD ,BE ,CE ,线段AD 分别与BE 和CE 相交于点M ,N ,给出下列结论:①∠AME =108°;②2AN
AM AD =?;③MN
=3;④1BE =.其中正确结论的序号是 ▲ .
三、解答题:本大题共11小题,共76分.把解答过程写在答题卡相应位置上,解答时应写出必要的计算过程、推演步骤或文字说明,作图时用2B 铅笔或黑色墨水签字笔.
19.(本题满分5
分)计算:202(π--+.
20.(本题满分5分)解不等式组:()
12221x x x ->???+≥-??
21.(本题满分6分)
21111x x x ?
?÷+ ?
--??
,其中x 1. 22.(本题满分6分)某校学生利用双休时间去距学校10 km 的天平山社会实践活动,一部
分学生骑电瓶车先走,过了20 min 后,其余学生乘公交车沿相同路线出发,结果他们同时到达.已知公交车的速度是电瓶车学生速度的2倍,求骑电瓶车学生的速度和公交车的速度?
23.(本题满分8分)如图,四边形ABCD 为平行四边形,∠BAD 的角平分线AE 交CD 于点F ,交BC 的延长线于点E .
(
1)求证:BE =CD ;
(2)连接BF ,若BF ⊥AE ,∠BEA =60°,AB =4,求平行四边形ABCD 的面积.
24.(本题满分8分)为庆祝建军90周年,某校计划在五月份举行“唱响军歌”歌咏比赛,要确定一首喜欢人数最多的歌曲为每班必唱歌曲.为此提供代号为A ,B ,C ,D 四首备选曲目让学生选择,经过抽样调查,并将采集的数据绘制如下两幅不完整的统计图.请根据图①,图②所提供的信息, 解答下列问题:
(1)本次抽样调查中,选择曲目代号为A 的学生占抽样总数的百分比为 ▲ ; (2)请将图②补充完整;
(3)若该校共有1260名学生,根据抽样调查的结果估计全校共有多少学生选择喜欢人数最多的歌曲?(要有解答过程)
25.(本题满分8分)如图,在平面直角坐标系中,O 为坐标原点,△ABO 的边AB 垂直于x 轴,垂足为点B ,反比例函数y =(x >0)的图象经过AO 的中点C ,且与AB 相交于点D ,OB =4,AD =3,
(1)求反比例函数y
=的解析式; (2)求cos ∠OAB 的值;
(3)求经过C 、D 两点的一次函数解析式.
26(本题满分10分)如图,点P 是⊙O 外一点,P A 切⊙O 于点A ,AB 是⊙O 的直径,连接OP ,过点B 作BC ∥OP 交⊙O 于点C ,连接AC 交OP 于点D .
(1)求证:PC 是⊙O 的切线; (2)若PD =
3
16
cm ,AC =8cm ,求图中阴影部分的面积; (3)在(2)的条件下,若点E 是AB ︵
的中点,连接CE ,求CE 的长.
27.(本题满分10分)在△ABC 中,∠BAC =90°,AB =AC ,点D 为直线BC 上一动点(点D 不与B ,C 重合),以AD 为边在AD 右侧作正方形ADEF ,连接CF . (1)如图1,当点D 在线段BC 上时,①BC 与CF 的位置关系为: ▲ .
第26题图
B
A
E P
O D
C
②BC,CD,CF之间的数量关系为:▲;(将结论直接写在横线上)
(2)如图2,当点D在线段CB的延长线上时,结论①,②是否仍然成立?若成立,请给予证明;若不成立,请你写出正确结论,再给予证明.
(3)如图3,当点D在线段BC的延长线上时,延长BA交CF于点G,连接GE.若已知
AB=2,CD=BC,请求出GE的长.
28.(本题满分10分)如图平面直角坐标系中,抛物线y=ax2+bx+c(a≠0)经过△ABC
的三个顶点,与y轴相交于(0,),点A坐标为(﹣1,2),点B是点A关于y轴的对称点,点C在x轴的正半轴上.
(1)求该抛物线的函数关系表达式.
(2)点F为线段AC上一动点,过F作FE⊥x轴,FG⊥y轴,垂足分别为E、G,当四边形OEFG为正方形时,求出F点的坐标.
(3)将(2)中的正方形OEFG沿OC向右平移,记平移中的正方形OEFG 为正方形DEFG,当点E和点C重合时停止运动,设平移的距离为t,正方形的边EF与AC交于点M,DG所在的直线与AC交于点N,连接DM,是否存在这样的t,使△DMN是等腰三角形?若存在,求t的值;若不存在请说明理由.
苏州市胥江实验中学校2017届初三二模试卷
数学参考答案及评分标准
一、选择题(每小题3分,共30分)
11.(a + 3)(a - 3) 12.4.51×107 13.8 14.29
15.600
16.
10
3
∏ 17.4
18.①、②、③
三、解答题(共11大题,共76分) 19.(本题共5分)
解:原式= 3-2 + 1 ·············································································· 3分
=2 ························································································· 5分
20.(本题共5分)
解:由①式得:x>3. ············································································ 2分
由②式得:x 4≤. ·········································································· 4分
∴不等式组的解集为: 34x <≤
. ····················································· 5分
21.(本题共6分) 解:原式=
2
1
1x x x x ÷-- ··········································································· 1分 =1
(1)(1)x x x x x
-?+- ····································································· 2分
=
11
x + ···················································································· 4分
当x 1时,原式
··································································· 5分
·················································································· 6分 22.(本题满分6分)
解:设骑电瓶车学生的速度为x km /h ,汽车的速度为2x km /h ,可得:··········1分
10x =102x +2060
, ···············································································3分
解得x =15,······················································································4分 经检验,x =15是原方程的解,······························································5分 2x =2×15=30.
答:骑车学生的速度和汽车的速度分别是15 km /h ,30 km /h .·························6分
23.(本题共8分)
···············1分
················2分
·················3分
···············4分
················5分
···············6分
,·················7分
1
AEBF······8分
2
24.(本题共8分)
1)由题意可得,本次抽样调查中,选择曲目代号为A的学生占抽样总数的百
分比为:×100%=20%.··················································2分
(2)由题意可得,选择C的人数有:30÷﹣36﹣30﹣44=70(人)补全的图②柱状图正确·········································5分
(3)由题意可得,全校选择此必唱歌曲共有:1260×=490(人),答:全校共有490名学生选择此必唱歌曲.········································8分25.(本题共8分)
解:(1)设点D的坐标为(4,m)(m>0),则点A的坐标为(4,3+m),
∵点C为线段AO的中点,∴点C的坐标为(2,).∵点C、点D均在反比例函数y=
的函数图象上,∴,···························1分解得:.·········2分
∴反比例函数的解析式为y=.········································3分
(2)∵m=1,∴点A的坐标为(4,4),········································4分
∴OB=4,AB=4.在Rt△ABO中,OB=4,AB=4,∠ABO=90°,
∴OA==4,cos∠OAB===.········································5分
(3))∵m =1,∴点C 的坐标为(2,2),点D 的坐标为(4,1). 设经过点C 、D 的一次函数的解析式为y =ax +b
,则有
,解得:
.·
····7分 ∴经过C 、D 两点的一次函数解析式为y =
﹣x +3. ········································8分 26.(本题共10分)
证明: ⑴如图,连接OC ,∵P A 切⊙O 于A .
∴∠P AO =90o. ····································································································· 1分 ∵OP ∥BC ,∴∠AOP =∠OBC ,∠COP =∠OCB .∵OC =OB ,∴∠OBC =∠OCB ,
∴∠AOP =∠COP . ······························································································· 2分 又∵OA =OC ,OP =OP , ∴△P AO ≌△PCO (SAS ).∴∠P AO =∠PCO =90 o, 又∵OC 是⊙O 的半径,
∴PC 是⊙O 的切线. ······························································································ 3分 ⑵解法不唯一. 解:由(1)得P A ,PC 都为圆的切线,
∴P A =PC ,OP 平分∠APC ,∠ADO =∠P AO =90 o,∴∠P AD+∠DAO =∠DAO+∠AOD , ∴∠P AD =∠AOD ,
∴△ADO ∽△PDA . ······························································································ 4分 ∴
AD DO PD AD =
,∴2AD PD DO =?,∵AC =8, PD =16
3
, ∴AD =
1
2
AC =4,OD =3,AO =5, 5分 由题意知OD 为△ABC 的中位线,∴BC =2OD =6,AB =10.
∴S 阴=S 半⊙O -S △ACB =()2
21101254868=cm 2222
ππ-??-?? ???. 答:阴影部分的面积为
2
2548cm 2
π-. ·
······································································ 6分 (3)如图,连接AE ,BE ,过点B 作BM ⊥CE 于点M . ················································· 7分 ∴∠CMB =∠EMB =∠AEB =90o,又∵点E 是AB ︵
的中点,
∴∠ECB =∠CBM =∠ABE =45o,CM =MB
=,BE =AB cos450
= ···························· 8分 ∴ EM
CE =CM +EM
=()cm .
·······················9分
答:CE
的长为. ······················································································· 10分
第23题答图
B
27.(本题共10分)
解:(1)①垂直; ·································································································1分
②BC=CF+CD;···························2分
(2)成立,∵正方形ADEF中,AD=AF,∵∠BAC=∠DAF=90°,∴∠BAD=∠CAF,
在△DAB与△F AC中,,∴△DAB≌△F AC,···························4分
∴∠B=∠ACF,CF=BD∴∠ACB+∠ACF=90°,即CF⊥BD;∵BC=BD+CD,
∴BC=CF+CD;···························6分
(3)解:过A作AH⊥BC于H,过E作EM⊥BD于M,EN⊥CF于N,
∵∠BAC=90°,AB=AC,∴BC=AB=4,AH=BC=2,∴CD=BC=1,CH=BC=2,
∴DH=3,···························7分
由(2)证得BC⊥CF,CF=BD=5,∵四边形ADEF是正方形,∴AD=DE,∠ADE=90°,
∵BC⊥CF,EM⊥BD,EN⊥CF,∴四边形CMEN是矩形,···························8分
∴NE=CM,EM=CN,∵∠AHD=∠ADC=∠EMD=90°,
∴∠ADH+∠EDM=∠EDM+∠DEM=90°,
∴∠ADH=∠DEM,
在△ADH与△DEM中,,
∴△ADH≌△DEM,∴EM=DH=3,DM=AH=2,
∴CN=EM=3,EN=CM=3,···························9分
∵∠ABC=45°,∴∠BGC=45°,∴△BCG是等腰直角三角形,∴CG=BC=4,∴GN=1,
∴EG==.··························10分
28.(本题共10分)
解:(1)∵点B是点A关于y轴的对称点,∴抛物线的对称轴为y轴,
∴抛物线的顶点为(0,),故抛物线的解析式可设为y=ax2+.
∵A(﹣1,2)在抛物线y=ax2+上,∴a+=2,解得a=﹣,
∴抛物线的函数关系表达式为y=﹣x2+;··························2分
(2)①当点F在第一象限时,如图1,令y=0得,﹣x2+=0,
解得:x 1=3,x 2=﹣3,∴点C 的坐标为(3,0).设直线AC 的解析式为y =mx +n ,
则有,解得,∴直线AC 的解析式为y =﹣x +.·········3分
设正方形OEFG 边长为p ,则F (p ,p ).∵点F (p ,p )在直线y =﹣x +上,
∴﹣
p +=p ,解得p =1,∴点F 的坐标为(1,1).·
························4分 ②当点F 在第二象限时,同理可得:点F 的坐标为(﹣3,3), 此时点F 不在线段AC 上,故舍去.·
·························5分 综上所述:点F 的坐标为(1,1);··························6分 (3)过点M 作MH ⊥DN 于H ,如图2,则OD =t ,OE =t +1. ∵点E 和点C 重合时停止运动,∴0≤t ≤2.
当x =t 时,y =﹣
t +,则N (t ,﹣t +),DN =﹣
t +
.
当x =t +1时,y =﹣(t +1)+=﹣
t +1,则M (t +1,﹣
t +1),ME =﹣
t +1.
在Rt △DEM 中,DM 2=12+(﹣t +1)2=t 2﹣t +2.
在Rt △NHM 中,MH =1,NH =(﹣t +
)﹣(﹣
t +1)=,
∴MN 2=12+(
)2=.·
·························7分
①当DN =DM 时,(﹣t +)2=t 2﹣t +2,
解得t =
;·
·························8分 ②当ND =NM 时,
﹣
t +
=
=
,
解得t =3﹣
;·
·························9分 ③当MN =MD 时,
=t 2﹣t +2, 解得t 1=1,t 2=3.
∵0≤t ≤2,∴t =1.·
·························10分
综上所述:当△DMN 是等腰三角形时,t 的值为
,3﹣
或1.