<h3>Answer:</h3>
56 m²
<h3>Explanation:</h3>
The altitude from point B to segment CP is the same for ∆BMP as for ∆BMC. Since both have the same base length (MP = MC), both have the same area, 21 m². Hence the area of ∆CPB is (21+21) m² = 42 m².
The altitude from point C to segment AB is the same for ∆CPA as for ∆CPB, so the areas of those triangles will have the same proportion as the base segments AP and BP. That is, ...
... AACPA : ACPB = PA : PB = 1 : 3
The ratio of ACPB to the total is then ...
... ACPB : (ACPA +ACPB) = 3 : (1+3) = 3 : 4
The area of ∆ABC is the total of the areas of the smaller triangles CPA and CPB, so we have
... ACPB : AABC = 3 : 4
... AABC/42 m² = 4/3 . . . . . rearranging slightly and substituting for ACPB
... AABC = (42 m²)×(4/3) . . . . multiply by the denominator
... AABC = 56 m²
