【答案】
分析:(1)由于三角形AHG和ACB相似,可通過相似比求出HG的值,然后根據(jù)三角形的面積計(jì)算公式即可求出三角形AHG的面積.
(2)①首先四邊形CDH′H是個(gè)矩形,如果使四邊形CDH′H成為正方形,那么需滿足的條件是CD=DH′,可先根據(jù)AH:AC的值,求出HC的長(zhǎng)即H′D的長(zhǎng),然后除以梯形的速度即可求出t的值.
②要分三種情況進(jìn)行討論:
一:當(dāng)E在三角形ABC內(nèi)部時(shí),即當(dāng)0≤t≤4時(shí),重合部分是整個(gè)直角梯形,因此可通過計(jì)算直角梯形的面積得出重合部分的面積.
二:當(dāng)E在三角形ABC外部,且H′在G點(diǎn)左側(cè)或G點(diǎn)上時(shí),即當(dāng)4<t≤5
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時(shí),重合部分是直角梯形,其面積可用:四邊形CBGH的面積一矩形CDH′H的面積來求得.
三:當(dāng)H′在G點(diǎn)右側(cè)一直到D與B重合的過程中,即當(dāng)5
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<t≤8時(shí),重合部分是個(gè)直角三角形.可通過計(jì)算這個(gè)直角三角形的面積來得出關(guān)于S,t的函數(shù)關(guān)系式.
解答:解:(1)∵AH:AC=2:3,AC=6
∴AH=
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AC=
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×6=4
又∵HF∥DE,
∴HG∥CB,
∴△AHG∽△ACB
∴
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=
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,即
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=
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,
∴HG=
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∴S
△AHG=
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AH•HG=
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×4×
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=
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.
(2)①能為正方形
∵HH′∥CD,HC∥H′D,
∴四邊形CDH′H為平行四邊形
又∠C=90°,
∴四邊形CDH′H為矩形
又CH=AC-AH=6-4=2
∴當(dāng)CD=CH=2時(shí),四邊形CDH′H為正方形
此時(shí)可得t=2秒時(shí),四邊形CDH′H為正方形.
②(Ⅰ)∵∠DEF=∠ABC,
∴EF∥AB
∴當(dāng)t=4秒時(shí),直角梯形的腰EF與BA重合.
當(dāng)0≤t≤4時(shí),重疊部分的面積為直角梯形DEFH′的面積.
過F作FM⊥DE于M,
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=tan∠DEF=tan∠ABC=
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=
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=
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∴ME=
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FM=
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×2=
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,HF=DM=DE-ME=4-
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=
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∴直角梯形DEFH′的面積為
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(4+
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)×2=
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∴y=
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.
(Ⅱ)∵當(dāng)4<t≤5
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時(shí),重疊部分的面積為四邊形CBGH的面積一矩形CDH′H的面積.
而S
邊形CBGH=S
△ABC-S
△AHG=
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×8×6-
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=
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S
矩形CDH′H?=2t
∴y=
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-2t.
(Ⅲ)當(dāng)5
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<t≤8時(shí),如圖,設(shè)H′D交AB于P,
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BD=8-t
又
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=tan∠ABC=
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∴PD=
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DB=
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(8-t)
∴重疊部分的面積y=S??
△PDB=
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PD•DB
=
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•
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(8-t)(8-t)
=
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(8-t)
2=
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t
2-6t+24.
∴重疊部分面積y與t的函數(shù)關(guān)系式:
y=
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.
點(diǎn)評(píng):本題著重考查了圖形平移變換、三角形相似以及二次函數(shù)的綜合應(yīng)用等重要知識(shí)點(diǎn),
要注意的是(2)中不確定直角梯形的位置時(shí),要根據(jù)不同的情況進(jìn)行分類討論,不要漏解.