【答案】
分析:(1)把BA,AD,DC它們的和求出來(lái)再除以速度每秒5個(gè)單位就可以求出t的值,然后也可以求出BQ的長(zhǎng);
(2)如圖1,若PQ∥DC,又AD∥BC,則四邊形PQCD為平行四邊形,從而PD=QC,用t分別表示QC,BA,AP,然后就可以得出關(guān)于t的方程,解方程就可以求出t;
(3)①當(dāng)點(diǎn)E在CD上運(yùn)動(dòng)時(shí),如圖2分別過(guò)點(diǎn)A、D作AF⊥BC于點(diǎn)F,DH⊥BC于點(diǎn)H,則四邊形ADHF為矩形,然后根據(jù)已知條件可以證明△ABF≌△DCH,根據(jù)全等三角形的性質(zhì)可以得到FH=AD=75,BF=CH=30,DH=AF=40,再求出tanC=
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,在Rt△CQE中,QE,QC就可以用t表示,這樣射線QK掃過(guò)梯形ABCD的面積為S也可以用t表示了;
②當(dāng)點(diǎn)E在DA上運(yùn)動(dòng)時(shí),如圖1.過(guò)點(diǎn)D作DH⊥BC于點(diǎn)H,由①知DH=40,CH=30,又QC=3t,從而ED=QH=QC-CH=3t-30,現(xiàn)在的射線QK掃過(guò)梯形ABCD的面積S就是梯形QCDE,可以用t表示了.
(4)△PQE能成為直角三角形.
①當(dāng)點(diǎn)P在BA(包括點(diǎn)A)上,即0<t≤10時(shí),如圖2.過(guò)點(diǎn)P作PG⊥BC于點(diǎn)G,則PG=PB•sinB=4t,又有QE=4t=PG,易得四邊形PGQE為矩形,此時(shí)△PQE總能成為直角三角形
②當(dāng)點(diǎn)P、E都在AD(不包括點(diǎn)A但包括點(diǎn)D)上,即10<t≤25時(shí),如圖1.由QK⊥BC和AD∥BC可知,此時(shí),△PQE為直角三角形,但點(diǎn)P、E不能重合,即5t-50+3t-30≠75,解得t≠
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.
③當(dāng)點(diǎn)P在DC上(不包括點(diǎn)D但包括點(diǎn)C),即25<t≤35時(shí),如圖3.由ED>25×3-30=45,
可知,點(diǎn)P在以QE=40為直徑的圓的外部,故∠EPQ不會(huì)是直角.由∠PEQ<∠DEQ,可知∠PEQ一定是銳角.對(duì)于∠PQE,
∠PQE≤∠CQE,只有當(dāng)點(diǎn)P與C重合,即t=35時(shí),如圖4,∠PQE=90°,△PQE為直角三角形.
解答:
解:(1)t=(50+75+50)÷5=35(秒)時(shí),點(diǎn)P到達(dá)終點(diǎn)C.(1分)
此時(shí),QC=35×3=105,
∴BQ的長(zhǎng)為135-105=30.(2分)
(2)如圖1,若PQ∥DC,
又∵AD∥BC,
∴四邊形PQCD為平行四邊形,
∴PD=QC,
由QC=3t,BA+AP=5t
得50+75-5t=3t,
解得t=
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.
經(jīng)檢驗(yàn),當(dāng)t=
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時(shí),有PQ∥DC.(4分)
(3)①當(dāng)點(diǎn)E在CD上運(yùn)動(dòng)時(shí),如圖2.分別過(guò)點(diǎn)A、D
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作AF⊥BC于點(diǎn)F,DH⊥BC于點(diǎn)H,則四邊形
ADHF為矩形,且△ABF≌△DCH,從而
FH=AD=75,于是BF=CH=30.
∴DH=AF=40.
又∵QC=3t,
從而QE=QC•tanC=3t•
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=4t.
(注:用相似三角形求解亦可)
∴S=S
△QCE=
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QE•QC=6t
2;(6分)
②當(dāng)點(diǎn)E在DA上運(yùn)動(dòng)時(shí),如圖1.過(guò)點(diǎn)D作DH⊥BC于點(diǎn)H,由①知DH=40,CH=30,又QC=3t,從而ED=QH=QC-CH=3t-30.
∴S=S
梯形QCDE=
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(ED+QC)DH=120t-600.(8分)
(4)△PQE能成為直角三角形.(9分)
當(dāng)△PQE為直角三角形時(shí),t的取值范圍是0<t≤25且t≠
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或t=35.(12分)根據(jù)全等三角形的性質(zhì)
(注:(4)問(wèn)中沒(méi)有答出t≠
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或t=35者各扣(1),其余寫(xiě)法酌情給分)
下面是第(4)問(wèn)的解法,僅供教師參考:
①當(dāng)點(diǎn)P在BA(包括點(diǎn)A)上,即0<t≤10時(shí),如圖2.
過(guò)點(diǎn)P作PG⊥BC于點(diǎn)G,則PG=PB•sinB=4t,
又有QE=4t=PG,易得四邊形PGQE為矩形,此時(shí)△PQE總能成為直角三角形.
②當(dāng)點(diǎn)P、E都在AD(不包括點(diǎn)A但包括點(diǎn)D)上,即10<t≤25時(shí),如圖1.
由QK⊥BC和AD∥BC可知,此時(shí),△PQE為直角三角形,但點(diǎn)P、E不能重合,
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即5t-50+3t-30≠75,解得t≠
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.
③當(dāng)點(diǎn)P在DC上(不包括點(diǎn)D但包括點(diǎn)C),
即25<t≤35時(shí),如圖3.由ED>25×3-30=45,
可知,點(diǎn)P在以QE=40為直徑的圓的外部,故
∠EPQ不會(huì)是直角.
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由∠PEQ<∠DEQ,可知∠PEQ一定是銳角.
對(duì)于∠PQE,∠PQE≤∠CQE,只有當(dāng)點(diǎn)P與C
重合,即t=35時(shí),如圖4,∠PQE=90°,△PQE
為直角三角形.
綜上所述,當(dāng)△PQE為直角三角形時(shí),t的取值范圍是0<t≤25且t≠
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或t=35.
點(diǎn)評(píng):此題綜合性很強(qiáng),把圖形的變換放在梯形的背景中,利用等腰梯形的性質(zhì)結(jié)合已知條件探究圖形的變換,根據(jù)變換的圖形的性質(zhì)求出運(yùn)動(dòng)時(shí)間.