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In triangle ABC, point P intersects AC such that AP:PC = 3:2. If the area of triangle BPC is 14 m², find the areas of triangles ABP and ABC.

Answer :

Final answer:

The area of triangle ABP is 21m² and the total area of triangle ABC is 35m². The areas are calculated based on the proportional base lengths and areas of constituent triangles.

Explanation:

In your question, you have a triangle ABC in which point P divides line AC such that AP:PC = 3:2. Also, you mentioned that the area of triangle BPC is 14m².

We know that in a triangle, the area is proportional to the length of its base. By extension, for two triangles with the same height, their areas will be in the proportion of their respective bases.

In this case, the height, perpendicular from point B to AC, is same for both triangles ABP and BPC. Hence, the areas of these two triangles will be directly proportional to their base lengths i.e., AP and PC. So, Area of ABP : Area of BPC = AP : PC = 3 : 2.

Let's denote the area of ABP as x. From the given proportion, we can write x/14 = 3/2. Solve this equation and you'll find that the area of ABP is 21m².

Adding these two areas together, we get the total area of triangle ABC. So, the area of triangle ABC is 35m² (21m² from ABP and 14m² from BPC).

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