梯形ABCD中,AD//BC,AB

1)在△DBC中,过点D作BC高DH,BH=1/2*BC=6,DH=根号下(10^2-6^2)=8
BE=12-2t
y=(1/2)(AD+BE)*DH
=(1/2)(4+12-2t)*8
=64-8t
0≤t≤6
2)三角形DEC面积=(1/2)EC*DH
=(1/2)2t*8
=8t
得64t-8t=8tt=4
此时BE=4=AD,又BE平行于AD,所以此时四边形ABED为平行四边形