,
则△APP′是等腰直角三角形,PP′=
,∠AP′P=45°,∵PP′ 2 +P′B 2 =
+ =4,PB 2 =4,∴PP′ 2 +P′B 2 =PB 2 ,
∴△PP′B是等腰直角三角形,
∴∠PP′B=90°,
过A作AN⊥BP′于N,
则∠AP′N=180°-90°-45°=45°,
即△ANP′是等腰直角三角形,
由勾股定理得:AN=NP′=
,由勾股定理得:AB 2 =AN 2 +BN 2 ,
=
+ ,=5,
∴正方形ABCD的面积是5.
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