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I could count to 12 (update! actually it was 11). But the answer given is 14. For a vertical rectangle, there are 6 rectangles in it as drawn by me. Same is the case for horizontal rectangle except the middle common rectangle. So I did

$6 * 2 - 1 = 12 - 1$ (common) $= 11$

Can you count 14 rectangles?

enter image description here

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    $\begingroup$ I see 11 rectangles. Not sure where the other 3 will come from. The diagonals surely don't help. $\endgroup$ – greenturtle3141 Oct 23 '16 at 7:30
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    $\begingroup$ Was 11 one of the lettered answers? The whole problem might have been sloppily translated from another language (and numbering system??), as seen in the grammar of "how many rectangle." $\endgroup$ – humn Oct 23 '16 at 9:44
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    $\begingroup$ Do the three "dots" of the letters (a. b. c.) count to the figure? if so, they would be TINY rectangles (depending on the font.) Otherwise, I don't see 14 neither... $\endgroup$ – BmyGuest Oct 23 '16 at 9:46
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    $\begingroup$ The options are labelled 25, 14 and 13 and their values are given in hexadecimal. The answer is option 14: 0x0b = 11 rectangles. (No, that wasn't a serious comment either. :)) $\endgroup$ – M Oehm Oct 23 '16 at 12:38
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    $\begingroup$ @MOehm It actually does work for octal: $13_8 = 11_{10}$ :P $\endgroup$ – Paul Evans Oct 23 '16 at 13:35
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Given the lines have non-zero thickness, I count 32.

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Possible Solution:

As pointed out in this answer to the related Math.SE question, the only way to count more than 11 rectangles in the image is to move the triangles around to make more rectangles.
If you use the four triangles to make two rectangles and then combine those rectangles, that gives you three more rectangles, making the total 14. I can't really see how anyone would expect people to come up with this solution though...

Image (Excuse my bad artistic skills)

enter image description here

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