The rectangle $A B C D$ with dimensions $A B=240 \mathrm{~cm}$ and $B C=288 \mathrm{~cm}$ represents a piece of paper that will be folded along the segment $E F$, where $E$ belongs to $A D$ and $F$ belongs to $B C$, such that point $C$ will be on the midpoint of $A B$.

![](https://cdn.mathpix.com/cropped/2024_05_01_4fd1b602786ad9981479g-06.jpg?height=597&width=973&top_left_y=542&top_left_x=423)

a) What is the length of $C C^{\prime}$?

b) What is the length of $E F$?
Solution

The operation of folding the paper is equivalent to producing a reflection along the fold. To see this, notice that corresponding segments have the same length, for example, $C^{\prime} F=C F$ and $C^{\prime} E=C E$. Thus, $E F$ bisects the segment $C C^{\prime}$ and is perpendicular to it.

![](https://cdn.mathpix.com/cropped/2024_05_01_4fd1b602786ad9981479g-06.jpg?height=581&width=507&top_left_y=1753&top_left_x=656)

a) Since $\triangle C B C^{\prime}$ is a right triangle at $B$, by the Pythagorean Theorem, we have $C C^{\prime}=\sqrt{120^{2}+288^{2}}=$ $312 \mathrm{~m}$.
b) Since $E F$ is perpendicular to $C C^{\prime}$, it follows that

$$
\angle C C^{\prime} B=90^{\circ}-\angle C^{\prime} C B=\angle E F C
$$

Taking $G \in B C$ such that $E G \perp B C$, we can conclude that $\triangle E F G$ is similar to $\triangle C C^{\prime} B$. Thus

$$
\begin{aligned}
\frac{E F}{E G} & =\frac{C C^{\prime}}{B C} \\
\frac{E F}{240} & =\frac{312}{288} \\
E F & =260 \mathrm{~m}
\end{aligned}
$$