1．下列在廚房中發(fā)生的變化是物理變化的是

A．榨取果汁　　　　　　 B．冬瓜腐爛　 　　　C．鐵鍋生銹　　　　 　D．煤氣燃燒

29．問題解決

解：方法一：如圖(1-1)，連接．

　　　 由題設，得四邊形和四邊形關于直線對稱．

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

　　　 ∵四邊形是正方形，∴

　　　 ∵設則

　　　　 在中，．

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

　　　 在和在中，

，

，

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

　　　 設則∴

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

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

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

　　　 如圖(1－2)，過點做交于點，連接

∵∴四邊形是平行四邊形．

　　　 ∴

　　　 同理，四邊形也是平行四邊形．∴

　　∵

　　在與中

　　∴····························· 5分

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

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

類比歸納

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

聯(lián)系拓廣

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

26．(1)解：由得點坐標為

由得點坐標為

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

由解得∴點的坐標為···································· (3分)

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

　 (2)解：∵點在上且

　　　　　　 ∴點坐標為······················································································ (5分)

又∵點在上且

∴點坐標為······················································································ (6分)

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

　 (3)解法一：當時，如圖1，矩形與重疊部分為五邊形(時，為四邊形)．過作于，則

∴即∴

∴

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

(2009年山西省太原市)29．(本小題滿分12分)

問題解決

如圖(1)，將正方形紙片折疊，使點落在邊上一點(不與點，重合)，壓平后得到折痕．當時，求的值．

類比歸納

在圖(1)中，若則的值等于　　 　　；若則的值等于　　 　　；若(為整數(shù))，則的值等于　　 　　．(用含的式子表示)

聯(lián)系拓廣

　 如圖(2)，將矩形紙片折疊，使點落在邊上一點(不與點重合)，壓平后得到折痕設則的值等于　　 　　．(用含的式子表示)

26．(2009年山西省)(本題14分)如圖，已知直線與直線相交于點分別交軸于兩點．矩形的頂點分別在直線上，頂點都在軸上，且點與點重合．

　　 (1)求的面積；

(2)求矩形的邊與的長；

(3)若矩形從原點出發(fā)，沿軸的反方向以每秒1個單位長度的速度平移，設

移動時間為秒，矩形與重疊部分的面積為，求關

的函數(shù)關系式，并寫出相應的的取值范圍．

23.(2009年河南省)(11分)如圖，在平面直角坐標系中，已知矩形ABCD的三個頂點B(4，0)、C(8，0)、D(8，8).拋物線y=ax²+bx過A、C兩點.　　

(1)直接寫出點A的坐標，并求出拋物線的解析式；

　　 (2)動點P從點A出發(fā)．沿線段AB向終點B運動，同時點Q從點C出發(fā)，沿線段CD

向終點D運動．速度均為每秒1個單位長度，運動時間為t秒.過點P作PE⊥AB交AC于點E

　　 ①過點E作EF⊥AD于點F，交拋物線于點G.當t為何值時，線段EG最長?

②連接EQ．在點P、Q運動的過程中，判斷有幾個時刻使得△CEQ是等腰三角形?

請直接寫出相應的t值.

解.(1)點A的坐標為(4，8)　　　　　　　　 …………………1分

將A　 (4,8)、C(8，0)兩點坐標分別代入y=ax²+bx

　　　　　　 8=16a+4b

　　　　 得　　　　　　　　　　　　　

　　　　 0=64a+8b

　　　　 解 得a=-,b=4

∴拋物線的解析式為：y=-x²+4x　 　　　　…………………3分

(2)①在Rt△APE和Rt△ABC中，tan∠PAE==,即=

∴PE=AP=t．PB=8-t．

∴點E的坐標為(4+t，8-t).

∴點G的縱坐標為：-(4+t)²+4(4+t)=-t²+8. …………………5分

∴EG=-t²+8-(8-t)

　　 =-t²+t.

∵-＜0，∴當t=4時，線段EG最長為2.　　　　　　 …………………7分

②共有三個時刻.　　　　　　　　　　　　　　　　　 …………………8分

t₁=， 　t₂=，t₃= ．　　　　　　　　　 …………………11分

26．解：(1)1，；

(2)作QF⊥AC于點F，如圖3， AQ = CP= t，∴．

由△AQF∽△ABC，，

得．∴．

∴，

即．

(3)能．

　 ①當DE∥QB時，如圖4．

　 ∵DE⊥PQ，∴PQ⊥QB，四邊形QBED是直角梯形．

　　 此時∠AQP=90°．

由△APQ　∽△ABC，得，

即． 解得．

②如圖5，當PQ∥BC時，DE⊥BC，四邊形QBED是直角梯形．

此時∠APQ =90°．

由△AQP　∽△ABC，得 ，

即． 解得．　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　

(4)或．

[注：①點P由C向A運動，DE經(jīng)過點C．

方法一、連接QC，作QG⊥BC于點G，如圖6．

，．

由，得，解得．

方法二、由，得，進而可得

，得，∴．∴．

②點P由A向C運動，DE經(jīng)過點C，如圖7．

，]

26．(2009年河北省)(本小題滿分12分)

如圖16，在Rt△ABC中，∠C=90°，AC = 3，AB = 5．點P從點C出發(fā)沿CA以每秒1個單位長的速度向點A勻速運動，到達點A后立刻以原來的速度沿AC返回；點Q從點A出發(fā)沿AB以每秒1個單位長的速度向點B勻速運動．伴隨著P、Q的運動，DE保持垂直平分PQ，且交PQ于點D，交折線QB-BC-CP于點E．點P、Q同時出發(fā)，當點Q到達點B時停止運動，點P也隨之停止．設點P、Q運動的時間是t秒(t＞0)．

(1)當t = 2時，AP =　　　 ，點Q到AC的距離是　　　 ；

(2)在點P從C向A運動的過程中，求△APQ的面積S與

t的函數(shù)關系式；(不必寫出t的取值范圍)

(3)在點E從B向C運動的過程中，四邊形QBED能否成

為直角梯形？若能，求t的值．若不能，請說明理由；

(4)當DE經(jīng)過點C　時，請直接寫出t的值．

26．解：(1)由已知，得，，

，

．

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

設過點的拋物線的解析式為．

將點的坐標代入，得．

將和點的坐標分別代入，得

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

解這個方程組，得

故拋物線的解析式為．··························································· (3分)

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

點在該拋物線上，且它的橫坐標為，

點的縱坐標為．······················································································· (5分)

設的解析式為，

將點的坐標分別代入，得

　 解得

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

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

過點作于點，

則．

，

．

又，

．

．

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

．

(3)點在上，，，則設．

，，．

①若，則，

解得．，此時點與點重合．

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

②若，則，

解得 ，，此時軸．

與該拋物線在第一象限內(nèi)的交點的橫坐標為1，

點的縱坐標為．

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

③若，則，

解得，，此時，是等腰直角三角形．

過點作軸于點，

則，設，

．

．

解得(舍去)．

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

綜上所述，存在三個滿足條件的點，

即或或．

(2009年重慶綦江縣)26．(11分)如圖，已知拋物線經(jīng)過點，拋物線的頂點為，過作射線．過頂點平行于軸的直線交射線于點，在軸正半軸上，連結．

(1)求該拋物線的解析式；

(2)若動點從點出發(fā)，以每秒1個長度單位的速度沿射線運動，設點運動的時間為．問當為何值時，四邊形分別為平行四邊形？直角梯形？等腰梯形？

(3)若，動點和動點分別從點和點同時出發(fā)，分別以每秒1個長度單位和2個長度單位的速度沿和運動，當其中一個點停止運動時另一個點也隨之停止運動．設它們的運動的時間為，連接，當為何值時，四邊形的面積最��？并求出最小值及此時的長．

*26．解：(1)拋物線經(jīng)過點，

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

二次函數(shù)的解析式為：·················································· 3分

(2)為拋物線的頂點過作于，則，

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

當時，四邊形是平行四邊形

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

當時，四邊形是直角梯形

過作于，則

(如果沒求出可由求)

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

當時，四邊形是等腰梯形

綜上所述：當、5、4時，對應四邊形分別是平行四邊形、直角梯形、等腰梯形．·· 7分

(3)由(2)及已知，是等邊三角形

則

過作于，則········································································· 8分

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

當時，的面積最小值為··································································· 10分

此時

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

26．(2009年重慶市)已知：如圖，在平面直角坐標系中，矩形OABC的邊OA在y軸的正半軸上，OC在x軸的正半軸上，OA=2，OC=3．過原點O作∠AOC的平分線交AB于點D，連接DC，過點D作DE⊥DC，交OA于點E．

(1)求過點E、D、C的拋物線的解析式；

(2)將∠EDC繞點D按順時針方向旋轉(zhuǎn)后，角的一邊與y軸的正半軸交于點F，另一邊與線段OC交于點G．如果DF與(1)中的拋物線交于另一點M，點M的橫坐標為，那么EF=2GO是否成立？若成立，請給予證明；若不成立，請說明理由；

(3)對于(2)中的點G，在位于第一象限內(nèi)的該拋物線上是否存在點Q，使得直線GQ與AB的交點P與點C、G構成的△PCG是等腰三角形？若存在，請求出點Q的坐標；若不存在，請說明理由．

25.(2009年北京)如圖，在平面直角坐標系中，三個機戰(zhàn)的坐標分別為

，，，延長AC到點D,使CD=,過點D作DE∥AB交BC的延長線于點E.

(1)求D點的坐標；

(2)作C點關于直線DE的對稱點F,分別連結DF、EF，若過B點的直線將四邊形CDFE分成周長相等的兩個四邊形，確定此直線的解析式；

(3)設G為y軸上一點，點P從直線與y軸的交點出發(fā)，先沿y軸到達G點，再沿GA到達A點，若P點在y軸上運動的速度是它在直線GA上運動速度的2倍，試確定G點的位置，使P點按照上述要求到達A點所用的時間最短。(要求：簡述確定G點位置的方法，但不要求證明)