中考數(shù)學(xué)二輪復(fù)習(xí)培優(yōu)專(zhuān)題03 折疊存在性及最值大全(填空壓軸)（教師版）

專(zhuān)項(xiàng)03折疊存在性及最值大全(填空壓軸)1．如圖，在菱形SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，點(diǎn)SKIPIF1<0為邊SKIPIF1<0的中點(diǎn)，SKIPIF1<0為射線SKIPIF1<0上一動(dòng)點(diǎn)，連接SKIPIF1<0，把SKIPIF1<0沿SKIPIF1<0折疊，得到SKIPIF1<0，當(dāng)SKIPIF1<0與菱形的邊垂直時(shí)，線段SKIPIF1<0的長(zhǎng)為_(kāi)_____．【答案】SKIPIF1<0或SKIPIF1<0【分析】存在兩種情況①當(dāng)點(diǎn)F在線段AB上時(shí)，由題意得出AE的長(zhǎng)，在SKIPIF1<0中可求出AG的長(zhǎng)，由SKIPIF1<0根據(jù)折疊的性質(zhì)，可知SKIPIF1<0在SKIPIF1<0中，可求出GF的長(zhǎng)，即可得出AF的長(zhǎng)．②當(dāng)點(diǎn)F在線段AB延長(zhǎng)線上時(shí)，由SKIPIF1<0得出SKIPIF1<0由SKIPIF1<0SKIPIF1<0中，求出SKIPIF1<0由SKIPIF1<0SKIPIF1<0得出SKIPIF1<0即可得出結(jié)果．【詳解】解：如圖1所示：當(dāng)點(diǎn)F在線段AB上時(shí)，過(guò)點(diǎn)E作SKIPIF1<0于G，∵四邊形SKIPIF1<0是菱形，SKIPIF1<0SKIPIF1<0∵點(diǎn)E是AD的中點(diǎn)，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0如圖2所示：當(dāng)點(diǎn)F在線段AB延長(zhǎng)線上時(shí)，過(guò)點(diǎn)E作SKIPIF1<0SKIPIF1<0交AD于點(diǎn)H，∵四邊形SKIPIF1<0是菱形，SKIPIF1<0SKIPIF1<0∵點(diǎn)E是AD的中點(diǎn)，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0又SKIPIF1<0SKIPIF1<0SKIPIF1<0故答案為：SKIPIF1<0或SKIPIF1<0【我思故我在】本題主要考查了菱形的性質(zhì)，折疊的性質(zhì)，銳角三角函數(shù)的知識(shí)，區(qū)分點(diǎn)F的位置在線段AB上和在線段AB的延長(zhǎng)線上是解本題的關(guān)鍵．2．如圖，菱形SKIPIF1<0的邊長(zhǎng)SKIPIF1<0，M是SKIPIF1<0邊上一點(diǎn)，SKIPIF1<0，N是SKIPIF1<0邊上一動(dòng)點(diǎn)，將梯形SKIPIF1<0沿直線SKIPIF1<0折疊，C對(duì)應(yīng)點(diǎn)SKIPIF1<0．當(dāng)SKIPIF1<0的長(zhǎng)度最小時(shí)，SKIPIF1<0的長(zhǎng)為_(kāi)_________．【答案】14【分析】作SKIPIF1<0于H，如圖，根據(jù)菱形的性質(zhì)可求得SKIPIF1<0，SKIPIF1<0，在SKIPIF1<0中，利用勾股定理計(jì)算出SKIPIF1<0，再根據(jù)兩點(diǎn)間線段最短得到當(dāng)點(diǎn)SKIPIF1<0在SKIPIF1<0上時(shí)，SKIPIF1<0的值最小，然后證明SKIPIF1<0即可．【詳解】解：作SKIPIF1<0于H，如圖，∵菱形SKIPIF1<0的邊SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，在SKIPIF1<0中，SKIPIF1<0，∵梯形SKIPIF1<0沿直線SKIPIF1<0折疊，C對(duì)應(yīng)點(diǎn)SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，∴當(dāng)點(diǎn)SKIPIF1<0在SKIPIF1<0上時(shí)，SKIPIF1<0的值最小，由折疊的性質(zhì)得SKIPIF1<0，而SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0．故答案為：14．【我思故我在】本題考查了菱形的性質(zhì)，折疊的性質(zhì)，勾股定理等知識(shí)，解決本題的關(guān)鍵是確定點(diǎn)SKIPIF1<0在SKIPIF1<0上時(shí)，SKIPIF1<0的值最?。?．如圖，在四邊形紙片ABCD中，ADSKIPIF1<0BC，AB＝10，∠B＝60°，將紙片折疊，使點(diǎn)B落在AD邊上的點(diǎn)G處，折痕為EF，若∠BFE＝45°，則BF的長(zhǎng)為_(kāi)_____．【答案】SKIPIF1<0【分析】由折疊的性質(zhì)知SKIPIF1<0，SKIPIF1<0，再由∠BFE＝45°得到SKIPIF1<0，過(guò)點(diǎn)A作SKIPIF1<0于點(diǎn)H，在SKIPIF1<0中求出SKIPIF1<0的長(zhǎng)度，再證明四邊形SKIPIF1<0是矩形，從而得出SKIPIF1<0，即可解決問(wèn)題．【詳解】解：如圖，過(guò)點(diǎn)A作SKIPIF1<0于點(diǎn)H，由折疊的性質(zhì)知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，在SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0四邊形SKIPIF1<0是矩形，SKIPIF1<0，SKIPIF1<0，故答案為：SKIPIF1<0．【我思故我在】本題考查折疊的性質(zhì)、解直角三角形、矩形的判定與性質(zhì)，根據(jù)已知角度和折疊的性質(zhì)得出SKIPIF1<0是解題的關(guān)鍵．4．如圖，在SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，點(diǎn)SKIPIF1<0在邊SKIPIF1<0上，并且SKIPIF1<0，點(diǎn)SKIPIF1<0為邊SKIPIF1<0上的動(dòng)點(diǎn)，將SKIPIF1<0沿直線SKIPIF1<0翻折，點(diǎn)SKIPIF1<0落在點(diǎn)SKIPIF1<0處，則點(diǎn)SKIPIF1<0到邊SKIPIF1<0距離的最小值是________.【答案】1.2【分析】過(guò)點(diǎn)F作FG⊥AB，垂足為G，過(guò)點(diǎn)P作PD⊥AB，垂足為D，根據(jù)垂線段最短，得當(dāng)PD與FG重合時(shí)PD最小，利用相似求解即可．【詳解】∵SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，∴AB=10，∵SKIPIF1<0，將SKIPIF1<0沿直線SKIPIF1<0翻折，點(diǎn)SKIPIF1<0落在點(diǎn)SKIPIF1<0處，∴CF=PF=2，AF=AC-CF=6-2=4，過(guò)點(diǎn)F作FG⊥AB，垂足為G，過(guò)點(diǎn)P作PD⊥AB，垂足為D，根據(jù)垂線段最短，得當(dāng)PD與FG重合時(shí)PD最小，∵∠A=∠A，∠AGF=∠ACB，∴△AGF∽△ACB，∴SKIPIF1<0，∴SKIPIF1<0，∴FG=3.2，∴PD=FG-PF=3.2-2=1.2，故答案為：1.2．【我思故我在】本題考查了勾股定理，折疊的性質(zhì)，三角形相似，垂線段最短，準(zhǔn)確找到最短位置，并利用相似求解是解題的關(guān)鍵．5．如圖，在矩形SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，點(diǎn)SKIPIF1<0是線段SKIPIF1<0上的一點(diǎn)(不與點(diǎn)SKIPIF1<0，SKIPIF1<0重合)，將△SKIPIF1<0沿SKIPIF1<0折疊，使得點(diǎn)SKIPIF1<0落在SKIPIF1<0處，當(dāng)△SKIPIF1<0為等腰三角形時(shí)，SKIPIF1<0的長(zhǎng)為_(kāi)__________．【答案】SKIPIF1<0或SKIPIF1<0【分析】根據(jù)題意分SKIPIF1<0，SKIPIF1<0，SKIPIF1<0三種情況討論，構(gòu)造直角三角形，利用勾股定理解決問(wèn)題．【詳解】解：∵四邊形SKIPIF1<0是矩形∴SKIPIF1<0，SKIPIF1<0∵將△SKIPIF1<0沿SKIPIF1<0折疊，使得點(diǎn)SKIPIF1<0落在SKIPIF1<0處，∴SKIPIF1<0SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0SKIPIF1<0，則SKIPIF1<0①當(dāng)SKIPIF1<0時(shí)，如圖過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0，則四邊形SKIPIF1<0為矩形SKIPIF1<0SKIPIF1<0，SKIPIF1<0在SKIPIF1<0中SKIPIF1<0SKIPIF1<0在SKIPIF1<0中SKIPIF1<0即SKIPIF1<0解得SKIPIF1<0SKIPIF1<0②當(dāng)SKIPIF1<0時(shí)，如圖，設(shè)SKIPIF1<0交于點(diǎn)SKIPIF1<0，設(shè)SKIPIF1<0SKIPIF1<0SKIPIF1<0垂直平分SKIPIF1<0SKIPIF1<0SKIPIF1<0在SKIPIF1<0中SKIPIF1<0即SKIPIF1<0在SKIPIF1<0中，SKIPIF1<0即SKIPIF1<0聯(lián)立SKIPIF1<0，解得SKIPIF1<0SKIPIF1<0③當(dāng)SKIPIF1<0時(shí)，如圖，又SKIPIF1<0SKIPIF1<0垂直平分SKIPIF1<0SKIPIF1<0SKIPIF1<0垂直平分SKIPIF1<0此時(shí)SKIPIF1<0重合，不符合題意綜上所述，SKIPIF1<0或SKIPIF1<0故答案為：SKIPIF1<0或SKIPIF1<0【我思故我在】本題考查了矩形的性質(zhì)，勾股定理，等腰三角形的性質(zhì)與判定，垂直平分線的性質(zhì)，分類(lèi)討論是解題的關(guān)鍵．6．如圖，在矩形SKIPIF1<0中，SKIPIF1<0，對(duì)角線SKIPIF1<0，點(diǎn)SKIPIF1<0，SKIPIF1<0分別是線段SKIPIF1<0，SKIPIF1<0上的點(diǎn)，將SKIPIF1<0沿直線SKIPIF1<0折疊，點(diǎn)SKIPIF1<0，SKIPIF1<0分別落在點(diǎn)SKIPIF1<0，SKIPIF1<0處．當(dāng)點(diǎn)SKIPIF1<0落在折線SKIPIF1<0上，且SKIPIF1<0時(shí)，SKIPIF1<0的長(zhǎng)為_(kāi)_____．【答案】2或SKIPIF1<0【分析】分兩種情況討論，由折疊的性質(zhì)和勾股定理可求解．【詳解】解：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，當(dāng)點(diǎn)SKIPIF1<0落在SKIPIF1<0上時(shí)，SKIPIF1<0將SKIPIF1<0沿直線SKIPIF1<0折疊，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0；當(dāng)點(diǎn)SKIPIF1<0落在SKIPIF1<0上時(shí)，如圖2，連接SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0于SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0將SKIPIF1<0沿直線SKIPIF1<0折疊，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，綜上所述：SKIPIF1<0的長(zhǎng)為2或SKIPIF1<0．【我思故我在】本題考查了矩形的性質(zhì)，折疊的性質(zhì)，勾股定理等知識(shí)，利用勾股定理列出方程是解題的關(guān)鍵．7．在數(shù)學(xué)探究活動(dòng)中，小美將矩形ABCD紙片先對(duì)折，展開(kāi)后折痕是EF，點(diǎn)M為BC邊上一動(dòng)點(diǎn)，連接AM，過(guò)點(diǎn)M作SKIPIF1<0交CD于點(diǎn)N．將SKIPIF1<0沿MN翻折，點(diǎn)C恰好落在線段EF上，已知矩形ABCD中SKIPIF1<0，SKIPIF1<0，那么BM的長(zhǎng)為_(kāi)______．【答案】4或SKIPIF1<0【分析】設(shè)BM=x，則CM=BC-BM=6-x，根據(jù)三角函數(shù)可得tan∠CMN=tan∠BAM=SKIPIF1<0，tan∠CMN=SKIPIF1<0，F(xiàn)N=CF-CN=SKIPIF1<0，由折疊可知∶C"N=CN=SKIPIF1<0，tanSKIPIF1<0=tan∠CMN=SKIPIF1<0，由tanSKIPIF1<0=SKIPIF1<0，可求SKIPIF1<0，在Rt△SKIPIF1<0中，由勾股定理，SKIPIF1<0，代入相關(guān)數(shù)據(jù)求解即可．【詳解】解：矩形ABCD中，AB=DC=4，BC=6，∠B=∠BCD=90°∴∠BAM+∠AMB=90°，∵M(jìn)N⊥AM，∴∠AMN=90°，∴∠CMN+∠AMB=90°，∴∠CMN=∠BAM，∵小美將矩形ABCD紙片先對(duì)折，展開(kāi)后折痕是EF，∴CF=SKIPIF1<0DC=2，設(shè)BM=x，則CM=BC-BM=6-x，在Rt△ABM中，tan∠BAMSKIPIF1<0∴tan∠CMN=tan∠BAM=SKIPIF1<0在Rt△CMN中，∴tan∠CMN=SKIPIF1<0CN=SKIPIF1<0∴FN=CF-CN=2-SKIPIF1<0由折疊可知∶C"N=CN=SKIPIF1<0連接SKIPIF1<0，如圖∶由折疊知∶MN垂直平分SKIPIF1<0，∴SKIPIF1<0+∠CMN=90°，而SKIPIF1<0=90°，∴SKIPIF1<0=∠CMN，∴tanSKIPIF1<0=tan∠CMN=SKIPIF1<0在Rt△CFC'中，tanSKIPIF1<0=SKIPIF1<0∴SKIPIF1<0在Rt△SKIPIF1<0中，由勾股定理，得SKIPIF1<0，即SKIPIF1<0∴SKIPIF1<0整理，得SKIPIF1<0，解得SKIPIF1<0∴BM的長(zhǎng)為4或SKIPIF1<0故答案為：4或SKIPIF1<0．【我思故我在】本題考查了矩形的性質(zhì)，折疊的性質(zhì)，解直角三角形，勾股定理，解一元二次方程等知識(shí)，運(yùn)用三角函數(shù)將邊長(zhǎng)表示出來(lái)，借助勾股定理建立方程是解題的關(guān)鍵．8．如圖，矩形ABCD中，AB=4，AD=6，點(diǎn)E為AD中點(diǎn)，點(diǎn)P為線段AB上一個(gè)動(dòng)點(diǎn)，連接EP，將△APE沿PE折疊得到△FPE，連接CE，DF，當(dāng)線段DF被CE垂直平分時(shí)，AF則線的長(zhǎng)為_(kāi)______．【分析】連接AF交PE于O，連接DF,先由矩形的性質(zhì)可得BC=AD=6、CD=AB=4，再由折疊的性質(zhì)和垂直平分線的性質(zhì)可得AF=2OA，AE=ED=EF=3；設(shè)AP=x，則PF=AP=x,BP=4-x，PC=PF+FC=x+4，運(yùn)用勾股定理可求得x，然后再運(yùn)用勾股定理求得PE的長(zhǎng)，再運(yùn)用等面積法求得AO的長(zhǎng)，最后根據(jù)AF=2AO解答即可．【詳解】解：連接AF交PE于O，連接DF，∵矩形ABCD，∴BC=AD=6,CD=AB=4，∵線段DF被CE垂直平分時(shí)，∴CF=CD=4,ED=EF，∵將△APE沿PE折疊得到△FPE，∴PE是線段AF的垂直平分線，∴AE=EF,AF=2OA，∴AE=ED=EF，∵AD=AE+ED=6，∴AE=ED=EF=3，設(shè)AP=x，則PF=AP=x，BP=4-x，PC=PF+FC=x+4，∵PC2=BP2+BC2,即(x+4)2=(4-x)2+62∴x=SKIPIF1<0，∵PE=SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，解得：AO=SKIPIF1<0，∴AF=2AO=SKIPIF1<0．故答案為SKIPIF1<0．【我思故我在】本題主要考查了矩形的性質(zhì)、折疊的性質(zhì)、線段垂直平分線的性質(zhì)、勾股定理等知識(shí)點(diǎn)，靈活應(yīng)用相關(guān)知識(shí)成為解答本題的關(guān)鍵．9．如圖，在矩形ABCD中，AB＝2，AD＝1，E是AB上一個(gè)動(dòng)點(diǎn)，F(xiàn)是AD上一個(gè)動(dòng)點(diǎn)(點(diǎn)F不與點(diǎn)D重合)，連接EF，把△AEF沿EF折疊，使點(diǎn)A的對(duì)應(yīng)點(diǎn)A′總落在DC邊上．若△A′EC是以A′E為腰的等腰三角形，則A′D的長(zhǎng)為_(kāi)_____．【答案】SKIPIF1<0或SKIPIF1<0【分析】分兩種情形分別畫(huà)出圖形，利用勾股定理構(gòu)建方程求解即可．【詳解】解：如圖1中，當(dāng)EA′＝CE時(shí)，過(guò)點(diǎn)E作EH⊥CD于H．∵四邊形ABCD是矩形，∴AD＝BC＝1，∠B＝90°，設(shè)AE＝EA′＝EC＝x，則BE＝2﹣x，在Rt△EBC中，則有x2＝12+(2﹣x)2，解得x＝SKIPIF1<0，∴EB＝2﹣x＝SKIPIF1<0，∵∠B＝∠BCH＝∠CHE＝90°，∴四邊形CBEH是矩形，∴CH＝BE＝SKIPIF1<0，∵EC＝EA′，EH⊥CA′，∴HA′＝CH＝SKIPIF1<0，∴DA′＝CD﹣CA′＝2﹣SKIPIF1<0＝SKIPIF1<0．如圖2中，當(dāng)A′E＝A′C時(shí)，設(shè)AE＝EA′＝CA′＝y(tǒng)．則CH＝EB＝2﹣y，A′H＝CA′﹣CH＝y(tǒng)﹣(2﹣y)＝2y﹣2，在Rt△A′EH中，則有y2＝12+(2y﹣2)2，解得y＝SKIPIF1<0或1(舍棄)，∴CA′＝SKIPIF1<0，∴DA′＝2﹣SKIPIF1<0＝SKIPIF1<0，∴DA′為SKIPIF1<0或SKIPIF1<0，故答案為SKIPIF1<0或SKIPIF1<0．【我思故我在】本題考查翻折變換，矩形的性質(zhì)，等腰三角形的判定和性質(zhì)，解直角三角形等知識(shí)，解題的關(guān)鍵是學(xué)會(huì)用分類(lèi)討論的思想思考問(wèn)題，屬于中考?？碱}型．10．如圖，長(zhǎng)方形SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，點(diǎn)E為射線SKIPIF1<0上一動(dòng)點(diǎn)(不與D重合)，將SKIPIF1<0沿AE折疊得到SKIPIF1<0，連接SKIPIF1<0，若SKIPIF1<0為直角三角形，則SKIPIF1<0________【分析】分兩種情況討論：①當(dāng)點(diǎn)E在線段CD上時(shí)，SKIPIF1<0三點(diǎn)共線，根據(jù)SKIPIF1<0可求得SKIPIF1<0，再由勾股定理可得SKIPIF1<0，進(jìn)而可計(jì)算SKIPIF1<0，在SKIPIF1<0中，由勾股定理計(jì)算SKIPIF1<0的值；②當(dāng)點(diǎn)E在射線CD上時(shí)，設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，由勾股定理可解得SKIPIF1<0，進(jìn)而可計(jì)算SKIPIF1<0，在SKIPIF1<0中，由勾股定理計(jì)算SKIPIF1<0的值即可．【詳解】解：根據(jù)題意，四邊形ABCD為長(zhǎng)方形，SKIPIF1<0，SKIPIF1<0，將SKIPIF1<0沿AE折疊得到SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，①如圖1，當(dāng)點(diǎn)E在線段CD上時(shí)，∵SKIPIF1<0，∴SKIPIF1<0三點(diǎn)共線，∵SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0；∴在SKIPIF1<0中，SKIPIF1<0；②如圖2，當(dāng)點(diǎn)E在射線CD上時(shí)，∵SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，∴SKIPIF1<0，∴在SKIPIF1<0中，SKIPIF1<0．綜上所述，AE的值為SKIPIF1<0或SKIPIF1<0．故答案為：SKIPIF1<0或SKIPIF1<0．【我思故我在】本題主要考查了折疊的性質(zhì)以及勾股定理等知識(shí)，運(yùn)用分類(lèi)討論的思想分析問(wèn)題是解題關(guān)鍵．11．如圖，已知SKIPIF1<0中，SKIPIF1<0，點(diǎn)SKIPIF1<0、SKIPIF1<0分別在線段SKIPIF1<0、SKIPIF1<0上，將SKIPIF1<0沿直線SKIPIF1<0折疊，使點(diǎn)SKIPIF1<0的對(duì)應(yīng)點(diǎn)SKIPIF1<0恰好落在線段SKIPIF1<0上，當(dāng)SKIPIF1<0為直角三角形時(shí)，折痕SKIPIF1<0的長(zhǎng)為_(kāi)__________．【答案】SKIPIF1<0或SKIPIF1<0【分析】由SKIPIF1<0為直角三角形，分兩種情況進(jìn)行討論：SKIPIF1<0分別依據(jù)含SKIPIF1<0角的直角三角形的性質(zhì)以及等腰直角三角形的性質(zhì)，即可得到折痕SKIPIF1<0的長(zhǎng)．【詳解】解：分兩種情況：SKIPIF1<0如圖，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0是直角三角形，SKIPIF1<0在SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，由折疊可得，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0；SKIPIF1<0如圖，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0是直角三角形，由題可得，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，過(guò)SKIPIF1<0作SKIPIF1<0于SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，由折疊可得，SKIPIF1<0，SKIPIF1<0是等腰直角三角形，SKIPIF1<0，SKIPIF1<0．故答案為：SKIPIF1<0或SKIPIF1<0．【我思故我在】本題考查了翻折變換SKIPIF1<0折疊問(wèn)題，勾股定理，含SKIPIF1<0角的直角三角形的性質(zhì)，等腰直角三角形的性質(zhì)，正確的作出圖形是解題的關(guān)鍵．折疊是一種對(duì)稱(chēng)變換，它屬于軸對(duì)稱(chēng)，折疊前后圖形的形狀和大小不變，位置變化，對(duì)應(yīng)邊和對(duì)應(yīng)角相等．12．如圖，在SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，點(diǎn)SKIPIF1<0、SKIPIF1<0分別是邊SKIPIF1<0、SKIPIF1<0上的點(diǎn)，且SKIPIF1<0，將SKIPIF1<0沿SKIPIF1<0對(duì)折，若點(diǎn)SKIPIF1<0恰好落到了SKIPIF1<0的外部，則折痕SKIPIF1<0的長(zhǎng)度范圍是______．【答案】SKIPIF1<0【分析】把SKIPIF1<0沿SKIPIF1<0對(duì)折，當(dāng)點(diǎn)SKIPIF1<0恰好落在SKIPIF1<0的SKIPIF1<0點(diǎn)處，SKIPIF1<0與SKIPIF1<0相交于SKIPIF1<0點(diǎn)，根據(jù)折疊的性質(zhì)得到SKIPIF1<0，SKIPIF1<0，證明SKIPIF1<0，同理可得SKIPIF1<0，于是可得SKIPIF1<0的長(zhǎng)，然后根據(jù)勾股定理計(jì)算SKIPIF1<0的長(zhǎng)，由正切的定義可得SKIPIF1<0和SKIPIF1<0的長(zhǎng)，計(jì)算SKIPIF1<0的長(zhǎng)，再計(jì)算當(dāng)SKIPIF1<0與SKIPIF1<0重合時(shí)SKIPIF1<0的長(zhǎng)，從而得結(jié)論．【詳解】解：把SKIPIF1<0沿SKIPIF1<0對(duì)折，當(dāng)點(diǎn)SKIPIF1<0恰好落在SKIPIF1<0的SKIPIF1<0點(diǎn)處，SKIPIF1<0與SKIPIF1<0相交于SKIPIF1<0點(diǎn)，如圖1，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，而SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，同理可得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，在SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，在SKIPIF1<0中，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，在SKIPIF1<0中，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，SKIPIF1<0；如圖2，當(dāng)SKIPIF1<0與SKIPIF1<0重合時(shí)，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0折痕SKIPIF1<0的長(zhǎng)度范圍是：SKIPIF1<0．故答案為：SKIPIF1<0．【我思故我在】本題考查了折疊的性質(zhì)：折疊是一種對(duì)稱(chēng)變換，它屬于軸對(duì)稱(chēng)，折疊前后圖形的形狀和大小不變，位置變化，對(duì)應(yīng)邊和對(duì)應(yīng)角相等．也考查了勾股定理和銳角三角函數(shù)．13．如圖，在?ABCD中，點(diǎn)E，F(xiàn)分別在邊AB、AD上，將△AEF沿EF折疊，點(diǎn)A恰好落在BC邊上的點(diǎn)G處．若∠A=45°，AB=6SKIPIF1<0，5BE=AE．則AF長(zhǎng)度為_(kāi)____．【答案】SKIPIF1<0【分析】過(guò)點(diǎn)B作BM⊥AD于點(diǎn)M，過(guò)點(diǎn)F作FH⊥BC于點(diǎn)H，過(guò)點(diǎn)E作EN⊥CB延長(zhǎng)線于點(diǎn)N，得矩形BHFM，可得△BEN和△ABM是等腰直角三角形，然后利用勾股定理即可解決問(wèn)題．【詳解】解：如圖，過(guò)點(diǎn)B作BM⊥AD于點(diǎn)M，過(guò)點(diǎn)F作FH⊥BC于點(diǎn)H，過(guò)點(diǎn)E作EN⊥CB延長(zhǎng)線于點(diǎn)N，得矩形BHFM，∴∠MBC=90°，MB=FH，F(xiàn)M=BH，∵AB=6SKIPIF1<0，5BE=AE，∴AE=5SKIPIF1<0，BE=SKIPIF1<0，由折疊的性質(zhì)可知：GE=AE=5SKIPIF1<0，GF=AF，∵四邊形ABCD是平行四邊形，∴∠ABN=∠A=45°，∴△BEN和△ABM是等腰直角三角形，∴EN=BN=SKIPIF1<0BE=1，AM=BM=SKIPIF1<0AB=6，∴FH=BM=6，在Rt△GEN中，根據(jù)勾股定理，得SKIPIF1<0，∴SKIPIF1<0，解得GN=±7(負(fù)值舍去)，∴GN=7，設(shè)MF=BH=x,則GH=GN-BN-BH=7-1-x=6-x，GF=AF=AM+FM=6+x，在Rt△GFH中，根據(jù)勾股定理，得SKIPIF1<0，∴SKIPIF1<0，解得x=SKIPIF1<0，∴AF=AM+FM=6+SKIPIF1<0=SKIPIF1<0．∴AF長(zhǎng)度為SKIPIF1<0．故答案為：SKIPIF1<0．【我思故我在】本題考查了翻折變換，平行四邊形的性質(zhì)，等腰直角三角形的性質(zhì)，勾股定理，解決本題的關(guān)鍵是掌握翻折的性質(zhì)．14．如圖，矩形SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0是SKIPIF1<0邊上的一個(gè)動(dòng)點(diǎn)，將SKIPIF1<0沿SKIPIF1<0折疊，得到SKIPIF1<0，則當(dāng)SKIPIF1<0最小時(shí)，折痕SKIPIF1<0長(zhǎng)為_(kāi)_____．【答案】SKIPIF1<0【分析】根據(jù)三角形的三邊關(guān)系得出：當(dāng)SKIPIF1<0最小時(shí)的圖形，利用勾股定理列出方程，求出SKIPIF1<0的長(zhǎng)度，進(jìn)行解答即可．【詳解】連接AC，依題意可知：SKIPIF1<0，如圖，當(dāng)A、C、F三點(diǎn)共線時(shí)，SKIPIF1<0取得最小值，在矩形SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0,∴SKIPIF1<0，由折疊可知：SKIPIF1<0,設(shè)SKIPIF1<0，∴SKIPIF1<0,SKIPIF1<0，在SKIPIF1<0中，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0．故答案為：SKIPIF1<0．【我思故我在】本題考查了矩形與折疊，勾股定理，二次根式的運(yùn)算，掌握勾股定理進(jìn)行求線段長(zhǎng)度是解題的關(guān)鍵．15．如圖，在正方形ABCD中，AB＝8，E是CD上一點(diǎn)，且DE＝2，F(xiàn)是AD上一動(dòng)點(diǎn)，連接EF，若將△DEF沿EF翻折后，點(diǎn)D落在點(diǎn)SKIPIF1<0處，則點(diǎn)SKIPIF1<0到點(diǎn)B的最短距離為_(kāi)_____．【答案】8【分析】連接SKIPIF1<0、SKIPIF1<0，當(dāng)B、SKIPIF1<0、E三點(diǎn)共線的時(shí)候點(diǎn)SKIPIF1<0到B點(diǎn)的距離最短，根據(jù)DE求出CE，再利用勾股定理求出BE，即可求解．【詳解】如圖，連接SKIPIF1<0、SKIPIF1<0，當(dāng)B、SKIPIF1<0、E三點(diǎn)共線的時(shí)候點(diǎn)SKIPIF1<0到B點(diǎn)的距離最短，在正方形ABCD中，AB=8，E是CD上一點(diǎn)，且DE=2，∴CE=CD-DE=8-2=6，BC=AB=8，∴SKIPIF1<0，根據(jù)折疊的性質(zhì)有SKIPIF1<0，∵B、SKIPIF1<0、E三點(diǎn)共線∴SKIPIF1<0，即點(diǎn)SKIPIF1<0到B點(diǎn)的距離最短為8，故答案為：8．【我思故我在】本題考查了正方形的性質(zhì)、翻折的性質(zhì)、勾股定理以及兩點(diǎn)之間線段最短的知識(shí)，找到B、SKIPIF1<0、E三點(diǎn)共線的時(shí)候點(diǎn)SKIPIF1<0到B點(diǎn)的距離最短是解答本題的關(guān)鍵．16．如圖，已知在矩形紙片SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，點(diǎn)E是SKIPIF1<0的中點(diǎn)，點(diǎn)F是SKIPIF1<0邊上的一個(gè)動(dòng)點(diǎn)，將SKIPIF1<0沿SKIPIF1<0所在直線翻折，得到SKIPIF1<0，連接SKIPIF1<0，SKIPIF1<0，則當(dāng)SKIPIF1<0是以SKIPIF1<0為腰的等腰三角形時(shí)，SKIPIF1<0的長(zhǎng)是_______________．【答案】1或SKIPIF1<0【分析】存在三種情況：當(dāng)SKIPIF1<0時(shí)，連接ED，利用勾股定理可以求得ED的長(zhǎng)，可判斷SKIPIF1<0三點(diǎn)共線，根據(jù)勾股定理即可求解；當(dāng)SKIPIF1<0時(shí)，可以證得四邊形SKIPIF1<0是正方形，即可求解；當(dāng)SKIPIF1<0時(shí)，連接EC，F(xiàn)C，證明SKIPIF1<0三點(diǎn)共線，再用勾股定理，即可求解．【詳解】解：①當(dāng)SKIPIF1<0時(shí)，連接ED，如圖，∵點(diǎn)E是SKIPIF1<0的中點(diǎn)，SKIPIF1<0，SKIPIF1<0，四邊形SKIPIF1<0是矩形，∴SKIPIF1<0，由勾股定理可得，SKIPIF1<0，∵將SKIPIF1<0沿SKIPIF1<0所在直線翻折，得到SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0三點(diǎn)共線，∵SKIPIF1<0，∴SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，在SKIPIF1<0中，SKIPIF1<0，∴SKIPIF1<0，解得SKIPIF1<0，∴SKIPIF1<0；②當(dāng)SKIPIF1<0時(shí)，如圖，∵SKIPIF1<0，∴點(diǎn)SKIPIF1<0在線段CD的垂直平分線上，∴點(diǎn)SKIPIF1<0在線段AB的垂直平分線上，∵點(diǎn)E是SKIPIF1<0的中點(diǎn)，∴SKIPIF1<0是AB的垂直平分線，∴SKIPIF1<0，∵將SKIPIF1<0沿SKIPIF1<0所在直線翻折，得到SKIPIF1<0，∴SKIPIF1<0，∴四邊形SKIPIF1<0是正方形，∴SKIPIF1<0；綜上所述，AF的長(zhǎng)為1或SKIPIF1<0．故答案為：1或SKIPIF1<0．【我思故我在】本題考查矩形中的翻折問(wèn)題，涉及矩形的性質(zhì)、等腰三角形的性質(zhì)、正方形的判定和性質(zhì)、勾股定理，分類(lèi)討論思想的運(yùn)用是解題的關(guān)鍵．17．如圖，在SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為邊SKIPIF1<0的中點(diǎn)，點(diǎn)SKIPIF1<0是SKIPIF1<0邊上的動(dòng)點(diǎn)，把SKIPIF1<0沿SKIPIF1<0翻折，點(diǎn)SKIPIF1<0落在SKIPIF1<0處，若SKIPIF1<0是直角三角形，則SKIPIF1<0的長(zhǎng)為_(kāi)_____．【答案】SKIPIF1<0或SKIPIF1<0【分析】在圖SKIPIF1<0中構(gòu)造正方形SKIPIF1<0，在SKIPIF1<0中即可解決問(wèn)題，在圖SKIPIF1<0中也要證明四邊形SKIPIF1<0是正方形解決問(wèn)題．【詳解】解：如圖SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，作SKIPIF1<0垂足為SKIPIF1<0，作SKIPIF1<0于SKIPIF1<0．SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1