【答案】
分析:本題的關(guān)鍵是求出重合部分的面積與x的函數(shù)關(guān)系式,可分三種情況進(jìn)行討論:
①當(dāng)B′在△ABC內(nèi)部時(shí),即當(dāng)0≤x≤
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時(shí),此時(shí)重合部分是平行四邊形,以MB′為底,以AM•sin45°為高.據(jù)此可求出此時(shí)y,x的函數(shù)關(guān)系式.
②當(dāng)A′,B′都在△ABC外部時(shí),即當(dāng)
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≤x≤
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時(shí),此時(shí)重合部分是個(gè)六邊形,可用△A′B′C′的面積-△A′ME的面積-△B′ND的面積-△GC′F的面積來求解.
③當(dāng)A′在△ABC內(nèi)部時(shí),即當(dāng)
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≤x≤
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時(shí),此時(shí)重合部分是平行四邊形,求法和①相同.
根據(jù)上述三種情況可得出三個(gè)不同的函數(shù)解析式,根據(jù)函數(shù)式即可求出當(dāng)y為
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平方厘米時(shí)x的值,以及y的最大值及對應(yīng)的x的值.
解答:
解:(1)①如圖1,當(dāng)B′在△ABC內(nèi)時(shí),重疊部分是平行四邊形,
由題意得:
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x=
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,
解得x=
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.
②如圖3,當(dāng)A′在△ABC內(nèi)時(shí),重疊部分是平行四邊形,
由題意得:A′N=6
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-x,y=(6
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-x)×
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=
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,
解得x=6
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-
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,
綜上所述,當(dāng)△A′B′C′與△ABC重疊部分面積為
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平方厘米時(shí),△A′B′C′移動(dòng)的時(shí)間為
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或(6
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-
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)秒.
(2)①如圖1,當(dāng)0≤x≤
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時(shí)y=
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x,
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②如圖2,當(dāng)
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≤x≤
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時(shí),如圖,△DB′N,△A′ME,△C′FG是等腰直角三角形,
由B′N=x-2
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,GF=MN=
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,A′M=4
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-x,
則y=
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×4×4-
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×2×2-
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×(x-2
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)
2-
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(4
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-x)
2
即y=-
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x
2+3
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x-4;
③如圖3,當(dāng)
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≤x≤
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時(shí),A′M=x-4
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,A′N=2
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-x+4
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=6
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-x,
y=A′N×
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=-
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x+12.
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(3)①當(dāng)0≤x≤
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時(shí),最大值=4,
②當(dāng)
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≤x≤
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時(shí),最大值=5,
③當(dāng)
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≤x≤
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時(shí),最大值=4,
所以,△A′B′C′與△ABC重疊部分面積的最大值為5.
點(diǎn)評:本題主要考查了等腰直角三角形的性質(zhì)、圖形面積的求法以及一次函數(shù)與二次函數(shù)的綜合應(yīng)用等知識點(diǎn).