Squares in the grid

If we add all the numbers $16 + 9 + 4 + 1 + 9 + 4 +2 +1 + 4 = 50$ squares.

General solution for a $p \times q$ grid, with $p \leq q$ . A coordinate system can be defined with $1 \leq x \leq p$ and $1 \leq y \leq q$ .

For any square $S$ , consider the minimum and maximum coordinates of the vertices, $x_{\text{min}}, x_{\text{max}}, y_{\text{min}}, y_{\text{min}}$ . It is not difficult to see that $d := x_{\text{max}} - x_{\text{min}} = y_{\text{max}} - y_{\text{min}} > 0$ . Define $x_\star$ such that $(x_\star, y_{\text{min}})$ is a vertex of the square; if there are two such vertices, choose $x_\star$ as large as possible. Any valid square is now fully specified by

• $1 \leq d \leq p-1$ ;

• $x_{\text{min}} + 1 \leq x_\star \leq x_{\text{min}} + d$ ( $d$ possible choices);

• $1 \leq x_{\text{min}} \leq p - d$ ( $p - d$ possible choices);

• $1 \leq y_{\text{min}} \leq q - d$ ( $q - d$ possible choices).

Therefore the number of squares is $\begin{aligned}N_{p,q} & = \sum_{d = 1}^{p-1} d(p-d)(q-d) \\ & = \sum_{d = 1}^{p-1} (d^3 - (p+q)d^2 + pqd) \\ & = \frac14p^2(p-1)^2 - \frac16(p+q)(p-1)p(2p-1) + \frac12qp^2(p-1) \\ & = \boxed{\dfrac{(p-1)p(p+1)(2q-p)}{12}}\end{aligned}.$ In this case, $p = q = 5$ so that we find $N_{5,5} = \frac{4\cdot 5\cdot 6\cdot 5}{12} = \boxed{50}.$

Each square takes up the same horizontal and vertical length, even if the square is tilted.

There are $4 \times 4 = 16$ spaces that are $1 \times 1$ , with only one possible type of square that can fit in this space.

In general, an $n \times n$ space has $n$ different squares that fully occupy that space, as all the n first points on the top row can be chosen as the starting point, and each successive point is going to be the same distance in from the corner on the next face going clockwise, and starting from the rightmost point is equivalent to starting from the leftmost.

So, we have $4 \times 4 \times 1 + 3 \times 3 \times 2 + 2 \times 2 \times 3 + 1 \times 1 \times 4 = 50$ distinct squares.

My strategy was to consider side leght/orientation combinations from vector forming the sides. In the first figure, the vectors are (1,0), (2,0), (3,0), (4,0). In the second, (1,1), (1,2), (1,3) ((1,4) is not possible). In the third, the reflections of (1,2) and (1,3). Finally in the fourth, (2,2), which is the largest without leaving the grid. Now you just have to count how many translations each square can have. Grouping by figure, we have $(16+9+4+1)+(9+4+1)+(4+1)+(1)=50.$

(4^2 + 3^2 + 2^2 + 1^2) + (3^2 + 2^2 + 1^2) + (2^2 + 1^2) + 1^2 = 50 Brackets contain different orientations.

To make a square select any 2 dots from 25 dots of the grid =25C2
The other two dots are automatically arranged so as to form a square . But all the squares counted are repeated $^5C_2$ times.
Therefore the required answer= 25C2/4C2 = 50
For a general n×n grid the number of such squares = (n×n)C2 / 4C2

I counted the small squares in the puzzle then multiplied it by the number of dots on one side and got the answer

formula might be

(n-1)(n-1)+(2)(n-2)(n-2)+3(n-3)(n-3)+4(n-4)(n-4)......till it becomes zero

$1^2 \times 4 + 2^2 \times 3 + 3^2 \times 2 + 4^2 \times 1$

it is 50 bc i said it was 50 so here's the answer have a nice day