6. (6 points) There are two squares, the side length of the larger square is 4 decimeters longer than that of the smaller square, and the area of the larger square is 80 square decimeters larger than that of the smaller square. The sum of the areas of the larger and smaller squares is $\qquad$ square decimeters.
6. (6 points) There are two squares, the side length of the larger square is 4 decimeters longer than that of the smaller square, and the area of the larger square is 80 square decimeters larger than that of the smaller square. The sum of the areas of the large and small squares is $\qquad$ square decimeters.

【Solution】Solution: According to the problem, draw a diagram and add auxiliary lines as follows:
It can be seen that the blank part is composed of two equal rectangles and a small square with a side length of 4 decimeters, so the area of one rectangle is:
$(80-4 \times 4) \div 2=32$ (square decimeters),
Then, according to the area of the rectangle $=$ length $\times$ width, the side length of the smaller square is: $32 \div 4=8$ (decimeters),
The side length of the larger square is: $8+4=12$ (decimeters),
Therefore, the sum of the areas of the large and small squares is: $8 \times 8+12 \times 12=208$ (square decimeters);
Answer: The sum of the areas of the large and small squares is 208 square decimeters.
So the answer is: 208.