2006年福建省泉州市数学试卷

2006年福建省泉州市初中毕业、升学考试
（Ａ卷：供课改实验区使用）
数学试题
一、填空题（每小题3分，共36分） 1．2-的相反数是 ． 2．分解因式：2
3x x +=
．
3．去年泉州市林业用地面积约为10200000 亩，用科学记数表示约为 亩． 4．甲、乙两人比赛射击，两人所得平均环数相同，其中甲所得环数的方差为12，乙所得环数的方差为8，那么成绩较为稳定的是 （填“甲”或“乙”）． 5．某商品每件进价200元，现加价10%出售，则每件商品可获利润 元． 6．计算：
2
22
x x x +=++ ．
7．如图，ABC △为
O 的内接三角形，AB 是直径，20A ∠=， 则B ∠=
度．
8．函数4y x =的图象经过原点、第一象限与第
象
限．
9．抛掷一个质地均匀的正方体骰子、骰子的六个面上分别刻有1至6的点数，则掷得点数是2的概率是 ．
10．只用同一种正多边形铺满地面，请你写出一种这样的正多边形： ． 11．如图，圆锥的高AO 与母线AB 的夹角30α=，2cm AB =，则该圆锥侧面展开扇形的弧长等于
cm ．
12．菱形ABCD 的一条对角线长为6，边AB 的长是方程2
7120
x x -+=的一个根，则菱形ABCD 的周长为 ． 二、选择题（每小题4分，共24分）
每小题有四个答案，其中有且只有一个答案是正确的，请把正确答案
的代号写在题后的括号内，答对的得4分，答错、不答或答案超过一个的一律得0分．
13．计算：2
4
a a 的结果是（ ） Ａ．2
a
Ｂ．6
a
Ｃ．8
a
Ｄ．16
a
Ａ第7题图
α
Ａ
Ｏ Ｂ 第11题图
14．下列事件中，是必然事件的为（ ）
Ａ．我市夏季的平均气温比冬季的平均气温高； Ｂ．每周的星期日一定是晴天； Ｃ．打开电视机，正在播放动画片； Ｄ．掷一枚均匀硬币，正面一定朝上． 15．右边物体的正视图是（ ）
16．已知两圆半径分别为1与5，圆心距为4，则这两圆的位置关系是（ ） Ａ．外离 Ｂ．外切 Ｃ．相交 Ｄ．内切
17．某校篮球队五名主力队员的身高是174，179，180，174，178（单位：cm ），则这组数据的中位数是（ ） Ａ．174cm Ｂ．177cm Ｃ．178cm Ｄ．180cm 18．如右图，在Rt ABC △中，90C ∠=，2AC =，BC 的长为常数，点P 从起点C 出发，沿CB 向终点B 运动，设点P 所走过路程
CP 的长为x ，APB △的面积为y ，则下列图象能大致反映y 与x
之间的函数关系的是（ ）
三、解答题（共90分）
19．（8分）计算：1
322006--+-． 20．（8分）先化简下面的代数式，再求值：
(1)(1)(1)a a a a -+-+
，其中1a ．
第15题图
Ａ．
Ｂ．
Ｃ．
Ｄ．
Ａ． Ｂ． Ｃ．
Ｄ．
21．（8分）如图，在矩形ABCD 中，E F ，分别是BC ，
AD 上的点，且BE DF =． 求证：ABE CDF △≌△．
22．（8分）《泉州晚报》2006年6月5日报道：去年我市空气质量状况良好．泉州市各县（市、区）空气质量API 指数年际比较图如下（API 指数越高，空气质量越差）：
根据上图信息，解答下列问题： （1）有哪些县（市、区）连续两年．．．．的空气质量API 指数小于或等于50？ （2）哪个县（市、区）2005年比2004年空气质量API 指数下降最多？下降多少？ 23．（8分）如图，小王在操场上放风筝，已知风筝线AB 长100米，风筝线与水平线的夹角36α=，小王拿风筝线的手
离地面的高度AD 为1.5米，求风筝离地面的高度BE （精确
到0.1米）．
24．（8分）在两个布袋中分别装有三个小球，这三个小球的颜色分别为红色、白色、绿色，其他没有区别．把两袋小球都搅匀后，再分别从两袋中各取出一个小球，试求取出两个相同．．颜色．．小球的概率（要求用树状图或列表方法求解）．
25．（8分）在左图的方格纸中有一个Rt ABC △（A B C ，，三点均为格点），90C ∠=．
（1）请你画出将Rt ABC △绕点C 顺时针旋转90后所得到的Rt A B C ''△，其中A B ，
的
α
Ｂ
Ａ
Ｃ
Ｅ Ｄ 泉州市区 洛江区
API /指数 县（市、区） 年API
年API
对应点分别是A B ''，（不必写画法）；
（2）设（1）中AB 的延长线与A B ''相交于D 点，方格纸中每
一个小正方形的边长为1，试求BD 的长（精确到0.1）． 26．（8分）某校的一间阶梯教室，第1排的座位数为a ，从第2排开始，每一排都比前一排增加b 个座位．
座位？ 27．（13分）一条隧道的截面如图所示，它的上部是一个以AD 为直径的半圆O ，下部是一个矩形ABCD ．
（1）当4AD =米时，求隧道截面上部半圆O 的面积；
（2）已知矩形ABCD 相邻两边之和为8米，半圆O 的半径为r 米． ①求隧道截面的面积S （米2
）关于半径r （米）的函数关系式（不
要求写出r 的取值范围）；
②若2米CD ≤≤3米，利用函数图象求隧道截面的面积S 的最大值（π取3.14，结果精确到0.1米）．
28．（13分）如图，在直角坐标系中，O 为原点，(412)A ，为双曲线(0)k
y x x
=
>上的一点． （1）求k 的值；
（2）过双曲线上的点P 作PB x ⊥轴于B ，连接OP ，若
Rt OPB
△两直角边的比值为1
4
，试求点P 的坐标；
（3）分别过双曲线上的两点1P ，2
P ，作11PB x ⊥轴于1B ，22P B x ⊥轴于2B ，连结1OP ，
2OP ．设11Rt OPB △，22Rt OP B △
的周长分别为1l ，2l ，内切圆的半径分别为1r ，2r ，若Ａ
Ｃ Ｂ Ａ
122l l =，试求12
r
r 的值． 四、附加题（共10分）
友情提示：请同学们做完上面考题后，再认真检查一遍，估计一下你的得分情况．如果你全卷得分低于90分（及格线），则本题的得分将计入全卷总分，但计入后全卷总分最多不超过90分；如果你全卷得分已达到或超过90分，则本题的得分不计入全卷总分． 1．（5分）将有理数1，2-，0按从小到大的顺序排列，用“<”号连接起来．
2．（5分）如图，在ABC △，AB AC =，50B ∠=．求A ∠的度数．
2006年福建省泉州市初中毕业、升学考试
（Ａ卷：供课改实验区使用）
数学试题参考答案及评分标准
说明：
（一）考生的正确解法与“参考答案”不同时，可参照“参考答案及评分标准”的精神进行评分．
（二）如解答的某一步出现错误，这一错误没有改变后续部分的考查目的，可酌情给分，但原则上不超过后面应得的分数的二分之一；如属严重的概念性错误，就不给分．
（三）以下解答各行右端所注分数表示正确做完该步应得的累计分数． （四）评分最小单位是1分，得分或扣分都不出现小数． 一、填空题（每小题3分，共36分） 1．2； 2．(3)x x +； 3．7
1.0210⨯ ； 4．乙； 5．20；
6．1；
7．70；
8．三；
9．
1
6
；
10．正三角形（或正四边形，正六边形）； 11．2π； 12．16． 二、选择题（每小题4分，共24分） 13．Ｂ； 14．Ａ； 15．Ｂ； 16．Ｄ； 17．Ｃ； 18．Ｃ．
三、解答题（共90分） 19．（本小题8分） 解：原式1
312
=+- ········································································································ 6分 12
2
= ················································································································ 8分 20．（本小题8分）
解：原式2
2
1a a a =-+- ······························································································ 4分 1a =- ·············································································································· 5分
当1a =时
原式11=- ······································································································ 7分
=··············································································································· 8分 21．（本小题8分）
证明：∵四边形ABCD 是矩形
A B C D =∴，90B D ∠=∠= ··········· 4分
在ABE △和CDF △中AB CD
B D BE DF =⎧⎪
∠=∠⎨⎪=⎩
A B E C D ∴△≌△ ·································································································· 8分
22．（本小题8分） 解：（1）永春县和惠安县连续两年的空气质量API 指数小于或等于50 ···························· 4分 （2）安溪县2005年比2004年空气质量API 指数下降最多，下降16．··························· 8分 23．（本小题8分）
解：在Rt ABC △中，36BAC α∠=∠=，100AB =米
sin BC
AB
α=
∵ ··························································· 4分 sin 100sin361000.587858.78BC AB α==⨯≈⨯=∴
（米） ······ 6分
又 1.5CE AD ==∵米
58.78 1.560.2860.3BE BC CE =+=+=≈∴（米） 答：风筝离地面的高度BE 约为60.3米 ·························· 8分
24．（本小题8分） 解：（解法一）
列举所有等可能结果，画树状图：
α
Ｂ
Ａ Ｃ Ｅ
Ｄ
··························· 4分
由上图可知，所有等可能结果共有9种，两个相同颜色小球的结果共3种，
P ∴（相同颜色）3193
=
= ··································································································· 8分 （解法二）列表如下： ········································································································································· 4分 由上表可知，所有等可能的结果共有9种，两个相同颜色小球的结果共3种，
P ∴（相同颜色）31
93
== ···················································· ·············································· 8分
25．（本小题8分） 解：（1）方格纸中Rt A B C ''△为所画的三角形 ············· 4分 （2）由（1）得A A '∠=∠ 又12∠=∠∵
ABC A BD '∴△∽△ ····················································· 5分
BC AB
BD A B
='∴
························································································································· 6分
1BC =∵，2A B '=，AB = =··········································································································· 7分
12
BD =∴
即0.6BD =
≈ BD ∴的长约为0.6 ················································································································ 8分
26．（本小题8分） 解：（1）3a b + ······················································································································ 3分
红 白 绿 红 白 绿 红 白 绿 红 白 绿
布袋1 布袋2 Ａ Ｃ Ｂ
Ｄ
A '
B '
1 2
（2）依题意得318
142(4)
a b a b a b +=⎧⎨
+=+⎩ ··············································································· 5分
解得12
2
a b =⎧⎨
=⎩ ············································································································ 7分
1220252+⨯=∴
答：第21排有52个座位． ···················································································· 8分 27．（本小题13分）
解：（1）当4AD =米时，2
2112222
AD S ⎛⎫=π⨯=π⨯ ⎪⎝⎭半圆
2=π（米2
） ···························································· 3分
（2）①2AD r =∵，8AD CD +=
882CD AD r =-=-∴ ········································· 4分
211
2(82)22
S r AD CD r r r 2=
π+=π+-∴ 21162r r ⎛⎫
=π-4+ ⎪⎝⎭
·
····································································································· 8分 ②由①知82CD r =- 又2∵米3CD ≤≤米 2823r -∴≤≤ 2.53r ∴≤≤ ················································································································ 9分 由①知21162S r r ⎛⎫
=π-4+
⎪⎝⎭
21 3.144162r r ⎛⎫
≈⨯-+ ⎪⎝⎭
22.4316r r =-+ ··········································································································· 10分
2
8642.43 2.43 2.43r ⎛
⎫=--+ ⎪
⎝
⎭ ∵ 2.430-<，∴函数图象为开口向下的抛物线．
∵函数图象对称轴8
3.32.43
r =≈ ··············································································· 11分
又2.53 3.3r <≤≤
由函数图象知，在对称轴左侧S 随r 的增大而增大， 故当3r =时，S 有最大值． ························································································ 12分
2131632S
⎛⎫
=π-4⨯+⨯ ⎪⎝⎭
最大值
Ａ
1 3.1449482⎛⎫
≈⨯-⨯+ ⎪⎝⎭ 26.13= 26.1≈（米2）
答：隧道截面面积S 的最大值约为26.1米2
． ··························································· 13分 28．（本小题13分）
解：（1）依题意，得124
k =，48k = ···························· 3分
（2）由（1）得双曲线解析式为48
y x
= ························· 4分
设48
()P m n n m
=，∴ 即48mn =···························· 5分
当
14OB PB =时，即1
4m n = 可设m z =，4n z =． 448z z =∴
，解得z =
m =∴
，n =
P ∴ ······················································································································ 7分 当
1
4
PB OB =
时，同理可求得P ············································································· 8分 （3）在11Rt OPB △中，设1OB a 1=，141PB b =，11OP c =，则111()P a b ，，由（2）得1148a b =； 在22Rt OP B △中，设22OB a =，222P B b =，22OP c =，则222()P a b ，，由（2）得2248a b =． ················································································································································· 9分
111111
()2422a b c r a b 11++==∵ 2222211
()2422
a b c r a b 2++== ························································································ 10分 11112222()()a b c r a b c r ++=++∴ ··················································································· 11分 即1122l r l r = 故
12
21
l r l r = ······························································································································ 12分 又1
2
2l l =∵
2
1
2r r =∴
即得
121
2
r r = ·
·························································································································· 13分 四、附加题（共10分） 1．解：201-<< ·················································································································· 5分 2．解：AB AC =∵
50C B ∠=∠=∴ ···································································································· 2分
180A B C ∠=-∠-∠∴ 1805050=--
80= ·························································· 5分。