SOLUTION:

Let the area of ΔCEG be x cm². Since  ΔCEG and ΔEFG share a common vertex i.e. E, they have the same height.

Since G is the midpoint of FC, the base of ΔCEG and ΔEFG are equal.

Thus, the area of ΔEFG is also x cm².

Likewise, we have area of ΔDEF = area of ΔCEF = area of ΔCEG + area of ΔEFG = 2x cm².

Using the same concept once again, since F is the midpoint of BC, and ΔBDF and ΔCDF share a common vertex,

area of ΔBDF = area of ΔCDF =  area of ΔCEG + area of ΔEFG + area of ΔCEF = x + x + 2x cm² = 4x cm².

Since, D and E are trisection points of line segment AC (as you've said), the base of ΔABD = 1/2 base of ΔBDC.

Since ΔABD and ΔBDC share a common vertex, the height of the two triangles are equal.

Hence, area of ΔABD = 1/2 area of ΔBDC.

Note that:

area ΔBDC = area ΔCEG + area ΔEFG + area ΔDEF + area ΔBDF

Substituting the areas of the triangles,

area ΔBDC = x + x + 2x + 4x = 8x cm²

Thus, area of ΔABD = 4x cm²

It follows that ΔABC = ΔABD + ΔBDC = 4x + 8x cm² = 12x cm².

From the given information, the area of ΔABC = 20 cm²,

so we have x = 5/3 cm².

Therefore, the area of the shaded region is

x + 2x + 4x = 7x = 7(5/3) = 35/3 cm² ≈ 11.67 cm²

ANSWER:

11.67 cm²