Florinda went for a walk. She left point $A$ and walked $1 \mathrm{~m}$. At this point, she turned left at an angle of $90^{\circ}$ and walked $2 \mathrm{~m}$. At the last point, she turned left again and walked $3 \mathrm{~m}$. She continued walking in this manner until, on the last segment, she walked $30 \mathrm{~m}$ and reached her final point, which we will call point $B$. The following figure illustrates the first seven segments of Florinda's walk (note that point $B$ is not illustrated).

![](https://cdn.mathpix.com/cropped/2024_05_01_463eeeabc25d438954b1g-13.jpg?height=523&width=536&top_left_y=1470&top_left_x=837)

Calculate the distance between points $A$ and $B$.

#
Solution

In this exercise, all distances are given in meters. Observe in the following figure that the horizontal straight-line segments have odd lengths while the vertical ones have even lengths.

![](https://cdn.mathpix.com/cropped/2024_05_01_463eeeabc25d438954b1g-43.jpg?height=354&width=355&top_left_y=526&top_left_x=732)

If we consider only the horizontal straight-line segments, we obtain that the total horizontal displacement is equal to

$$
1-3+5-7+9-11+13-\cdots+25-27+29
$$

To calculate this value, we can group the terms as follows:

$$
1 \underbrace{-3+5}_{+2} \underbrace{-7+9}_{+2} \underbrace{-11+13}_{+2}-\ldots \underbrace{-27+29}_{+2}
$$

The total horizontal displacement is, then, equal to $1+2 \times 7=15$.

Now considering the vertical displacements and calculating similarly, we obtain

$$
2 \underbrace{-4+6}_{+2} \underbrace{-8+10}_{+2} \underbrace{-12+14}_{+2}-\ldots \underbrace{-28+30}_{+2}=2+2 \times 7=16
$$

![](https://cdn.mathpix.com/cropped/2024_05_01_463eeeabc25d438954b1g-43.jpg?height=460&width=471&top_left_y=1821&top_left_x=671)

Finally, by the Pythagorean Theorem, we conclude that the total displacement is equal to

$$
\sqrt{15^{2}+16^{2}}=\sqrt{481}
$$