发布时间 : 星期三 文章数学f1初中数学200862522335526991(1)更新完毕开始阅读26b1ef6df5335a8102d220af
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?x?1或x?3．
可得y??3或y??1．
于是??x?1，?x?3，或?
y??3y??1．??而点A的坐标是(1，?3)，
···································································································· 2分 ?1)． ·?点B的坐标为(3，五、（每小题10分，共20分） 19．解：（1）从纸箱中随机地一次取出两个小球，所标数字的所有可能结果有：
(1，，2)(13)，，(1，，，，，，，4)(23)(24)(34)，共6种；
而所标数字一个是奇数另一个是偶数的有4种． ································································· 3分
?P?42······················································································································ 2分 ?． ·
63
（2）画树状图：
开始
第一次 第二次
组成的两位数
1 2 3 4 1 2 3 4 1 2 3 4 1 2 （（ 34）（ 33）（ 32）（ 14）（（ 24） 31）（ 44） （ （13）（ 23）（ （43）（11） 12）（ 21） 22）（ 41） 42）
3 4 1
2 3 4
或用列表法：
第 一 次 1 2 3 4 （11） （21） （31） （41） （12） （22） （32） （42） （13） （23） （33） （43） （14） （24） （34） （44） 第 二 次 1 2 3 4 ··································································· 3分
所有可能出现的结果共有16种，其中能被3整除的有5种．
5． ···························································································································· 2分 1620．（1）解：由题意，有△BEF≌△DEF．
································································ 1分 ?BF?DF． ·D A 如图，过点A作AG?BC于点G． E ?P?
B G F C 知识决定命运 百度提升自我
则四边形AGFD是矩形．
?AG?DF，GF?AD?4． 在Rt△ABG和Rt△DCF中， ?AB?DC，AG?DF，
（HL） ?Rt△ABG≌Rt△DCF．
······················································································································· 2分 ?BG?CF． ·
11?BG?(BC?GF)?(8?4)?2．
22··············································································· 2分 ?DF?BF?BG?GF?2?4?6． ·
11··················································· 1分 ?S梯形ABCD?(AD?BC)?DF??(4?8)?6?36． ·
221（2）猜想：CG?k?BE（或BE??． ·································································· 1分 CG）
k证明：如图，过点E作EH∥CG，交BC于点H． 则?FEH??FGC． 又?EFH??GFC， ?△EFH∽△GFC． EFEH． ??GFGCGF而FG?k??k． EF，即
EFEH1······················· 2分 ??．即CG?k?EH． ·
D A GCk?EH∥CG，??EHB??DCB．
E 而ABCD是等腰梯形，??B??DCB．
C B ??B??EHB．?BE?EH． H F ··················································· 1分 ?CG?k?BE．·G
B卷（共50分）
一、填空题：（每小题4分，共20分） 21．1； 22．4； 23．分别作点A关于OM，ON的对称点A?，A??；连结A?，A??，分别交OM，ON于点B、点C，则点B、点C即为所求．（2分）如图所示（2分）；
A? M B A 24．
3； 425．
1113． 5O C N
A?? 226．解：（1）设乙队单独完成这项工程需要x天，则甲队单独完成这项工程需要x天．
3二、（共8分）
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???11?10根据题意，得?30????1．
22?xx?x3?3?解得x?90．
经检验，x?90是原方程的根． ··························································································· 3分 22?x?90?60． 33答：甲、乙两队单独完成这项工程各需要60天和90天． ·················································· 1分 （2）设甲、乙两队合作完成这项工程需要y天．则有y?1??1???1． ?6090?解得y?36． ························································································································· 2分 需要施工费用：36?(0.84?0.56)?50.4（万元）． ·························································· 1分
?50.4?50，
····················································· 1分 ?工程预算的施工费用不够用，需追加预算0.4万元． ·
三、（共10分）
27．解：（1）连结OB，OM． 则在Rt△OMB中，
C E O B ?OB?2，MB?3，
?OM?1．
1?OM?OB，??OBM?30?．
2??MOB?60?．
连结OA．则?AOB?120．
?D A M 1····································································································· 3分 ??C??AOB?60?．
2[或：延长BO与?O相交于点F，连结AF．
则有?ACB??AFB，且?FAB?90．
在Rt△ABF中，?BO?2，?BF?2BO?2?2?4． 又sin?AFB??AB233??， BF42??AFB?60?．
??AFB??ACB，??C?60?．]
（2）在△CDE和△CBA中，
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??CDE??CBA，?ECD??ACB， ?△CDE∽△CBA． DEDC． ??ABBC连结BD．则?BDC??ADB?90． 在Rt△BCD中，
???BCD?60?，??CBD?30?．
?BC?2DC． DC1DE1??．即?． BC2AB211························································································ 3分 ?DE?AB??23?3． ·
22[或：?点C在?AB上移动，??C恒为60，DE长始终不变．当点C移动到BO延长线与?O交点处时，
?可求得DE?AB?sin30?23?（3）连结AE．
?1?3．] 2?AB是?M的直径，??AEB??AEC?90?．
AD?x，可得AD?x?DC，AC?AD?DC?(x?1)?DC． DC在Rt△ACE中，
CEAE，sin?ACE?， ?cos?ACE?ACAC1?CE?AC?cos?ACE?(x?1)?DC?cos60??(x?1)?DC；
2由
AE?AC?sin?ACE?(x?1)?DC?sin60??又由（2），知BC?2DC．
3(x?1)?DC． 211············································ 3分 ?BE?BC?CE?2DC?(x?1)?DC?(3?x)?DC． ·
22在Rt△ABE中，
3(x?1)?DCAE3(x?1)， ?tan?ABC??2?1BE3?x(3?x)?DC2?y?3(x?1)(0?x?3)． ······························································································· 1分
3?x[或：由（2），知△CDE∽△CBA，