In △ABC, point P∈
1 answer:
<h3>Answer:</h3>
56 m²
<h3>Explanation:</h3>A diagram can be helpful.
Triangles with the same altitude will have areas proportional to the length of their bases.
The altitude from B to PC is the same for triangles BMP and BMC, so they have areas that are in the same proportion as MP : MC. Since M is the midpoint of CP, MP = MC and ABMP = ABMC = 21 m². Then ...
... ACPB = 21 m² + 21 m² = 42 m²
The altitude from C to AB is the same for triangles CPA and CPB, so those triangles have areas in the sampe proportion as AP : BP = 1 : 3. Then ...
... ACPA : ACPB = PA : PB = 1 : 3
... ACPA : 42 m² = 1 : 3
So, the area of ∆CPA is 1/3 of 42 m², or 14 m². The area of ABC is the sum of the areas of CPA and CPB, so is ...
... AABC = ACPA + ACPB = 14 m² + 42 m²
... AABC = 56 m²
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