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The red square is placed on top of a blue square The goal is to cut the red square into 4 pieces and assemble them with the blue square to create a larger square. The area of the resulting square is the sum of the area of the red and blue squares.

enter image description here

There are at least two solutions I am aware of. The first one to show at least two solutions will be granted the 15 points:)

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    $\begingroup$ What is the significance of "The red square is placed on top of a blue square", as the first sentence? Or is this a red herring? $\endgroup$ – Stilez Jun 17 at 13:39
  • $\begingroup$ What is the blue triangular thing at right hand side of red cube? $\endgroup$ – Always Confused Sep 8 at 9:29
  • $\begingroup$ My old-syllabus school math textbook contained a bit comlex version of this puzzle: testing the Pythagoras theorem. "A right-angled triangle given. Draw a square using base as arm and another square using perpendicular as arm. Now find such an way to cut the 2 squares that fuse perfectly into a square with an arm same as the hypoteneuse." $\endgroup$ – Always Confused Sep 8 at 9:39
  • $\begingroup$ "Placed on top" does it mean perfectly overlapping? $\endgroup$ – Always Confused Sep 8 at 9:45
  • $\begingroup$ No - a portion of the blue is hidden. $\endgroup$ – Moti Sep 9 at 2:57
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I think you can dissect the red square (on the left here) as follows

enter image description here

And then rearrange to form the larger square

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This is always possible when the red square is larger than the blue.

Construction

If the length of the side of the larger square is $A$ and the smaller square $B$ then we find the point along each side of the large square which is a distance $\frac{A+B}{2}$ from each vertex going in a clockwise direction. Then join opposing points in this construction.

The lines traversing the square intersect at right angles due to symmetry. Also, the length of each line $d$ can be determined with Pythagoras theorem by constructing a line from one of the vertices to the opposite side parallel to the side of the square. That is, $$ d^2 = A^2 + \left(\frac{A+B}{2} - \frac{A-B}{2}\right)^2 = A^2 + B^2$$ It follows that the five pieces can be rearranged to form the combined square as shown.

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    $\begingroup$ This is a great solution! Much better than what I had in mind. Though it seems like a "single" solution it holds for an infinite "blue" squares. $\endgroup$ – Moti Jun 17 at 19:10

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