10. A $15 \times 15$ grid composed of unit squares, using the grid points as vertices to form squares with integer side lengths, then the number of squares with side lengths greater than 5 is $\qquad$.
【Answer】 393
【Question Type】Counting: Geometric Counting
【Analysis】
Comment: When looking for squares in a grid, one must consider both upright and tilted directions. This problem only requires finding squares with integer side lengths, which is not particularly nasty.
(1) Upright:

The smallest is a $6 \times 6$ square, as shown in the figure, with a total of $10 \times 10$; then $7 \times 7$ has $9 \times 9$; and so on.
Thus, the total number of upright squares is $10^{2}+9^{2}+8^{2}+\cdots+1^{2}=\frac{10 \times 11 \times 21}{6}=385$.
(2) Tilted: To have integer side lengths, we can try special Pythagorean triples.
We can find a square with a side length of 10, as shown in the figure, with a total of 4 (which can be moved to the right and down). Constructing a gnomon, the side length of the larger square is 14, and within each larger square of side length 14, 2 squares with a side length of 10 can be formed.
In summary, we can find $385+4 \times 2=393$ squares with side lengths greater than 5.