Area of a trapezoidal

Since the red square has an area of $4$ , then its sides are $2$ .

Since the yellow triangle has an area of 4, and the side sharing the red square and yellow triangle is $2$ , and the area of a triangle is $A = \frac{1}{2}bh$ then $X = 4$ .

Since $X = 4$ and the side sharing the red square and green trapezoid is $2$ , the parallel sides of the green trapezoid are $2$ and $10$ , and the area $A$ of the green trapezoid is $A = \frac{1}{2}(2 + 10)Y = 6Y$ .

Since $Y$ is directly proportional to the area of the green trapezoid, the maximum area of the green trapezoid occurs when $Y$ is a maximum. Since $Y \leq X$ , and $X = 4$ , the maximum value of $Y$ is $4$ , so the maximum area of the green trapezoid is $A = 6 \cdot 4 = \boxed{24}$ .