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(shè)，得四邊形和四邊形關(guān)于直線對(duì)稱．

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

　　　 ∵四邊形是正方形，∴

　　　 ∵設(shè)則

　　　　 在中，．

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

　　　 在和在中，

，

，

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

　　　 設(shè)則∴

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

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

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

　　　 如圖(1－2)，過點(diǎn)做交于點(diǎn)，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(diǎn)坐標(biāo)為

由得點(diǎn)坐標(biāo)為

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

由解得∴點(diǎn)的坐標(biāo)為···································· (3分)

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

　 (2)解：∵點(diǎn)在上且

　　　　　　 ∴點(diǎn)坐標(biāo)為······················································································ (5分)

又∵點(diǎn)在上且

∴點(diǎn)坐標(biāo)為······················································································ (6分)

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

　 (3)解法一：當(dāng)時(shí)，如圖1，矩形與重疊部分為五邊形(時(shí)，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(diǎn)落在邊上一點(diǎn)(不與點(diǎn)，重合)，壓平后得到折痕．當(dāng)時(shí)，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(diǎn)落在邊上一點(diǎn)(不與點(diǎn)重合)，壓平后得到折痕設(shè)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(diǎn)分別交軸于兩點(diǎn)．矩形的頂點(diǎn)分別在直線上，頂點(diǎn)都在軸上，且點(diǎn)與點(diǎn)重合．

　　 (1)求的面積；

(2)求矩形的邊與的長；

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

移動(dòng)時(shí)間為秒，矩形與重疊部分的面積為，求關(guān)

的函數(shù)關(guān)系式，并寫出相應(yīng)的的取值范圍．

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

(1)直接寫出點(diǎn)A的坐標(biāo)，并求出拋物線的解析式；

　　 (2)動(dòng)點(diǎn)P從點(diǎn)A出發(fā)．沿線段AB向終點(diǎn)B運(yùn)動(dòng)，同時(shí)點(diǎn)Q從點(diǎn)C出發(fā)，沿線段CD

向終點(diǎn)D運(yùn)動(dòng)．速度均為每秒1個(gè)單位長度，運(yùn)動(dòng)時(shí)間為t秒.過點(diǎn)P作PE⊥AB交AC于點(diǎn)E

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

②連接EQ．在點(diǎn)P、Q運(yùn)動(dòng)的過程中，判斷有幾個(gè)時(shí)刻使得△CEQ是等腰三角形?

請(qǐng)直接寫出相應(yīng)的t值.

解.(1)點(diǎn)A的坐標(biāo)為(4，8)　　　　　　　　 …………………1分

將A　 (4,8)、C(8，0)兩點(diǎn)坐標(biāo)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(diǎn)E的坐標(biāo)為(4+t，8-t).

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

∴EG=-t²+8-(8-t)

　　 =-t²+t.

∵-＜0，∴當(dāng)t=4時(shí)，線段EG最長為2.　　　　　　 …………………7分

②共有三個(gè)時(shí)刻.　　　　　　　　　　　　　　　　　 …………………8分

t₁=， 　t₂=，t₃= ．　　　　　　　　　 …………………11分

26．解：(1)1，；

(2)作QF⊥AC于點(diǎn)F，如圖3， AQ = CP= t，∴．

由△AQF∽△ABC，，

得．∴．

∴，

即．

(3)能．

　 ①當(dāng)DE∥QB時(shí)，如圖4．

　 ∵DE⊥PQ，∴PQ⊥QB，四邊形QBED是直角梯形．

　　 此時(shí)∠AQP=90°．

由△APQ　∽△ABC，得，

即． 解得．

②如圖5，當(dāng)PQ∥BC時(shí)，DE⊥BC，四邊形QBED是直角梯形．

此時(shí)∠APQ =90°．

由△AQP　∽△ABC，得 ，

即． 解得．　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　

(4)或．

[注：①點(diǎn)P由C向A運(yùn)動(dòng)，DE經(jīng)過點(diǎn)C．

方法一、連接QC，作QG⊥BC于點(diǎn)G，如圖6．

，．

由，得，解得．

方法二、由，得，進(jìn)而可得

，得，∴．∴．

②點(diǎn)P由A向C運(yùn)動(dòng)，DE經(jīng)過點(diǎn)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(diǎn)P從點(diǎn)C出發(fā)沿CA以每秒1個(gè)單位長的速度向點(diǎn)A勻速運(yùn)動(dòng)，到達(dá)點(diǎn)A后立刻以原來的速度沿AC返回；點(diǎn)Q從點(diǎn)A出發(fā)沿AB以每秒1個(gè)單位長的速度向點(diǎn)B勻速運(yùn)動(dòng)．伴隨著P、Q的運(yùn)動(dòng)，DE保持垂直平分PQ，且交PQ于點(diǎn)D，交折線QB-BC-CP于點(diǎn)E．點(diǎn)P、Q同時(shí)出發(fā)，當(dāng)點(diǎn)Q到達(dá)點(diǎn)B時(shí)停止運(yùn)動(dòng)，點(diǎn)P也隨之停止．設(shè)點(diǎn)P、Q運(yùn)動(dòng)的時(shí)間是t秒(t＞0)．

(1)當(dāng)t = 2時(shí)，AP =　　　 ，點(diǎn)Q到AC的距離是　　　 ；

(2)在點(diǎn)P從C向A運(yùn)動(dòng)的過程中，求△APQ的面積S與

t的函數(shù)關(guān)系式；(不必寫出t的取值范圍)

(3)在點(diǎn)E從B向C運(yùn)動(dòng)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(qǐng)說明理由；

(4)當(dāng)DE經(jīng)過點(diǎn)C　時(shí)，請(qǐng)直接寫出t的值．

26．解：(1)由已知，得，，

，

．

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

設(shè)過點(diǎn)的拋物線的解析式為．

將點(diǎn)的坐標(biāo)代入，得．

將和點(diǎn)的坐標(biāo)分別代入，得

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

解這個(gè)方程組，得

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

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

點(diǎn)在該拋物線上，且它的橫坐標(biāo)為，

點(diǎn)的縱坐標(biāo)為．······················································································· (5分)

設(shè)的解析式為，

將點(diǎn)的坐標(biāo)分別代入，得

　 解得

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

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

過點(diǎn)作于點(diǎn)，

則．

，

．

又，

．

．

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

．

(3)點(diǎn)在上，，，則設(shè)．

，，．

①若，則，

解得．，此時(shí)點(diǎn)與點(diǎn)重合．

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

②若，則，

解得 ，，此時(shí)軸．

與該拋物線在第一象限內(nèi)的交點(diǎn)的橫坐標(biāo)為1，

點(diǎn)的縱坐標(biāo)為．

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

③若，則，

解得，，此時(shí)，是等腰直角三角形．

過點(diǎn)作軸于點(diǎn)，

則，設(shè)，

．

．

解得(舍去)．

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

綜上所述，存在三個(gè)滿足條件的點(diǎn)，

即或或．

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

(1)求該拋物線的解析式；

(2)若動(dòng)點(diǎn)從點(diǎn)出發(fā)，以每秒1個(gè)長度單位的速度沿射線運(yùn)動(dòng)，設(shè)點(diǎn)運(yùn)動(dòng)的時(shí)間為．問當(dāng)為何值時(shí)，四邊形分別為平行四邊形？直角梯形？等腰梯形？

(3)若，動(dòng)點(diǎn)和動(dòng)點(diǎn)分別從點(diǎn)和點(diǎn)同時(shí)出發(fā)，分別以每秒1個(gè)長度單位和2個(gè)長度單位的速度沿和運(yùn)動(dòng)，當(dāng)其中一個(gè)點(diǎn)停止運(yùn)動(dòng)時(shí)另一個(gè)點(diǎn)也隨之停止運(yùn)動(dòng)．設(shè)它們的運(yùn)動(dòng)的時(shí)間為，連接，當(dāng)為何值時(shí)，四邊形的面積最��？并求出最小值及此時(shí)的長．

*26．解：(1)拋物線經(jīng)過點(diǎn)，

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

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

(2)為拋物線的頂點(diǎn)過作于，則，

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

當(dāng)時(shí)，四邊形是平行四邊形

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

當(dāng)時(shí)，四邊形是直角梯形

過作于，則

(如果沒求出可由求)

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

當(dāng)時(shí)，四邊形是等腰梯形

綜上所述：當(dāng)、5、4時(shí)，對(duì)應(yīng)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(dāng)時(shí)，的面積最小值為··································································· 10分

此時(shí)

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

26．(2009年重慶市)已知：如圖，在平面直角坐標(biāo)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(diǎn)O作∠AOC的平分線交AB于點(diǎn)D，連接DC，過點(diǎn)D作DE⊥DC，交OA于點(diǎn)E．

(1)求過點(diǎn)E、D、C的拋物線的解析式；

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

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

25.(2009年北京)如圖，在平面直角坐標(biāo)系中，三個(gè)機(jī)戰(zhàn)的坐標(biāo)分別為

，，，延長AC到點(diǎn)D,使CD=,過點(diǎn)D作DE∥AB交BC的延長線于點(diǎn)E.

(1)求D點(diǎn)的坐標(biāo)；

(2)作C點(diǎn)關(guān)于直線DE的對(duì)稱點(diǎn)F,分別連結(jié)DF、EF，若過B點(diǎn)的直線將四邊形CDFE分成周長相等的兩個(gè)四邊形，確定此直線的解析式；

(3)設(shè)G為y軸上一點(diǎn)，點(diǎn)P從直線與y軸的交點(diǎn)出發(fā)，先沿y軸到達(dá)G點(diǎn)，再沿GA到達(dá)A點(diǎn)，若P點(diǎn)在y軸上運(yùn)動(dòng)的速度是它在直線GA上運(yùn)動(dòng)速度的2倍，試確定G點(diǎn)的位置，使P點(diǎn)按照上述要求到達(dá)A點(diǎn)所用的時(shí)間最短。(要求：簡述確定G點(diǎn)位置的方法，但不要求證明)