2012初三数学质量检测试题及答案

2012年福州市初中毕业班质量检查 数学试卷参考答案及评分标准

一、选择题

1．A 2．B 3．C 4．C 5．D 6．A 7．D 8．A 9．B 10．C 二、填空题：

11．(x?3)(x?3) 12．8 13．三、解答题：

2 14.1 15．3 9116．（1）解：?1?2 =1? =

?1?8?(?3.14)???

?2?0?11?22?1?2 ····································································4分 2·······················································································7分 2.

2（2）解：?x?1??x?x?2?

=x2?2x?1?x2?2x ·····································································4分

=2x2?1, ····················································································5分

当x?2时，原式=2??2?2?1=5． ············································7分

17．（1）证明：∵AB与CD是平行四边形ABCD的对边，

∴AB∥CD， ······································································2分 ∴∠F=∠FAB． ···································································4分 ∵E是BC的中点， ∴BE=CE, ············································5分 又∵ ∠AEB=∠FEC， ·······················································6分 ∴ △ABE≌△FCE． ··························································7分

（2）解：如图，a = 45°，β= 60°， AD＝80． 在Rt△ADB中,

BD， AD∴BD?AD?tan??80?tan45?=80．······· 2分 ∵tan??在Rt△ADC中, ∵tan??CD, AD ∴CD?AD?tan??80?tan60?=803． ··· 5分

∴BC?BD?CD?80?803?218.6． 答：这栋楼高约为218.6m． ······· 7分

18．（1）a= 25 %， 90o ． ··························································2分

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补全条形图. ···················································································4分

（2）众数是5，中位数是5． ························································8分 （3）该市初一学生第一学期社会实践活动时间不少于5天的人数约是：

． ····································· 12分 20000?(30%?25%?20%)?15000（人）19．（1）证法一：连接AE, ······················· 1分

∵AC为⊙O的直径,

∴∠AEC=90o,即AE⊥BC. ················· 4分 ∵AB=AC,

∴BE=CE，即点E为BC的中点． ····· 6分 证法二：连接OE， ························· 1分 ∵OE=OC, ∴∠C=∠OEC. ∵AB=AC, ∴∠C=∠B, ∴∠B=∠OEC,

∴OE∥AB. ····································· 4分 ∴

ECOC??1, BEAO∴EC=BE,即点E为BC的中点. ······· 6分 ⑵∵∠COD=80o,

∴∠DAC=40o . ······························· 8分 ∵∠DAC+∠DEC=180o,∠BED+∠DEC=180o, ∴∠BED=∠DAC=40o． ················· 11分

20．解：（1）设A型计算器进价是x元，B型计算器进价是y元， ·········1分

x?y8?880?10得 ? ·································································3分

2x?5y?380.??x?40,解得? ······································································5分

y?60.?答：每只A型计算器进价是40元，每只B型计算器进价是60元． ·6分

（2）设购进A型计算器为z只，则购进B型计算器为（50-z）只，得 ?40z?60(50?z)?2520, ·························································9分 ?10z?15(50?z)?620.?解得24≤z≤26,

因为z是正整数，所以z=24,25,26. ·································· 11分

答：该经销商有3种进货方案：①进24只A型计算器，26只B型计算器；②进25只A型计算器，25只B型计算器；③进26只A型计算器，24只B型计算器. ·········································································································· 12分

21．解：

（1）BE??t?4?cm， ········ 1分

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EF?5?t?4?cm. 8 ····················································4分

（2）分三种情况讨论： ①当DF?EF时，

有?EDF??DEF??B,

∴点B与点D重合， ∴t?0. ······························ 5分 ②当DE?EF时, ∴4?5?t?4?, 8解得：t?12···················· 7分 . ·5③当DE?DF时,

有?DFE??DEF??B??C, ∴△DEF∽△ABC.

5t?4?48?DEEF?∴, 即, ?1016ABBC解得：t?156. ················· 9分 2512156或秒时， 525综上所述，当t=0、

△DEF为等腰三角形．

（3）设P是AC的中点，连接BP， ∵EF∥AC, ∴△FBE∽△ABC.

EFBEENBE?, ∴?. ACBCCPBC又?BEN??C, ∴△NBE∽△PBC, ∴

∴?NBE??PBC.············································································· 10分 ∴点N沿直线BP运动，MN也随之平移.

如图，设MN从ST位置运动到PQ位置，则四边形PQST是平行四边形.11分 ∵M、N分别是DF、EF的中点，∴MN∥DE，且ST=MN=

1DE?2. 2分别过点T、P作TK⊥BC，垂足为K，PL⊥BC，垂足为L，延长ST交PL于点R，则四边形TKLR是矩形， 当t=0时，EF=

1551533（0+4）=,TK=EF·sin?DEF?··?;

2222548113AC·10·?3. sinC?·225

当t=12时，EF=AC=10，PL=∴PR=PL-RL=PL-TK=3-

39?. 44 7

99∴S?PQST?ST·PR=2×?.

42∴整个运动过程中，MN所扫过的面积为

?c?4,??4?3?9?3b?c?0,?22．解：（1）由题意

∴抛物线解析式

（2）令y?0，得

9cm2. ······························ 13分 216?b??,?………3 1分 得： 解得：??c?4,?………2分 4216为y?·····················································x?x?4. ·334216x?x?4?0. 33解得：x1?1，x2=3.

∴C点坐标为（1，0）. ··························· 4分 作CQ⊥AB，垂足为Q，延长CQ，使CQ=C'Q，

则点

C'

就是点C关于直线AB的对称点． 由△ABC的面积得：

11CQ?AB?CA?OB, 22∵AB?OA2?OB2?5,CA=2，

816，CC'=. ·································································6分 55作C'T⊥x轴，垂足为T，则△CTC'∽△BOA. ∴CQ=∴

C'TCC'CT4864 ∴C'T=，CT= ??OAABOB2525∴OT=1+

64898948= ∴C'点的坐标为（，） ····················8分 25252525（3）设⊙D的半径为

r，∴AE=r+3，BF=4－r，HB=BF=4－r.

∵AB=5，且AE=AH, ∴r+3=5+4－r,

∴r=3． ·····························10分 HB=4－3=1.

作HN⊥y轴，垂足为N，

HNHBBNHB，, ??OAABOBAB34∴HN=，BN=,

55则

∴H点坐标为（?324，）. ······12分 55根据抛物线的对称性，得PA=PC,

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∵PH?PA?PH?PC?HC,

∴当H、C、P三点共线时，PH?PC最大.

832242)?()=55585∵HC=(1?10, ∴PH?PA的最大值为

10. ··················································· 14分

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