【答案】
分析:(1)根據(jù)題意,AD=8,B點(diǎn)在y=
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x上,則y=6,即B點(diǎn)坐標(biāo)為(8,6),AB=6,可求得矩形的周長(zhǎng)為28.
(2)由(1)可知AB+BC=14,點(diǎn)P的速度為每秒1個(gè)單位.可求得D點(diǎn)坐標(biāo)為(4,3),點(diǎn)P坐標(biāo)為(12,8);
(3)設(shè)線段所在直線為y=kx+b,把點(diǎn)(8,0),(12,8),代入解析式利用待定系數(shù)法求解得:函數(shù)關(guān)系式為y=2x-16;
(4)①當(dāng)點(diǎn)P在AB邊運(yùn)動(dòng)時(shí),即0≤t≤6,點(diǎn)D的坐標(biāo)表示為
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,點(diǎn)P的坐標(biāo)為(8+
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t,
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),根據(jù)相似三角形的相似比,解得t=6;當(dāng)點(diǎn)P與點(diǎn)B重合,此時(shí)矩形PEOF與矩形BADC是位似形.則根據(jù)相似三角形的相似比可解得t=20,因?yàn)?0>6,所以此時(shí)點(diǎn)P不在AB邊上,舍去.②當(dāng)點(diǎn)P在BC邊運(yùn)動(dòng)時(shí),即6≤t≤14,則點(diǎn)D的坐標(biāo)為
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,點(diǎn)P的坐標(biāo)為(14-
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t,
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t+6)
同樣利用三角形的相似比可求得
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,因?yàn)?img src="http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192240607009378/SYS201311011922406070093023_DA/8.png">,此時(shí)點(diǎn)P不在BC邊上,舍去.
綜合可知,當(dāng)t=6時(shí),點(diǎn)P到達(dá)點(diǎn)B,矩形PEOF與矩形BADC是位似形.
解答:解:(1)AD=8,B點(diǎn)在y=
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x上,則y=6,B點(diǎn)坐標(biāo)為(8,6),AB=6,矩形的周長(zhǎng)為28.
(2)由(1)可知AB+BC=14,P點(diǎn)走過AB、BC的時(shí)間為14秒,因此點(diǎn)P的速度為每秒1個(gè)單位.
∵矩形沿DB方向以每秒1個(gè)單位長(zhǎng)運(yùn)動(dòng),出發(fā)5秒后,OD=5,此時(shí)D點(diǎn)坐標(biāo)為(4,3),
同時(shí)點(diǎn)P沿AB方向運(yùn)動(dòng)了5個(gè)單位,則點(diǎn)P坐標(biāo)為(12,8).
(3)點(diǎn)P運(yùn)動(dòng)前的位置為(8,0),5秒后運(yùn)動(dòng)到(12,8),
已知它運(yùn)動(dòng)路線是一條線段,設(shè)線段所在直線為y=kx+b
∴
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,
解得:
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函數(shù)關(guān)系式為y=2x-16.
(4)方法一:①當(dāng)點(diǎn)P在AB邊運(yùn)動(dòng)時(shí),即0≤t≤6,
點(diǎn)D的坐標(biāo)為
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,
∴點(diǎn)P的坐標(biāo)為(8+
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t,
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).
若
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,則
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,解得t=6.
當(dāng)t=6時(shí),點(diǎn)P與點(diǎn)B重合,此時(shí)矩形PEOF與矩形BADC是位似形.
若
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,則
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,
解得t=20.
因?yàn)?0>6,所以此時(shí)點(diǎn)P不在AB邊上,舍去.
②當(dāng)點(diǎn)P在BC邊運(yùn)動(dòng)時(shí),即6≤t≤14,
點(diǎn)D的坐標(biāo)為
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,
∴點(diǎn)P的坐標(biāo)為(14-
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t,
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t+6).
若
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,則
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,
解得t=6.
因?yàn)?img src="http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192240607009378/SYS201311011922406070093023_DA/24.png"><14,此時(shí)點(diǎn)P不在BC邊上,舍去.
綜上,當(dāng)t=6時(shí),點(diǎn)P到達(dá)點(diǎn)B,矩形PEOF與矩形BADC是位似形.
方法二:當(dāng)點(diǎn)P在AB上沒有到達(dá)點(diǎn)B時(shí),
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,
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更不能等于
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.
則點(diǎn)P在AB上沒到達(dá)點(diǎn)B時(shí),兩個(gè)矩形不能構(gòu)成相似形
當(dāng)點(diǎn)P到達(dá)點(diǎn)B時(shí),矩形PEOF與矩形BADC是位似形,此時(shí)t=6.
當(dāng)點(diǎn)P越過點(diǎn)B在BC上時(shí),
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若
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時(shí),由點(diǎn)P在BC上時(shí),坐標(biāo)為:
(14-
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t,
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t+6),(6≤t≤14),
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,
解得t=
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,但
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<14.
因此當(dāng)P在BC上(不包括B點(diǎn))時(shí),矩形PEOF與矩形BCDA不相似.
綜上,當(dāng)t=6時(shí),點(diǎn)P到達(dá)點(diǎn)B,矩形PEOF與矩形BADC是位似形.
點(diǎn)評(píng):主要考查了函數(shù)和幾何圖形的綜合運(yùn)用.解題的關(guān)鍵是會(huì)靈活的運(yùn)用函數(shù)圖象的性質(zhì)和交點(diǎn)的意義求出相應(yīng)的線段的長(zhǎng)度或表示線段的長(zhǎng)度,再結(jié)合具體圖形的性質(zhì)求解.