2009年甘肃省武威、金昌、定西、白银、酒泉、嘉峪关市中考试卷答案

2009年甘肃省武威、金昌、定西、白银、酒泉、嘉峪关市中考试卷
数学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 12．34
x y =⎧⎨=⎩， 13．60o 14．二、四 15．1->x 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-x 2+bx +c =0的两个根为-3，1；⑦y >0时，-3<x <1；或y <0时，x <-3或x >1；⑧当x >-1时，y 随x 的增大而减小；或当x <-1时，y 随x 的增大而增大等等
三、解答题（一）：本大题共5小题，共38分．
19．本小题满分6分
解：∵a =2007200920082009⨯⨯(20081)(20081)20082009-⨯+=⨯22
2008120082009
-=⨯， ··································· 3分 b 2
200820082009
=⨯， ···································································································· 4分 222200812008-<， ······························································································· 5分 ∴ a <b ． ··························································································································· 6分 说明：求差通分作，参考此标准给分．若只写结论a <b ，给1分．
20．本小题满分6分
解：∵22a b a b ⊕=-，∴2222(43)(43)77x x x x ⊕⊕=-⊕=⊕=-． ···················· 3分
∴22724x -=．∴225x =． ······················································································ 4分 ∴5x =±． ····················································································································· 6分
21．本小题满分8分
解：∵随机闭合开关1S 、2S 、3S 中的两个，共有3种情况：12S S ，13S S ，23S S ．能让灯泡发光的有13S S 、23S S 两种情况． ·············································································· 4分 ∴ 能让灯泡发光的概率为
23
． ····················································································· 8分 22．本小题满分8分
解：从图中可以看出，在室内厚为a cm 的墙面、宽为4cm 的门框及开成120°的门之间构成了
一个直角三角形，且其中有一个角为60°． ···································································· 3分
从而 a =4×tan60° ················································· 6分
． ····································· 8分
即室内露出的墙的厚度约为6.9cm ．
23．本小题满分10分
解：（1）一次函数． ··············································································································· 2分
（2）设y kx b =+． ········································································································ 3分
由题意，得22162819k b k b =+⎧⎨=+⎩
，． ····························································································· 5分 解得210k b =⎧⎨=-⎩
，． ·············································································································· 7分 ∴210y x =-．（x 是一些不连续的值．一般情况下，x 取16、16.5、17、17.5、…、26、26.5、27等） ··············································································································· 8分 说明：只要求对k 、b 的值，不写最后一步不扣分．
（3）44y =时，27x =．
答：此人的鞋长为27cm ． ···························································································· 10分 说明：只要求对x =27cm ，不答不扣分．
四、解答题（二）：本大题共5小题，共50分 (不含附加4分) ．
24．本小题满分8分
解：（1）如图：
············································································ 4分
（2）∵参加足球运动项目的学生占所有运动项目学生的比例为15
=1050， ······················· 6分 ∴扇形统计图中表示“足球”项目扇形圆心角的度数为1
360725
⨯= ． ····················· 8分
25．本小题满分10分
解法1：设第一天捐款x 人，则第二天捐款（x +50）人， ··················································· 1分 由题意列方程：x
4800=506000+x ． ··············································································· 5分 解得 x =200．·················································································································· 7分 检验：当x =200时，x （x +50）≠0，
∴ x =200是原方程的解．······························································································· 8分 两天捐款人数x +（x +50）=450，人均捐款x
4800=24（元）． 答：两天共参加捐款的有450人，人均捐款24元． ·················································· 10分 说明：只要求对两天捐款人数为450， 人均捐款为24元，不答不扣分．
解法2：设人均捐款x 元， ···································································································· 1分 由题意列方程：6000x －4800x
＝50 ． ········································································ 5分 解得x =24． ······················································································································ 7分 以下略．
26．本小题满分10分
解：（1）如图，过A 作AO ⊥AC ，过B 作BO ⊥BD ，AO 与BO 相交于O ，O 即圆心． ············ 3分
说明：若不写作法，必须保留作图痕迹．其它作法略．
（2）∵ AO 、BO 都是圆弧AmB 的半径，O 是其圆心，
∴ ∠OBA =∠OAB =150°-90°=60°． ········································ 5分
∴ △AOB 为等边三角形．∴ AO =BO =AB =180． ·················· 7分
∴ 弧)(60180180
60m AB ππ=⨯⨯=
∴ A 到B 这段弧形公路的长为60πm ． ········································································· 10分
27．本小题满分10分
b
证明：（1）∵ACB ECD ∠=∠，
∴ ACE ACD BCD ACD ∠+∠=∠+∠．
即 ACE BCD ∠=∠． ·················································· 2分
∵ EC DC AC BC ==,，
∴ △ACE ≌△BCD ． ······················································ 4分
（2）∵ ACB ∆是等腰直角三角形，
∴ ︒=∠=∠45BAC B ． ··············································· 5分
∵ △ACE ≌△BCD ， ∴ ︒=∠=∠45CAE B ． ········ 6分
∴ ︒=︒+︒=∠+∠=∠904545BAC CAE DAE ． ················································ 7分 ∴ 2
22DE AE AD =+． ························································································· 9分 由（1）知AE =DB ，
∴ 222AD DB DE +=． ························································································ 10分
28．本小题满分16分(含附加4分)
解：（1）3k =-， ········································································ 1分
A （-1，0）， ··································································· 2分
B （3，0）． ···································································· 3分
（2）如图，抛物线的顶点为M （1，-4），连结OM ． 4分
则 △AOC 的面积=23，△MOC 的面积=2
3， △MOB 的面积=6， ························································· 5分
∴ 四边形 ABMC 的面积
=△AOC 的面积+△MOC 的面积+△MOB 的面积=9． ············································· 6分 说明：也可过点M 作抛物线的对称轴，将四边形ABMC 的面积转化为求1个梯形与2
个直角三角形面积的和．
（3）如图，设D （m ，322
--m m ），连结OD ．
则 0＜m ＜3，322--m m ＜0．
且 △AOC 的面积=23，△DOC 的面积=m 2
3， △DOB 的面积=-
23（322--m m ）， ·································································· 8分 ∴ 四边形 ABDC 的面积=△AOC 的面积+△DOC 的面积+△DOB 的面积 =629232++-
m m =8
75)23(232+--m ． ··························································································· 9分 ∴ 存在点D 3
15()24-，
，使四边形ABDC 的面积最大为875． ································ 10分 （4）有两种情况：
如图，过点B 作BQ 1⊥BC ，交抛物线于点Q 1、交y 轴于点E ，连接Q 1C ．
∵ ∠CBO =45°，∴∠EBO =45°，BO =OE =3．
∴ 点E 的坐标为（0，3）．
∴ 直线BE 的解析式为3y x =-+． ·········································································· 12分
由2323
y x y x x =-+⎧⎨=--⎩， 解得1125x y ，；ì=-ïïíï=ïî 2230.x y ，ì=ïïíï=ïî ∴ 点Q 1的坐标为（-2，5）． ······················································································· 13分 如图，过点C 作CF ⊥CB ，交抛物线于点Q 2、交x 轴于点F ，连接BQ 2．
∵ ∠CBO =45°，∴∠CFB =45°，OF =OC =3．
∴ 点F 的坐标为（-3，0）．
∴ 直线CF 的解析式为3y x =--． ·········································································· 14分 由2323y x y x x =--⎧⎨=--⎩，
解得⎩⎨⎧-==3011y x ；⎩⎨⎧-==4
122y x ∴点Q 2的坐标为（1，-4）．························································································· 15分 综上，在抛物线上存在点Q 1（-2，5）、Q 2（1，-4），使△BCQ 1、△BCQ 2是以BC 为直角边的直角三角形． ····································································································· 16分 说明：如图（4），点Q 2即抛物线顶点M ，直接证明△BCM 为直角三角形同样得2分． 附加题：如果你的全卷得分不足150分，则本题与28题附加的4分的得分将记入总分，但记入总分后全卷得分不得超过150分，超过按150分算．
29．本小题满分7分
解：学生可能写出不同程度的一般的结论，由一般化程度不同得不同分．
若m 、n 是任意正整数，且m ＞n ，则11
n n m m +<+． ······················································ 4分 若m 、n 是任意正实数，且m ＞n ，则
11n n m m +<+． ······················································ 5分 若m 、n 、r 是任意正整数，且m ＞n ；或m 、n 是任意正整数，r 是任意正实数，且m ＞n ，则n n r m m r
+<+． ··············································································································· 6分 若m 、n 是任意正实数，r 是任意正整数，且m ＞n ；或m 、n 、r 是任意正实数，且m ＞n ，则n n r m m r +<+． ······································································································· 7分。