2009年天津中考数学试题答案

参考答案及评分标准
评分说明：
1．各题均按参考答案及评分标准评分．
2．若考生的非选择题答案与参考答案不完全相同但言之有理，可酌情评分，但不得超过该题所分配的分数．
一、选择题：本大题共10小题，每小题3分，共30分.
1．A 2．B 3．B 4．C 5．D 6．A 7．A 8．B 9．D 10．C 二、填空题：本大题共8小题，每小题3分，共24分.
11
12．2
13．正方形（对角线互相垂直的四边形均可） 14．()01-，
15．56，80，156.8 16．60；13 17．21
18．①3，4（提示：答案不惟一）；
②裁剪线及拼接方法如图所示：图中的点E 可以是以BC 为直径的半圆上的任意一点（点B C ，除外）.BE CE ，的长分别为两个小正方形的边长. 三、解答题：本大题共8小题，共66分 19．本小题满分6分
解：5125431x x x x ->+⎧⎨-<+⎩，①②
由①得2x >， ······························································································· 2分
由②得，5
2
x >-
···························································································· 4分 ∴原不等式组的解集为2x > ··········································································· 6分
20．本小题满分8分. 解：（Ⅰ）这个反比例函数图象的另一支在第三象限. ············································· 1分 因为这个反比例函数的图象分布在第一、第三象限， 所以50m ->，解得5m >. ············································································ 3分
（Ⅱ）如图，由第一象限内的点A 在正比例函数2y x =的图象上，
设点A 的坐标为()()00020x x x >，，则点B 的坐标为()00x ，，
0014242
OAB S x x =∴=△，·，解得02x =（负值舍去）.
∴点A 的坐标为()24，. ·
··················································································· 6分 D
C
A
E 2 3
1 2
3
又
点A 在反比例函数5
m y x
-=
的图象上， 5
42
m -∴=
，即58m -=. ∴反比例函数的解析式为8y x
=
. ········································································ 8分 21．本小题满分8分.
解（Ⅰ）法一：根据题意，可以画出如下的树形图：
从树形图可以看出，摸出两球出现的所有可能结果共有6种； 法二：根据题意，可以列出下表：
从上表中可以看出，摸出两球出现的所有可能结果共有6种. ··································· 4分 （Ⅱ）设两个球号码之和等于5为事件A .
摸出的两个球号码之和等于5的结果有2种，它们是：()()2332，，，.
()21
63
P A ∴=
=. ·
·························································································· 8分 22．本小题满分8分.
解（Ⅰ）PA 是O ⊙的切线，AB 为O ⊙的直径， PA AB ∴⊥.
90BAP ∴∠=°.
30BAC ∠=°，
9060CAP BAC ∴∠=-∠=°°． ·
··································
····························· 2分 又PA 、PC 切O ⊙于点A C ，. PA PC ∴=. PAC ∴△为等边三角形.
60P ∴∠=°. ·
································································································ 5分
（Ⅱ）如图，连接BC ， 则90ACB ∠=°．
在Rt ACB △中，230AB BAC =∠=，°，
AC AB ∴=·cos 2BAC ∠=cos 30°=. PAC △为等边三角形， PA AC ∴=.
1 2 3
2 1
3 3 1 2 第一个球 第二个球 第二个球 第一个球 （1，3） （2，3） （1，2） （3，2）
（3，1） （2，1） 3 2 1 1 2 3 P C B A O
PA ∴=································································································· 8分
23．本小题满分8分
解：如图，过C 点作CD 垂直于AB 交BA 的延长线于点D . ··································· 1分 在Rt CDA △中，3018018012060AC CAD CAB =∠=-∠=︒-︒=︒，°. ················ 2分
CD AC ∴=·sin 30CAD ∠=·
sin 60=°AD AC =·cos 30CAD ∠=·cos 60°=15. 又在Rt CDB △中，
22270BC BD BC CD ==，-，
65BD ∴==. ········································································ 7分
651550AB BD AD ∴=-=-=，
答：A B ，两个凉亭之间的距离为50m. ······························································· 8分
24．本小题满分8分.
解（Ⅰ）220630424260600x x x x ---+，，； ···················································· 3分 （Ⅱ）根据题意，得2
124260*********x x ⎛
⎫-+=-⨯⨯ ⎪⎝⎭
. ···································· 5分 整理，得2665500x x -+=.
解方程，得125106
x x ==，（不合题意，舍去）.
则55
2332
x x ==，．
答：每个横、竖彩条的宽度分别为
53cm ，5
2
cm. ···················································· 8分 25．本小题满分10分.
解（Ⅰ）如图①，折叠后点B 与点A 重合， 则ACD BCD △≌△.
设点C 的坐标为()()00m m >，. 则4BC OB OC m =-=-. 于是4AC BC m ==-.
在Rt AOC △中，由勾股定理，得222
AC OC OA =+，
即()2
2242m m -=+，解得32
m =
. ∴点C 的坐标为302⎛⎫
⎪⎝⎭
，. ··················································································· 4分
图①
图②
图③
（Ⅱ）如图②，折叠后点B 落在OA 边上的点为B ', 则B CD BCD '△≌△. 由题设OB x OC y '==，， 则4B C BC OB OC y '==-=-，
在Rt B OC '△中，由勾股定理，得222B C OC OB ''=+.
()2
224y y x ∴-=+，
即2
128
y x =-
+ ·
··························································································· 6分 由点B '在边OA 上，有02x ≤≤，
∴ 解析式21
28
y x =-+()02x ≤≤为所求.
∴ 当02x ≤≤时，y 随x 的增大而减小，
y ∴的取值范围为3
22
y ≤≤. ·
···································································· 7分 （Ⅲ）如图③，折叠后点B 落在OA 边上的点为B ''，且B D OB ''∥. 则OCB CB D ''''∠=∠. 又CBD CB D OCB CBD ''''∠=∠∴∠=∠，，有CB BA ''∥. Rt Rt COB BOA ''∴△∽△. 有OB OC OA OB
''=，得2OC OB ''=. ·································································· 9分 在Rt B OC ''△中，
设()00OB x x ''=>，则02OC x =. 由（Ⅱ）的结论，得2
001228
x x =-
+，
解得000808x x x =-±>∴=-+，
∴点C 的坐标为()
016. ··································································· 10分
26．本小题满分10分. 解（Ⅰ）
212120y x y x bx c y y ==++-=，，，
()210x b x c ∴+-+=. ·············································································· 1分 将1
132
αβ==
，分别代入()2
10x b x c +-+=，得 ()()2
2
111110103322b c b c ⎛⎫⎛⎫
+-⨯+=+-⨯+= ⎪ ⎪⎝⎭⎝⎭
，，
解得1166
b c =
=，. ∴函数2y 的解析式为2y 251
66
x x =-
+． ·
····················································· 3分
（Ⅱ）由已知，得6
AB =
，设ABM △的高为h ，
31121212ABM S AB h h ∴=
==△·1144
=.
根据题意，t T -=，
由211
66
T t t =++，得251166144t t -+-=
. 当251166144t t -
+=-
时，解得12512
t t ==；
当2511
66144
t t -
+=
时，解得34551212t t -==.
t ∴的值为
512··································································· 6分 （Ⅲ）由已知，得
222b c b c T t bt c αααβββ=++=++=++，，. ()()T t t b ααα∴-=-++， ()()T t t b βββ-=-++，
()()22b c b c αβααββ-=++-++，化简得()()10b αβαβ-++-=.
01αβ<<<，得0αβ-≠， 10b αβ∴++-=.
有1010b b αββα+=->+=->，. 又01t <<，0t b α∴++>，0t b β++>，
∴当0t a <≤时，T αβ≤≤；
当t αβ<≤时，T αβ<≤；
当1t β<<时，T αβ<<. ············································································ 10分。