福州市2013—2014学年第一学期九年级 期末质量检测数学试卷+答案

∴∠ACP =∠B = 45°，∠CPB = 90°. ······ 5分 ∴∠BPE = 90°-∠CPE. 又∵∠DPC = 90°-∠CPE,

∴∠DPC = ∠EPB. ·································· 6分 ∴△PCD ≌ △PBE.

∴DC = EB， ·············································· 7分 ∴DC+CE = EB+CE = BC = 2. ··················· 8分

（3）△CMN的周长为定值，且周长为2. ·········································· 9分 在EB上截取EF = DM，如图， ···················································· 10分 由（2）可知：PD = PE，∠PDC = ∠PEB，

∴△PDM ≌ △PEF. ··································································· 11分 ∴∠DPM = ∠EPF，PM = PF.

∵∠NPF = ∠NPE + ∠EPF = ∠NPE + ∠DPM = ∠DPE - ∠MPN

=45°= ∠NPM. ∴△PMN ≌ △PFN，

∴MN = NF. ·········································· 12分 ∴MC + CN + NM = MC + CN + NE + EF = MC + CE + DM = DC + CE = 2.

∴△CMN的周长是2. ····························· 13分 22．解：

（1）令y?0，得：x?4x?1?0， ··········· 1分 解得：x1?2?3，x2?2?3. ·············· 3分 ∴点A的坐标为（2?3，0），

点B的坐标为（2?3，0）． ················· 4分 ∴AB的长为23． ···································· 5分 （由韦达定理求出AB也可）

（2）由已知得点C的坐标为（0，1）， 由y?x?4x?1?(x?2)?3，可知抛物线的对

称轴为直线x?2， ················································································· 6分 设△ABC的外接圆圆心D的坐标为（2，n），连接AD、CD，

九年级数学答案 — 9 — （共11页）

22APDMCNEFB2?(?2∴DC=DA，即2?(n?1)?[22232)]?n2, ································ 8分

解得：n?1， ············································································· 9分

∴点D的坐标为（2，1）， ∴△ABC的外接圆⊙D半径为2． ··············································· 10分 （3）解法一：由（2）知,C是弧MN的中点.

在半径DN上截取EN= MG， ·························································· 11分 又∵DM=DN，∴DG=DE.

则点G与点E关于点D对称，连接CD、CE、PD、PE．由圆的对称性可得：图形PMC的面积与图形PECN的面积相等． ······················ 12分

?由PC把图形PMCN（指圆弧MCN和线段

PM、PN组成的图形）分成两部分，这两部分面积之差为4．可知△PCE的面积为4．设点P坐标为（m，n）

∴S?CEP?2S?CDP?2?1?CD?n?1?4， 22∴n1?3，n2??1. ······································································· 13分 由点P在抛物线y?x?4x?1上，得：

； x2?4x?1?3，解得：x1?2?6，x2?2?6（舍去）

2或x?4x?1??1，解得：x3?2?2，x4?2?2（舍去）.

∴点P的坐标为（2?2，?1）或（2?6，3）． ························· 14分 解法二：设点P坐标为（m，n），点G坐标为（2，c），直线PC的解析式为y?kx?b，

n?1?k??b?1?得：?，解得：?m ，

?n?km?b??b?1∴直线PC的解析式为y?当x?2时，c?n?1··········································· 11分 x?1. ·

m2(n?1)?1． m九年级数学答案 — 10 — （共11页）

由（2）知,C是弧MN的中点.连接CD, 图形PCN的面积与图形PMC的面积差为：

………12分

=

=2(c?1)?1(2c?2)(m?2) 2=(c?1)(2?m?2) =[2(n?1)?1?1]m m=2(n?1)?4．

∴n1?3，n2??1． ························· 13分 由点P在抛物线y?x?4x?1上，得：

； x2?4x?1?3，解得：x1?2?6，x2?2?6（舍去）

2或x?4x?1??1，解得：x3?2?2，x4?2?2（舍去）.

2∴点P的坐标为（2?2，?1）或（2?6，3）． ························· 14分

九年级数学答案 — 11 — （共11页）