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解:(1)過D作DH⊥BC,DH與EF、BC分別相交于點G、H,
∵梯形ABCD中,∠B=90°,
∴DH∥AB,
又∵AD∥BC,
∴四邊形ABHD是矩形,
∵∠C=45°,
∴∠CDH=45°,
∴CH=DH=AB=8,
∴AD=BH=BC-CH=6.
(2)∵DH⊥EF,∠DFE=∠C=∠FDG=45°,
∴FG=DG=AE=x,
∵EG=AD=6,
∴EF=x+6,
∵PE=PF,EF∥BC,
∴∠PFE=∠PEF=∠PMN=∠PNM,
∴PM=PN,
過點P作QR⊥EF,QR與EF、MN分別相交于Q、R,
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∵∠MPN=∠EPF=90°,QR⊥MN,
∴PQ=
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EF=
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,PR=
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MN=
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,
∵QR=BE=8-x,
∴
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,
∴y關(guān)于x的函數(shù)解析式為y=-3x+10.定義域為1≤x<
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.
(3)當點P在梯形ABCD內(nèi)部時,由MN=2及(2)的結(jié)論得2=-3x+10,AE=
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,
∴
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(AD+EF)•AE=
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,
當點P在梯形ABCD外部時,由MN=2及與(2)相同的方法得:
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,AE=x=4,
∴
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(AD+EF)•AE=
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.
分析:(1)過D作DH⊥BC,DH與EF、BC分別相交于點G、H,從而判定四邊形ABHD是矩形,在RT△DHC中求出CH的長,利用AD=BH=BC-CH可得出AD的長.
(2)首先確定PM=PN,過點P作QR⊥EF,QR與EF、MN分別相交于Q、R,根據(jù)∠MPN=∠EPF=90°,QR⊥MN,可表示出PQ、PR,繼而可得出y關(guān)于x的函數(shù)解析式,也能得出定義域.
(3)①當點P在梯形ABCD內(nèi)部時,由MN=2及(2)的結(jié)論得2=-3x+10,AE=
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,可求得梯形的面積,②當點P在梯形ABCD外部時,由MN=2及與(2)相同的方法得:
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,AE=x=4,可求得梯形的面積.
點評:本題考查梯形及有實際問題列一次函數(shù)關(guān)系式的知識,屬于綜合性較強的題目,難度較大,對于此類題目要學會由小及大,將所求的問題縮小,一步一步求解.