2007—2014年陇南市中考题（含答案）(11)

在Rt△DEC中，∵ CD=5， ································· 7分 ∴ BC=DC，又∠C=∠C， ···································· 8分 ∴ Rt△BFC≌Rt△DEC． ······································ 9分 ∴ BF= DE=4． ················································ 10分 图②

方法二：

在Rt△DEC中，∵ CD=5， ····································································· 7分

∴ sinC=

45． ······················································································ 8分 在Rt△BFC中，BF=BC·sinC=5?45=4． ·················································· 10分

27． 本小题满分10分

解：(1) 树状图为：

共有12种可能结果． ············································································· 4分 说明：无最后一步不扣分．

（2）游戏公平． ·············································································· 6分 ∵ 两张牌的数字都是偶数有6种结果：

（6，10），（6，12），（10，6），（10，12），（12，6），（12，10）．

∴ 小明获胜的概率P=

612=12． ····························································· 8分 小慧获胜的概率也为12．

∴ 游戏公平． ················································································ 10分

解：(1)（4，0），（0，3）； ······································································ 2分 (2) 2，6； ··························································································· 4分 (3) 当0＜t≤4时，OM=t． 由△OMN∽△OAC，得OMOA?ONOC， ∴ ON=

334t，S=8t2． ·

······························ 6分 当4＜t＜8时，

如图，∵ OD=t，∴ AD= t-4． 方法一：

由△DAM∽△AOC，可得AM=

34(t?4)，∴ BM=6-34t． ·

······················· 7分 由△BMN∽△BAC，可得BN=43BM=8-t，∴ CN=t-4． ······························ 8分

S=矩形OABC的面积-Rt△OAM的面积- Rt△MBN的面积- Rt△NCO的面积

=12-32(t?4)-12（8-t）

（6-34t）-32(t?4) =?38t2?3t． ···············································································10分

方法二：

易知四边形ADNC是平行四边形，∴ CN=AD=t-4，BN=8-t． ····························· 7分 由△BMN∽△BAC，可得BM=3334BN=6-4t，∴ AM=4(t?4)． ·

············· 8分 以下同方法一． (4) 有最大值． 方法一： 当0＜t≤4时，

∵ 抛物线S=38t2的开口向上，在对称轴t=0的右边， S随t的增大而增大， ∴ 当t=4时，S可取到最大值38?42=6； ················································ 11分

当4＜t＜8时， ∵ 抛物线S=?328t?3t的开口向下，它的顶点是（4，6）

，∴ S＜6． 综上，当t=4时，S有最大值6． ····························································· 12分 方法二：

?3t2，0?t≤∵ S=???84

????38t2?3t，4?t?8∴ 当0＜t＜8时，画出S与t的函数关系图像，如图所示． ························· 11分 显然，当t=4时，S有最大值6． ·························································· 12分

说明：只有当第（3）问解答正确时，第（4）问只回答“有最大值”无其它步骤，可给1分；否则，不给分．

附加题 (12分)

C 1． （1）三条中线交于一点； ······················· 2分 B （2）在同一条中线上，这个点到边中点的距离等

A 于它到顶点距离的一半． ······················· 5分

2． 能消去AC、BC、CD，得到sin(α+β)= sinα·cosβ+cosα·sinβ． ··························· 2分 解：给AC·BC·sin(α+β)= AC·CD·sinα+BC·CD·sinβ两边同除以AC·BC，得

sin(α+β)= CDBC·sinα+CDAC·sinβ， ···················································· 4分

∵ CDBC=cosβ， CDAC=cosα． ······················································ 6分

∴ sin(α+β)= sinα·cosβ+cosα·sinβ． ······················································ 7分

说明：如果上边解法没有第1个步骤的采分点，则后边三个采分点得分分别改为2分、6分、7分． 全卷说明：对于以上各解答题学生试卷中出现的不同解法，请参考本标准给分

2009年初中毕业、高中招生考试

数学试卷参考答案与评分标准

一、选择题：本大题共10小题，每小题3分，共30分．

题号

1 2 3 4 5 6 7 8 9 10 答案

B D A B D B C

A

A

C

二、填空题：本大题共8小题，每小题4分，共32分．

11．9 12． ??x?3，?y?4 13．60o 14．二、四

15．x??1 16．答案不唯一，如AC=BD，∠BAD=90o，等 17． 5

18．答案不唯一．如：①c=3；②b+c=1；③c-3b=9；④b=-2；⑤抛物线的顶点为（-1，4），或二次函数的最大值为4；⑥方程-x2+bx+c=0的两个根为-3，1； …… 此处隐藏：913字，全部文档内容请下载后查看。喜欢就下载吧 ……