Determining Areas of Triangles in Hexagon ABCDEF

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The discussion revolves around finding the areas of two unknown triangles within hexagon ABCDEF, given the total area of the hexagon is 30 cm² and the areas of four triangles formed with point M inside. The known areas are ABM (3 cm²), BCM (2 cm²), DEM (7 cm²), and FEM (8 cm²), totaling 20 cm², leaving 10 cm² for the two unknown triangles. Participants note that while the total area can be calculated, the lack of symmetry in the hexagon means the individual areas of the remaining triangles cannot be determined uniquely. The conversation highlights the challenge of the problem due to the absence of regularity in the hexagon's shape. Ultimately, the problem is deemed unfair if it does not allow for a definitive solution.
Numeriprimi
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I have hexagon ABCDEF (30 cm2) and point M inside.
True: ABM = 3 cm2; BCM = 2 cm2; DEM = 7 cm2 ; FEM = 8cm2

How can I determine area of others two triangles? I know their total area, but how individually?

Thanks very much and if you don't understand, write, I will try to write better.
Poor Czech Numeriprimi
 
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You exploit the symmetry of the hexagon ... did you sketch it out?
Point M is closest to B and C, closer to B than C - right?

Can you find the length of the line segments radiating from M in terms of the areas you know?
 
How can I exploit symmetry?
And yes, it is right, but but what good is it useful?
I don't understand your third question... What length from M?
 
Simon Bridge said:
You exploit the symmetry of the hexagon ... did you sketch it out?
Point M is closest to B and C, closer to B than C - right?

Can you find the length of the line segments radiating from M in terms of the areas you know?
The problem, as stated, does not suggest that this is a "regular" hexagon and so does not imply any "symmetry".

Numeri Primi, it is easy, as you say, to see that the total area of the two remaining triangles is 30- (3+ 2+ 7+ 8)= 30- 20= 10. But there is NO way to determine the area of the two triangles separately. It is possible to construct many different (non-symmetric) hexagons having the given information but different areas for the last two triangles.
 
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The problem, as stated, does not suggest that this is a "regular" hexagon and so does not imply any "symmetry".
That's a good point... though the question would seem somewhat unfair if it were not.
 

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