Area of trapezoid, did I count right? Top and bottom lines are parallel.
Trapezoid:

I got ~$6397 m^2$
 A: It's $$\frac{(100+77)76\sin72^{\circ}}{2}$$ and we get your answer.
In another hand, we obtain:
$$\cos72^{\circ}=\frac{23^2+76^2-72^2}{2\cdot23\cdot76},$$ which is wrong, which says that this trapezoid does not exist.
A: Let's use:
$$A=\frac12(76\sin 72°)(100+77)=6397$$
Note that the result seems to be numerically exact 76*sin(72°)*177/2.
However note also that the left side should be $\geq 76 \sin 72°\approx72.28$ (maybe it's a typo?).
In case the bottom and top lines were not parallel you could consider a diagonal and solve for triangles or use Bretschneider's formula.
Notably, with reference to the following figure:

we obtain from the Law of cosines
$$AB^2=AC^2+BC^2-2AC\cdot BC \cos 72°\implies AB=115.9$$
and thus
$$S_{ABC}=\frac12AC\cdot BC \sin 72° = 3614$$
and from Heron's formula
$$S_{DBC}=\sqrt{p(p-a)(p-b)(p-c)} \approx 2710.65$$
and finally
$$S_{ABCD}\approx 6324.65$$
A: Not sure what I'm missing, but isn't the area just this?
$$A = \frac{(B_1 + B_2)h}{2} = \frac{(100m+77m)72m}{2} = 6372m^2$$
($B_1,B_2$ are the two bases of the trapezoid.)
It looks like the lengths of the non-parallel sides were rounded to the nearer integer, so you have to pick one non-parallel side to be the "right" one. I also assumed that the $72$-m side was perpendicular to the bases (it appears to be so).
