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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?



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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  • 4
    $\begingroup$ 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
  • 1
    $\begingroup$ According to the figure on the right, $\frac25=\frac38$. Who knew? $\endgroup$ – G Tony Jacobs Sep 2 '17 at 1:41

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

  • $\begingroup$ Nicely done (+1). $\endgroup$ – dxiv Sep 2 '17 at 0:47

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".


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 $\,90^\circ - \arctan 3/8 - \arctan 5/2 \simeq 1.25 ^\circ$.

  • $\begingroup$ 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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