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（2）经过点A、B、C三点的抛物线的解析式.
23．（10分）汶川地震后，抢险队派一架直升飞机去A、B两个村庄抢险，飞机在距地面450
米上空的P点，测得A村的俯角为30?，B村的俯角为60?（．如图7）．求A、B两个
村庄间的距离．（结果精确到米，参考数据2?1.414，3?1.732）
AB图7
Q60?30?P450C
24.（本题14分）如图所示，在直角梯形ABCD中，AD//BC，
AD=21。动点P从点D出发，沿射线DA的方
向以每秒2个单位长的速度运动，动点Q从点C出发，在线段CB上以每秒1个单位长的速度向点B运动，点P，Q分别从点D，C同时出发，当点Q运动到点B时，点P随之停止运动。设运动的时间为t（秒）。 （1）设△BPQ的面积为S，求S与t之间的函数关系式； （2）当线段PQ与线段AB相交于点O，且BO=2AO时，求的正切值；
（3）当t为何值时，以B、P、Q三点为顶点的三角形是等腰三角形？
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参考答案
一、选择题（本题共10小题，每小题4分，共40分） 题号 答案 评分标准 1 B 2 A 3 B 4 A 5 C 6 A 7 C 8 D 9 D 10 B 选对一个得4分，不选、多选、错选均不给分 8
二、填空题（本题有6小题，每小题5分，共30分）
11．a（x+2y）（x-2y）；12．2.46×10 ；13．2/5；14．X﹥3； 15．1000；16．30． 三、解答题（本题有8小题，共80分） 17.（1）（本题6分）解：原式?1?3?3?2?注：上面的计算每错一处扣1分.
3＝－2 2a2?1a?1?2（2）（本题6分）原式 = 2????????2′
a?2aa?2a(a2?1)?(a?1) = 2a?2aa2?1?a?1a2?aa(a?1) =??????2′ ??22a?2aa?2aa(a?2)a?1???????????????????2′ a?218.解：方程两边同乘(x?1)(x?1)，得
=
2(x?1)?x?0． ········································································································· 3分
解这个方程，得 x?2． ···························································································································· 3分
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检验：当x?2时，(x?1)(x?1)?0．
所以x?2是原方程的解． ······························································································ 2分 19．（本题8分）解：连结CF，猜想CF=EA。证明：略
20. （本题9分）解：（1）全班共１５人； ???????2分 (2) 补图如右，a?20,b?30?? 4分 (3)估计全校大约能捐22824元. ???3分 21、（本题9分）解：略 22．（本题9分）
（1）∵连结BM
则BM?BO?OM 即5?BO?4
∴BO?3，BO??3（舍去）
∴B（0，3）?????????????????????2′ ∵AO?5?4?1 OC?4?5?9
∴A（-1，0）?????????????????????1′ C（9，0）?????????????????????1′
（2）因为抛物线经过A（-1，0），B（9，0）
所以设y?a(x?1)(x?9)??????????????1′
把（0，3）代入得
222222人数 捐款人数条形统计· 20 15 · 10 · 5 · · · · · · O 10 15 20 25 30 金额
图5
3?a?1?(?9)??????????????2′ 1??????????????2′ 31∴y??(x?1)(x?9)
3128即y??x?x?3??????????????1′
33∴a??(其它正确解法都可得分)
23、（本题10分）
解：解：根据题意得: ?A?30? ， ?PBC?60? 所以?APB?60??30?，所以?APB??A ，
所以AB=PB ······································· 3分
在Rt?BCP中，?C?90?,?PBC?60?，PC=450， ····································· 2分
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所以PB =
450900??3003
sin60?3所以AB?PB?3003?520(米)
答：略 ······························································································································· 5分 24.解：（1）如图1所示，过点P作，垂足为M，则四边形PDCM为矩形。
··········································································································································· 5分 （2）如图2所示，由 得： 图2
··············································································································································· 6分
过点Q作，垂足为E
在中， ················································································································································· 8分
（3）由图1可知：CM=PD=2t，CQ=t。
若以B、P、Q三点为顶点的三角形是等腰三角形，可以分三种情况： ①若PQ=BQ。在 由中， 解得 ··········································· 10分
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②若 由 即。在，得： ················································································································································· 5分 ③若 由 得 整理，得解得 综合上面的讨论可知：当 （不合题意，舍去）
秒或秒时，以B、P、Q三点为顶点的三角形是等腰
三角形。 ································································································································ 14分
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