（3）如图4，在直角梯形ABCD中，?D??C?90，对角线AC、BD交于点F，延长AB、DC交于点E，连接EF交梯形上、下底于G、H两点，请问直线GH是不是直角梯形ABCD的黄金分割线，并证明你的结论. · A
C C
A
· C 图1
· B
A
D 图2
B
图3
D
A
B
H B D
F ?C 图4
E
25.（本小题满分10分）如图1所示，已知直线y?kx?m与x轴、y轴分别交于A、C两
点，抛物线y??x?bx?c经过A、C两点，点B是抛物线与x轴的另一个交点，当
2x??12时，y取最大值
254.
（1）求抛物线和直线的解析式；
（2）设点P是直线AC上一点，且S?ABP ：S?BPC ?1:3，求点P的坐标； （3）若直线y?12x?a与（1）中所求的抛物线交于M、N两点，问:
0①是否存在a的值，使得?MON?90？若存在，求出a的值；若不存在，请说明理由；
②猜想当?MON?90时，a的取值范围（不写过程，直接写结论）. （参考公式：在平面直角坐标系中，若M(x1,y1)，N(x2,y2)，则M，N两点间的距离为MN?
0(x2?x1)2?(y2?y1)2） y C A O 图1
B x
黄石市2013年初中毕业生学业考试
数学答案及评分标准
一、选择题（每小题3分，共30分） 题号 答案 1 A 2 C 3 D 4 B 5 A 6 D 7 C 8 C 9 B 10 A
二、填空题（每小题3分，共18分）
11．3(x?3)(x?3) 12．k?0或k??1 13．
58
14．6?32 15．y??x?3 16．170
三、解答题（9小题，共72分） 17．(7分)解：原式?3?3?33?2?1?3 ······························································· （5分）
?4 ·································································································· （2分）
18．（7分）解：原式?ab?a2?ab?b2ab(a?b)(a?b)2ab(a?b)? ···································································· （2分）
?a?bab ·································································· （2分）
当a?5?12，b?5?12时，原式的值为5。 （ 3分）
19．（1）证明：连接OE，AM是⊙O的切线，OA是⊙O的半径
∴?DAO?90° ∵AD∥BC
∴?AOD??OBE，?DOE??OEB ∵OB?OE ∴?OEB??OBE 在△AOD和△DOE中
[来源学科网ZXXK]
?OA?OE???AOD??DOE ?OD?OD?∴△AOD≌△DOE
∴?DAO??DEO?90° ∴DE与⊙O相切 ·························································································· （3分）
（2）∵AM和BN是⊙O的两切线 ∴MA?AB，NB?AB ∴AD∥BC
∵O是AB的中点，OF∥BN ∴OF∥
12(AD?BC)且OF?12(AD?BC)
∵DE切⊙O于点E ∴DA?DE，CB?CE ∴DC?AD?CB ∴OF?12·································································· （4分） CD?ADE??CBF
1① ?222x?y???2 20．(8分)解：依题意?···································································· （2分）
?2x?25y?3② ? 由①得 4x?2y??1 ③
由②得 2x?2225? 3 ④
2将④代入③化简得9y?65y?5?0 ·················································· （4分）
即 y1?y2??53 代入②得 x1?x2??16
1?x?x??12?6?∴原方程组的解为?······················································· （4分）
?y?y??512?3?21．（8分）解：（1） 分 组 50.5～60.5 60.5～70.5 70.5～80.5 80.5～90.5 90.5～100.5 合 计 频数 频率 4 14 16 6 10 50 0.08 0.28 0.32 0.12 0.20 1.00 y 频率 组距 O 50.5 60.5 70.5 80.5 90.5 100.5 x ········································································································································· （6分）
（2）0.32?0.12?0.20?0.64?0.70说明该校的学生心理健康状况不正常，需要加强
心理辅导 ··················································································································· （2分） 22．（8分）解：过点A作AH?CF交CF于H点，由图可知
∵?ACH?75?15?60 ··················································································· （3分）
0000sin60?125?∴AH?AC?32?125?1.7322?108.25(m) ····························· （3分）
∵AH?100米 ∴不需要改道行驶 ··································································································· （２分）
23．（8分）解：（1）y1?60x （0≤x?10）
y2??100x?600 （0≤x?6） ···································· （2分）
15??160x?600(0?x?)4??15?（2）∴S??160x?600 (?x?6)
4????60x(6?x?10)（3）由题意得：S?200
①当0?x?154时，?160x?600?200 ∴x?∴y1?60x?150（km）
52
②当
154?x?6时，160x?600?200 ∴x?5
∴y1?60x?300（km）
③当6?x?10时，60x?360（舍） ······························ （3分）
24．（9分）解：（1）点D是AB边上的黄金分割点，理由如下：
∵?A?36°，AB?AC ∴?B??ACB?72° ∵CD平分?ACB ∴?DCB?36°
∴?BDC??B?72°
∵?A??BCD，?B??B ∴△BCD ∽△BAC
∴
BCABBC又∵BC?CD?AD
?BD
ABAB∴D是AB边上的黄金分割点 ················································ （3分）
（2）直线CD是△ABC的黄金分割线，理由如下：
设△ABC的边AB上的高为h，则
111S?ADC?AD?h，S?DBC?BD?h，S?ABC?AB?h
222∴S?ADC:S?ABC?AD:AB，S?DBC:S?ADC?BD:AD ∵D是AB的黄金分割点 ∴
∴
AD?BD
ADAB?BDAD
∴S?ADC:S?ABC?S?DBC:S?ADC
∴CD是△ABC的黄金分割线 ················································· （3分）
（3）GH不是直角梯形ABCD的黄金分割线
∵BC∥AD
∴△EBG ∽△EAH，△EGC ∽△EHD
∴
BGAHGCHD??EGEHEGEH ① ②
由①、 ②得
BGAHHDGCHD同理，由△BGF ∽△DHF，△CGF ∽△AHF得 BGGCBGHD 即 ④ ??HDAHGCAHAHHD由③、④得 ?HDAH∴AH?HD ∴BG?GC
∴ 梯形ABGH与梯形GCDH上下底分别相等，高也相等
1∴S梯形ABGH?S梯形GCDH?S梯形ABCD
2∴GH不是直角梯形ABCD的黄金分割线 ······························ （3分）
?GC 即
BG?AH ③
b1?????2?(?1)2b??125．（10分）解：（1）由题意得? 解得 2c?64?(?1)c?b25??4??4?(?1)?∴抛物线的解析式为y??x?x?6 ∴A(?3,0)，B(2,0)
2