【答案】
分析:(1)由點(diǎn)C的坐標(biāo)為(3,4),利用勾股定理得到OC的長(zhǎng),利用菱形的性質(zhì)得到OA的長(zhǎng),即可確定點(diǎn)A的坐標(biāo);
(2)先利用待定系數(shù)法確定直線AC的解析式,得到DO=10,利用勾股定理得到DC、DA的長(zhǎng).然后分類討論:①當(dāng)點(diǎn)P在線段AB上,Q在線段CD上;②當(dāng)點(diǎn)P在線段BC上,Q在線段CD上;③當(dāng)點(diǎn)P在線段BC上,Q在線段CA上;過點(diǎn)P作PH⊥AD于H,利用三角形相似比表示出PH,在根據(jù)三角形的面積公式分別表示出S;
(3)分類討論:當(dāng)0<t≤2.5;當(dāng)2.5<t<3;當(dāng)3<t<5,由(2)知道PH,然后分別表示出對(duì)應(yīng)的QH,再根據(jù)正切的定義得到PH:QH=
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,解關(guān)于t的方程,得到滿足條件的t的值即可.
解答:解:(1)∵點(diǎn)C的坐標(biāo)為(3,4),
∴OC=
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=5,
又∵四邊形OABC為菱形,
∴OA=OC=5,
∴點(diǎn)A的坐標(biāo)為(5,0);
(2)設(shè)直線AC的解析式為y=kx+b,
把A(5,0)和點(diǎn)C(3,4)分別代入得,5k+b=0,3k+b=4,解得k=-2,b=10,
∴OD=10,
∴AD=
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=5
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,
延長(zhǎng)BC交OD于M,則CM=3,OM=4,
∴DM=10-4=6,
DC=
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=3
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;
①當(dāng)點(diǎn)P在線段AB上,Q在線段CD上,
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PA=2t,DQ=
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t,CQ=3
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-
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t,
過點(diǎn)P作PH⊥AD于H,如圖,
∵四邊形OABC為菱形,
∴∠PAH=∠OAD,
∴Rt△PHA∽R(shí)t△DOA,
∴PH:OD=AH:OA=PA:AD,即PH:10=AH:5=2t:5
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,
∴PH=
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t,AH=
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t,
∴S=
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QC•PH=
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•(3
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-
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t)•
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t=-2t2+6t(0<t≤2.5)
②當(dāng)點(diǎn)P在線段BC上,Q在線段CD上,
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PC=10-2t,
過點(diǎn)P作PH⊥AD于H,如圖,
易證Rt△PCH∽R(shí)t△DAO,
∴PH:OD=CH:OA=PC:AD,即PH:10=CH:5=(10-2t):5
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,
∴PH=4
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-
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t,CH=2
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-
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t,
∴S=
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PH•CQ=
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•(4
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-
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t)•(3
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-
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t)=2t
2-16t+30(2.5<t<3);
③當(dāng)點(diǎn)P在線段BC上,Q在線段CA上,
過點(diǎn)P作PH⊥AD于H,如圖,
由②知PH=4
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-
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t,
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∴S=
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PH•CQ=
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•(4
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-
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t)•(
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t-3
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)=-2t
2+16t-30(3<t<5);
(3)當(dāng)0<t≤2.5,
∵PH=
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t,AH=
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t,
∴QH=5
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-
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t-
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t=5
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-
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t,
∵tan∠PQH=
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,
∴PH:QH=
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=
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,解得t=
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;
當(dāng)2.5<t<3,
∵PH=4
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-
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t,CH=2
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-
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t,
∴QH=3
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-
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t+2
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-
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t=5
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-
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t,
∵tan∠PQH=
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,
∴PH:QH=(4
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-
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t):(5
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-
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t)=
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,解得t=
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(舍去);
當(dāng)3<t<5,
∵PH=4
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-
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t,CH=2
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-
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t,
∴QH=
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t-3
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-(2
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-
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t)=
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t-5
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,
∵tan∠PQH=
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,
∴PH:QH=(4
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-
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t):(
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t-5
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)=
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,解得t=
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;
∴點(diǎn)P在運(yùn)動(dòng)的過程中t為
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或
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時(shí),tan∠PQH=
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.
點(diǎn)評(píng):本題考查了一次函數(shù)的綜合題:利用待定系數(shù)法確定一次函數(shù)的解析式,然后根據(jù)解析式確定線段的長(zhǎng)或根據(jù)相似比求線段的長(zhǎng).也考查了菱形的性質(zhì)、三角函數(shù)的定義以及分類討論思想的運(yùn)用.