In the given diagram, all the small squares have a side length of $1 \mathrm{~cm}$. Which of the given segments is the longest?

(a) $A E$

(b) $C D+C F$

(c) $A C+C F$

(d) $F D$

(e) $A C+C E$

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-014.jpg?height=503&width=326&top_left_y=748&top_left_x=1502)
The correct option is (e).

Note that

- AE is the hypotenuse of a triangle with legs of $5 \mathrm{~cm}$ and $9 \mathrm{~cm}$;
- CF is the hypotenuse of a triangle with legs of $2 \mathrm{~cm}$ and $4 \mathrm{~cm}$;
- $\mathrm{AC}$ is the hypotenuse of a triangle with legs of $3 \mathrm{~cm}$ and $4 \mathrm{~cm}$;
- FD is the hypotenuse of a triangle with legs of $2 \mathrm{~cm}$ and $9 \mathrm{~cm}$;
- CE is the hypotenuse of a triangle with legs of $2 \mathrm{~cm}$ and $5 \mathrm{~cm}$.

Using the Pythagorean Theorem, we calculate these hypotenuses.

$$
\begin{aligned}
& \mathrm{AE}=\sqrt{5^{2}+9^{2}}=\sqrt{106} \approx 10.3 \\
& \mathrm{CF}=\sqrt{2^{2}+4^{2}}=\sqrt{20} \approx 4.47 \\
& \mathrm{CF}=\sqrt{3^{2}+4^{2}}=\sqrt{25}=5 \\
& \mathrm{FD}=\sqrt{2^{2}+9^{2}}=\sqrt{85} \approx 9.22 \\
& \mathrm{CE}=\sqrt{2^{2}+5^{2}}=\sqrt{29} \approx 5.39
\end{aligned}
$$

Since $\mathrm{CD}=5 \mathrm{~cm}$, we get $\mathrm{AE} \approx 10.3, \mathrm{CD}+\mathrm{CF} \approx 5+4.47=9.47$, $\mathrm{AC}+\mathrm{CF} \approx 5+4.47=9.47, \mathrm{FD} \approx 9.22$ and $\mathrm{AC}+\mathrm{CE} \approx 5+5.39=10.39$. Therefore, the longest segment is $\mathrm{AC}+\mathrm{CE}$, which measures $10.39 \mathrm{~cm}$.