2008 年新疆兵团中考数学真题及答案
)
)
D.4
C. 4
一、精心选择(本大题共 10 题,每题所给四个选项中,只有一个是正确的.每题 5 分,共
50 分.)
1.| 4 |等于(
A. 2
B.2
2.2008 年 5 月 12 日,四川省汶川县发生了里氏 8.0 级大地震.新
疆各族群众积极捐款捐物,还紧急烤制了 2×104 个饱含新疆各
族人民深情的特色食品——馕(náng),运往灾区.每个馕厚度
约为 2cm,若将这批馕摞成一摞,其高度大约相当于(
A.160 层楼房的高度(每层高约 2.5m)
B.一棵大树的高度
C.一个足球场的长度
D.2000m 的高度
3.如图,下列推理不正确...的是(
A.∵AB∥CD ∴∠ABC+∠C=180°
B.∵∠1=∠2 ∴AD∥BC
C.∵AD∥BC ∴∠3=∠4
D.∵∠A+∠ADC=180° ∴AB∥CD
4.下列事件属于必然事件的是(
A.打开电视,正在播放新闻
C.实数 a<0,则 2a<0
5.下列调查方式中,合适的是(
A.要了解约 90 万顶救灾帐蓬的质量,采用普查的方式
B.要了解外地游客对旅游景点“新疆民街”的满意程度,采用抽样调查的方式
C.要保证“神舟七号”飞船成功发射,对主要零部件的检查采用抽样调查的方式
D.要了解全疆初中学生的业余爱好,采用普查的方式
B.我们班的同学将会有人成为航天员
D.新疆的冬天不下雪
)
)
)
6.在函数
y
的图象上有三个点的坐标分别为(1, 1y )、(
1
x
1
2
, 2y )、( 3 , 3y ),函数
B.y3<y2<y1 C.y2<y1<y3
中 BC 边上的高为 1h , DEF△
D.y3<y1<y2
中 DE 边上的高为 2h ,下列结论正确的
值 y1、y2、y3 的大小关系是(
A.y1<y2<y3
)
7.如图, ABC△
是(
)
h
A. 1
h
2
h
C. 1
h
2
h
B. 1
h
2
D.无法确定
8.傍晚,小明陪妈妈在路灯下散步,当他们经过路灯时,身体的影长(
A.先由长变短,再由短变长
C.保持不变
B.先由短变长,再由长变短
D.无法确定
)
9.如图,圆内接四边形 ABCD是由四个全等的等腰梯形组成,AD是⊙O的直径,则∠BEC的
度数为(
A.15°
)
B.30°
C.45°
D.60°
10.古尔邦节,6 位朋友均匀地围坐在圆桌旁共度佳节.圆桌半径为 60cm,每人离圆桌的距
离均为 10cm,现又来了两名客人,每人向后挪动了相同的距离,再左右调整位置,使 8 人
都坐下,并且 8 人之间的距离与原来 6 人之间的距离(即在圆周上两人之间的圆弧的长)相
等.设每人向后挪动的距离为 x,根据题意,可列方程(
)
2π(60 10)
2π(60 10
)
x
A.
B.
6
2π(60
8
x
)
8
2π 60
6
2π(60
C. 2π(60 10) 6
x
) 8
D. 2π(60
x
) 8 2π(60
x
) 6
二、合理填空(本大题共 4 题,每题 5 分,共 20 分)
11.根据下列图形的排列规律,第 2008 个图形是福娃
(填写福娃名称即可).
12.如图,在平面直角坐标系中,线段 1 1A B 是由线段 AB 平移得到的,已知 A B, 两点的坐
标分别为 ( 2 3)
A , , ( 31)
B , ,若 1A 的坐标为 (3 4), ,则 1B 的坐标为
.
13.已知一元二次方程有一个根是 2,那么这个方程可以是
方程即可).
14.如图,一束光线从 y轴上点 A(0,1)发出,经过 x轴上点 C反射后,经过点 B(6,2),
则光线从 A点到 B点经过的路线的长度为
(填上一个符合条件的
.(精确到 0.01)
三、准确解答(本大题共 10 题,共 80 分)
15.(6 分)计算: 2
2
1
4
0
18 (π 6)
.
16(6 分)化简分式
的数代入求值.
2
x
1
2
x
1
x
x
1
1
2
x
,并从 2 、 1 、0、1、2 中选一个能使分式有意义
17.(6 分)城区某中学要从自愿报名的张、王、李、赵 4 名老师中选派 2 人下乡支教,请
用画树状图(或列表)的方法求出张、王两位老师同时被选中的概率.
18.(8 分)如图,⊙O的半径
AB
16cm
,直线 l平移多少厘米时能与⊙O相切?
OC
10cm
,直线 l⊥CO,垂足为 H,交⊙O于 A、B两点,
19.(9 分)某水果销售公司去年 3 至 8 月销售吐鲁番葡萄、哈密大枣的情况见下表:
3 月
4 月
5 月
6 月
吐鲁番葡萄(吨)
哈密大枣(吨)
4
8
8
7
5
9
8
7
7 月
10
10
8 月
13
7
(1)请你根据以上数据填写下表:
平均数
8
方差
9
吐鲁番葡萄
哈密大枣
(2)补全折线统计图.
(3)请你从以下两个不同的方面对这两种水果在去年 3 月份至 8 月份的销售情况进行分析:
①根据平均数和方差分析;
②根据折线图上两种水果销售量的趋势分析.
20.(8 分)如图,某市区南北走向的北京路与东西走向的喀什路相交于点 O处.甲沿着喀
什路以 4m/s 的速度由西向东走,乙沿着北京路以 3m/s 的速度由南向北走.当乙走到 O点以
北 50m 处时,甲恰好到点 O处.若两人继续向前行走,求两个人相距 85m 时各自的位置.
21.(8 分)如图,在△ABC中,∠C=2∠B,AD是△ABC的角平分线,∠1=∠B.
求证:AB=AC+CD.
22.(9 分)某社区计划购买甲、乙两种树苗共 600 棵,甲、乙两种树苗单价及成活率见下
表:
种类
单价(元)
成活率
甲
乙
60
80
88%
96%
(1)若购买树苗资金不超过 44000 元,则最多可购买乙树苗多少棵?
(2)若希望这批树苗成活率不低于 90%,并使购买树苗的费用最低,应如何选购树苗?购
买树苗的最低费用为多少?
23.(10 分)(1)请用两种不同的方法,用尺规在所给的两个矩形中各作一个不为正方形的
菱形,且菱形的四个顶点都在矩形的边上.(保留作图痕迹)
(2)写出你的作法.
24.(10 分)某工厂要赶制一批抗震救灾用的大型活动板房.如图,板房一面的形状是由矩
形和抛物线的一部分组成,矩形长为 12m,抛物线拱高为 5.6m.
(1)在如图所示的平面直角坐标系中,求抛物线的表达式.
(2)现需在抛物线 AOB的区域内安装几扇窗户,窗户的底边在 AB上,每扇窗户宽 1.5m,
高 1.6m,相邻窗户之间的间距均为 0.8m,左右两边窗户的窗角所在的点到抛物线的水平距
离至少为 0.8m.请计算最多可安装几扇这样的窗户?
参考答案
一、选择题(本大题共 10 题,每题 5 分,共 50 分)
题号
选项
1
D
2
A
3
C
4
C
5
B
6
D
7
C
8
A
9
B
10
A
二、填空题(本大题共 4 题,每题 5 分,共 20 分)
11.欢欢
12.(2,2)
13. 2
x (答案不惟一)
4
14.6.71
三、解答题(本大题共 10 题,共 80 分)
15.(6 分)解:原式
······························································· 4 分
1 3 2 1
1
4
4
3 2
············································································· 6 分
16.(6 分)解:原式
(
x
1)(
(
x
x
1)
2
1)
x
x
1
1
························································· 1 分
2
x
x
(
x
x
x
2
1)(
1
1
1)
(
x
1)
x
························································································· 3 分
································································································ 2 分
········································································································ 5 分
1
1
1)
(
x
4
x
2
1
x
把 0
x 代入
原式 0 ········································································································ 6 分
或把 2
4 2
原式 2
2
1
2
x 代入
4 ( 2)
2
( 2)
1
··························································································6 分
.······································ 6 分 17.(6 分)解:方法 1:画树状图
x 代入
原式
或把
8
3
8
3
张、王两位老师同时被选中的概率是
方法 2:列表
·······························································4 分
.····························································· 6 分
1
6
张
王
李
赵
张
王张
李张
赵张
王
张王
李王
赵王
李
张李
王李
赵李
赵
张赵
王赵
李赵
·················································· 4 分
张、王两位老师同时被选中的概率是
1
6
.····························································· 6 分
18.(8 分)解法 1:如图,连结 OA,延长 CO交⊙O于 D,
∵l⊥OC,
∴OC平分 AB.
∴AH=8.······································································ 3 分
在 Rt△AHO中,
CH
4cm
2
AO AH
2
2
10
2
8
,······· 6 分
6
,
OH
DH
16cm
.
∴
答:直线 AB向左移 4cm,或向右平移 16cm 时与圆相切.·········································8 分
解法 2:设直线 AB平移 cmx 时能与圆相切,
(10
························································································· 3 分
x
10
8
2
)
2
2
1 16
x
x
2
4
.·········································································· 8 分
CH
DH
4cm
16cm
,
∴
答:略.
(只答一个方向的平移扣 2 分)
19.(9 分)
解:(1)
平均数
方差
吐鲁番葡萄
哈密大枣
8
8
9
4
3
……(4 分)
(2)
·············································· (7 分)
(3)①由于平均数相同, 2
S
S大枣
2
葡萄 ,所以大枣的销售情况相对比较稳定.·············8 分
②从图上看,葡萄的月销售量呈上升趋势.·························································· 9 分
(答案不惟一,合理均可得分)
20.(8 分)解法 1:设经过 x秒时两人相距 85m·····················································1 分
根据题意得:
2
(4 )
x
(50 3 )
x
2
2
85
································································ 4 分
化简得: 2 12
x
x
189 0
21
x
2
9
,
x 时, 4
(不符合实际情况,舍去)···············································6 分
x
解得: 1
当 9
∴当两人相距 85m 时,甲在 O点以东 36m 处,乙在 O点以北 77m 处.·························8 分
解法 2:设甲与 O处的距离为 xm 时,两人相距 85m
36 50 3
77
,
x
x
则乙与 O处的距离为
3
4
x
50 m
······································································ 1 分
2
x
3
4
x
50
2
2
85
·····················································································4 分
x
解得: 1
336
,
4
当
x
36
,
x
2
84
(不符合实际情况,舍去 )··········································· 6 分
x
50 77
···················································································· 7 分
CAD
答:当两人相距 85 米时,甲在 O点以东 36 米处,乙在 O点以北 77 米处.················· 8 分
21.(8 分)证明:
∵∠1=∠B
∴∠AED=2∠B,DE=BE······················································································· 2 分
∴∠C=∠AED··································································································· 3 分
在△ACD和△AED中
EAD
AD AD
∴△ACD≌△AED·······························································································5 分
∴AC=AE,CD=DE,∴CD=BE.·············································································· 6 分
∴AB=AE+EB=AC+CD.·························································································8 分
22.(9 分)解:(1)设最多可购买乙树苗 x棵,则购买甲树苗( 600
x )棵·············1 分
60(600
··············································································3 分
) 80
x
≤
44000
AED
C
x
400
x ≤ .
答:最多可购买乙树苗 400 棵.········································································· 5 分
(2)设购买树苗的费用为 y
则 60(600
y
x
) 80
x
y
20
x
36000
························································································· 6 分
根据题意 0.88(600
) 0.96
x
x
≥
0.9 600