It was claimed that there are 204 squares on an ordinary chessboard. Can you justify this claim?

Answer 1

There is one square with 64 smaller squares
there are 4 with 49 squares .... to there are 64 squares with one square, find them all and add them up.

It is a good rainy day activity with a bunch of grade school kids that can't go out and play in the rain.

Answer 2

http://mathcentral.uregina.ca/QQ/database/QQ.09.00/tom3.html has this description:

The 204 is correct. What is best to do is to think of the chessboard as being drawn in the plane by 9 vertical and 9 horizontal equally spaced line segments so that the lines meet at (0,0), (0,1), ... (8,8) (that's 9x9 points altogether).

To count the number of squares think of placing a 1x1 square on the grid to cover one of the unit squares. Where can the upper right hand corner of this square be? At (1,1), (1,2), ... , (8,8) giving 8x8 choices. How about 2x2 squares? The upper right hand corner must appear at (2,2), (2,3), ... , (8,8) giving 7x7 choices. If you continue this until you place an 8x8 square you will end up with just 1x1 choice for the upper right hand corner. Thus the total number of (integer sided) squares that exist on the board is 8x8 + 7x7 + ... + 2x2 + 1x1 = 204.

Answer 3

The 204 is correct. What is best to do is to think of the chessboard as being drawn in the plane by 9 vertical and 9 horizontal equally spaced line segments so that the lines meet at (0,0), (0,1), ... (8,8) (that's 9x9 points altogether).

To count the number of squares think of placing a 1x1 square on the grid to cover one of the unit squares. Where can the upper right hand corner of this square be? At (1,1), (1,2), ... , (8,8) giving 8x8 choices. How about 2x2 squares? The upper right hand corner must appear at (2,2), (2,3), ... , (8,8) giving 7x7 choices. If you continue this until you place an 8x8 square you will end up with just 1x1 choice for the upper right hand corner. Thus the total number of (integer sided) squares that exist on the board is 8x8 + 7x7 + ... + 2x2 + 1x1 = 204.

For an nxn board you get nxn + (n-1)x(n-1) + ... + 1x1 = n(n+1)(2n+1)/2 in all.

To find how many rectangles you need the same idea but this time you need to look at

1x1 rectangles, 1x2's, ..., 1x8's and then
2x1 rectangles, 2x2's , ... 2x8's and then
.
.
.
8x1 rectangles, 8x2's , ..., 8x8's

and how they can be placed. I think you might get 1296 (which is 9C2 x 9C2).

Answer 4

multiply the length by the width and ull get the answer...i dont know the answer right off the bat i need to find a chess board but if u find one just do that.

Answer 5

there's 64

Answer 6

I'm sorry -- what??? A chess board is 8 x 8. YOU do the math and figure it out.