To create a 4-sided figure in this grid, we have to pick 2 vertical lines and 2 horizontal lines. The figure is created by the intersection points of these lines.
This grid is a 9 x 8 grid. There 10 horizontal lines and 9 vertical lines.
Number of ways to choose 2 vertical lines from 9 choices = 9C2 = 9! ÷ 2! ÷ (9-2)! = 9! ÷ 2! ÷ 7! = 36
Number of ways to choose 2 horizontal lines from 10 choices = 10C2 = 10! ÷ 2! ÷ (10-2)! = 10! ÷ 2! ÷ 8! = 45
Number of 4-sided figures possible = 36 x 45 = 1620
But this includes both squares and rectangles. So we need to remove the number of possibilities that result in squares.
No. of 1-unit squares = 9 x 8 = 72
Logic here is :
The squares are 1 unit long and wide. Since the grid is 9 units wide (vertical length) and 8 units long(horizontal length) ,there are 9 choices to pick the 1 unit for the width and 8 choices for the length.
There are 9 and 8 degrees of freedom respectively.
No. of 2-unit squares = 8 x 7 = 56
Logic here is the same. Since each square is 2 units long and wide now, there are only 8 possible positions for a 2-unit width (from top to bottom/other way round) and 7 possible positions for the length (left to right/other way round). There is one less degree of freedom than before for each dimension.
Likewise in a similar fashion :
No. of 3-unit squares = 7 x 6 = 42
No. of 4-unit squares = 6 x 5 = 30
No. of 5-unit squares = 5 x 4 = 20
No. of 6-unit squares = 4 x 3 = 12
No. of 7-unit squares = 3 x 2 = 6
No. of 8-unit squares = 2 x 1 = 2
The total number of squares = 2+6+12+20+30+42+56+72 = 240
This can be easily summed since it's the summation of n(n+1) or n² + n from n = 1 to n = 8, a total of 8 terms
∑(n²+n) = 1/6 n(n+1)(2n+1) + ½n(n+1)
= 1/6 (8)(8+1)(2(8)+1) + ½(8)(8+1)
= 1/6 x 8 x 9 x 17 + ½ x 8 x 9
= 204 + 36
= 240
(For this question, not really necessarily but if they ask for things like 99 x 100 grid it will come in handy.)
So number of rectangles = 1620 - 240 = 1380