How to find the length of an unfolded piece of paper which has been folded by its diagonal?

Question

The problem is as follows:

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&\textrm{68 cm}\\ 2.&\textrm{130 cm}\\ 3.&\textrm{63 cm}\\ 4.&\textrm{112 cm}\\ \end{array}$

For this particular problem. I'm really stuck. It seems that there isn't any way to go further because the only thing which I'm getting is that as a consequence of the lenght of the diagonal the sum of the other sides must add up to 90 hence to be 50 cm.

But I cannot say for sure which side is greater or what?. Am I missing something?. Can someone help me here?. For reference this problem was obtained from my Reason and logic book from 2000's. And it seems to be a reprinted copy of a Martin Gardner's 70's book Puzzle Carnival with some modifications.

Since I'm not good at this it could help me a lot if it would include some drawing to see what's happening?.

Answer 1

Draw a vertical from the apex of the blue region to the diagonal fold. You should be able to see that the two right angled triangles you create are congruent and can be solved using what you have. Furthermore, those triangles are similar to the triangle which is half the page.

I get the fourth option as my answer, and Its at least possible ie it doesn't seem crazy when you look at all the other lengths you know.

Answer 2

We can see that a $3-4-5$ right triangle is created here.

Knowing the hypotenuse of $3-4-5$, the lengths of the other two sides are $\frac{3+4}5=1.4$ times the length of the hypotenuse.

Since the triangle above and the triangle equal to half the piece of paper are similar, we can conclude that the length and width of the paper are $1.4$ times the length of the diagonal. Thus, we can calculate: $$40\times1.4\times2=112.$$

Answer 3

First, note that because of symmetry, the triangle $AMB$, whose perimeter is 90cm, is isosceles. Therefore $AM=BM=\frac12(90-AB)=25cm$

Next, draw $MH$ perpendicular to $AB$ and note again by symmetry we have $AH=\frac12AB=20cm$. The sides of the right triangle $AHM$ are $25, 20\; \& MH=15$. You could calculate $MH$ using Pythagoras' theorem, or you could simply note that the lengths of the sides are 5 times those of the famous 3-4-5 right triangle.

Now, note the similarity of triangles $ADB$ and $AHM$: $$\frac{AB}{AM} = \frac{AD}{AH} = \frac{DB}{HM}$$ $$AD=AH \frac{AB}{AM}=20 \times \frac{40}{25}=32 cm$$ $$DB=HM \frac{AB}{AM}=15 \times \frac{40}{25}=24 cm$$ Therefore the perimeter of the unfolded rectangle is $2 \times (32+24) = 112cm.$