【答案】
分析:(1)根據(jù)勾股定理求出即可;
(2)要使△PCQ與△ACB相似,必須有∠PQC=∠B或∠PQC=∠A成立.當(dāng)∠PQC=∠A時(shí),△PCQ∽△BCA,得出
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,代入求出即可;當(dāng)∠PQC=∠B時(shí),△PCQ∽△ACB,得出
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,代入求出即可;
(3)分為兩種情況:畫出圖形,當(dāng)0<t<5時(shí),過(guò)點(diǎn)E作HE⊥CE交AC于H,求出∠HEP=∠CEQ,∠QCE=∠PCE=45°,PE=QE,證△QCE≌△PHE,推出QC=PH,根據(jù)勾股定理求出即可;當(dāng)t≥5時(shí),過(guò)點(diǎn)E作ME⊥CE交AC于M,同法可證△QCE≌△PME,根據(jù)勾股定理求出即可.
解答:解:(1)在Rt△ABC中,∠ACB=90°,AB=10cm,BC=5cm,由勾股定理得:AB=
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=5
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(cm)
(2)如圖1,由題意可知:PC=2t,QB=t,QC=5-t.
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∵∠PCQ=∠ACB,
∴要使△PCQ與△ACB相似,必須有∠PQC=∠B或∠PQC=∠A成立.
當(dāng)∠PQC=∠A時(shí),△PCQ∽△BCA,
由
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可得
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,
解得:t=1,
當(dāng)∠PQC=∠B時(shí),△PCQ∽△ACB,
由
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可得
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,
解得
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,
∴當(dāng)t=1或
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秒時(shí),△PCQ與△ACB相似;
(3)當(dāng)0<t<5時(shí),如圖2,
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過(guò)點(diǎn)E作HE⊥CE交AC于H,則∠HEP+∠PEC=90°,
∵∠ACB=90°,
∴PQ為△PCQ的外接圓的直徑,
∴∠QEP=90°,即∠QEC+∠PEC=90°,
∴∠HEP=∠CEQ,
又∵CE平分∠ACB且∠ACB=90°,
∴∠QCE=∠PCE=45°,
∴
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,
∴PE=QE,
∴∠QCE=∠PHE=45°,
∵在△QCE和△PHE中
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∴△QCE≌△PHE(AAS)
∴QC=PH,
在Rt△HEC中,EC
2+EH
2=HC
2,EC=EH,
即2EC
2=(CP+CQ)
2∴
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;
當(dāng)t≥5時(shí),如圖3,
過(guò)點(diǎn)E作ME⊥CE交AC于M,同法可證△QCE≌△PME,
∴
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,
綜上所述,當(dāng)0<t<5時(shí),
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;當(dāng)t≥5時(shí),
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.
故答案為:5
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;.
點(diǎn)評(píng):本題考查了等腰直角三角形,三角形的外接圓,相似三角形的性質(zhì)和判定的應(yīng)用,主要考查學(xué)生綜合運(yùn)用性質(zhì)進(jìn)行計(jì)算的能力,題目綜合性比較強(qiáng),難度偏大.