解答:解:(1)當(dāng)點(diǎn)E運(yùn)動(dòng)到DB邊上時(shí),點(diǎn)H才可能落在AD邊上.
在直角△ABD中,AD=8cm,BD=4cm.
由勾股定理,得AD=4
cm.
從點(diǎn)A運(yùn)動(dòng)到點(diǎn)D用時(shí)間t=4秒,
則DE=t-4,
∵四邊形HEFG是正方形,
∴BE=HE=BD-DE=4-(t-4)=8-t.
∵△DEH∽△DBA,
∴
=
,
=
,
解得t=
秒
(2)設(shè)HG與AD相交于點(diǎn)I,tan∠A=
=
=
=
,EF=
AF,IG=
AG
①當(dāng)0<t<4時(shí),∵△AFE∽△ABD,
=
,AE=
t,解得EF=t,AF=2t,GF=t,AG=2t-t=t,IG=
t.HI=
t.S=t
2-
×
×t=
t
2
②當(dāng)4<t<
時(shí),設(shè)HE與AD相交于點(diǎn)M,tan∠ADF=
==2,ME=2DE=2(t-4)=2t-8,GF=EF=4-(t-4)=8-t,AG=AF-GF=8-(8-t)=t,GI=
t.S=(8-t)
2-
××(16-3t)=
t2-20t+64
③當(dāng)
<t<8時(shí),正方形EFGH與△ABD重合部分的面積為正方形EFGH,S=(8-t)
2=t
2-16t+64.
(3)設(shè)點(diǎn)E到過點(diǎn)D后,運(yùn)動(dòng)的時(shí)間為a,∵PE∥BC,∴△DEP∽△DBC,
=①P在D→C運(yùn)動(dòng)時(shí),DP=8-8a,
=,解得a=
②P在D→C→D運(yùn)動(dòng)時(shí),DP=8a-8,
=,解得a=
③P在D→C→D→C運(yùn)動(dòng)時(shí),DP=24-8a,
=,解得a=
④P在D→C→D→C→D運(yùn)動(dòng)時(shí),DP=8a-24,
=,解得a=4.此時(shí)PE與BC重合.
因此滿足PE∥BC時(shí)t的取值分別為
秒,
秒,
秒.