青海省西宁市2008年高中招生考试

青海省西宁市2008年高中招生考试

数学试卷参及评分意见

一、填空题（本题共12个小题，每空2分，共30分） 1．2，3a； 4．运； 8．

2．x(x1)(x1)，

1

（答案不惟一）； 2

x1

3．4.010；

105．180； 9．10；

6．(3，4)，二、四；

7．相似变换； 12．121．

1； 310．35；

11．

3； 2二、选择题（本题共8个小题，每小题3分，共24分） DCCB DBAB

三、解答题（本题共3个小题，每小题7分，共21分） 21．解：原式322·········································································· 3分 1 ·

2································································································· 5分 321 ·

····································································································· 7分 22 ·

22．（1）只要度量残留的三角形模具片的B，C的度数和边BC的长，················· 1分

因为两角及其夹边对应相等的两个三角形全等． ··················································· 3分 （2）按尺规作图的要求，正确作出ABC的图形． ··········································· 7分 23．证明：四边形ABCD是平行四边形（已知），

AD∥BC，ABCD（平行四边形的对边平行，对边相等） ····························· 1分 GBCBGA，BCECED（两直线平行，内错角相等） ······················· 2分 又BG平分ABC，CE平分BCD（已知），

ABGGBC，BCEECD（角平分线定义） ······································ 3分 ABGGBA，ECDCED． ··························································· 4分 ABAG，CEDE（在同一个三角形中，等角对等边） ································ 5分 AGDE ·································································································· 6分 AGEGDEEG，即AEDG． ·························································· 7分 四、（本题共3个小题，每小题8分，共24分） 24．解：（1）根据题意，得mn50(412171)16；

17m100％％．2分 50则mn1617m32①② ····················································································· 4分

解之，得m15 ··························································································· 5分

n148100％96％ ························· 7分 50（2）7~8分数段的学生最多 ·············································································· 6分 及格人数412171548（人），及格率答：这次1分钟跳绳测试的及格率为96％． ························································ 8分 25．解：（1）由题意知，B场地宽为(30x)m ···················································· 1分 ··········································································· 2分 yx(30x)x230x ·

当y0时，即x30x0，x10，x230 ··············································· 3分 ······················································· 4分 0)，(30，0)． ·函数与x轴的交点坐标为(0，自变量x的取值范围为0x30． ························· 5分 （2）yx230x(x15)2225，

当x15m时，种植菊米的面积最大，······················ 6分 最大面积为225m2 ················································· 7分 草图（如右图所示）． ············································ 8分 26．解：由题意可得1≤2x7≤6，化为不等式组解得

225 x

30 （长：m）

y （面积：m2）

2O 2x7≤6 ························ 2分

2x7≥11≤x≤3 ····························································································· 3分 21≤x≤6，且x为正整数，x1，2，3． ························································ 4分

要使点P落在直线y2x7图象上，则对应的y5，3，1 ································ 5分 ，（2，3），（3，1） ··················································· 6分 满足条件的点P有（1，5）

抛掷骰子所得P点的总个数为36．

31········································· 7分 点P落在直线y2x7图象上的概率P36121答：点P落在直线y2x7图象上的概率是． ············································· 8分

12五、（本题共2个小题，第27题9分，第28题12分，共21分） 27．方案一（计算过程）

cot ·解：在Rt△ACD中，ACDC····························································· 1分

Rt△BCD中，BCDCcot ······································································· 2分

ABACBC．

33DC10，解得DC53(m) ··············· 3分 (cot30cot60)DC10，3AMCN，DNDCCNDCAM(531.5)(m)

（测量结果：）DN(531.5)m ···································································· 4分 方案二（计算过程）

cot ·解：在Rt△ACD中，ACDC····························································· 1分

Rt△BCD中，BCDCcot ······································································· 2分

3ABACBC，(cot30cot60)DC20，3DC20， 3解得DC53(m) ···················································· 3分

D AMCN，DNDCCNDCAM(531.5)(m)

（测量结果：）DN(531.5)m ································ 4分

A a M  b C N

方案三（不惟一） 能正确画出示意图 ······················································· 6分 （测量工具）：皮尺、测角仪；（测量数据）：AMa，ACb，DAC ·········· 7分 （计算过程）解：在Rt△ACD中，CDbtan， DNDCCN，AMCN，DNbtana ········································· 8分 （测量结果）：DNbtana ······································································ 9分 28．解：（1）圆心O1的坐标为(2，·············· 1分 0)，O1半径为1，A(1，0)，B(3，0) ·

二次函数yx2bxc的图象经过点A，B，

1bc0 ·········································································· 2分 可得方程组93bc0b4解得：··········································· 3分 二次函数解析式为yx24x3 ·

c3（2）过点M作MFx轴，垂足为F． ···························································· 4分

OM是O1的切线，M为切点，O1MOM（圆的切线垂直于经过切点的半径）．

在Rt△OO1M中，sinO1OMO1M1 OO12y P P2 1O M B x ··························· 5分 O1OM为锐角，OOM30 ·13OMOO1cos3023，

2H A F O1 在Rt△MOF中，OFOMcos30333． 2213． MFOMsin3032233················································································· 6分 点M坐标为2，2 ·

设切线OM的函数解析式为ykx(k0)，由题意可知333 ·········· 7分 k，k223切线OM的函数解析式为y3x ································································ 8分 3（3）存在． ·································································································· 9分

①过点A作AP与OM交于点P可得Rt△APO1x轴，1∽Rt△MOO1（两角对应相等两三角形相1．似）

33PtanAOP，P········································· 10分 11AOA1tan301，3 ·3

H． ②过点A作AP2OM，垂足为P2，过P2点作P2HOA，垂足为

可得Rt△APO∽Rt△O1MO（两角对应相等两三角开相似） 2OA1，OP2OA在Rt△OPcos302A中，

3， 2在Rt△OPcosAOP222H中，OHOP333， 224P2HOP2sinAOP233313，P2， ······································ 11分 22444333··················································· 12分 符合条件的P点坐标有1，3，4，4 ·

（注：用不同于述方法解答正确的相应给分）