Little clue.

Let the side length of square $ABCD$ be $a$ . Then $AE = \dfrac {16}{25}a$ and $AF = \dfrac 9{25}a$ . Then $AE+AF = a$ .

Let $FG = EH = b$ . We note that $\triangle AFG$ and $\triangle AEH$ are similar. Therefore,

$\begin{aligned} \frac {FG}{AF} & = \frac {AE}{EH} \\ \frac {b}{AF} & = \frac {AE}{b} \\ \implies b^2 & = AE \cdot AF \\ & = \frac {16}{25} a \cdot \frac 9{25} a \\ \implies b & = \frac {12}{25} a \end{aligned}$

Now, we note that the area of the orange rectangle is $ab = \dfrac {12}{25} a^2 = 48$ . Then the area of square $ABCD$ , $a^2 = 48 \times \dfrac {25}{12} = 100$ . Then the area of the green part:

$\begin{aligned} \color{#20A900} [GBCDH] & = [ABCD] - \color{#EC7300} [AGH] & \small \color{#EC7300} \text{Note that } [AGH] = \frac 12 [EFGH] \\ & = 100 - \color{#EC7300} 24 \\ & = \boxed{76} \end{aligned}$