# How do you find the perimeter of a set of odd looking squares and triangles?

The problem is as follows: 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.

• Could you transcribe the image? Images are non-accessible and non-searchable, while text is more usable and screen-reader-friendly. – bobble Jan 6 at 0:39
• @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. – Deusovi Jan 6 at 0:41
• 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. – Deusovi Jan 6 at 0:42
• Maybe there is a way to exploit $80+56−60=76$. – RobPratt Jan 6 at 4:05
• @JaapScherphuis Use $a^2+b^2=c^2$, $6a+2c+2b=80$, $6a+2b=60$, $a, b, c > 0$ can calculate all $a$, $b$, $c$. – tsh Jan 6 at 7:43

## 2 Answers

Are you looking for something like this?

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.