2009 AMC 12B Problem 25

$G$ splits into four $5 \times 5$ blocks, one per quadrant; a side-$\ge 6$ square takes exactly one vertex from each block

Slide the four blocks together onto a single $5 \times 5$ grid $G';$ each valid square maps to a point of $G'$ or a square in $G'$

$G$ consists of four $5 \times 5$ blocks $G_1, \ldots, G_4,$ one in each quadrant. Any square of side $\ge 6$ uses exactly one vertex in each block, since two points in one block are less than $6$ apart while points in different blocks are at least $6$ apart.

Sliding each block inward by $(\pm 5, \pm 5)$ superimposes them on one $5 \times 5$ grid $G'$ (points with $|x|, |y| \le 2$). Each such square maps to either a single point of $G'$ or a square in $G'.$ So the count equals the number of points of $G'$ plus $4$ times the number of squares with vertices in $G'.$

The grid has $4^2+3^2+2^2+1^2=30$ axis-parallel squares. For a tilted square, let one side move $a$ units horizontally and $b$ units vertically, where $a,b\gt0$ and $a+b\le4.$ For each ordered pair $(a,b),$ there are $(5-a-b)^2$ placements. The totals for $a+b=2,3,4$ are $9,8,3,$ respectively, giving $20$ tilted squares and $50$ squares altogether. Therefore the required count is $25+4\cdot50=225.$

Thus, the correct answer is E.