济南市2007年高中阶段学校招生考试 数学试题参考答案及评分标准
一、选择题 1．C 2．D 8．A 9．B 二、填空题 13．x??
3．B 10．C
4．B 11．C
45．C 12．B
6．A 7．D
1 214．y(y?2)2
15．1.3?10
16．
2 317．(123?36)
三、解答题 18．（1）解：
x2??2 x?33?x去分母得：x?2?2(x?3) ··································································································· 1分 解得：x?4 ··························································································································· 2分 经检验x?4是原方程的根． ································································································· 3分 （2）解法一：①?2?②得5x?10 ··················································································· 4分 解得：x?2 ··························································································································· 5分 将x?2代入①得y??2 ······································································································· 6分
?x?2 ······································································································· 7分 ?方程组的解为??y??2解法二：由①得y?2x?6 ③ ·························································································· 4分 将③代入②得x?2(2x?6)??2
解得：x?2 ··························································································································· 5分 将x?2代入③得y??2 ······································································································· 6分
?x?2 ······································································································· 7分 ?方程组的解为?y??2?19．（1）证明：?AF?BE，EF?EF，?AE?BF ················································· 1分
?四边形ABCD是矩形，
?∠A?∠B?90?，AD?BC，
O ?△DAE≌△CBF ························································ 2分 ?DE?CF······································································ 3分 C B A （2）解：过点O作OC?AB，垂足为C，
第19题图2
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
?40x?30(8?x)≥290由题意得：? ····················································································· 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，
DM?CD?sinC?10?4?8 ········································· 1分 5B F
························· 2分 CM?CD2?DM2?102?82?6 ·
?BM?BC?CM?10?6?4，?AD?4 ··············· 3分
E M
第22题图
N
C
?S梯形ABCD?11(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 第23题图
x ················································································································· 6分 ?OD?OA?3
在Rt△DOE中，∠DOE?180?120?60
???313sin60?? ?OE?OD?cos60??OD?，ED?OD?222?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，································ 1分 3) ·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
44?3?k?b39x? ··············································································· 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? ·4?4?（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
则
m?133?43?13?m1254，解得m? ················································································ 9分
365
2008数学中考答案
一、选择题
1．A 2．B 3．C 4．B 5．B 6．C 7．C 8．A 9．B 10．D 11．B 12．C 二、填空题
13．9 14．(x?3)(x?1) 15． BD=CD，OE=OF，DE∥AC等 16．4 17．15 三、解答题 18．
（1）解：2x?2?1?0 ................................................................................................. 1分
2x?1 ............................................................................................................. 2分
1x? ............................................................................................................... 3分
2（2）解：解①得x＞－2 ................................................................................................ 4分
解②得x＜3 .................................................................................................... 5分 ∴此不等式组的解集是－2＜x＜3 ...................................................................... 6分 解集在数轴上表示正确 ............................................................................................ 7分 19．
（1）证明：∵AB∥DE，∴∠B=∠DEF
∵AC∥DF，∴∠F=∠ACB ................................................................................... 1分 ∵BE=CF，∴BE+EC= CF + EC即BC=EF ....................................................... 2分 ∴△ABC≌△DEF
∴AB=DE ....................................................... 3分 （2）解：过点O作OG⊥AP于点G
连接OF ..................................................... 4分 ∵ DB=10，∴ OD=5 ∴ AO=AD+OD=3+5=8
∵∠PAC=30°
11∴ OG=AO=?8?4cm ........................... 5分
22∵ OG⊥EF，∴ EG=GF
∵ GF=OF2?OG2?52?42?3
∴ EF=6cm ................................................. 7分
20．解：组成的所有坐标列树状图为：
第19题图2
A F Ｇ E D O B C P