得分 评卷人
七、（本题12分）
25．有两张完全重合的矩形纸片，小亮同学将其中一张绕点A顺时针旋转90°后得到矩形
AMEF（如图12），连结BD、MF，若此时他测得BD=8cm，∠ADB=30°． （1）试探究线段BD与线段MF的关系，并简要说明理由；
C
D M E
B
A 图12
F
（2）小红同学用剪刀将△BCD与△MEF剪去，与小亮同学继续探究．他们将△ABD绕点A顺时针旋转得△AB1D1，AD1交FM于点K（如图13），设旋转角为?（0°＜），当△AFK为等腰三角形时，请直接写出旋转角?的度数； ?＜90°
D M B1 B K F D1
A 图13
（3）若将△AFM沿AB方向平移得到△A2F2M2（如图14），F2M2与AD交于点P，A2M2
与BD交于点N，当NP∥AB时，求平移的距离是多少？
D M2 N B
A2
M P A 图14
F2
F
得分 评卷人 八、（本题14分）
226 ．已知：在平面直角坐标系中，抛物线y?ax?x?3（a?0）交x轴于A、B两点，
交y轴于点C，且对称轴为直线x??2．
（1）求该抛物线的解析式及顶点D的坐标；
（2）若点P（0，t）是y轴上的一个动点，请进行如下探究：
探究一：如图15，设△PAD的面积为S，令W＝t·S，当0＜t＜4时，W是否有最大
值？如果有，求出W的最大值和此时t的值；如果没有，说明理由；
探究二：如图16，是否存在以P、A、D为顶点的三角形与Rt△AOC相似？如果存
在，求点P的坐标；如果不存在，请说明理由． （参考资料：抛物线y?ax?bx?c(a?0)对称轴是直线x??
D A O y C B x A O D y C B x 2b） 2a图15
图16
2009年丹东市初中毕业生毕业升学考试数学试题
参考答案及评分标准
一、选择题 题号 得分 二、填空题
9．3(a?3)(a?3) 10．4.7 11．（?2，?1） 12．5 13．302 14．53 15．m?2且m?0 16．8 三、（每小题8分，共16分）
1 C 2 D 3 B 4 B 5 D 6 A 7 B 8 C ?1?17．计算：32?(π?1)?4sin45°???
?3?0?1 ?42?1?4?2?3 ························································································· 4分 2 ?42?1?22?3 ····························································································· 6分 ?62?2 ············································································································· 8分 18．（1）如图，线段BC就是小芳能看到的那段公路． ·· 2分 （2）过点A作AM⊥BC，垂足为M，交DE于点N． ∵DE∥BC，∴?3??4，?1??2?90°， P ∴AN⊥DE． ···································································· 3分 又∵?DAE??BAC， ∴△ADE∽△ABC． ······················································· 4分 ∴
B 4 M C 2 N 3 1 Q
DEAN． ··································································· 5分 ?A BCAM根据题意得：BC?1.2?10?12（米）． ························· 6分
34又∵AN?4米，DE?3米，∴，∴AM?16（米）． ································ 7分 ?12AM答：点A到公路的距离为16米． ························································································· 8分
19．解：（1）（每空1分，共4分） 命中环数 命中次数 10 9 8 7 10环 40% 7环
10% 8环 20%
D E 4 3 2 1 （2）应该派甲去． ······································································· 5分 理由：x甲?（10?4?9?3?8?2?7?1······ 7分 ）?9（环）． ·
2S甲?1101[4?(10?9)2?3?(9?9)2?2?(8?9)2?1?(7?9)2]?1． 109环 30%
················································ 9分
因为甲、乙两人的平均成绩相同，而S甲?S乙，说明甲的成绩比乙稳定．
所以应派甲去． ···················································································································· 10分
20．解：设每个中国结的原价为x元， ················································································ 1分 根据题意得
22160160····················································································································· 5分 ??2 ·
0.8xx解得 x?20． ····················································································································· 8分 经检验，x?20是原方程的根． ··························································································· 9分
答：每个中国结的原价为20元． ························································································ 10分 21．
G F B 60°A 30° E
D C
过点C作CG⊥AE，垂足为G，过点D作DF⊥AE，垂足为F，得矩形CDFG． ∴CD?GF，CG?DF?900（米） ··············································································· 2分 在Rt△AGC中，∵?A?30°，∴?ACG?60°．
∴AG?CG?tan60°?9003（米）． ·············································································· 4分 同理，在Rt△BFD中，BF?DF?tan30°?3003（米）． ········································· 6分 ∵AB?1503?20?30003（米）． ············································································· 7分 ∴CD?GF?AB?BF?AG?24003（米）． ···························································· 8分 ∴搜寻的平均速度为24003?20?1203≈208（米/分）． ······································· 9分 答：搜救船搜寻的平均速度为208米/分． ········································································· 10分 （其它方法可参照此答案给分） 22．（1）解法一：
开始
第一天 A B
第二天 C D E C D E
所有可能出现的结果（A，C）（A，D）（A，E）（B，C）（B，D）（B，E） ∴小刚所有可能选择的方式有6种． ···················································································· 7分 解法二： 第二天 第一天 A B C （A，C） （B，C） D （A，D） （B，D） E （A，E） （B，E） ∴小刚所有可能选择的方式有6种． ···················································································· 7分 （2）∵一共有六种等可能的结果，而恰好选中A、D两处的可能性只有一种， ∴小刚恰好选中A和D这两处的概率为
1． ····································································· 10分 623．解：（1）四边形EFPG是平行四边形． ··························· 1分
M
理由：∵点E、F分别是BC、PC的中点， ∴EF∥BP． ·············································································· 2分
A D 同理可证EG∥PC． ································································· 3分 ∴四边形EFPG是平行四边形． ················································ 4分
P （2）方法一：当PC?3时，四边形EFPG是矩形． ·············· 5分 G F 证明：延长BA、CD交于点M． B C E ∵AD∥BC，AB?CD，?BAD?120°，∴?ABC??C?60°． ∴?M?60°，∴△BCM是等边三角形． ······························ 7分 ∵?MAD?180°?120°?60°，∴AD?DM?2． ∴CM?DM?CD?2?4?6． ·············································· 8分
∵PC?3，∴MP?3，∴MP?PC，∴BP⊥CM即?BPC?90°． 由（1）可知，四边形EFPG是平行四边形， ∴四边形EFPG是矩形． ···································································································· 10分 方法二：当PC?3时，四边形EFPG是矩形．·································································· 5分 证明：延长BA、CD交于点M．由（1）可知，四边形EFPG是平行四边形． 当四边形EFPG是矩形时，?BPC?90°．
M ∵AD∥BC，?BAD?120°，∴?ABC?60°．
∵AB?CD，∴?C??ABC?60°．
A D ∴?PBC?30°且△BCM是等边三角形．······························ 7分
1·············· 8分 CM． ·
P 2G F 同方法一，可得CM?DM?CD?2?4?6，
B C 1E
∴PC?6??3．
2即当PC?3时，四边形EFPG是矩形． ··········································································· 10分
∴?ABP??PBC?30°，∴PC?PM?（其它方法可参照此答案给分）
24．解：（1）y?20x?28(10?x)??8x?280．
∴y与x的函数关系式为y??8x?280． ·········································································· 3分
（2）??4x?6(10?x)≥51 ···························································································· 5分
?20x?28(10?x)≤260解得2.5≤x≤4.5． ············································································································ 6分 ∵x为非负整数，∴x?3或4． ∴有两种购买方案，
第一种：买甲种水壶3个，乙种水壶7个； 第二种：买甲种水壶4个，乙种水壶6个． ········································································· 8分 ∵y??8x?280，?8?0，