【答案】
分析:(1)根據(jù)矩形的性質(zhì)可知OB=AC,根據(jù)直角三角形的性質(zhì)可知
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=100,即OB=AC=100.判定Rt△OPT∽R(shí)t△OBC則可得出
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,即可得出函數(shù)解析式,根據(jù)P的運(yùn)動(dòng)情況求出t的取值范圍即可.
(2)當(dāng)O點(diǎn)關(guān)于直線AP的對(duì)稱點(diǎn)O′恰好在對(duì)角線OB上時(shí),A,T,P三點(diǎn)在一條直線上.判定Rt△AOP∽R(shí)t△OCB,則可得出
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,點(diǎn)P的坐標(biāo)為(45,0).列出AP的函數(shù)解析式將點(diǎn)A(0,60)和點(diǎn)P(45,0)代入解析式,解出即可.
(3)由(2)知,當(dāng)t=9時(shí),A,T,P三點(diǎn)在一條直線上,此時(shí)點(diǎn)A,T,P不構(gòu)成三角形.所以分兩種情況:1、當(dāng)0<t<9時(shí),列出方程求解看有無實(shí)數(shù)根即可.2、當(dāng)9<t≤16時(shí),根據(jù)圖(3)列出方程求解看有無實(shí)數(shù)根即可.
解答:解:(1)在矩形OABC中,
因?yàn)镺A=60,OC=80,
所以O(shè)B=AC=
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=100.
因?yàn)镻T⊥OB,
所以Rt△OPT∽R(shí)t△OBC.
因?yàn)?img src="http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231743464078357/SYS201310212317434640783028_DA/4.png">,即
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,
所以y=PT=3t.
當(dāng)點(diǎn)P運(yùn)動(dòng)到C點(diǎn)時(shí)即停止運(yùn)動(dòng),此時(shí)t的最大值為
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.
(2)(如圖2)當(dāng)O點(diǎn)關(guān)于直線AP的對(duì)稱點(diǎn)O'恰好在對(duì)角線OB上時(shí),A,T,P三點(diǎn)在
一條直線上.
所以AP⊥OB,∠1=∠2.
所以Rt△AOP∽R(shí)t△OCB,
所以
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.
所以O(shè)P=45.
所以點(diǎn)P的坐標(biāo)為(45,0).
設(shè)直線AP的函數(shù)解析式為y=kx+b.
將點(diǎn)A(0,60)和點(diǎn)P(45,0)代入解析式,
得
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,
解這個(gè)方程組得
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.
所以此時(shí)直線AP的函數(shù)解析式是
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.
(3)由(2)知,當(dāng)
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時(shí),A,T,P三點(diǎn)在一條直線上,此時(shí)點(diǎn)A,T,P不構(gòu)
成三角形.
所以分兩種情況:
1、當(dāng)0<t<9時(shí),點(diǎn)T位于△AOP的內(nèi)部(如圖1),過A點(diǎn)作AE⊥OB,垂足為點(diǎn)E,
由AO•AB=OB•AE可得AE=48.
所以S
△APT=S
△AOP-S
△ATO-S
△OTP=
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×60×5t-
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×4t×48-
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×4t×3t=-6t
2+54t.
若S
△APT=
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S
矩形OABC,
則-6t
2+54t=1200,即t
2-9t+200=0.
此時(shí),△=(-9)
2-4×1×200<0,
所以該方程無實(shí)數(shù)根.
所以當(dāng)0<t<9時(shí),以A,P,T為頂點(diǎn)的△APT的面積不能達(dá)到矩形OABC面積的
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.
2、當(dāng)9<t≤16時(shí),點(diǎn)T位于△AOP的外部.
此時(shí)S
△APT=S
△ATO+S
△OTP-S
△AOP=6t
2-54t.
若S
△APT=
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S
矩OABC,
則6t
2-54t=1200,即t
2-9t-200=0.
解得
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,
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(舍去).
由于881>625=25
2,
所以
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.
而此時(shí)9<t≤16,
所以
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也不符合題意,應(yīng)舍去.
所以當(dāng)9<t≤16時(shí),以A,P,T為頂點(diǎn)的△APT的面積也不能達(dá)到矩形OABC面積的
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.
綜上所述,以A,P,T為頂點(diǎn)的△APT的面積不能達(dá)到矩形OABC面積的
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.
點(diǎn)評(píng):本題要注意利用待定系數(shù)法求一次函數(shù)解析式的方法,列出方程,得出未知數(shù).同學(xué)們需熟練掌握.