Medians And Areas

In $\triangle CGM$ , since $S$ is the midpoint of $\overline{CG}$ , $[CGM]=2[GSM]$ . This same argument can be applied to the other parts of the triangles. Theoretically, just by the nature of the centroid and proportionality, this will logically imply that the final proportion is $\boxed{2}$ . (Sorry I half-@$$ed it)

Well we need to construct $GP, GN, GM$ . So considering $Δ BPG$ we see that the median divides the triangle into two equal areas. So $S(Δ BPG) =2 S(Δ GPR)$ In this way on dividing the whole hexagon we get, ( S(Δ ABC)) = 2. S(PQNSMR)

Just think, one triangle APG, here Q is the middle point of AG. So that

triangle APQ= triangle PGQ All is same, so the hexagonal is half to the hole triangle.

Just keep on guessing knowing that it cannot most likely be 1.

The relation is too simple! mid point=1÷2 the whole side so every side of this polygon is half the other so the whole polygon is half the triangle😎

Add R&Q,Q&S,R&S. You can see that the triangles ABG, AGC & BGC are divided into four equal parts, of which 2 are in the hexagon and all 4 in triangle. Adding them all, we get the value of x, 2.