2011年中考数学复习高经典习题1-5(6)

19． （本题满分7分）

某中学对全校学生60秒跳绳的次数进行了统计，全校平均次数是100次．某班体育委员统计了全班50名学生60秒跳绳的成绩，列出的频数分布直方图如下（每个分组包括左端点，不包括右端点）：

求：（1）该班60秒跳绳的平均次数至少是多少？是否超过全校平均次数？

（2）该班一个学生说：“我的跳绳成绩在我班是中位数”，请你给出该生跳绳成绩的所在范围．

（3）从该班中任选一人，其跳绳次数达到或超过校平均次数的概率是多少？

20． （本题满分7分）

如图，⊙O的直径AB=4，C为圆周上一点，AC=2，过点C作⊙O的切线l，过点B作l的垂线BD，垂足为D，BD与⊙O交于点 E．

D （1） 求∠AEC的度数； （2）求证：四边形OBEC是菱形．

（第20题图）

A

O l B

C E 13 频数 19 7 5 4 2 O 60 80 100 120 140 160 180 次数

（第19题图）

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参考答案

一、选择题：（本大题共12小题，每小题3分，共36分） 题号 答案 1 D 2 D 3 C 4 A 5 A 6 B 7 A 8 B 9 B 10 A 11 D 12 C 二、填空题：（本大题共5小题，每小题4分，共20分） 13．4.834×103； 14．乙；

15．∠DAC＝∠ADB，∠BAD＝∠CDA，∠DBC＝∠ACB，∠ABC＝∠DCB，OB＝OC，OA＝OD；（任选其一） 16．

127n?1或2； 17．?2n?1，2?．

三、解答题：（本大题共7小题， 共64分） 18．（本小题满分6分）

解：原式=

x?yx?3yx?yx?3y?

x?6xy?9yx?y2222?2yx?y2yx?y ··································································· 1分 ···································································· 4分

= = =

?x?3y??

?x?y??x?y??2yx?y2?x?3yx?yx?yx?y ······························································································· 6分

=1． ······························································································ 7分

19．（本小题满分9分）

解：（1）该班60秒跳绳的平均次数至少是：

60?4?80?13?100?19?120?7?140?5?160?250＝100.8．

因为100.8>100，所以一定超过全校平均次数． ························································· 3分

（2）这个学生的跳绳成绩在该班是中位数，由4+13+19=36，所以中位数一定在100～120范围内． ··························································································································· 6分

（3）该班60秒跳绳成绩大于或等于100次的有：19+7+5+2=33（人）， ·················· 8分

3350?0.66．

所以，从该班任选一人，跳绳成绩达到或超过校平均次数的概率为0.66． ·········· 9分 20．（本题满分9分）

（1）解：在△AOC中，AC=2， ∵ AO＝OC＝2，

∴ △AOC是等边三角形． ··································· 2分 ∴ ∠AOC＝60°，

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∴∠AEC＝30°． ···················································· 4分 （2）证明：∵OC⊥l，BD⊥l．

∴ OC∥BD． ························································ 5分 ∴ ∠ABD＝∠AOC＝60°． ∵ AB为⊙O的直径，

∴ △AEB为直角三角形，∠EAB＝30°． ············································································· 7分 ∴∠EAB＝∠AEC．

∴ 四边形OBEC 为平行四边形． ············································································· 8分 又∵ OB＝OC＝2．

∴ 四边形OBEC是菱形． ························································································· 9分

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