I think it would be 2x2 squares
3 squares and 4 triangles will have 24 vertices. There will be 4 vertices for each of the 3 squares. Since there are 3 squares, that will be 12 vertices. There are 3 vertices for each of the 4 triangles. Since there are 4 triangles, that will be 12 vertices. 12+12=24.
GFYM
There are 36 unique quadrilaterals in a 3x3 square grid: 14 squares = 9 (1x1) 4 (2x2) 1 (3x3) 22 rectangles = 6 (1x2) 6 (2x1) 6 (3x3) 2 (2x3) 2 (3x2) (the total number of quadrilaterals formed by 3 x 3 pin sets will be larger, i.e. 78)
3-4
4 squares in a 2 by 2 grid 9 squares in a 3 by 3 grid 16 squares in a 4 by 4 grid 25 squares in a 5 by 5 grid 36 squares in a 6 by 6 grid 49 squares in a 7by 7 grid 64 squares in a 8 by 8 grid 81 squares in a 9 by 9 grid 100 squares in a 10 by 10 grid
In a 4 by 3 grid, there are a total of 20 squares. To calculate this, you can start by counting the individual squares of each size within the grid. There are 12 one-by-one squares, 6 two-by-two squares, and 2 three-by-three squares. Adding these together gives a total of 20 squares in a 4 by 3 grid.
30 squares within a 1 unit grid. 30 squares in all: 4*4 square: 1 3*3 squares: 4 2*2 squares: 9 1*1 squares: 16
There are 14 squares in a 3x3 grid. 9 for the separates squares, 4 made up of the upper left 4 squares, upper right, lower right, lower left. 1 Last square is the entire grid. 9 + 4 + 1 = 14
25 Squares * * * * * 30 squares A 5*5 grid offers squares of sides 4, 3, 2 and 1 - as follows: 1 of 4*4 4 of 3*3 9 of 2*2 16 of 1*1
25 or something * * * * * 30 squares A 5*5 grid offers squares of sides 4, 3, 2 and 1 - as follows: 1 of 4*4 4 of 3*3 9 of 2*2 16 of 1*1
Infinitely many, but only 30 squares within a 1 unit grid. 4*4 square: 1 3*3 squares: 4 2*2 squares: 9 1*1 squares: 16
22
There are 4 squares in a 2 x 2 grid.
-3
25 I think cos 5 x 5 = 25 * * * * * Correction. 30 squares A 5*5 grid offers squares of sides 4, 3, 2 and 1 - as follows: 1 of 4*4 4 of 3*3 9 of 2*2 16 of 1*1
4X4