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The problem is as follows:

Sketch of the problem

The alternatives given in my book are:

  1. 76 cm
  2. 80 cm
  3. 92 cm
  4. 100 cm

Upon the first inspection. I'm getting the idea that I have to make a system of equations.

Assuming that the edge of the squares is $a$ and the smallest edge of the triangle is $b$ and the diagonal of the right triangle is $c$.

I'm getting for each set: (for purposes of brevity I'm omitting the units but you get the idea)

$A:$

$6a+2c+2b=80$

$B:$

$6a+2b=60$

$C:$

$5a+b+c=56$

At this point it is possible to solve the system:

$a=8$

$b=6$

$c=10$

The for $D:$

It is kind of tricky because one square is shifted a little bit to right. But I understood it as it will make that the whole length in that side makes it equal to $6$ because the same amount which is shifted to the right is to the left when you add up these quantities they cancel and you end up with $6$.

Therefore:

$8\times 8+10+(8-6)=76\,cm$

To which appears in the first alternative. And I believe its right. But to me, this process was more mathematical in nature other than solving a puzzle with some intuition or something along those lines.

Therefore, does it exist a way to solve this more intuitively?. Perhaps faster?. Solving a system of three unknowns isn't that quick. For reference, this riddle was obtained from my book Reason and logic from the 2000s and it appears to be a reprinted version of the 70's book from Martin Gardner's Puzzle Carnival's with some modifications.

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  • $\begingroup$ Could you transcribe the image? Images are non-accessible and non-searchable, while text is more usable and screen-reader-friendly. $\endgroup$
    – bobble
    Commented Jan 6, 2021 at 0:39
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    $\begingroup$ @bobble No, there is no way to transcribe the image. I understand wanting things to be accessible, but sometimes images are genuinely the best way to present something. $\endgroup$
    – Deusovi ♦
    Commented Jan 6, 2021 at 0:41
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    $\begingroup$ As for the question, your method seems to be the natural way to me. You can get $c$ pretty easily, because the perimeters in (A) and (B) only differ by adding 2c. $\endgroup$
    – Deusovi ♦
    Commented Jan 6, 2021 at 0:42
  • $\begingroup$ Maybe there is a way to exploit $80+56−60=76$. $\endgroup$
    – RobPratt
    Commented Jan 6, 2021 at 4:05
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    $\begingroup$ @JaapScherphuis Use $ a^2+b^2=c^2 $, $ 6a+2c+2b=80 $, $ 6a+2b=60 $, $ a, b, c > 0 $ can calculate all $a$, $b$, $c$. $\endgroup$
    – tsh
    Commented Jan 6, 2021 at 7:43

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Figure A has the side of the square equal to 8. The right angle triangle must have one side equal to 6 so the hypotenuse is equal to 10. If we add them all we have (6x8)+(2x6)+(2x10)=80. Figure D has 7 sides whose sum is equal to 56, and the hypotenuse is equal to 10. 8-6=2 and 16-8=8 and if we add all we have (7x8)+10+(8-6)+(16-8)=76. So figure D has perimeter equal to 76.

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