摘要：26．(1)解:由得点坐标为 由得点坐标为 ∴··················································································· 由解得∴点的坐标为···································· ∴··························································· (2)解:∵点在上且 ∴点坐标为······················································································ 又∵点在上且 ∴点坐标为······················································································ ∴··········································································· (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··································································· 29． 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) 联系拓广 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)

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