题目内容
如下图,折叠矩形纸片ABCD,先折出折痕BD,再折叠使AD边与对角线BD重合,得折痕DG,若AB=2,BC=1,则AG的长是__________。
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试题分析:已知AB=2,BC=1,可知AD=BC=1,在Rt△ABD中根据勾股定理求得BD的长;设AG=x,由折叠的性质可知,GH=x,BH=BD-DH=BD-AD=
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由题意得AB=2,AD=BC=1,
在Rt△ABD中,

过点G作GH⊥BD,垂足为H,
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由折叠可知:△AGD≌△HGD,
∴AD=DH=1,设AG的长为x,HG=AG=x,BG=2-x,BH=
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在Rt△BGH中,由勾股定理得

即
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解得

则AG的长是
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点评:解答本题的关键是熟练掌握折叠的性质,折叠前后图形的形状和大小不变.
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