Question 815188
<pre>
There are many different solutions to this problem,
depending on how you cut the rectangle

If you cut it into equal rectangles like this:

{{{drawing(400,400,-5,9,-7,7,
rectangle(-4,0,0,6),
rectangle(0,0,4,6),
rectangle(4,0,8,6),
rectangle(-4,-6,0,0),
rectangle(0,-6,4,0),
rectangle(4,-6,8,0),
locate(-2,0,W), locate(.1,3,L),
locate(-2+4,0,W), locate(.1+4,3,L),
locate(-2+4+4,0,W),
locate(.1,-3,L),locate(.1+4,-3,L)



)}}}

where each has perimeter 2L+2W,

then the sum of the perimeters is 6(2L+2W)
6(2L+2W) = 120
 12L+12W = 120
     L+W = 10
       L = 10-W

      3W = 2L
      3W = 2(10-W)
      3W = 20-2W
      5W = 20
       W = 4 
       L = 10-W
       L = 10-4
       L = 6

So the a side of the square is 3W = 2L = 3(4) = 2(6) = 12

So the area of a 12×12 square is 144cm²

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However if you cut it into 6 equal pieces lke this,
which is what the other tutor did, you get a smaller area.

{{{drawing(300,300,-5,9,-7,7,
rectangle(-4,-6,-2,6),
rectangle(-2,-6,0,6),
rectangle(0,-6,2,6),
rectangle(2,-6,4,6),
rectangle(4,-6,6,6),
rectangle(6,-6,8,6),
locate(-3,-6,W),locate(-3+2,-6,W),locate(-3+4,-6,W),
locate(-3+6,-6,W),locate(-3+8,-6,W),locate(-3+10,-6,W),
locate(-3.9,0,L),locate(-3.9+2,0,L),locate(-3.9+4,0,L),
locate(-3.9+6,0,L),locate(-3.9+8,0,L),locate(-3.9+10,0,L)


)}}}

Then up to a certain point, it's like the above:

where each has perimeter 2L+2W,

then the sum of the perimeters is 6(2L+2W)
6(2L+2W) = 120
 12L+12W = 120
     L+W = 10
       L = 10-W

But this time

      6W = L  

      6W = 10-W
      7W = 10
       W = 10/7

And the side of the square is L = 6W = 6(10/7) = 60/7 cm

And the area of the square is (60/7)² = 3600/49 = 73.46938776 cm². 
 
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Now if you cut into 6 UNequal rectangles like this


{{{drawing(400,400,-5,9,-7,7,
rectangle(-4,-1,-1,6),
rectangle(-1,-1,3,6),
rectangle(3,-1,5,6),
rectangle(5,-6,8,6),
rectangle(-4,-6,0,-1),
rectangle(0,-6,5,-1))}}}

You would get still a totally different answer.

Edwin</pre>