A sheet of paper with side lengths of $1 \mathrm{~cm}$ and $\sqrt{2} \mathrm{~cm}$ was folded, as shown in the figure below, so that one vertex lies on the opposite side. What is the value of the length $d$ in centimeters?

![](https://cdn.mathpix.com/cropped/2024_05_01_a50b5476db779c3986d5g-29.jpg?height=386&width=449&top_left_y=854&top_left_x=883)

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Solution

Unfolding the sheet of paper, we obtain the rectangle $B C D F$. Thus, $B C=D F=D E$ and $A F=$ $A E=x \text{ cm}$. Applying the Pythagorean Theorem in the triangle $C D E$, we have

$$
C E=\sqrt{D E^{2}-C D^{2}}=\sqrt{2-1}=1
$$

Consequently, $B E=(\sqrt{2}-1) \text{ cm}$ and the triangle $C E D$ is a right isosceles triangle. Since $\angle A E D=90^{\circ}$, this leads to $\angle B E A=180^{\circ}-90^{\circ}-45^{\circ}=45^{\circ}$. Therefore, $A B E$ is also a right isosceles triangle and $d=B E=(\sqrt{2}-1) \text{ cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_a50b5476db779c3986d5g-29.jpg?height=380&width=452&top_left_y=1999&top_left_x=879)

Note: Since $C D=B F=1 \text{ cm}$, we can conclude that $x=(1-d) \text{ cm}$. Using the Pythagorean Theorem again, this time in the triangle $A B E$, we get

$$
\begin{aligned}
A B^{2}+B E^{2} & =A E^{2} \\
d^{2}+(\sqrt{2}-1)^{2} & =(1-d)^{2} \\
d^{2}+2-2 \sqrt{2}+1 & =1-2 d+d^{2} \\
d & =\sqrt{2}-1 \text{ cm}
\end{aligned}
$$