∵QH=HD=DQ=(12﹣2t)=6﹣t, 由AH=BP, ∴6﹣t+2t=3t 解得:t=3秒; 当PQ=DQ时,
QH=AH﹣AQ=BP﹣AQ=3t﹣2t=t,DQ=12﹣2t,
∵DQ2=PQ2=t2+102, ∴(12﹣2t)2=102+t2, 整理得:3t2﹣48t+44=0, 解得:t=∵0<t<5, ∴t=
当DQ=PD时,
(秒),
(秒);
DH=AD﹣AH=AD﹣BP=12﹣3t,
∵DQ=PD=PH+HD=10+(12﹣3t) ∴(12﹣2t)2=102+(12﹣3t)2 即5t2﹣24t+100=0, ∵△<0, ∴方程无实根,
综上可知,当t=3秒或t=
秒时,△PQD是等腰三角形.
2
2
2
2
2
2
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