Solution: Let QD be perpendicular to AB, let the time be t seconds, then BP is 6-t, BQ is 2t, and △BQD is similar to △BCA, then QB/CB=QD/CA, and QD=2/3t, then the area of △PBQ is expressed as: S = BP * QD * 1/.

Divide by angle

1, acute triangle: all three internal angles of the triangle are less than 90 degrees.

2. Right triangle: One of the three internal angles of the triangle is equal to 90 degrees, which can be recorded as Rt△.

3. obtuse triangle: one of the three internal angles of the triangle is greater than 90 degrees.