作DE⊥BC,AF⊥BC,垂足分别为E、F,角C=180°-角ADC=45°,
△DEC是等腰RT△,
DE=CD*sin45°=4√2,
CE=DE=4√2,
EF=AD=6,
BF=BC-CE-EF=30-6-4√2=24-4√2,
tanB=AF/BF=4√2/(24-4√2)=(3√2+1)/17,角B=arctan[(3√2+1)/17],
∴角ABC≈17°8'
S梯形ABCD=(AD+BC)*DE/2=(6+30)*4√2/2=72√2,
∴土石料的体积V=72√2*100=7200√2≈10182.338