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This question already has an answer here:

Using 2 triangles each with base of 8 and height of 3, and 2 trapezoids with heights of 3 on top, 5 on bottom and height of 5, these four figures can create an area with 64 units squared. However, when rearranged as a rectangle with 13 x 5=65, one additional unit squared seemed to have been created. How is this possible?

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marked as duplicate by David K, dxiv, G Tony Jacobs, J. M. is a poor mathematician, Lord Shark the Unknown Sep 2 '17 at 3:18

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    $\begingroup$ This is the missing square puzzle. $\endgroup$ – José Carlos Santos Sep 2 '17 at 0:04
  • $\begingroup$ so the second figure has triangles slightly bigger than the triangles in figure one? $\endgroup$ – Goodwin Lu Sep 2 '17 at 0:06
  • $\begingroup$ also, unlike the wikipedia article, the hypothenuse is completely and utterly straight: prntscr.com/gg199b $\endgroup$ – Goodwin Lu Sep 2 '17 at 0:10
  • $\begingroup$ @GoodwinLu then it doesn't work, the truth is it's not possible it's an illusion. $\endgroup$ – user451844 Sep 2 '17 at 0:14
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    $\begingroup$ According to the figure on the right, $\frac25=\frac38$. Who knew? $\endgroup$ – G Tony Jacobs Sep 2 '17 at 1:41
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Here's a slightly less subtle "demonstration" that a rectangle with area 5 can be rearranged into a rectangle with area 6. enter image description here enter image description here

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  • $\begingroup$ Nicely done (+1). $\endgroup$ – dxiv Sep 2 '17 at 0:47
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This is a classic illusion based on the Fibonacci number identity $$ 13 \times 5 = 1 + 8 \times 8 . $$

The "diagonal" of the rectangle isn't one. The slopes on each segment don't agree. There's one unit of area between the "diagonals".

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unlike the wikipedia article, the hypothenuse is completely and utterly straight

No, it's not. Consider the bottom-left corner of the rectangle.

  • Let $\alpha$ be the angle in the yellow triangle, then $\tan \alpha = 3/8\,$.
  • Let $\beta$ be the angle in the green trapezoid, then $\tan \beta = 5/(5-3)=5/2\,$.

But then $\,\tan \alpha \tan \beta = 15 / 16 \ne 1\,$, so $\,\alpha+\beta \ne 90^\circ\,$ i.e. the two angles do not add up to a right angle. The slopes of the two hypotenuses differ by $\,90^\circ - \arctan 3/8 - \arctan 5/2 \simeq 1.25 ^\circ$.

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  • $\begingroup$ In fact, not only is the "hypotenuse" not straight, it is bent in exactly the same way as in the Wikipedia article: part of the "hypotenuse" has slope 2/5, and the rest of it has slope 3/8. $\endgroup$ – David K Sep 2 '17 at 0:57
  • $\begingroup$ @DavidK Indeed, and also thanks for the pointer to that cute animation in the duplicate link. Somewhat related: the perpetual chocolate. $\endgroup$ – dxiv Sep 2 '17 at 1:08

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