Draw the perpendicular CD from the vertex C of the triangle ABC to its side AB (so the point D lies on the straight line AB).



Then CD is the common height in triangles ABC, APC and BPC.



Since the triangle area is half the product of the base length and the height drawn to the base, it is clear that

    - the area of the triangle ABC is the sum of areas of triangles APC and BPC,  and

    - the ratio of the area of triangle APC to the area of triangle BPC  is  the same as the ratio of the segment 
      lengths |AP| to |BP|, i.e. 3:4.


Then it is clear that area of the triangle APC = 3x and the area of the triangle BPC = 4x, where 3x + 4x = 231.



Hence,  7x = 231,  x =  = 33,  

area of the triangle APC = 3x = 3*33 = 99 square inches and area of the triangle BPC = 4*33 = 132 square inches.     ANSWER