发布时间 : 星期四 文章陕西省2010年中考数学试题（word版附带详细答案）更新完毕开始阅读15ffeeee172ded630b1cb63b
···························································································································· （4分） 从上表可以看出，一次游戏共有20种等可能结果，其中两数和为偶数的共有8种.将参 加联欢会的某位同学即兴表演节目记为事件A，
?P(A)?P（两数和为偶数）=
820?25． ························································· （6
分）
（2）?50?25?20（人）.
?估计本次联欢会上有20名同学即兴表演节目. ········································ （8分）
23.解：（1）?DE垂直平分AC，
??DEC?90°．
?DC为△DEC?DC外接圆的直径.
的中点O即为圆心.
·················································································（1分） 连接OE.
又知BE是⊙O的切线，
??EBO??BOE?90°． ··················································································· （2
分）
在Rt△ABC中，E是斜边AC的中点，
?BE?EC． ??EBC??C.
又??BOE?2?C,
??C?2?C?90°.
??C?30°. ······································································································· （4
分）
（2）在Rt△ABC中，AC??EC?
AB?BC22?5. 12AC?52． ··························································································· （6
9
分）
??ABC??DEC?90°，
?△ABC∽△DEC． ?AC?DC54BC ．EC ?DC?.
58C ?△DE外接圆的半径为. ···································································· （8分）
24.解：（1）设该抛物线的表达式为y?ax2?bx?c．根据题意，得
1?a?，?3?a?b?c?0，?2??b??，9a?3b?c?0，解之，得 ································································· （2分） ??3??c??1．??c??1．???所求抛物线的表达式为y?13x?223x?1．
（2）①当AB为边时，只要PQ∥AB，且PQ?AB?4即可. 又知点Q在y轴上，?点P的横坐标为4或?4. 这时，符合条件的点P有两个，分别记为P1，P2． 而当x?4时，y???5?3?53；当x??4时，y?7．
7)．此时P1?4，?，P2(?4， ·················································································· （7分）
②
当AB为对角线时，只要线段PQ与线段AB互相平分即可.
又知点Q在y轴上，且线段AB中点的横坐标为1，
?点P的横坐标为2.
这时，符合条件的点P只有一个，记为P3.
?1). 而当x?2时，y??1．此时P3(2，7)，P3(2，．1) 综上，满足条件的点P为P1?4，?，P2(?4，······························· （10分）
??5?3?25.解：（1）如图①，作直线DB，直线DB即为所求.（所求直线不唯一，只要过矩形 对称中心的直线均可） ···················································································· （2分）
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（2）如图②，连接AC、DB交于点P，则点P为矩形ABCD的对称中心.作直线MP， 直线MP即为所求. ·························································································· （5分） （3）如图③，存在符合条件的直线l. ·························································· （6分） 过点D作DA⊥OB于点A，
则点P(4，························································· （6分） 2)为矩形ABCD的对称中心. ·
?过点P的直线只要平分△DOA的面积即可.
易知，在OD边上必存在点H，使得直线PH将△DOA面积平分. 从而，直线PH平分梯形OBCD的面积.
即直线PH为所求直线l. ·················································································· （9分） 设直线PH的表达式为y?kx?b，且点P(4， 2)，?2?4k?b．即b?2?4k． ?y?kx?2?4k．?直线OD的表达式为y?2x．
2?4k?x?，??y?kx?2?4k，?2?k解之，得? ???y?2x．?y?4?8k．?2?k??点H的坐标为??2?4k4?8，2?k2?k?? ?．??PH与线段AD的交点F的坐标为(2，2?2k)，
2k?2．??4?k?1． ?0??
?S△DHF?12?4?2?k4???2k·?2?2???2?k???211 4??2．2 解之，得k?8 ?b??13?3??13?3．k?不合题意，舍去? ???22???2．13
?直线l的表达式为y?13?32x?8?213． ················································ （12分）
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