中考數(shù)學(xué)二輪專題復(fù)習(xí)：三大幾何變換 (含答案)

PAGE2PAGE1PAGE1三大幾何變換三大幾何變換知識(shí)互聯(lián)網(wǎng)知識(shí)互聯(lián)網(wǎng)題型一：平移變換題型一：平移變換思路導(dǎo)航思路導(dǎo)航平移一般是在需要同時(shí)移動(dòng)兩條線段或元素的時(shí)候，才考慮的方法．典題精練典題精練已知：如圖，正方形SKIPIF1<0中，SKIPIF1<0是SKIPIF1<0上一點(diǎn)，SKIPIF1<0于點(diǎn)SKIPIF1<0．⑴求證：SKIPIF1<0．⑵求證：SKIPIF1<0． 延長(zhǎng)SKIPIF1<0到點(diǎn)SKIPIF1<0，使得SKIPIF1<0，連接SKIPIF1<0、SKIPIF1<0．⑴∵SKIPIF1<0，SKIPIF1<0∴四邊形SKIPIF1<0為平行四邊形∴SKIPIF1<0，SKIPIF1<0又∵SKIPIF1<0，∴SKIPIF1<0∴SKIPIF1<0在SKIPIF1<0和SKIPIF1<0中SKIPIF1<0∴SKIPIF1<0∴SKIPIF1<0⑵由⑴知道SKIPIF1<0為等腰直角三角形∴SKIPIF1<0在SKIPIF1<0中，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，取到等號(hào)．在Rt△ABC中，∠A=90°，D、E分別為AB、AC上的點(diǎn)．⑴如圖1，CE=AB，BD=AE，過(guò)點(diǎn)C作CF∥EB，且CF=EB，連接DF交EB于點(diǎn)G，連接BF，請(qǐng)你直接寫(xiě)出SKIPIF1<0的值；⑵如圖2，CE=kAB，BD=kAE，SKIPIF1<0，求k的值． 圖2圖2圖1【解析】(1)SKIPIF1<0.(2)過(guò)點(diǎn)C作CF∥EB且CF=EB，連接DF交EB于點(diǎn)G,連接BF.∴四邊形EBFC是平行四邊形.∴CE∥BF且CE=BF.∴∠ABF=∠A=90°.∵BF=CE=kAB.∴SKIPIF1<0.∵BD=kAE，∴SKIPIF1<0.∴SKIPIF1<0.∴SKIPIF1<0∽SKIPIF1<0.∴SKIPIF1<0,∠GDB=∠AEB.∴∠DGB=∠A=90°.∴∠GFC=∠BGF=90°.∵SKIPIF1<0.∴SKIPIF1<0.∴k=SKIPIF1<0.題型二：軸對(duì)稱變換題型二：軸對(duì)稱變換典題精練典題精練⑴如圖，已知正方形紙片SKIPIF1<0的邊長(zhǎng)為SKIPIF1<0，SKIPIF1<0的半徑為SKIPIF1<0，圓心在正方形的中心上，將紙片按圖示方式折疊，使SKIPIF1<0恰好與SKIPIF1<0相切于點(diǎn)SKIPIF1<0（SKIPIF1<0與SKIPIF1<0除切點(diǎn)外無(wú)重疊部分），延長(zhǎng)SKIPIF1<0交SKIPIF1<0邊于點(diǎn)SKIPIF1<0，則SKIPIF1<0的長(zhǎng)是．⑵將弧SKIPIF1<0沿弦SKIPIF1<0折疊交直徑SKIPIF1<0于點(diǎn)SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0的長(zhǎng)是______________．⑴過(guò)SKIPIF1<0點(diǎn)作SKIPIF1<0于SKIPIF1<0．則四邊形SKIPIF1<0是矩形，∴SKIPIF1<0，設(shè)SKIPIF1<0，則根據(jù)對(duì)稱性可知SKIPIF1<0∴SKIPIF1<0，在SKIPIF1<0中，SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，∴SKIPIF1<0．⑵將半圓還原，點(diǎn)SKIPIF1<0關(guān)于SKIPIF1<0的對(duì)稱點(diǎn)為SKIPIF1<0，作SKIPIF1<0于SKIPIF1<0．根據(jù)“翻折”的性質(zhì)可知SKIPIF1<0，則SKIPIF1<0∵SKIPIF1<0，則SKIPIF1<0，BC2=BH·AB∴SKIPIF1<0．把正方形沿著SKIPIF1<0折疊使點(diǎn)SKIPIF1<0落在SKIPIF1<0上，SKIPIF1<0交SKIPIF1<0于點(diǎn)SKIPIF1<0，已知正方形的邊長(zhǎng)為SKIPIF1<0，求SKIPIF1<0的周長(zhǎng)．在SKIPIF1<0上取點(diǎn)SKIPIF1<0，使SKIPIF1<0，連接SKIPIF1<0．∵SKIPIF1<0，∴SKIPIF1<0由翻折得對(duì)稱性可知SKIPIF1<0∴SKIPIF1<0在SKIPIF1<0和SKIPIF1<0中SKIPIF1<0∴SKIPIF1<0∴SKIPIF1<0，SKIPIF1<0在SKIPIF1<0和SKIPIF1<0中SKIPIF1<0∴SKIPIF1<0∴SKIPIF1<0∴SKIPIF1<0的周長(zhǎng)為SKIPIF1<0．題型三：旋轉(zhuǎn)變換題型三：旋轉(zhuǎn)變換典題精練典題精練在Rt△ABC中，AB=BC，∠B=90°，將一塊等腰直角三角板的直角頂點(diǎn)O放在斜邊AC上，將三角板繞點(diǎn)O旋轉(zhuǎn)．⑴當(dāng)點(diǎn)O為AC中點(diǎn)時(shí)，①如圖1，三角板的兩直角邊分別交AB，BC于E、F兩點(diǎn)，連接EF，猜想線段AE、CF與EF之間存在的等量關(guān)系（無(wú)需證明）；②如圖2，三角板的兩直角邊分別交AB，BC延長(zhǎng)線于E、F兩點(diǎn)，連接EF，判斷①中的猜想是否成立．若成立，請(qǐng)證明；若不成立，請(qǐng)說(shuō)明理由；⑵當(dāng)點(diǎn)O不是AC中點(diǎn)時(shí)，如圖3,，三角板的兩直角邊分別交AB，BC于E、F兩點(diǎn)，若SKIPIF1<0，求SKIPIF1<0的值． CCOBAOE圖1FBAOCEFABCEF圖2圖3CBACBAOEF猜想：SKIPIF1<0.成立.證明：連結(jié)OB.∵AB=BC,∠ABC=90°,O點(diǎn)為AC的中點(diǎn)，∴SKIPIF1<0,∠BOC=90°,∠ABO=∠BCO=45°.∵∠EOF=90°，∴∠EOB=∠FOC.又∵∠EBO=∠FCO，∴△OEB≌△OFC（ASA）.∴BE=CF.又∵BA=BC,∴AE=BF.在RtΔEBF中，∵∠EBF=90°,SKIPIF1<0.SKIPIF1<0.（2）解：如圖，過(guò)點(diǎn)O作OM⊥AB于M，ON⊥BC于N.AOBCEFMN∵∠AOBCEFMN∵∠EOF=90°,∴∠EOM=∠FON.∵∠EMO=∠FNO=90°，∴△OME∽△ONF.∴SKIPIF1<0∵△AOM和△OCN為等腰直角三角形，∴△AOM∽△OCN∴SKIPIF1<0.∵SKIPIF1<0，∴SKIPIF1<0.SKIPIF1<0和SKIPIF1<0是繞點(diǎn)SKIPIF1<0旋轉(zhuǎn)的兩個(gè)相似三角形，其中SKIPIF1<0與SKIPIF1<0、SKIPIF1<0與SKIPIF1<0為對(duì)應(yīng)角．⑴如圖1，若SKIPIF1<0和SKIPIF1<0分別是以SKIPIF1<0與SKIPIF1<0為頂角的等腰直角三角形，且兩三角形旋轉(zhuǎn)到使點(diǎn)SKIPIF1<0、SKIPIF1<0、SKIPIF1<0在同一條直線上的位置時(shí)，請(qǐng)直接寫(xiě)出線段SKIPIF1<0與線段SKIPIF1<0的關(guān)系；⑵若SKIPIF1<0和SKIPIF1<0為含有SKIPIF1<0角的直角三角形，且兩個(gè)三角形旋轉(zhuǎn)到如圖2的位置時(shí)，試確定線段SKIPIF1<0與線段SKIPIF1<0的關(guān)系，并說(shuō)明理由；⑶若SKIPIF1<0和SKIPIF1<0為如圖3的兩個(gè)三角形，且SKIPIF1<0SKIPIF1<0，SKIPIF1<0，在繞點(diǎn)SKIPIF1<0旋轉(zhuǎn)的過(guò)程中，直線SKIPIF1<0與SKIPIF1<0夾角的度數(shù)是否改變？若不改變，直接用含SKIPIF1<0、SKIPIF1<0的式子表示夾角的度數(shù)；若改變，請(qǐng)說(shuō)明理由．⑴線段SKIPIF1<0與線段SKIPIF1<0的關(guān)系是SKIPIF1<0．⑵如圖2，連接SKIPIF1<0、SKIPIF1<0并延長(zhǎng)，設(shè)交點(diǎn)為點(diǎn)SKIPIF1<0．∵SKIPIF1<0SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0．∵SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0．∴SKIPIF1<0．∴SKIPIF1<0．∴SKIPIF1<0．在SKIPIF1<0中，SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0．又∵SKIPIF1<0∴SKIPIF1<0，∴SKIPIF1<0．∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0．即SKIPIF1<0．⑶在繞點(diǎn)SKIPIF1<0旋轉(zhuǎn)的過(guò)程中，直線SKIPIF1<0與SKIPIF1<0夾角度數(shù)不改變，SKIPIF1<0度．

復(fù)習(xí)鞏固復(fù)習(xí)鞏固題型一平移變換鞏固練習(xí)如圖，已知SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0的度數(shù)為_(kāi)_____．SKIPIF1<0.通過(guò)作平行線平移角，使角與角之間聯(lián)系起來(lái)．A如下圖，兩條長(zhǎng)度為SKIPIF1<0的線段SKIPIF1<0和SKIPIF1<0相交于SKIPIF1<0點(diǎn)，且SKIPIF1<0，求證：SKIPIF1<0．A考慮將SKIPIF1<0、SKIPIF1<0和SKIPIF1<0集中到同一個(gè)三角形中，以便運(yùn)用三角形的不等關(guān)系．作SKIPIF1<0且SKIPIF1<0，則四邊形SKIPIF1<0是平行四邊形，從而SKIPIF1<0．（教師可告訴學(xué)生：一組對(duì)邊平行且相等的四邊形是平行四邊形），在SKIPIF1<0中可得SKIPIF1<0，即SKIPIF1<0．由于SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0是等邊三角形，故SKIPIF1<0，所以SKIPIF1<0．題型二軸對(duì)稱變換鞏固練習(xí)如圖矩形紙片SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0上有一點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0上有一點(diǎn)SKIPIF1<0，SKIPIF1<0，過(guò)SKIPIF1<0作SKIPIF1<0交SKIPIF1<0于SKIPIF1<0，將紙片折疊，使SKIPIF1<0點(diǎn)與SKIPIF1<0點(diǎn)重合，折痕與SKIPIF1<0交于SKIPIF1<0點(diǎn)，則SKIPIF1<0的長(zhǎng)是________cm．SKIPIF1<0.解法：過(guò)Q作QM⊥DC，設(shè)QP=x，∴QE=x，∵DE=2，∴SKIPIF1<0∴在Rt△QME中，SKIPIF1<0，∴SKIPIF1<0題型三旋轉(zhuǎn)變換鞏固練習(xí)已知正方形SKIPIF1<0中，點(diǎn)SKIPIF1<0在邊SKIPIF1<0上，SKIPIF1<0，SKIPIF1<0（如圖所示）把線段SKIPIF1<0繞 點(diǎn)SKIPIF1<0旋轉(zhuǎn)，使點(diǎn)SKIPIF1<0落在直線SKIPIF1<0上的點(diǎn)SKIPIF1<0處，則SKIPIF1<0、SKIPIF1<0兩點(diǎn)的距離為．SKIPIF1<0或SKIPIF1<0．題目里只說(shuō)“旋轉(zhuǎn)”，并沒(méi)有說(shuō)順時(shí)針還是逆時(shí)針，而且說(shuō)的是“直線SKIPIF1<0上的點(diǎn)”，所以有兩種情況如圖所示：順時(shí)針旋轉(zhuǎn)得到SKIPIF1<0點(diǎn)，則SKIPIF1<0，逆時(shí)針旋轉(zhuǎn)得到SKIPIF1<0點(diǎn)，則SKIPIF1<0，SKIPIF1<0．在平面直角坐標(biāo)系中，矩形SKIPIF1<0的頂點(diǎn)SKIPIF1<0、SKIPIF1<0的坐標(biāo)分別為SKIPIF1<0和SKIPIF1<0．將矩形SKIPIF1<0繞點(diǎn)SKIPIF1<0順時(shí)針旋轉(zhuǎn)SKIPIF1<0度，得到四邊形SKIPIF1<0，使得邊SKIPIF1<0與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0，此時(shí)邊SKIPIF1<0、SKIPIF1<0分別與SKIPIF1<0邊所在的直線相交于點(diǎn)SKIPIF1<0、SKIPIF1<0．⑴如圖1，當(dāng)點(diǎn)SKIPIF1<0與點(diǎn)SKIPIF1<0重合時(shí)，求點(diǎn)SKIPIF1<0的坐標(biāo)；⑵在⑴的條件下，求SKIPIF1<0的值；⑶如圖2，若點(diǎn)SKIPIF1<0與點(diǎn)SKIPIF1<0不重合，則SKIPIF1<0的值是否發(fā)生變化？若不變，試證明你的結(jié)論；若有變化，請(qǐng)說(shuō)明理由．（圖1）（圖2）（圖1）（圖2）⑴∵將矩形SKIPIF1<0繞點(diǎn)SKIPIF1<0順時(shí)針旋轉(zhuǎn)SKIPIF1<0度，得到四邊形SKIPIF1<0，且SKIPIF1<0、SKIPIF1<0的坐標(biāo)分別為SKIPIF1<0和SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0．（圖1）∴SKIPIF1<0．（圖1）∴點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0．⑵∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0．∵SKIPIF1<0，且SKIPIF1<0，∴SKIPIF1<0．同理SKIPIF1<0．∴SKIPIF1<0，∴SKIPIF1<0．（或：∵SKIPIF1<0．∴SKIPIF1<0．）⑶如圖2所示，作SKIPIF1<0SKIPIF1<0交SKIPIF1<0于點(diǎn)SKIPIF1<0，∵SKIPIF1<0SKIPIF1<0，且SKIPIF1<0SKIPIF1<0，∴四邊形SKIPIF1<0是平行四邊形．（圖2）∴SKIPIF1<0．（圖2）∵SKIPIF1<0，∴SKIPIF1<0．∴SKIPIF1<0．又∵SKIPIF1<0，∴SKIPIF1<0．∴SKIPIF1<0．∴SKIPIF1<0的值不會(huì)發(fā)生改變．課后測(cè)課后測(cè)【測(cè)試1】在四邊形SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，SKIPIF1<