When the triangular area PQC is equal to the quadrilateral area PABQ

1/2*ab= 1/2*3b？ /4= 1/2* 1/2*3*4

b=2√2，a=3√2/2，

(2) Let CQ=a, CP=b, PQ‖AB, point P on AC, and a=3b/4.

When the perimeter of the triangle PQC is equal to the perimeter of the quadrilateral PABQ

a+b==3-a+(4-b)+5

3b/4+b=3-3b/4+(4-b)+5

B = 24/7 = personal computer

(3) Let CQ=a, CP=b, PQ‖AB, point P on AC, and a=3b/4.

If there is a point m on AB, the triangle PQM becomes an isosceles right triangle.

It is more convenient to use the distance between two parallel lines in high school.

PQ？ =a？ +b？ =25b？ / 16，PQ=5b/4

Take the midpoint g of PQ, then GM=5b/8.

PQ‖AB, the distance from P to AB =PQ=5b/8.

Take advantage of similar triangles:

5b/24=(4-b)/5

b=96/49，a=72/49

PQ？ =(96? +72? )/49? = 14400/49?

PQ= 120/49

I wonder if the answer is right. Please refer to it.