Question

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

Answer 1

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．

Answer 2

26-3T=t
26=4t
t=6.5

Answer 3

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Answer 4

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。