（2007-2010年）济南中考数学试题及答案(2)

?△DAE≌△CBF ························································ 2分 ?DE?CF······································································ 3分 （2）解：过点O作OC?AB，垂足为C， 则有AC?BC ·································································· 4分 ?AB?4，?AC?2 ··········································································································· 5分 在Rt△AOC中，

················································································· 6分 OC?OA2?AC2?32?22?5 ·

sinA?OC5 ················································································································· 7分 ?OA320．解：（1）?在7张卡片中共有两张卡片写有数字1 ···················································· 1分

2················································ 2分 ?从中任意抽取一张卡片，卡片上写有数字1的概率是 ·7（2）组成的所有两位数列表为： 十位数 个位数 1 2 3 或列树状图为：

1 2 3 4 十位数

1 1 1 2 3 2 3 2 3 2 3 个位数 1

（11） （12） （13）（（（ 21） （22） （23） 31） （32） （33） 41） （42） （43）

········································································· 6分

1 11 12 13 2 21 22 23 3 31 32 33 4 41 42 43 ?这个两位数大于22的概率为

7． ···················································································· 8分 1221．解：（1）由租用甲种汽车x辆，则租用乙种汽车(8?x)辆 ········································ 1分

由题意得：??40x?30(8?x)≥290 ····················································································· 4分

10x?20(8?x)≥100?解得：5≤x≤6 ··················································································································· 5分

即共有2种租车方案：

第一种是租用甲种汽车5辆，乙种汽车3辆； 第二种是租用甲种汽车6辆，乙种汽车2辆． ····································································· 6分 （2）第一种租车方案的费用为5?2000?3?1800?15400元； 第二种租车方案的费用为6?2000?2?1800?15600元 ··················································· 7分

······························································································ 8分 ?第一种租车方案更省费用． ·

22．解：（1）过点D作DM?BC，垂足为M，

D A 在Rt△DMC中，

6

F

B E M

第22题图

N

C

DM?CD?sinC?10?4?8 ········································· 1分 5························· 2分 CM?CD2?DM2?102?82?6 ·

?BM?BC?CM?10?6?4，?AD?4 ··············· 3分

11?S梯形ABCD?(AD?BC)DM?(4?10)?8?56························································· 4分

22（2）设运动时间为x秒，则有BE?CF?x，EC?10?x ············································ 5分 过点F作FN?BC，垂足为N，

4sinC?x 在Rt△FNC中，FN?CF?············································································· 6分

51142?S△EFC?EC?FN?(10?x)?x??x2?4x ······················································· 7分

2255当x??24?5时，S△EFC???52?4?5?10

5?2?2?????5?即△EFC面积的最大值为10 ······························································································· 8分 此时，点E，F分别在BC，CD的中点处 ·········································································· 9分 23．解：（1）?∠AOC?90，

??AC是?B的直径，?AC?2 ·························································································· 1分

又?点A的坐标为(?3，0)，?OA?3 ?OC?AC2?OA2?22?(3)2?1 ············································································ 2分

OC1?，?∠CAO?30? ········································································ 3分 AC2（2）如图，连接OB，过点D作DE?x轴于点E ··························································· 4分 ?OD为?B的切线， ?sin∠CAO??OB?OD，?∠BOD?90? ·························································································· 5分

y ?AB?OB，?∠AOB?∠OAB?30?，

CD A B ?∠AOD?∠AOB?∠BOD?30??90??120?，

在△AOD中，∠ODA?180?120?30?30?∠OAD

????O E x 第23题图

················································································································· 6分 ?OD?OA?3

在Rt△DOE中，∠DOE?180?120?60

???313sin60?? ?OE?OD?cos60??OD?，ED?OD?222

7

?33? ····································································································· 7分 ?点D的坐标为???2，?2??设过D点的反比例函数的表达式为y?k x?k?3333 ············································································································· 8分 ??22433 ····························································································································· 9分 4x?y?0)，C(1，0) 24．解：（1）?点A(?3，?AC?4，BC?tan∠BAC?AC?3?4?3，B点坐标为(1，3) ································· 1分 4设过点A，B的直线的函数表达式为y?kx?b，

由??0?k?(?3)?b39 得k?，b? ················································································· 2分

443?k?b?39x? ··············································································· 3分 44y （2）如图1，过点B作BD?AB，交x轴于点D， B 在Rt△ABC和Rt△ADB中，

P ?∠BAC?∠DAB ?Rt△ABC∽R△tAD，B

?D点为所求 ··································································· 4分

4O QC D x A 又tan∠ADB?tan∠ABC?， 3第24题图1

49?CD?BC?tan∠ADB?3??·················································································· 5分

34?直线AB的函数表达式为y??OD?OC?CD?13?13?，?D?，··············································································· 6分 0? ·44??（3）这样的m存在 ················································································································ 7分

在Rt△ABC中，由勾股定理得AB?5 如图1，当PQ∥BD时，△APQ∽△ABD

y P B m则?53?13?m254，解得m? ·································· 8分 1393?4A Q O C 第24题图2

D x

如图2，当PQ?AD时，△APQ∽△ADB

8

则

m?133?43?13?m1254，解得m? ················································································ 9分

365

济南市2008年高中阶段学校招生考试

1．－2的绝对值是（ ） A．2

B．－2

C．

1 21D．?

22．下列计算正确的是（ ）

A．a3?a4?a7

B．a3?a4?a7

C．(a3)4?a7

D．a6?a3?a2

3．下面简单几何体的主视图是（ ） ．

A． B． C． D． 正面

4．国家游泳中心——“水立方”是2008年北京奥运会标志性建筑物之一，其工程占地面积为

62828平方米，将62828用科学记数法表示是（保留三个有效数字） （ ） A．62.8?103 B．6.28?104

y C．6.2828?104 D．0.62828?105

C 5．已知?ABC在平面直角坐标系中的位置如图所示，将?ABC向右平移6个单位，则平移后A点的坐标是（ ）

A．（?2，1）

B．（2，1）

A B 1 O 1 x C．（2，?1） D．（?2，?1） 6．四川省汶川发生大地震后，全国人民“众志成城，抗震救灾”，积极开展捐款捐物献爱心活动．下表是我市某中学初一·八班50名同学捐款情况统计表：

捐款数（元） 10 人 数（人） 3 15 10 20 10 30 15 50 5 60 2 70 1 80 1 第5题图

90 1 100 2 根据表中提供的信息，这50名同学捐款数的众数是（ ） A．15 B．20 C．30 D．100 7．如图：点A、B、C都在⊙O上，且点C在弦AB所对的优弧上， 若?AOB?72?，则?ACB的度数是（ ） A．18° B．30° C．36° D．72°

18．如果xa?2y3与?3x3y2b?1是同类项，那么a、b的值分别是（ ） A 3O C

B

第7题图

9

?a?1A．?

b?2??a?0B．?

b?2??a?2C．?

b?1??a?1D．?

b?1?9．“迎奥运，我为先”联欢会上，班长准备了若干张相同的卡片，上面写的是联欢会上同学

们要回答的问题．联欢会开始后，班长问小明：你能设计一个方案，估计联欢会共准备了多少张卡片？小明用20张空白卡片（与写有问题的卡片相同），和全部写有问题的卡片洗匀，从中随机抽取10张，发现有2张空白卡片，马上正确估计出了写有问题卡片的数目，小明估计的数目是（ ） A．60张 B．80张 C．90张 D．110张 10．关于x的一元二次方程2x2?3x?a2?1?0的一个根为2，则a的值是（ ）

A．1

B．3 C．?3 D．?3

S(吨) 11．济南市某储运部紧急调拨一批物资，调进物资共用４小时，调

30 进物资2小时后开始调出物资（调进物资与调出物资的速度均

保持不变）．储运部库存物资S(吨)与时间t(小时)之间的函数关

系如图所示，这批物资从开始调进到全部调出需要的时间是（ ） A．4小时 B．4.4小时 10 C．4.8小时 D．5小时

12．如图：等腰直角三角形ABC位于第一象限，AB=AC=2，直

角顶点A在直线y=x上，其中A点的横坐标为1，且两条直

k

角边AB、AC分别平行于x轴、y轴，若双曲线y?(k≠0)

x 得 分 评卷人 二、填空题：本大题共5个小题．每小题3分，共15分．把答案填在题中横线上．

A 与?ABC有交点，则k的取值范围是（ ） A．1?k?2 B．1≤k≤3 C．1≤k≤4 D．1≤k?4

O O 4 t(时)

第11题图

2 y C A B 1 x 第12题图 13．当x?3,y?1时，代数式(x?y)(x?y)?2y的值是 ．

14．分解因式：x2?2x?3= ．

15．如图，在?ABC中，EF为?ABC的中位线，Ｄ为BC

边上一点(不与B、C重合)，AD与EF交于点Ｏ，连接DE、DF，要使四边形AEDF为平行四边形，需要添加条件 ．(只添加一个条件)

16．如图：矩形纸片ABCD，AB=2，点E在BC上，且

10

E B

O

F C

D 第15题图 A

D

B

E

第16题图

C

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