（2）解法一：如图1，连结DE．
?AE是?O的直径， ??ADE?90?．
?AD:AO?8:5，
AD4?cosA??． ············································································································ 3分
AE5??C?90?，?CBD??A，
BC4?． ··································································································· 4分 BD55?BC?2， ?BD?．························································································· 5分
21解法二：如图2，过点O作OH?AD于点H． ?AH?DH?AD．
2?AD:AO?8:5，
C AH4?cosA??． ························· 3分
AO5D?cos?CBD? ??C?90?，?CBD??A，
A
H B
O BC4?cos?CBD??． ········································· 4分
BD5图2 ?BC?2，
5?BD?． ··························································································································· 5分
2五、解答题（本题满分6分） 解：（1）补全图1见下图． ··································································································· 1分 “限塑令”实施前，平均一次购物使 用不同数量塑料购物袋的人数统计图 ．． 人数／位
40 37 35 30 26
25 20 15 11 10 9 10 4 3 5
9?1?37?2?26?3?110 ?4?10?6?76 3007 塑料袋数／个 1 ?52 ?43 4 3?5 ??3（个）．
图1 100100这100位顾客平均一次购物使用塑料购物袋的平均数为3个． ·········································· 3分
2000?3?6000．
估计这个超市每天需要为顾客提供6000个塑料购物袋． ···················································· 4分 （2）图2中，使用收费塑料购物袋的人数所占百分比为25%． ········································ 5分 根据图表回答正确给1分，例如：由图2和统计表可知，购物时应尽量使用自备袋和押金式环保袋，少用塑料购物袋；塑料购物袋应尽量循环使用，以便减少塑料购物袋的使用量，为环保做贡献． ·························································································································· 6分
六、解答题（共2道小题，共9分）
21．解：设这次试车时，由北京到天津的平均速度是每小时x千米，则由天津返回北京的平均速度是每小时(x?40)千米． ···························································································· 1分
30?61x?(x?40)． ······················································································ 3分 602解得x?200． ······················································································································· 4分
依题意，得
答：这次试车时，由北京到天津的平均速度是每小时200千米． ······································ 5分 22．解：（1）重叠三角形A?B?C?的面积为3． ································································ 1分 （2）用含m的代数式表示重叠三角形A?B?C?的面积为3(4?m)2； ····························· 2分
m的取值范围为
8≤m?4． ······························································································· 4分 3七、解答题（本题满分7分）
23．（1）证明：?mx2?(3m?2)x?2m?2?0是关于x的一元二次方程，
???[?(3m?2)]2?4m(2m?2)?m2?4m?4?(m?2)2．
?当m?0时，(m?2)2?0，即??0．
·························································································· 2分 ?方程有两个不相等的实数根． ·
(3m?2)?(m?2)（2）解：由求根公式，得x?．
2m2m?2?x?或x?1． ········································································································ 3分
m?m?0，
2m?22(m?1)???1．
mm?x1?x2，
2m?2． ······································································································ 4分 m2m?22?y?x2?2x1??2?1?．
mm2y 即y?(m?0)为所求．······························· 5分
4 my?2m(m?0)
?x1?1，x2?（3）解：在同一平面直角坐标系中分别画出
2y?(m?0)与y?2m(m?0)的图象．
m ············································································· 6分 由图象可得，当m≥1时，y≤2m． ·············· 7分 八、解答题（本题满分7分）
2(m?0) mx -4 -3 -2 -1 O 1 2 3 4 -1 y?3 2 1 -2 -3 -4
24．解：（1）?y?kx沿y轴向上平移3个单位长度后经过y轴上的点C，
?C(0，3)．
设直线BC的解析式为y?kx?3．
?B(3，0)在直线BC上， ?3k?3?0． 解得k??1．
··················································································· 1分 ?直线BC的解析式为y??x?3． ·
?抛物线y?x2?bx?c过点B，C， ?9?3b?c?0， ???c?3．解得??b??4，
c?3．?··············································································· 2分 ?抛物线的解析式为y?x2?4x?3． ·（2）由y?x2?4x?3．
y 4 3 C 2 1 -2 -1 O -1 -2 A P E B 1 2 F 3 4 D P? 图1
x
?1)，A(1，0)． 可得D(2，?OB?3，OC?3，OA?1，AB?2．
可得△OBC是等腰直角三角形．
??OBC?45，CB?32．
如图1，设抛物线对称轴与x轴交于点F，
??AF?1AB?1． 2过点A作AE?BC于点E．
??AEB?90?．
可得BE?AE?2，CE?22．
?在△AEC与△AFP中，?AEC??AFP?90，?ACE??APF，
?△AEC∽△AFP．
?AECE222??，． AFPF1PF
解得PF?2．
?点P在抛物线的对称轴上，
2)或(2，?2)． ······················································································· 5分 ?点P的坐标为(2，0)关于y轴的对称点A?，则A?(?1，（3）解法一：如图2，作点A(1，0)．
连结A?C，A?D，
可得A?C?AC?10，?OCA???OCA． 由勾股定理可得CD?20，A?D?10． 又A?C?10，
222y 4 3 C 2 1 A? A B -1 O 1 2 F 3 4 -1 D -2 图2
x
?A?D?A?C?CD．
?△A?DC是等腰直角三角形，?CA?D?90?，
222??DCA??45?．
??OCA???OCD?45?． ??OCA??OCD?45?．
即?OCA与?OCD两角和的度数为45． ········································································· 7分
y 解法二：如图3，连结BD． 同解法一可得CD??20，AC?10．
4 3 C 2 1 -2 -1 O -1 -2 A B 1 2 F 3 4 D 图3
x
?在Rt△DBF中，?DFB?90，BF?DF?1，
?DB?DF?BF?2．
在△CBD和△COA中，
22BC32DB2CD20??2，??2，??2． OC3AO1CA10?DBBCCD??． AOOCCA?△CBD∽△COA． ??BCD??OCA．
??OCB?45?，
??OCA??OCD?45?．
即?OCA与?OCD两角和的度数为45． ········································································· 7分 九、解答题（本题满分8分）
25．解：（1）线段PG与PC的位置关系是PG?PC；
?PG?PC·························································································································· 2分 3． ·
（2）猜想：（1）中的结论没有发生变化．
证明：如图，延长GP交AD于点H，连结CH，CG． ?P是线段DF的中点， ?FP?DP．
由题意可知AD∥FG． ??GFP??HDP．
??GPF??HPD， ?△GFP≌△HDP．
?GP?HP，GF?HD． ?四边形ABCD是菱形，
D H
A
P
C
G
B
E
F
?CD?CB，?HDC??ABC?60?．
由?ABC??BEF?60，且菱形BEFG的对角线BF恰好与菱形ABCD的边AB在同一条直线上，
可得?GBC?60．
????HDC??GBC． ?四边形BEFG是菱形， ?GF?GB． ?HD?GB．
?△HDC≌△GBC．
?CH?CG，?DCH??BCG．
??DCH??HCB??BCG??HCB?120?．
即?HCG?120．
??CH?CG，PH?PG，
?PG?PC，?GCP??HCP?60?．
?PG?3． ························································································································ 6分 PCPG?tan(90???)． ·（3）································································································· 8分 PC