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.