The Purple Triangle of Jeffrey and Trevor

Position the given square on an $xy$ -grid with coordinates $A(0,0), B(12,0), C(12,12)$ and $D(0,12)$ .

With $|GD| = \frac{1}{4}|DA| = 3$ we then have that the coordinates of $G$ are $(0,9)$ .

Next, since $AH = \frac{1}{3}AE$ and the coordinates of $E$ , being the midpoint of $CB)$ , are $(12,6)$ , we find that the coordinates of $H$ are $\frac{1}{3}(12,6) = (4,2)$ .

Now since $|FC| = \frac{1}{3}|DC| = 4$ we see that the coordinates of $F$ are $(8,12)$ . The coordinates of $J$ , being the midpoint of $EF$ , are then $\left(\dfrac{8 + 12}{2}, \dfrac{12 + 6}{2}\right) = (10,9)$ .

With the base of $\Delta GHJ$ being the horizontal segment $GJ$ , length $J_{x} - G_{x} = 10 - 0 = 10$ , the height will then be $J_{y} - H_{y} = 9 - 2 = 7$ , resulting in an area value of $\dfrac{10*7}{2} = \boxed{35}$ .

Let the coordinates of $G,H,J$ be, which can be figured out from inspection (see Brian's solution)

$G = ( {x}_{g},{y}_{g} ) = (0,9)$
$H = ( {x}_{h},{y}_{h} ) = (4,2)$
$J = ( {x}_{j},{y}_{j} ) = (10,9)$

then the area of the triangle $\Delta GHJ$ is

$\dfrac { 1 }{ 2 } \left( \; (x_{ g }{ y }_{ h }-{ y }_{ h }x_{ g })+({ x }_{ h }{ y }_{ j }-{ y }_{ j }{ x }_{ h })+({ x }_{ j }{ y }_{ g }-{ y }_{ g }{ x }_{ j })\; \right) =35$

This is one of the variations of the "shoelace formula" for finding the area of a general irregular polygon, given the coordinates of its vertices. The derivation of this formula comes from the fact that the vector cross product yields the area between the two vectors. The interesting detail about doing a vector cross product of successive pairs of vectors, the area yielded can sometimes be negative, so that for a closed loop, the net area is what's inside the loop.

Let A(0,0),....D(0,12)......
So H(AB/3,BE/3)=H(4,2),....................G(0,AD-GD)=G(0,9),...................J(DC - FC/2, BC - CE/2)=J(10,9). GJ=10 horizontal.
Vertical from H to GJ=9 - 2=7.
So area JGH= 1/2 * 10 * 7= 35.