Final Answer:
The areas are:
Area(∆APC) = 70 m²
Area(∆PMC) = 35 m²
Step-by-step explanation:
To find the areas of ∆APC and ∆PMC when given that Area(∆APM) = 35 m², let's consider the properties of a median and the fact that M is the midpoint of BC.
Since AM is a median, it divides triangle ABC into two triangles of equal area. Therefore, the area of ∆AMB is equal to the area of ∆AMC.
Now let's focus on line MP, which is given to intersect AB at its midpoint P. This implies that AP = PB. As a result, triangle APM is congruent to triangle BPM by the Side-Side-Side (SSS) congruency rule (since AP = PB, PM is a common side, and AM = MB by definition of a median).
Since ∆APM is congruent to ∆BPM, they have the same area. We are given that Area(∆APM) = 35 m², so Area(∆BPM) must also be 35 m².
Now, let's consider the larger triangles. Since M is the midpoint of BC, and P is the midpoint of AB, we have established that:
Area(∆AMB) = Area(∆AMC) (due to AM being a median)
Area(∆APM) = Area(∆BPM) (due to P being the midpoint of AB)
Adding the areas of ∆APM and ∆BPM gives us the area of the entire ∆ABM (because they are two non-overlapping triangles that together form ∆ABM):
Area(∆ABM) = Area(∆APM) + Area(∆BPM) = 35 m² + 35 m² = 70 m²
Since ∆AMB is equal in area to ∆AMC (because AM is a median), we also find that:
Area(∆AMC) = 70 m²
Now, to find Area(∆APC), we need to add the areas of ∆APM and ∆PMC. However, remember that ∆PMC is part of ∆AMC, which we already know has an area of 70 m².
Since ∆PMC and ∆APM both share side PM, we can express the area of ∆APC as the sum of the areas of ∆APM and ∆PMC:
Area(∆APC) = Area(∆APM) + Area(∆PMC)
We know Area(∆APM) is 35 m², and because ∆APM and ∆BPM are congruent and combine to form the same area as ∆AMC, Area(∆PMC) must also be 35 m².
Therefore:
Area(∆APC) = 35 m² + 35 m² = 70 m²
Complete question:
AM is a median in △ABC (M∈ BC ). A line drawn through point M intersects AB at its midpoint P. Find areas of △APC and △PMC, if Area of APM=35m².