(1)解:∵A(-3,4),
∴AH=3,OH=4,
由勾股定理得:AO=
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=5,
答:OA的長是5.
(2)解:∵菱形OABC,
∴OA=OC=BC=AB=5,
5-3=2,
∴B(2,4),C(5,0),
設(shè)直線AC的解析式是y=kx+b,
把A(-3,4),C(5,0)代入得:
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,
解得:
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,
∴直線AC的解析式為
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,
當x=0時,y=2.5
∴M(0,2.5),
答:直線AC的解析式是
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,點M的坐標是(0,2.5).
(3)①解:過M作MN⊥BC于N,
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∵菱形OABC,
∴∠BAC=∠OCA,
∵MO⊥CO,MN⊥BC,
∴OM=MN,
當0≤t<2.5時,P在AB上,MH=4-2.5=
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,
S=
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×BP×MH=
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×(5-2t)×
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=-
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t+
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,
∴
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,
當2.5<t≤5時,P在BC上,S=
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×PB×MN=
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×(2t-5)×
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=
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t-
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,
∴
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,
答:S與t的函數(shù)關(guān)系式是
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(0≤t<2.5)或
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(2.5<t≤5).
②解:當P在AB上時,高MH一定,只有BP取最大值即可,即P與A重合,S最大是
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×5×
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=
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,
同理在BC上時,P與C重合時,S最大是
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×5×
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=
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,
∴S的最大值是
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,
答:S的最大值是
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.
分析:(1)根據(jù)A的坐標求出AH、OH,根據(jù)勾股定理求出即可;
(2)根據(jù)菱形性質(zhì)求出B、C的坐標,設(shè)直線AC的解析式是y=kx+b,把A(-3,4),C(5,0)代入得到方程組,求出即可;
(3)①過M作MN⊥BC于N,根據(jù)角平分線性質(zhì)求出MN,P在AB上,根據(jù)三角形面積公式求出即可;P在BC上,根據(jù)三角形面積公式求出即可;②求出P在AB的最大值和P在BC上的最大值比較即可得到答案.
點評:本題主要考查對勾股定理,三角形的面積,菱形的性質(zhì),角平分線性質(zhì),解二元一次方程組,用待定系數(shù)法求一次函數(shù)的解析式等知識點的理解和掌握,綜合運用這些性質(zhì)進行計算是解此題的關(guān)鍵.