As shown in the figure, in △ABC, ∠ b = 90, BC=8cm, and AB=6cm. Point P moves from point A to point B at a speed of 1cm/s, and point Q moves from point B to point C at a speed of 2cm/s along BC. (p, q are on the side of the triangle)

(1) if p and q start from a and b at the same time, after a few seconds, the area of △PBQ is equal to 8cm2?

Solution: Let x seconds pass, and the area of △PBQ is equal to 8cm2. Pb = 6-x.bq = 2x。 S △PBQ = (6-x) 2x/2, s △ pbq = 6x-x 2 = 8, X=2, or X=4. Therefore, the area of △ pbq is equal to 8cm2 in 4 seconds and 2 seconds.

(2) If P and Q start from A and B at the same time, P will continue to advance on BC side after arriving at B, and Q will continue to advance on CA side after arriving at C.. After a few seconds, the area of △PCQ is equal to 12.6cm2?

Solution: When P reaches point B, it takes 6 seconds. Therefore, according to the speed of point Q, it has passed 12cm and point C, and CQ is already on the AC line, BC = 8 cm, and CQ is not equal to the length of 12CM-BC. CQ = 4cm. CQ =12-8 = 4cm. So suppose that the area of △PCQ after y seconds is equal to 12.6cm2, and according to similar triangles's theorem, the area of △PCQ = CP * QH/2. CP = 8-y。 CQ =(4+2Y)* 8 * 3/5。 △ area of pcq = CP * CQ * 1/2 = 12. Therefore, when Y= 1 or Y=5, the area of △PCQ is equal to12.6cm2.