1．下列在厨房中发生的变化是物理变化的是

A．榨取果汁　　　　　　 B．冬瓜腐烂　 　　　C．铁锅生锈　　　　 　D．煤气燃烧

29．问题解决

解：方法一：如图(1-1)，连接．

　　　 由题设，得四边形和四边形关于直线对称．

　　　 ∴垂直平分．∴··········································· 1分

　　　 ∵四边形是正方形，∴

　　　 ∵设则

　　　　 在中，．

　　　 ∴解得，即················································ 3分

　　　 在和在中，

，

，

······································································· 5分

　　　 设则∴

　　　 解得即················································································· 6分

　　　 ∴··································································································· 7分

　　　 方法二：同方法一，········································································· 3分

　　　 如图(1－2)，过点做交于点，连接

∵∴四边形是平行四边形．

　　　 ∴

　　　 同理，四边形也是平行四边形．∴

　　∵

　　在与中

　　∴····························· 5分

∵······························································ 6分

∴································································································· 7分

类比归纳

(或)；； ·········································································· 10分

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···································································································· 12分

26．(1)解：由得点坐标为

由得点坐标为

∴··················································································· (2分)

由解得∴点的坐标为···································· (3分)

∴··························································· (4分)

　 (2)解：∵点在上且

　　　　　　 ∴点坐标为······················································································ (5分)

又∵点在上且

∴点坐标为······················································································ (6分)

∴··········································································· (7分)

　 (3)解法一：当时，如图1，矩形与重叠部分为五边形(时，为四边形)．过作于，则

∴即∴

∴

即··································································· (10分)

(2009年山西省太原市)29．(本小题满分12分)

问题解决

如图(1)，将正方形纸片折叠，使点落在边上一点(不与点，重合)，压平后得到折痕．当时，求的值．

类比归纳

在图(1)中，若则的值等于　　 　　；若则的值等于　　 　　；若(为整数)，则的值等于　　 　　．(用含的式子表示)

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　 如图(2)，将矩形纸片折叠，使点落在边上一点(不与点重合)，压平后得到折痕设则的值等于　　 　　．(用含的式子表示)

26．(2009年山西省)(本题14分)如图，已知直线与直线相交于点分别交轴于两点．矩形的顶点分别在直线上，顶点都在轴上，且点与点重合．

　　 (1)求的面积；

(2)求矩形的边与的长；

(3)若矩形从原点出发，沿轴的反方向以每秒1个单位长度的速度平移，设

移动时间为秒，矩形与重叠部分的面积为，求关

的函数关系式，并写出相应的的取值范围．

23.(2009年河南省)(11分)如图，在平面直角坐标系中，已知矩形ABCD的三个顶点B(4，0)、C(8，0)、D(8，8).抛物线y=ax²+bx过A、C两点.　　

(1)直接写出点A的坐标，并求出抛物线的解析式；

　　 (2)动点P从点A出发．沿线段AB向终点B运动，同时点Q从点C出发，沿线段CD

向终点D运动．速度均为每秒1个单位长度，运动时间为t秒.过点P作PE⊥AB交AC于点E

　　 ①过点E作EF⊥AD于点F，交抛物线于点G.当t为何值时，线段EG最长?

②连接EQ．在点P、Q运动的过程中，判断有几个时刻使得△CEQ是等腰三角形?

请直接写出相应的t值.

解.(1)点A的坐标为(4，8)　　　　　　　　 …………………1分

将A　 (4,8)、C(8，0)两点坐标分别代入y=ax²+bx

　　　　　　 8=16a+4b

　　　　 得　　　　　　　　　　　　　

　　　　 0=64a+8b

　　　　 解 得a=-,b=4

∴抛物线的解析式为：y=-x²+4x　 　　　　…………………3分

(2)①在Rt△APE和Rt△ABC中，tan∠PAE==,即=

∴PE=AP=t．PB=8-t．

∴点E的坐标为(4+t，8-t).

∴点G的纵坐标为：-(4+t)²+4(4+t)=-t²+8. …………………5分

∴EG=-t²+8-(8-t)

　　 =-t²+t.

∵-＜0，∴当t=4时，线段EG最长为2.　　　　　　 …………………7分

②共有三个时刻.　　　　　　　　　　　　　　　　　 …………………8分

t₁=， 　t₂=，t₃= ．　　　　　　　　　 …………………11分

26．解：(1)1，；

(2)作QF⊥AC于点F，如图3， AQ = CP= t，∴．

由△AQF∽△ABC，，

得．∴．

∴，

即．

(3)能．

　 ①当DE∥QB时，如图4．

　 ∵DE⊥PQ，∴PQ⊥QB，四边形QBED是直角梯形．

　　 此时∠AQP=90°．

由△APQ　∽△ABC，得，

即． 解得．

②如图5，当PQ∥BC时，DE⊥BC，四边形QBED是直角梯形．

此时∠APQ =90°．

由△AQP　∽△ABC，得 ，

即． 解得．　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　

(4)或．

[注：①点P由C向A运动，DE经过点C．

方法一、连接QC，作QG⊥BC于点G，如图6．

，．

由，得，解得．

方法二、由，得，进而可得

，得，∴．∴．

②点P由A向C运动，DE经过点C，如图7．

，]

26．(2009年河北省)(本小题满分12分)

如图16，在Rt△ABC中，∠C=90°，AC = 3，AB = 5．点P从点C出发沿CA以每秒1个单位长的速度向点A匀速运动，到达点A后立刻以原来的速度沿AC返回；点Q从点A出发沿AB以每秒1个单位长的速度向点B匀速运动．伴随着P、Q的运动，DE保持垂直平分PQ，且交PQ于点D，交折线QB-BC-CP于点E．点P、Q同时出发，当点Q到达点B时停止运动，点P也随之停止．设点P、Q运动的时间是t秒(t＞0)．

(1)当t = 2时，AP =　　　 ，点Q到AC的距离是　　　 ；

(2)在点P从C向A运动的过程中，求△APQ的面积S与

t的函数关系式；(不必写出t的取值范围)

(3)在点E从B向C运动的过程中，四边形QBED能否成

为直角梯形？若能，求t的值．若不能，请说明理由；

(4)当DE经过点C　时，请直接写出t的值．

26．解：(1)由已知，得，，

，

．

．············································································································ (1分)

设过点的抛物线的解析式为．

将点的坐标代入，得．

将和点的坐标分别代入，得

··································································································· (2分)

解这个方程组，得

故抛物线的解析式为．··························································· (3分)

(2)成立．························································································· (4分)

点在该抛物线上，且它的横坐标为，

点的纵坐标为．······················································································· (5分)

设的解析式为，

将点的坐标分别代入，得

　 解得

的解析式为．········································································ (6分)

，．··························································································· (7分)

过点作于点，

则．

，

．

又，

．

．

．··········································································································· (8分)

．

(3)点在上，，，则设．

，，．

①若，则，

解得．，此时点与点重合．

．··········································································································· (9分)

②若，则，

解得 ，，此时轴．

与该抛物线在第一象限内的交点的横坐标为1，

点的纵坐标为．

．······································································································· (10分)

③若，则，

解得，，此时，是等腰直角三角形．

过点作轴于点，

则，设，

．

．

解得(舍去)．

．··········································· (12分)

综上所述，存在三个满足条件的点，

即或或．

(2009年重庆綦江县)26．(11分)如图，已知抛物线经过点，抛物线的顶点为，过作射线．过顶点平行于轴的直线交射线于点，在轴正半轴上，连结．

(1)求该抛物线的解析式；

(2)若动点从点出发，以每秒1个长度单位的速度沿射线运动，设点运动的时间为．问当为何值时，四边形分别为平行四边形？直角梯形？等腰梯形？

(3)若，动点和动点分别从点和点同时出发，分别以每秒1个长度单位和2个长度单位的速度沿和运动，当其中一个点停止运动时另一个点也随之停止运动．设它们的运动的时间为，连接，当为何值时，四边形的面积最小？并求出最小值及此时的长．

*26．解：(1)抛物线经过点，

·························································································· 1分

二次函数的解析式为：·················································· 3分

(2)为抛物线的顶点过作于，则，

··················································· 4分

当时，四边形是平行四边形

················································ 5分

当时，四边形是直角梯形

过作于，则

(如果没求出可由求)

····························································································· 6分

当时，四边形是等腰梯形

综上所述：当、5、4时，对应四边形分别是平行四边形、直角梯形、等腰梯形．·· 7分

(3)由(2)及已知，是等边三角形

则

过作于，则········································································· 8分

=·································································································· 9分

当时，的面积最小值为··································································· 10分

此时

······················································ 11分

26．(2009年重庆市)已知：如图，在平面直角坐标系中，矩形OABC的边OA在y轴的正半轴上，OC在x轴的正半轴上，OA=2，OC=3．过原点O作∠AOC的平分线交AB于点D，连接DC，过点D作DE⊥DC，交OA于点E．

(1)求过点E、D、C的抛物线的解析式；

(2)将∠EDC绕点D按顺时针方向旋转后，角的一边与y轴的正半轴交于点F，另一边与线段OC交于点G．如果DF与(1)中的抛物线交于另一点M，点M的横坐标为，那么EF=2GO是否成立？若成立，请给予证明；若不成立，请说明理由；

(3)对于(2)中的点G，在位于第一象限内的该抛物线上是否存在点Q，使得直线GQ与AB的交点P与点C、G构成的△PCG是等腰三角形？若存在，请求出点Q的坐标；若不存在，请说明理由．

25.(2009年北京)如图，在平面直角坐标系中，三个机战的坐标分别为

，，，延长AC到点D,使CD=,过点D作DE∥AB交BC的延长线于点E.

(1)求D点的坐标；

(2)作C点关于直线DE的对称点F,分别连结DF、EF，若过B点的直线将四边形CDFE分成周长相等的两个四边形，确定此直线的解析式；

(3)设G为y轴上一点，点P从直线与y轴的交点出发，先沿y轴到达G点，再沿GA到达A点，若P点在y轴上运动的速度是它在直线GA上运动速度的2倍，试确定G点的位置，使P点按照上述要求到达A点所用的时间最短。(要求：简述确定G点位置的方法，但不要求证明)