Posted October 12, 2013 What is the area of the smallest square that holds five unit squares without overlap? 0 Share this post Link to post Share on other sites

0 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 0 Share this post Link to post Share on other sites

0 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 ... 0 Share this post Link to post Share on other sites

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

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