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Using two - $ 4 \times 4$ squares, three - $3 \times 3 $ squares, four - $2 \times 2$ squares and four - $1 \times 1$ squares draw a diagram to show how you can make a square using some or all of these squares together without gaps or overlaps to make a square that is as large as possible.
Explain why you cannot make a square larger than this square.

problem-solving

Widawensen
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T.Bear
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  • Do you know the answer already? – Matthew Anderson Apr 10 '17 at 03:53
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    What have you tried? The total area of the squares gives an upper limit. Have you found that? Can you achieve that? – Ross Millikan Apr 10 '17 at 03:57
  • No- Hence the question. I am in Year 8 at school and i need help. 79 squares is the total area once you add all of the segments together. SO does that mean an 8 x 8 square is the maximum area, since 9 x9 would be too big – T.Bear Apr 10 '17 at 04:25
  • Yes, that is exactly what I was suggesting. That does not guarantee that you can make $8 \times 8$, but means you don't need to look at $9 \times 9$ Also note that if you don't use one $4 \times 4$ you only have $63$ available, so if you are going to make $8 \times 8$ you need to use both of the $4 \times 4$s. – Ross Millikan Apr 11 '17 at 15:46

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I can get $6 \times 6$ as shown in the first figure below. I can get close to $8 \times 8$ but would need six $1 \times 1$ squares to finish the second figure, not four. I haven't proven that $7 \times 7$ and $8 \times 8$ are impossible. The problem is that the $4 \times 4$ and $3 \times 3$ squares don't provide enough flexibility. To really prove that you would have to go through all the possible locations for the big squares and show they don't work. enter image description here enter image description here

Ross Millikan
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