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(1)解:
過G作GH⊥EF于H,
∵等邊三角形GEF,
∴EH=HF=1,
由勾股定理得:GH=
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=
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,
S=S
矩形ABCD-2S
△AIH=8×2
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-2×
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×1×
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=15
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,
故答案為:15
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.
(2)解:將△EFG移到四邊形ABCD的左上角(圖1),
則△AEG為△EFG無法掃到的一部分,
此時,由于AD、BC的距離為
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,△EFG的高為
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,
易得點(diǎn)E恰好是AB的中點(diǎn),
過點(diǎn)B、E分別作EJ⊥AD于J、BK⊥AD于K,作AD的中點(diǎn)H,BC的中點(diǎn)I,
∵
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,
∵AD=10,BC=8∴AH=5,BI=4,
∴AK=AH-BI=1,
∵E是線段AB的中點(diǎn),EJ⊥AD,BK⊥AD,
∴AJ=
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AK=
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,
∵∠JEG=30°,
∴JG=
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GE=1,
∴AG=AJ+JG=
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,
∴
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,
∴△EFG無法掃到的部分的總面積為
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,
∴S=
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,
答:S的值是
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.
(3)解:將△EFG移到四邊形ABCD的左下角(圖2),
則△BEG為△EFG無法掃到的一部分,
此時,由于AD、BC的距離為
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,△EFG的高為
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,
易得點(diǎn)G恰好是AB的中點(diǎn),
過點(diǎn)A、G分別作AK⊥BC于K、GJ⊥BC于J,作AD的中點(diǎn)H,BC的中點(diǎn)I
∵
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,
∵AD=2,BC=8,
∴AH=1,BI=4,
∴BK=BI-AH=3,
∵G是線段AB的中點(diǎn),AK⊥BC,GJ⊥BC,
∴BJ=
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BK=
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,
∵∠JGE=30°,
∴JE=
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GE=1,
∴BE=BJ-EJ=
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,
∴
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,
∴△EFG無法掃到的部分的總面積為
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,
∴S=
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,
答:S的值是
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.
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分析:(1)過G作GH⊥EF于H,求出等邊三角形GEF的高GH,關(guān)鍵面積公式求出即可;
(2)過點(diǎn)B、E分別作EJ⊥AD于J、BK⊥AD于K,作AD的中點(diǎn)H,BC的中點(diǎn)I,求出平行四邊形ABCD的面積和三角形AGE的面積,代入求出即可;
(3)過點(diǎn)A、G分別作AK⊥BC于K、GJ⊥BC于J,作AD的中點(diǎn)H,BC的中點(diǎn)I,求出平行四邊形ABCD和三角形BGE的面積,代入即可求出答案.
點(diǎn)評:本題主要考查對平行四邊形的性質(zhì),三角形的面積,等邊三角形的性質(zhì),勾股定理,矩形的性質(zhì),等腰梯形的性質(zhì)等知識點(diǎn)的理解和掌握,綜合運(yùn)用這些性質(zhì)進(jìn)行推理是解此題的關(guān)鍵.