25．（本题8分）在一次远足活动中，小聪由甲地步行到乙地后原路返回，小明由甲地步行到乙地后原路返回，到达途中的丙地时发现物品可能遗忘在乙地，于是从丙返回乙地，然后沿原路返回．两人同时出发，步行过程中保持匀速．设步行的时间为t（h），两人离甲地的距离分别为S1（km）和S2(km)，图中的折线分

S(km) 别表示S1、S2与t之间的函数关系．

A C 10 （1）甲、乙两地之间的距离为 km，乙、

丙两地之间的距离为 km； 8 B （2）分别求出小明由甲地出发首次到达乙地及由乙地到达丙地所用的时间． （3）求图中线段AB所表示的S2与t间的函数关系式，并写出自变量t的取值范围．

O 2 t(h) （第25题图）

26．（本题8分）已知抛物线C1：y??x2?2mx?1（m为常数，且m＞0）的顶点为A，与y轴交于点C；抛物线C2与抛物线C1关于y轴对称，其顶点为B，连接AC，BC，AB． （1）当m?1时，判定△ABC的形状，并说明理由； （2）抛物线C1上是否存在点P，使得四边形ABCP为菱形？如果存在，请求出m的值；如果不存在，请说明理由．

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27．（本题10分）在△ABC中， AB、BC、AC三边的长分别为5、10、13，求这个三角形的面积．小华同学在解答这道题时，先画一个正方形网格（每个小正方形的边长为1），再在网格中画出格点△ABC（即△ABC三个顶点都在小正方形的顶点处），如图1所示．这样不需求△ABC的高，而借用网格就能计算出它的面积．这种方法叫做构图法． ．．．（1）△ABC的面积为： ▲ ．

（2）若△DEF三边的长分别为5、请在图1的正方形网格中画出相应的△DEF，22、17，并利用构图法求出它的面积． ．．．

（3）如图2，一个六边形的花坛被分割成7个部分，其中正方形PRBA，RQDC，QPFE的

面积分别为13、10、17，且△PQR、△BCR、△DEQ、△AFP的面积相等，求六边形花坛ABCDEF的面积．

A

B

C

（第27题图1）

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28．（本题12分）如果一个点能与另外两个点能构成直角三角形，则称这个点为另外两个点的勾股点．例如：矩形ABCD中，点C与A，B两点可构成直角三角形ABC，则称点C为A，B两点的勾股点．同样，点D也是A，B两点的勾股点．

（1）如图1，矩形ABCD中，AB＝2，BC＝1，请在边CD上作出A，B两点的勾股点（点．C和点）（要求：尺规作图，保留作图痕迹，不要求写作法）． ．．．D．除外．．

C D

A B （第28题图1）

（2）矩形ABCD中，AB＝3，BC＝1，直接写出边CD上A， B两点的勾股点的个数． （3）如图2，矩形ABCD中，AB＝12cm，BC＝4 cm，DM＝8 cm，AN＝5 cm．动点P从D点出发沿着DC方向以1 cm／s的速度向右移动，过点P的直线l平行于BC，当点P运动到点M时停止运动．设运动时间为t(s) ，点H为M，N两点的勾股点，且点H在直线l上． ①当t＝4时，求PH的长．

②探究满足条件的点H的个数（直接写出点H的个数及相应t的取值范围，不必证明）．

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参考答案及评分标准

说明：本评分标准每题给出了一种或几种解法供参考，如果考生的解法与本解答不同，参照本评分标准的精神给分．

一、选择题（每小题2分，共计16分）

题号 答案 1 B 2 A 3 C 4 A 5 C 6 A 7 B 8 D 二、填空题（每小题2分，共计20分） 9．答案不唯一：如13 10．a（a＋b）（a－b） 11．x? 12．20

22??x?113．?

??y?114．

13 15．－7 16． 17．－3≤b≤0 18．2?32

三、解答题（本大题共10小题，共计84分） 19．（本题12分）

a2?aba2?b2?(1)解：原式=··························································································· 1分

aba2?a(a?b)(a?b)(a?b)? ·································································································· 3分

aba2??a(a?b)ab? ··································································································· 4分 2(a?b)(a?b)ab ··································································································································· 6分 a?b解不等式②，得x＜3． ····························································································· 4分

(2)解：解不等式①，得x≥－1． ···························································································· 2分

所以，不等式组的解集是－1≤x＜3． ················································································· 5分 所以，不等式组的整数解为－1，0，1，2． ········································································· 6分 20．（本题6分）

解：(1)抽样调查． ·················································································································· 2分 (2)25 ，图略 ··························································································································· 4分 (3)800?20?160（人） 100···································································· 6分 ?估计该校喜欢跳绳的学生人数约为160人． ·

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21．（本题7分）

解：（1）所有可能出现的结果如下：

A 1 1 1 0 0 0 B 1 0 0 0 1 1 C 0 1 0 1 0 1 D 0 0 1 1 1 0 结果 1100 1010 1001 0011 0101 0110 总共有6种结果，每种结果出现的可能性相同． ································································· 4分 （2）所有的结果中，满足A、B两个元件“开” “关”状态不同的结果有4种，所以A、B两个

2元件“开” “关”状态不同的概率是． ···················································································· 7分

322．（本题7分）

解：过点C作CE⊥AB，垂足为E，设CE长为x米． 在Rt△BEC中，tan∠CBE＝

CEx3，即tan60°＝，∴BE＝x ··································· 2分 BEBE3CEx，即tan30°＝，∴AE＝3x ··································· 4分 AEAE在Rt△AEC中，tan∠CAE＝

∵AB＝AE－BE，∴3x－3x＝4000． ············································································ 5分 3解得x＝20003，∴h＝20003＋500． ············································································ 6分 答：海底黑匣子C点处距离海面的深度为（20003＋500）米． ····································· 7分 23．（本题7分）

（1）解：∵四边形ABCD是正方形 ， ∴∠ABC＝∠DCB＝90°． ∵PB＝PC，∴∠PBC＝∠PCB． ······················································································· 1分 ∴∠ABC－∠PBC ＝∠DCB－∠PCB， 即∠ABP＝∠DCP． ············································ 2分 又∵AB＝ DC，PB＝PC，∴△APB≌△DPC． ··································································· 3分 （2）证明：∵四边形ABCD是正方形， ∴∠BAC＝∠DAC＝45°． ∵△APB≌△DPC， ∴AP＝ DP．

又∵AP＝AB＝AD ，∴DP ＝ AP ＝AD． ∴△APD是等边三角形． ∴∠DAP＝60°． 5分 ∴∠PAC＝∠DAP －∠DAC＝15°． ∴∠BAP＝∠BAC－∠PAC＝30°． ························································································ 6分 ∴∠BAP＝2∠PAC． ·············································································································· 7分 24．（本题7分） 解：设每个房间的定价增加x元． 根据题意得：(180＋x－20)(50－解得：x＝170．

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x)＝10890， ······························································ 4分 10