(2) Because DE is parallel to AB and CB⊥AB, the height of DE on the edge of △PDE is h=BE=20-0.8t, so S=0.5*0.6t*(20-0.8t)=6t-0.24t? (0 & ltt & lt25)
(3) Because △PAD is close to △PBE and point D is not on the line AO, angle APD = 90 and angle BED= angle PDA, so PE is parallel to AC, so S △ BPE = S △ APD = S △ DEC = 0.25 s △ ABC = 0.25 * 0.5 * 25 * 65433.
S = S△ABC-S△BPE-S△APD-S△DEC = 37.5 = 6t-0.24t? (0 & ltt & lt 16)。 T= do the math yourself. . . . The third question is not guaranteed to be correct. . . .