Current location - Training Enrollment Network - Mathematics courses - The problem of moving point in the second volume of mathematics in the second day of junior high school
The problem of moving point in the second volume of mathematics in the second day of junior high school
(1) Solution: When the distance between P and Q is t seconds, the area of quadrilateral PBCQ is 33cm2.

Then (3t+2t) × 6x1/2 = 33.

The solution is: t= 1 1/5s, and because 3t < 16cm, when p and q start from 1 1/5s, the area of quadrilateral PBCQ is 33cm2.

(2) Point Q is QT⊥AB, so QT=BC=6cm.

TQ= 10cm, so PT=8cm.

Because AB= 16cm, CQ+AP=8cm.

Let CQ=2t AP=3t.

So: 5t=8, t=8/5s.