如图,直角梯形ABCD中,AD∥BC,∠B=90°,AD=24cm,BC=26cm,动点P从A开

如图,直角梯形ABCD中,ADBC,∠B=90°,AD=24cmBC=26cm,动点PA开始沿AD边向D1cm/s的速度运动,动点Q从点C开始沿CB3cm/s的速度向点B运动.PQ同时出发,当其中一点到达顶点时,另一点也随之停止运动,设运动时间为tst为何值时.

    1)四边形PQCD是平行四边形.(2)当t为何值时,四边形PQCD为等腰梯形.

答案

解:(1)∵PDCQ,∴当PD=CQ时,四边形PQCD是平行四边形.

    PD=24-tCQ=3t

    24-t=3t,解得t=6

    t=6时,四边形PQCD是平行四边形.

    2)过点DDEBC,则CE=BC-AD=2cm

    CQ-PD=4时,四边形PQCD是等腰梯形.

    3t-24-t=4.∴t=7

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