bonanova 77 Report post Posted October 12, 2013 What is the area of the smallest square that holds five unit squares without overlap? Share this post Link to post Share on other sites

0 harey 7 Report post Posted October 13, 2013 (edited) Place four unit squares into the corners of the resulting square and the 5th one into the middle after a 45 degree rotation. The diagonal of the new square: d=sqrt(2)+1+sqrt(2) giving the surface 4.5+2*sqrt(2) (about 7.33). Edited October 13, 2013 by harey Share this post Link to post Share on other sites

0 dgreening 5 Report post Posted October 13, 2013 I think the smallest square would be a 3 x 3. If we start with the most compact area to contain 4 unit squares, it would be a 2 x 2 square. The addition of another unit square will force it to add at least one row or column - thus forcing you to a 2x3 configuration. To meet the criteria of the problem, we must expand that to a 3 x 3 There are multiple configurations [many are just reflections of others # # # # # - - - - or # # # # - - # - - or # - # - # - # - # or ... Share this post Link to post Share on other sites

What is the area of the smallest square that holds five unit squares without overlap?

## Share this post

## Link to post

## Share on other sites