在直角梯形ABCD中,AD‖BC,角B=90°,AD=24CM,AB=8CM,BC=26CM,动点P从A直角梯形

在直角梯形ABCD中,AD//BC,角B=90度,AB=8cm,AD=24cm,BC=26cm,动点P从点A开始沿AD向点D以1cm/s的速度移动,动点Q从点C开始沿CB... 在直角梯形ABCD中,AD//BC,角B=90度,AB=8cm,AD=24cm,BC=26cm,动点P从点A开始沿AD向点D以1cm/s的速度移动,动点Q从点C开始沿CB向B点以3cm/s的速度,如果P,Q分别从A,C同时出发,设移动时间为t(s).

当T为何值时,PQCD为直角梯形
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鑫哥nice
2012-07-01
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解:(1)设经过xs,四边形PQCD为平行四边形
即PD=CQ
所以24-x=3x,得x=6.(3分)
(2)设经过ys,四边形PQBA为矩形,
即AP=BQ,所以y=26-3y,得y=13 2 .(3分)
(3)设经过ts,四边形PQCD是等腰梯形.过Q点作QE⊥AD,过D点作DF⊥BC,
∵四边形PQCD是等腰梯形,
∴PQ=DC.
又∵AD∥BC,∠B=90°,
∴AB=QE=DF.
∴△EQP≌△FDC.
∴FC=EP=BC-AD=26-24=2.
又∵AE=BQ=26-3t EP=t-AE,
∴EP=AP-AE=t-(26-3t)=2.
得:t=7.
∴经过7s,四边形PQCD是等腰梯形.
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匿名用户
2010-08-09
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26-3T=t
26=4t
t=6.5
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haiyuh8hf
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因为要使四边形PQCD为直角梯形,则当连结PQ时,PQ要分别垂直于AD与BC,所以AP=BQ.
所以1t=26-3t
t=6.5
检验:当t=6.5时,
1t=6.5<24,3t=19.5<26
所以当t=6.5时,四边形PQCD为直角梯形。
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