28．（12分）如图20，在平面直角坐标系中，四边形OABC是矩形，点B的坐标为（4，3）．平行于对角线AC的直线m从原点O出发，沿x轴正方向以每秒1个单位长度的速度运动，设直线m与矩形OABC的两边分别交于点M、N，直线m运动的时间为t（秒）． ．．(1) 点A的坐标是__________，点C的坐标是__________； (2) 当t= 秒或 秒时，MN=
1AC； 2(3) 设△OMN的面积为S，求S与t的函数关系式；
(4) 探求(3)中得到的函数S有没有最大值？若有，求出最大值；若没有，要说明理由．
图20
附加题 (12分) 1．（5分）如图21，网格小正方形的边长都为1．在⊿ABC中，试画出三边的中线（顶点与对边中点连结的线段），然后探究三条中线位置及其有关线段之间的关系，你发现了什么有趣的结论？请说明理由． C B
A 图21 2．（7分）如图22（1），由直角三角形边角关系，可将三角形面积公式变形， 得 S△ABC=
1bc·sin∠A． ① 2A C b c 图22 (1) 即 三角形的面积等于两边之长与夹角正弦之积的一半． 如图22（2），在⊿ABC中，CD⊥AB于D，∠ACD=α， ∠DCB=β． ∵ S△ABC? S△ADC?S△BDC， 由公式①，得
B 111AC·BC·sin(α+β)= AC·CD·sinα+BC·CD·sinβ， 222即 AC·BC·sin(α+β)= AC·CD·sinα+BC·CD·sinβ． ②
你能利用直角三角形边角关系，消去②中的AC、BC、CD吗？不能， 说明理由；能，写出解决过程．
A C α β D 图22 (2) B 白银等九市试题答案
一、选择题：本大题共10小题，每小题3分，共30分．
1．A 2．C 3．B 4．D 5．B 6．C 7．A 8．A 9．D 10．B 二、填空题：本大题共8小题，每小题4分，共32分．
11． 3 12．（-2，-3） 13．4 14． (0，-4) 15． 90o 16． 150×80%－x＝20 17． y=?1 x18． 答案不唯一． 可供参考的有：①它内角的度数为60°、60°、120°、120°；②它的腰长
等于上底长；③它的上底等于下底长的一半． 三、解答题(一)：本大题共5小题，共38分． 19． 本小题满分6分
解法1：原式=(a+2)-(a-2) ··································································· 4分
=4． ··············································································· 6分
a(a?2)?a(a?2)a2?4?解法2：原式= ····································· 2分 2a?4a=(a?2)?(a?2) ······················································· 4分 =4． ·············································································· 6分
20． 本小题满分6分
答案不唯一． 可供参考的有：
相离：
······························· 1分
相切： ······························· 3分
相交： ······························· 5分
其它：
························································ 6分
21． 本小题满分8分
解：（1）7，
15． ·········································································· 4分 8（2）设所求的解析式为y?kx?b， ·························································· 5分 ∵ 点（0，15）、（1，7）在图像上，
?15?b, ……………………………………………………………………… 6分 ∴ ??7?k?b. 解得 k??8，b?15．
∴ 所求的解析式为y??8x?15． (0≤x≤
15) …………………………… 8分 8说明：只要求对k??8、b?15，不写最后一步，或者未注明x的取值范围，都不扣
分．
22． 本小题满分8分
证明：(1)
∵ 四边形ABCD是平行四边形，
∴ AD∥BF，∴ ∠D=∠ECF． ···························································· 3分
∵ E是CD的中点，∴ DE = CE．
又 ∠AED=∠FEC， ··································· 4分
∴ △ADE≌△FCE． ·································· 5分 (2) D．或填“平行四边形”． ······························ 8分 23． 本小题满分10分
解；（1）不及格，及格； ········································································ 4分 （2）抽到的考生培训后的及格与优秀率为（16+8）÷32=75%， ······················· 6分 由此，可以估计八年级320名学生培训后的及格与优秀率为75%． ················ 8分 所以，八年级320名学生培训后的及格与优秀人数为75%×320=240． ············ 10分
四、解答题(二)：本大题共5小题，共50分． 24． 本小题满分8分
解：(1)56； ··························································································· 3分 （2）如图，△O1 O2 O3是边长为8mm的正三角形， 作底边O2O3上的高O1 D． ······························· 4分 则 O1D=O1O3·sin60°=43≈6.92． ···················· 6分 ∴ AD=2(O1D+4)=2×10.92≈21.8（mm）． ··············· 8分 说明：(1)用勾股定理求O1D，参考本标准评分； （2）在如图大正三角形中求高后再求AD，
也参考本标准评分．
25． 本小题满分10分
解：设花边的宽为x分米， ································································ 1分 根据题意，得(2x?6)(2x?3)?40． ·············································· 5分 解得x1?1，x2??x2=?O2 D O3 O1 11． ··························································· 8分 411不合题意，舍去． ································································ 9分 4答: 花边的宽为1米． ····························································· 10分 说明：不答不扣分． 26． 本小题满分10分
解：（1）如图①，作DE⊥BC于E， ·················· 1分 ∵ AD∥BC，∠B=90°， ∴ ∠A=90°．又∠DEB=90°，
图①
∴ 四边形ABED是矩形． ·································································· 2分 ∴ BE=AD=2， ∴ EC=BC-BE=3． ······················································· 3分 在Rt△DEC中，DE= EC·tanC =3?4=4． ·········································· 5分 3（2）如图②，作BF⊥CD于F． ·························································· 6分 方法一：
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 方法二：
图②
在Rt△DEC中，∵ CD=5， ····································································· 7分
4． ······················································································ 8分 54在Rt△BFC中，BF=BC·sinC=5?=4． ·················································· 10分
5∴ sinC=
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）．
61=． ···························································· 8分 1221 小慧获胜的概率也为．
2∴ 小明获胜的概率P=
∴ 游戏公平． ··············································································· 10分 28． 本小题满分12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∴ ON=
OMON， ?OAOC33······························ 6分 t，S=t2． ·
48当4＜t＜8时，
如图，∵ OD=t，∴ AD= t-4． 方法一：
33······················· 7分 (t?4)，∴ BM=6-t． ·
444由△BMN∽△BAC，可得BN=BM=8-t，∴ CN=t-4． ····························· 8分
3由△DAM∽△AOC，可得AM=
2008年白银等九市中考数学试题及答案
