题目内容
如图1,梯形
中,
∥
,
,
.一个动点
从点
出发,以每秒
个单位长度的速度沿线段
方向运动,过点
作
,交折线段
于点
,以
为边向右作正方形
,点
在射线
上,当
点到达
点时,运动结束.设点
的运动时间为
秒(
).
(1)当正方形
的边
恰好经过点
时,求运动时间
的值;
(2)在整个运动过程中,设正方形
与△
的重合部分面积为
,请直接写出
与
之间的函数关系式和相应的自变量
的取值范围;
(3)如图2,当点
在线段
上运动时,线段
与对角线
交于点
,将△
沿
翻折,得到△
,连接
.是否存在这样的
,使△
是等腰三角形?若存在,求出对应的
的值;若不存在,请说明理由.
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(1)当正方形
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(2)在整个运动过程中,设正方形
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(3)如图2,当点
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(1)当t=4时,正方形PQMN的边MN恰好过点D
(2)
(3)当
时,∆PEF是等腰三角形
(2)
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(3)当
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试题分析:(1)作AG⊥BC,DH⊥BC,垂足分别为G、H,可以得出四边形AGHD为矩形,根据矩形的性质及相关条件可以得出△ABG≌△DCH,可以求出BG=CH的值,再由勾股定理就可以求出AG=DH的值,就可以求出BP的值,即可以求出结论t的值;
(2)运用求分段函数的方法,分四种情况,当0<t≤3,当3<t≤4,4<t≤7,7<t≤8时,运用梯形的面积公式和三角形的面积公式就可以求出S的值;
(3)先由条件可以求出EF=EQ=PQ-EP=4-
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试题解析:(1)如图2,作AG⊥BC于G,DH⊥BC于H,则四边形AGHD是矩形。
∵梯形ABCD中,AB=AD=DC=5,
∴∆ABG≌∆DCH,
∴
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∴当正方形PQMN的边MN恰好过点D时,点M与点D重合,此时MQ=4,
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∴当t=4时,正方形PQMN的边MN恰好过点D。
(2)
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如图1,当0<t≤3时,BP=t,∵
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∴
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如图3,当3<t≤4时,BP="t,"
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∴
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如图4,当4<t≤7时,BP="t,"
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∴
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如图5,当7<t≤8时,BP="t,"
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∴
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∴
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(3)∵
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∴
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∴
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由(1)可知
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如图6,当EF=EP时,
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∴
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如图7,当FE=FP时,作FR⊥EP于R,∴
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∴
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∴
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如图8,当PE=PF时,作PS⊥EF于S,∴
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∴
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∴
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∴当
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