The problem is as follows:
The alternatives given in my book are:
- 76 cm
- 80 cm
- 92 cm
- 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)
At this point it is possible to solve the system:
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 $6$.
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.