10. (10 points) 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 length greater than 5 is $\qquad$ .
【Analysis】Count the number of squares with side lengths of 6, 7, 8, 9, 10, 11, 12, 13, and 14, respectively, and add the 4 × 2 = 8 tilted squares, then sum them up to find the solution.

【Solution】Solution: The number of squares with a side length of 6 is 100,
The number of squares with a side length of 7 is 64,
The number of squares with a side length of 8 is 49,
The number of squares with a side length of 9 is 36,
The number of squares with a side length of 10 is 25,
The number of squares with a side length of 11 is 16,
The number of squares with a side length of 12 is 9,
The number of squares with a side length of 13 is 4,
The number of squares with a side length of 14 is 1,
The number of tilted squares is 4 × 2 = 8,
$$
100+64+49+36+25+16+9+4+1+8=393 \text { (squares) }
$$

Answer: There are 393 squares with a side length greater than 5.