7. (15 points) As shown in the figure, there are two small squares and one large square. The side length of the large square is twice the side length of the small squares. The area of the shaded triangular region is 240. What is the sum of the areas of the three squares? $\qquad$ .
7. (15 points) As shown in the figure, there are two small squares and one large square. The side length of the large square is twice the side length of the small square. The area of the shaded triangular region is 240. Please find the sum of the areas of the three squares, which is 360. Let the side length of the smaller square be \( x \), and the side length of the larger square be \( 2x \), and the side length of the largest square be \( x + 2x = 3x \).
$$
\begin{array}{l}
S \triangle A E B = S \triangle A D C = x \times (x + 2x) \div 2 = 1.5 x^{2} \\
S \triangle B F C = 2x \times 2x \div 2 = 2 x^{2} \\
S \triangle A E B + S \triangle A D C + \triangle B F C \\
= 1.5 x^{2} + 1.5 x^{2} + 2 x^{2} \\
= 5 x^{2} \\
S \square A D F E = (3x)^{2} = 9 x^{2} \\
S \triangle A B C = S \square A D F E - (S \triangle A E B + S \triangle A D C + S \triangle B F C) \\
= 9 x^{2} - 5 x^{2} \\
= 4 x^{2}
\end{array}
$$

Since \( S \triangle A B C = 240 \), then \( 4 x^{2} = 240 \), so \( x^{2} = 240 \div 4 = 60 \).
Therefore, the sum of the areas of the three squares is:
$$
x^{2} + (2x)^{2} + x^{2} = 60 + 240 + 60 = 360.
$$

Answer: The sum of the areas of the three squares is 360.
So the answer is: 360.