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分析:作HE⊥AB于E,CF⊥BC′于F,設(shè)BH=x,則DH=x,在Rt△BEH中,根據(jù)勾股定理得到BE
2+HE
2=BH
2,即(2-x)
2+1
2=x
2,可解得x=
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,即BH=
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;再根據(jù)旋轉(zhuǎn)的性質(zhì)得到∠ABA′=∠CBC′,BC′=BC=1,根據(jù)相似三角形的判定方法易得Rt△BEH∽Rt△BFC,則HE:FC=BH:BC,即1:FC=
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:1,可求出FC,然后利用三角形的面積公式計算△BCC′的面積.
解答:作HE⊥AB于E,CF⊥BC′于F,如圖,
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設(shè)BH=x,則DH=x,
∵矩形ABCD中,AB=2,AD=1,
∴AE=DH=x,HE=AD=1,
∴BE=AB-AE=2-x,
在Rt△BEH中,∵BE
2+HE
2=BH
2,即(2-x)
2+1
2=x
2,
∴x=
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,即BH=
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,
∵矩形ABCD繞點B按順時針方向旋轉(zhuǎn)后得到矩形A′BC′D,
∴∠ABA′=∠CBC′,BC′=BC=1,
∴Rt△BEH∽Rt△BFC,
∴HE:FC=BH:BC,即1:FC=
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:1,
∴FC=
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,
∴△BCC′的面積=
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BC′•FC=
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×1×
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=
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.
故答案為
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.
點評:本題考查了旋轉(zhuǎn)的性質(zhì):旋轉(zhuǎn)前后兩圖形全等;對應點與旋轉(zhuǎn)中心的連線段的夾角等于旋轉(zhuǎn)角;對應點到旋轉(zhuǎn)中心的距離相等.也考查了勾股定理、矩形的性質(zhì)以及相似三角形的判定與性質(zhì).