(1) because a (-9,0), B(0, 12), c (16,0); Therefore, AO=25, OB= 12, CO= 16, and x axis ⊥Y axis is 0; So AB= 15, CB=20, AC = 25 because15 *15+20 * 20 = 25 * 25, and the angle ABC=90. Because DE⊥CB is at point E with an angle of DEc = 90, DE is parallel to AB, so △CDE is similar to△ △CAB and CD=t, so DE: AB = CE: CB = CD: CA, so DE:15 = CE: 20 = T: 25; So DE=0.6t, CE=0.8t, so BE = 20-0.8t.

(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. . . .